A-level PhysicsOCR B (Advancing Physics)

Contents

1 AS 1

2 Optics 22.1 Lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

2.1.1 Curvature of Wavefronts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.2 Power of lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.3 The Lens Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.1.4 Types of Lens . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1.5 Magnification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1.6 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.2 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2.1 Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2.2 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.3 Total Internal Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2.5 See also . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Communications 53.1 Digital Storage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.1.1 Digital Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.1.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.2 Digital Processing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2.1 Mean Smoothing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2.2 Median Smoothing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2.3 Edge Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.3 Digitisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.3.1 Digitisation & Reconstruction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.3.2 Sampling Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3.3 Number of Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.4 Signal Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.4.1 Multiple Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

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3.4.2 Frequency Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.4.3 Fundamental Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.4.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.5 Bandwidth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.5.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Electricity 114.1 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.1.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114.3 Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.3.1 Electromotive Force (EMF) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3.2 Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.4 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.4.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.5 Resistance and Conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5.1 Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5.2 In Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5.3 In Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.5.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.6 Internal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.6.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

4.7 Potential Dividers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154.7.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.8 Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.8.1 Temperature Sensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.8.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.8.3 Signal Amplification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164.8.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.9 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.9.1 Symbols and Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.9.2 Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.9.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.10 Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.10.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.10.2 See Also . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5 Material Structure 195.1 Stress, Strain & the Young Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5.1.1 Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

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5.1.2 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.1.3 Young’s Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.1.4 Stress-Strain Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195.1.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

5.2 Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.3 Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3.1 Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.3.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.3.4 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

6 Waves 236.1 What is a wave? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

6.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.1.2 Types of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.1.3 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.1.4 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.1.5 Velocity, frequency and wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246.1.6 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

6.2 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

6.3 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.3.1 Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.3.2 Pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266.3.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6.4 Young’s Slits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.4.1 Calculating the angles at which fringes occur . . . . . . . . . . . . . . . . . . . . . . . . . 276.4.2 Calculating the distances angles correspond to on the screen . . . . . . . . . . . . . . . . . 286.4.3 Diffraction Grating . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.4.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.5 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.5.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.6 Finding the Distance of a Remote Object . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296.6.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

7 Quantum Physics 317.1 Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7.1.1 Evidence for the Quantum Behaviour of Light . . . . . . . . . . . . . . . . . . . . . . . . 317.1.2 The Relationship between Energy and Frequency . . . . . . . . . . . . . . . . . . . . . . . 327.1.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7.2 Quantum Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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7.2.1 Many Paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.2.2 Calculating Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

7.3 Electron Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.3.1 Frequency and Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347.3.2 De Broglie Wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357.3.3 Potential Difference and Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 357.3.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

8 Mechanics 368.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

8.1.1 What is a vector? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.2 Vector Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.3 Vector Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.4 Vector Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368.1.5 Predicting Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378.1.6 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

8.2 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.2.1 Distance-time Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.2.2 Position-time Graphs or Displacement - Time Graphs . . . . . . . . . . . . . . . . . . . . 388.2.3 Velocity-time Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

8.3 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398.3.1 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398.3.2 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408.3.3 Derivations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408.3.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

8.4 Forces and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408.4.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408.4.2 Work Done . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408.4.3 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418.4.4 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418.4.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

9 A2 42

10 Decay 4310.1 Exponential Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

10.1.1 Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.1.2 Exponential Relationships in the Real World . . . . . . . . . . . . . . . . . . . . . . . . . 4310.1.3 Mathematical Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.1.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

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10.2 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.2.1 Exponential Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.2.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.2.3 Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

10.3 Radioactive Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.3.1 Decay Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.3.2 Activity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.3.3 Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.3.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

10.4 Half-lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.4.1 Half Life of a Radioisotope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.4.2 Half-Life of a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.4.3 Time Constant of a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.4.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

11 Gravity 4811.1 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

11.1.1 Gravitational Force Inside an Object . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4811.1.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

11.2 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4911.2.1 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4911.2.2 Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4911.2.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

11.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5011.3.1 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5011.3.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

11.4 Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.4.1 Equipotentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.4.2 Summary of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5111.4.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

12 Mechanics 5212.1 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

12.1.1 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5212.1.2 Time Period . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5312.1.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5312.1.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

12.2 Energy in Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5312.2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

12.3 Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5412.3.1 Critical Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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12.3.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5412.4 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

12.4.1 Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.5 Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

12.5.1 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.5.2 Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5512.5.3 Explosions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612.5.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

12.6 Forces and Impulse in Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612.6.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

12.7 Rockets, Hoses and Machine Guns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.7.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

12.8 Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5712.8.1 Angular Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5812.8.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

13 Astrophysics 5913.1 Radar and Triangulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

13.1.1 Radar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5913.1.2 Triangulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5913.1.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

13.2 Large Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.2.1 Light Years . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.2.2 Astronomical Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.2.3 Parsecs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

13.3 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.3.1 Kepler’s Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6113.3.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

13.4 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6113.4.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

13.5 The Big Bang . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.5.1 Hubble’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.5.2 The Age of the Universe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.5.3 More Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6213.5.4 Evidence for the Big Bang . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6313.5.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6313.5.6 Footnotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

14 Thermodynamics 6414.1 Heat and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

14.1.1 Changes of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

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14.1.2 Activation Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6414.1.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

14.2 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6514.2.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

14.3 Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6514.3.1 Boyle’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6514.3.2 Charles’ Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6514.3.3 Amount Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6614.3.4 Pressure Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6614.3.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

14.4 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6614.4.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6614.4.2 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

14.5 Boltzmann Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6714.5.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6714.5.2 A Graph of this Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6814.5.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

15 Magnetic Fields 6915.1 Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

15.1.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6915.2 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

15.2.1 Flux Linkage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6915.2.2 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7015.2.3 Lenz’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7015.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

15.3 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7015.3.1 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7015.3.2 Direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7015.3.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

15.4 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7115.4.1 Ideal Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7215.4.2 Eddy Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7215.4.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

15.5 Motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7215.5.1 Simple DC Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7215.5.2 Three-phase Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7315.5.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

15.6 Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7415.6.1 Moving Coil . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7415.6.2 Moving Magnet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7415.6.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

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16 Electric Fields 7616.1 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

16.1.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7616.2 Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

16.2.1 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7616.2.2 Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7716.2.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

16.3 Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.3.1 Relationship to Electric Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.3.2 Relationship to Electric Field Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.3.3 Equipotentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.3.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

16.4 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.4.1 The Electronvolt . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7816.4.2 Summary of Electric Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7916.4.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

17 Particle Physics 8017.1 The Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

17.1.1 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.1.2 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.1.3 Generations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.1.4 Antiparticles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.1.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

17.2 Quarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.2.1 Generations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.2.2 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.2.3 Hadrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8117.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

17.3 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8217.3.1 Feynman Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8217.3.2 Photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8217.3.3 W and Z Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8217.3.4 Gluons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.3.5 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

17.4 Leptons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.4.1 Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.4.2 Neutrinos . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.4.3 Lepton Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.4.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

17.5 Millikan’s Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8317.5.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

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17.6 Pair Production and Annihilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8417.6.1 Pair Production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8417.6.2 Annihilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8417.6.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

17.7 Particle Accelerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8417.7.1 Linear Accelerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8517.7.2 Cyclotrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8517.7.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

17.8 Cloud Chambers and Mass Spectrometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8517.8.1 Cloud Chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8517.8.2 Mass Spectrometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8617.8.3 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

18 Nuclear Physics 8818.1 Quantum Principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

18.1.1 Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8818.1.2 Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

18.2 Radioactive Emissions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8818.2.1 α Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8918.2.2 β Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8918.2.3 γ Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8918.2.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

18.3 Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8918.3.1 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

18.4 Fission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9018.4.1 Chain Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9018.4.2 Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9018.4.3 Neutron Moderator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9018.4.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

18.5 Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.5.1 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.5.2 Uses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.5.3 Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.5.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

18.6 Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9118.6.1 The Unified Atomic Mass Unit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9218.6.2 Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9218.6.3 The Binding Energy Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9218.6.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

18.7 Risks, Doses and Dose Equivalents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9318.7.1 Risk . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9318.7.2 Absorbed Dose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

x CONTENTS

18.7.3 Dose Equivalent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9318.7.4 Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

19 Appendices 9419.1 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9419.2 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9419.3 Delta - 'difference in' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9419.4 Sigma - 'sum of' . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9519.5 Derivation of Equations for Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . 95

20 Worked Solutions 9620.1 Lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9620.2 Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9620.3 Digital Storage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9720.4 Digital Processing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9720.5 Digitisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9720.6 Signal Frequencies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9820.7 Bandwidth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9820.8 Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9920.9 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9920.10Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9920.11Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9920.12Resistance and Conductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10020.13Internal Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10020.14Potential Dividers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10120.15Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10120.16Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10220.17Semiconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10220.18Stress, Strain & the Young Modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10320.19Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10320.20Polymers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10320.21What is a wave? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10420.22Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10420.23Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10420.24Young’s Slits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10420.25Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10520.26Light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10520.27Electron Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10520.28Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10520.29Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10620.30Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10820.31Forces and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

CONTENTS xi

20.32Exponential Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10820.33Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10920.34Radioactive Decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11020.35Half-lives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11020.36Gravitational Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11120.37Gravitational Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11120.38Gravitational Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11220.39Gravitational Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11220.40Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11320.41Energy in Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11320.42Damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11420.43Conservation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11520.44Forces and Impulse in Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11620.45Rockets, Hoses and Machine Guns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11620.46Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11620.47Radar and Triangulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11720.48Large Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11820.49Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11820.50Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11820.51The Big Bang . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11820.52Heat and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11920.53Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11920.54Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11920.55Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11920.56Boltzmann Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12020.57Magnetic Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12020.58Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12020.59Magnetic Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12120.60Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12220.61Motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12220.62Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12220.63Electric Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12320.64Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12320.65Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12320.66Electric Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12420.67The Standard Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12420.68Quarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12520.69Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12520.70Leptons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12520.71Millikan’s Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12620.72Pair Production and Annihilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

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20.73Particle Accelerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12620.74Cloud Chambers and Mass Spectrometers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12720.75Radioactive Emissions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12720.76Energy Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12820.77Fission . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12820.78Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12920.79Binding Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12920.80Risks, Doses and Dose Equivalents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

21 Text and image sources, contributors, and licenses 13121.1 Text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13121.2 Images . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13721.3 Content license . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Chapter 1

AS

1

Chapter 2

Optics

2.1 Lenses

2.1.1 Curvature of Wavefronts

Light can be viewed as beams travelling between points.However, from most light sources, the light radiates out-wards as a series of wavefronts. Light from a light sourceis bent - wavefronts of light have a property known ascurvature.

Decreasing curvatures of wavefronts

As light travels further away from its source, its curvaturedecreases. Consider a sphere expanding gradually froma point, which represents a given wavefront of light. Asthe sphere expands, the curvature of its surface decreaseswhen we look at any part of the surface with a constantarea. It should be noted at this point that light from asource infinitely far away has 0 curvature - it is straight.This is useful, as ambient light (light from a source that isfar away) can be assumed to have a curvature of 0, as thedifference between this and its actual curvature is negli-gible.The curvature of a wavefront is given as:C = 1

v ,where v is the distance from the wavefront to the in-focusimage depicted by the light. Curvature is measured indioptres (D).

2.1.2 Power of lenses

Calculating the power of a lens

The function of a lens is to increase or decrease the cur-vature of a wavefront. Lenses have a 'power'. This is thecurvature which the lens adds to the wavefront. Power ismeasured in dioptres, D, and is given by the formula:P = 1

f ,where f equals the focal length of the lens. This is thedistance between the lens and the point where an imagewill be in focus, if the wavefronts entering the other sideof the lens are parallel.

2.1.3 The Lens Equation

Overall, then, the formula relating the curvature of thewavefronts leaving a lens to the curvature of the wave-fronts entering it is:1v = 1

u + 1f

where v is the distance between the lens (its centre) andthe in-focus image formed, u is the distance between thelens (its centre) and the object which the in-focus imageis of, and f is the focal length of the lens. The power ofthe lens can be substituted in for the reciprocal of f, asthey are the same thing.

2

2.2. REFRACTION 3

The lens equation, applied to a single pixel.

The Cartesian Convention

If we were to place a diagram of the lens on a grid, la-belled with cartesian co-ordinates, we would discover thatmeasuring the distance of the object distance is negative,in comparison to the image distance. As a result, thevalue for u must always be negative. This is known asthe Cartesian convention.This means that, if light enters the lens with a positivecurvature, it will leave with a negative curvature unlessthe lens is powerful enough to make the light leave with apositive curvature.

2.1.4 Types of Lens

Types of lens

There are two types of lens:Converging lenses add curvature to the wavefronts,causing them to converge more. These have a positivepower, and have a curved surface which is wider in themiddle than at the rim.Diverging lenses remove curvature from the wavefronts,causing them to diverge more. These have a negativepower, and have a curved surface with a dip in the middle.

2.1.5 Magnification

Magnification is a measure of how much an image hasbeen enlarged by a lens. It is given by the formula:M = h2

h1

where h1 and h2 are the heights of the image (or object)before and after being magnified, respectively. If an im-age is shrunk by a lens, the magnification is between 0and 1.Magnification can also be given as:M = v

u

where v and u are the image and object distances. There-fore:M = h2

h1= v

u

An easy way to remember this in the middle of an examis the formula:I = AM

where I is image size, A is actual size of the object M isthe magnification factor.

2.1.6 Questions

1. A lens has a focal length of 10cm. What is its power,in dioptres?2. Light reflected off a cactus 1.5m from a 20D lens formsan image. How many metres is it from the other side ofthe lens?3. A lens in an RGB projector causes an image to focuson a large screen. What sort of lens is it? Is its powerpositive or negative?4. What is the focal length of a 100D lens?5. The film in a camera is 5mm from a lens when au-tomatically focussed on someone’s face, 10m from thecamera. What is the power of the lens?6. The light from a candle is enlarged by a factor of 0.5by a lens, and produces an image of a candle, 0.05m high,on a wall. What is the height of the candle?Worked Solutions

2.2 Refraction

2.2.1 Reflection

Reflection is when light 'bounces’ off a material which isdifferent to the one in which it is travelling. You mayremember from GCSE (or equivalent) level that we cancalculate the direction the light will take if we consider aline known as the 'normal'. The normal is perpendicularto the boundary between the two materials, at the point atwhich the light is reflected. The angle between the normal

4 CHAPTER 2. OPTICS

Angles of reflection and incidence

and the ray of light is known as the angle of reflection (r).The ray of light will be reflected back at the same angle asit arrived at the normal, on the other side of the normal.

2.2.2 Refraction

Refraction is when light changes velocity when it travelsacross the boundary between two materials. This causesit to change direction. The angle between the normal andthe refracted ray of light is known as the angle of refrac-tion (r).

The Refractive Index

The refractive index is a measure of how much light willbe refracted on the boundary between a material and a'reference material'. This reference material is usuallyeither air or a vacuum. It is given by the following for-mula:n = c0

c1

where c0 is the speed of light in a vacuum (3 x 108 m/s)and c1 is the speed of light in the material.

Snell’s Law

We can relate the refractive index to the angles of inci-dence (i) and refraction (r) using the following formula,known as Snell’s Law:n = sin i

sin r = c0c1

2.2.3 Total Internal Reflection

Normally, when light passes through a non-opaque ma-terial, it is both reflected and refracted. However, some-times, rays of light are totally internally reflected; in otherwords, they are not refracted ( completely transmittedthrough that material and no reflection), so no light goesoutside the material. This is useful in optic fibres, which

allow a signal to be transmitted long distances at the speedof light because the light is totally internally reflected.

Critical Angle

The critical angle is the minimum angle of incidence, fora given material, at which rays of light are totally inter-nally reflected. At the critical angle (C), the angle of re-fraction is 90°, as any smaller angle of incidence will re-sult in refraction. Therefore:n = sin 90

sin r

Since sin 90° = 1:n = 1

sin r

sin r = 1n = sinC

In word form, in a material with refractive index n, lightwill be totally internally reflected at angles greater thanthe inverse sine of the reciprocal of the refractive index.

2.2.4 Questions

1. A ray of light is reflected from a mirror. Its angle tothe normal when it reaches the mirror is 70°. What is itsangle of reflection?2. The speed of light in diamond is 1.24 x 108 m/s. Whatis its refractive index?3. The refractive index of ice is 1.31. What is the speedof light in ice?4. A ray of light passes the boundary between air and atransparent material. The angle of refraction is 20°, andthe angle of incidence is 10°. What is the speed of lightin this material? Why is it impossible for this material toexist?5. What is the critical angle of a beam of light leaving atransparent material with a refractive index of 2?Worked Solutions

2.2.5 See also

• Optics/Refraction

Chapter 3

Communications

3.1 Digital Storage

3.1.1 Digital Data

There are two different types of data: analogue and digi-tal. Analogue data can, potentially, take on any value. Ex-amples include a page of handwritten text, a cassette, or apainting. Digital data can only take on a set range of val-ues. This enables it to be processed by a computer. Ex-amples include all files stored on computers, CDs, DVDs,etc.

Pixels

Enlarged image of a computer, showing individual pixels.

Digital images are made up of pixels. A pixel representsthe value of an individual square of the image, and it hasa value assigned to it. The total number of pixels in animage is just like the formula for the area of a rectangle:number of pixels across multiplied by number of pixelsdown. When representing text, each pixel is a componentof one character (for example, a letter, a number, a space,or a new line), it is not the entirety of a character. Forinstance if the letter 'E' was to be taken as an example anda section was to be taken through the three protrusions;a minimum of seven (7) pixels would be used, one whitepixel at the top, then one black (for the first protrusion),then one white for the gap, then a black one for the centre- and so on. A type face - such as Helvetica, or TimesNew Roman, maybe made up of a more complex patternof pixels to allow for serif details.

Bits

Each pixel’s value is digital: it takes on a definite value.In a higher quality image, each pixel can take on a greatervariety of values. Each pixel’s value is encoded as a num-ber of bits. A bit is a datum with a value of either 0 or 1.The more values a pixel can take on, the more bits mustbe used to represent its value. The number of values (N)that a pixel represented by I bits can take on is given bythe formula:N = 2I

Hence:I = logN

log 2 ≈ logN0.301029996 Log base 10 used here. For

ratios, the base of the log does not matter, now we haveevaluated log 2 using base 10 log N must be base 10 aswell.A pixel may be represented by values for red, green andblue, in which case each colour channel will have to beencoded separately. When dealing with text, the numberof values is equal to the number of possible characters.Overall, for an image:Amount of information in an image (bits) = number ofpixels x bits per pixel.

Bytes

A byte is equal to 8 bits. The major difference betweenbytes and SI units is that when prefixes (such as kilo-,mega-, etc.) are attached, we do not multiply by 103 asthe prefix increases. Instead, we multiply by 1024. So,1 kilobyte = 1024 bytes, 1 megabyte = 10242 bytes, 1gigabyte = 10243 bytes, and 1 terabyte = 10244 bytes.

3.1.2 Questions

1. An image transmitted down a SVGA video cable is800 pixels wide, and 600 pixels high. How many pixelsare there in the image?2. A grayscale image is encoded using 3 bits. How manypossible values can each pixel have?3. The characters in a text document are numbered from

5

6 CHAPTER 3. COMMUNICATIONS

0 - 255. Howmany bits should each character be encodedwith?4. A page contains 30 lines of text, with an average of15 characters on each line. Each character is representedby 4 bits. How many megabytes of uncompressed stor-age will a book consisting of 650 pages like this fill on acomputer’s hard disk?5. A 10cmwide square image is scanned into a computer.Each pixel is encoded using 3 channels (red, green andblue), and each channel can take on 256 possible values.One pixel is 0.01 mm wide. How much information doesthe scanned image contain? Express your answer usingan appropriate unit.Worked Solutions

3.2 Digital Processing

As we have already seen, a digital image consists of pix-els, with each pixel having a value which represents itscolour. For the purposes of understanding how digitalimages are manipulated, we are going to consider an 8-bit grayscale image, with pixel values ranging from 0 to255, giving us 256 (28) levels of grey. 0 represents white,and 255 represents black. This is the image we are goingto consider:The image consists of an edge, and some random noise.There are two methods of smoothing this image (i.e. re-moving noise) that you need to know about:

3.2.1 Mean Smoothing

In order to attempt to remove noise, we can take the meanaverage of all the pixels surrounding each pixel (and thepixel itself) as the value of the pixel in the smoothed im-age, as follows:This does remove the noise, but it blurs the image.

3.2.2 Median Smoothing

A far better method is, instead of taking the mean, to takethe median, as follows:For this image, this gives a perfect result. In more com-plicated images, however, data will still be lost, although,in general, less data will be lost by taking the median thanby taking the mean.

3.2.3 Edge Detection

We can detect the positioning of edges in an image usingthe 'Laplace rule', or 'Laplace kernel'. For each pixel inthe image, wemultiply its value by 4, and then subtract thevalues of the pixels above and below it, and on either side

of it. If the result is negative, we treat it as 0. So, takingthe median-smoothed image above, edge detection givesthe following result:

3.2.4 Questions

1. How could the above methods be applied to a digitalsound sample?2. Which of the above methods would be suitable forsmoothing sharp edges? Why?3. Use median smoothing to remove noise from the fol-lowing image of a white cat in a snowstorm (the blackpixels have a value of 255):4. Why would mean sampling not be appropriate forsmoothing the image given in question 3?5. Use mean smoothing to remove noise from the follow-ing image of a black cat in a coal cellar:Worked Solutions

3.3 Digitisation

Digitisation of a signal is the process by which an ana-logue signal is converted to a digital signal.

3.3.1 Digitisation & Reconstruction

Let us consider the voltage output from a microphone.The signal which enters themicrophone (sound) is an ana-logue signal - it can be any of a potentially infinite rangeof values, and may look something like this waveform(from an artificial (MIDI) piano):

When the microphone converts this signal to an electri-cal signal, it samples the signal a number of times, andtransmits the level of the signal at that point. The follow-ing diagram shows sample times (vertical black lines) andthe transmitted signal (the red line):

When we wish to listen to the sound, the digital signalhas to be reconstructed. The gaps between the samplesare filled in, but, as you can see, the reconstructed signalis not the same as the original sound:

3.4. SIGNAL FREQUENCIES 7

3.3.2 Sampling Rate

The sampling rate when digitising an analogue signal isdefined as the number of samples per. second, and ismeasured in Hertz (Hz), as it is a frequency. You cancalculate the sampling rate using the formula:

Sampling Rate (Hz) = No. of samplesNo. of seconds

The higher the sampling rate, the closer the reconstructedsignal is to the original signal, but, unfortunately, we arelimited by the bandwidth available. Theoretically, a sam-pling rate of twice the highest frequency of the originalsignal will result in a perfect reconstructed signal. In theexample given above, the sampling rate is far too low,hence the loss of information.

3.3.3 Number of Levels

Another factor which may limit the quality of the recon-structed signal is the number of bits with which the signalis encoded. For example, if we use 3 bits per. sample,we only have 8 (23) levels, so, when sampling, we musttake the nearest value represented by one of these levels.This leads to quantization errors - when a sample doesnot equal the value of the original signal at a given sam-ple point.

3.3.4 Questions

1. Take samples for the signal below every 0.1ms, andthen produce a reconstructed signal. How does it differfrom the original?

2. A signal is sampled for 5 seconds at a sampling rate of20 kHz. How many samples were taken?3. Most sounds created by human speech except for 'ss’and 'ff' have a maximum frequency of 4 kHz. What is asuitable sampling rate for a low-quality telephone?4. Using a sampling rate of 20 kHz and 3 bits, sample thefollowing signal, and then produce a reconstructed signal.What is the maximum frequency that can be perfectly re-constructed using this sampling rate?

Worked Solutions

3.4 Signal Frequencies

The frequency of a wave describes how many waves gopast a certain point in one second. Frequency is measuredin Hertz (usually abbreviated Hz), and can be calculatedusing the formula:V = fλwhere V is the velocity of the wave (in ms−1), f is thefrequency of the wave (in Hz), and λ (the Greek letterlambda) is the wavelength of the wave (distance from onepeak / trough to the next, in m).

3.4.1 Multiple Frequencies

Let us consider the following signal (time is in ms, andthe y-axis represents volts):

8 CHAPTER 3. COMMUNICATIONS

+1 +5 +10 x

-5

-1

+1

+5

This signal is constructed from a number of different sinewaves, with different frequencies, added together. Thesesine waves are as follows:

+1 +5 +10 x

-5

-1

+1

+5

3.4.2 Frequency Spectra

Each of these sine waves has a different frequency. Youcan see this, as they have different distances betweentheir peaks and troughs. These frequencies can be plot-ted against the amplitude of the wave, as in the table, andchart drawn from it, below:

3.5. BANDWIDTH 9

This chart is known as the frequency spectrum of a signal.

3.4.3 Fundamental Frequency

The fundamental freqency is the lowest frequency thatmakes up a signal. In the above example, the fundamen-tal frequency is 80 Hz. It is always the frequency farthestto the left of a frequency spectrum, ignoring noise. Otherfrequencies are known as overtones, or harmonics.

3.4.4 Questions

1. What is the frequency of an X-ray (wavelength0.5nm)?2. A sound wave, with a frequency of 44 kHz, has a wave-length of 7.7mm. What is the speed of sound?3. What is the fundamental frequency of the followingsignal?

4. Approximately how many harmonics does it contain?5. The three sine waves sin x°, 4sin(2x-50)° and0.5sin(3x+120)° are added together to form a signal.What are the frequencies of each of the waves? Whatis the signal’s fundamental frequency? Assume that thewaves are travelling at the speed of light, and that 60° =1mm.Worked Solutions

3.5 Bandwidth

Bandwidth is the frequency of a signal. Although originalsignals have varying frequencies, when these are trans-mitted, for example, as FM radio waves, they are mod-ulated so that they only use frequencies within a certainrange. FM radio modulates the frequency of a wave, soit needs some variation in the frequencies to allow fortransmission of multiple frequencies. Since bandwidthis a frequency, it is the number of bits per. second. Thebandwidth required to transmit a signal accurately can becalculated by using 1 as the number of bits, giving theformula:B = 1

t

where B is bandwidth (in Hz), and t is the time taken totransmit 1 bit of data (in s).The bandwidth of a signal regulates the bit rate of thesignal, as, with a higher frequency, more information canbe transmitted. This give us the formula (similar to theformula for lossless digital sampling):b = 2Bwhere b is the bit rate (in bits per. second), and B is the

10 CHAPTER 3. COMMUNICATIONS

bandwidth (in Hz).

3.5.1 Questions

1. A broadband internet connection has a bit rate of8Mbit s−1 when downloading information. What is theminimum bandwidth required to carry this bit rate?2. The same connection has a bandwidth of 100 kHz re-served for uploading information. What is the maximumbit rate that can be attained when uploading informationusing this connection?3. A lighthouse uses a flashing light and Morse Code tocommunicate with a nearby shore. A 'dash' consists of thelight being on for 2s. The light is left off for 1s betweendots and dashes. What is the bandwidth of the connec-tion?4. The broadband connection in question two is used toupload a 1Mbyte image to a website. How long does ittake to do this?Worked Solutions

Chapter 4

Electricity

4.1 Charge

Electrons, likemany other particles, have a charge. Whilesome particles have a positive charge, electrons have anegative charge. The charge on an electron is equalto approximately −1.6 x 10−19 coulombs. Coulombs(commonly abbreviated C) are the unit of charge. Onecoulomb is defined as the electric charge carried by 1 am-pere (amp) of current in 1 second. It is normal to ignorethe negative nature of this charge when considering elec-tricity.If we have n particles with the same charge Q ₐᵣ ᵢ ₑ, thenthe total charge Q ₒ ₐ is given by:Q ₒ ₐ = n Q ₐᵣ ᵢ ₑBy a simple rearrangement:n = Qtotal

Qparticle

4.1.1 Questions

1. How much charge do 1234 electrons carry?2. How many electrons does it take to carry 5 C ofcharge?3. The total charge on 1 mole of electrons (6 x 1023 parti-cles) is equal to 1 faraday of charge. Howmany coulombsof charge are equal to 1 faraday?4. The mass of a ball is 50mg. It is supplied 5C of charge.Will there be any change in the mass of the ball? If so,calculate the change of the mass.Worked Solutions

4.2 Current

Current is the amount of charge (on particles such as elec-trons) flowing through part of an electric circuit per sec-ond. Current is measured in amperes (usually abbreviatedA), where 1 ampere is 1 coulomb of charge per second.The formula for current is:I = ∆Q

∆t ([The triangle (Greek letter delta) means changein the quantity])

i1 + i4 = i2 + i3

where I is current (in A), Q is charge (in C) and t is thetime it took for the charge to flow (in seconds).In a series circuit, the current is the same everywhere inthe circuit, as the rate of flow of charged particles is con-stant throughout the circuit. In a parallel circuit, however,the current is split between the branches of the circuit, asthe number of charged particles flowing cannot change.This is Kirchoff’s First Law, stating that:In mathematical form:∑

Iin =∑

Iout (The character that resembles a side-ways M is the Greek letter sigma, meaning 'sum of'.)

4.2.1 Questions

1. 10 coulombs flow past a point in a wire in 1 minute.How much current is flowing through the point?2. How long does it take for a 2A current to carry 5C?3. In the diagram on the left, I = 9A, and I1 = 4.5A.Whatis the current at I2?4. What would I equal if I1 = 10A and I2 = 15A?

11

12 CHAPTER 4. ELECTRICITY

1R

1I 2I

2R

I

5. In the diagram on the left, in 5 seconds, 5C of chargedparticles flow past I1, and 6.7C flow past I2. How longdoes it take for 10C to flow past I?Worked Solutions

4.3 Voltage

Charge moves through a circuit, losing potential energy asit goes. This means that the charge travels as an electriccurrent. Voltage is defined as the difference in potentialenergy per. unit charge, i.e.V = E

Q

where V is voltage (in V), E is the difference in potentialenergy (in joules) and Q is charge (in coulombs).There are two electrical properties which are both mea-sured in volts (commonly abbreviated V), and so both areknown under the somewhat vague title of 'voltage'. Bothare so called because they change the potential energy ofthe charge.

4.3.1 Electromotive Force (EMF)

Keep in mind, that EMF as the name suggests is notan electrical force, it is basically the potential differenceacross the terminals when the key is open i.e. when nocurrent is drawn from the cell. EMF is named so by thescientists who performed faulty experiments and namedit so, hence, just a tribute to their contribution to physicsit is still called EMF but the definition has changed withtime.

4.3.2 Potential Difference

As charge travels around a circuit, each coulomb ofcharge has less potential energy, so the voltage (relativeto the power source) decreases. The difference betweenthe voltage at two points in a circuit is known as potentialdifference, and can be measured with a voltmeter.

Series Circuits

In a series circuit, the total voltage (EMF) is dividedacross the components, as each component causes thevoltage to decrease, so each one has a potential differ-ence. The sum of the potential differences across all thecomponents is equal to the potential difference (but bat-teries have their own 'internal resistances’, which compli-cates things slightly, as we will see).

Parallel Circuits

In a parallel circuit, the potential difference across eachbranch of the circuit is equal to the EMF, as the same'force' is pushing along each path of the circuit. Thenumber of charge carriers (current) differs, but the 'force'pushing them (voltage) does not.

4.3.3 Questions

1. A battery has an EMF of 5V.What is the total potentialdifference across all the components in the circuit?2. The voltages (relative to the voltage of the battery) oneither side of a resistor are −6V and −5V. What is thepotential difference across the resistor?3. At a given point in a circuit, 5C of charge have 10 kJof potential energy. What is the voltage at this point?4. Why do the electrons move to a point 1cm furtheralong the wire?Worked Solutions

4.4 Power

Power is a measure of how much potential energy is dis-sipated (i.e. converted into heat, light and other forms ofenergy) by a component or circuit in one second. This isdue to a drop in the potential energy, and so the voltage,of charge. Power is measured in Watts (commonly ab-breviated W), where 1 W is 1 Js−1. It can be calculatedby finding the product of the current flowing through acomponent / circuit and the potential difference acrossthe component / circuit. This gives us the equation:P = E

t = IV

4.5. RESISTANCE AND CONDUCTANCE 13

where P is the power dissipated (in W), E is the drop inpotential energy (in Joules, J), t is the time taken (in s), Iis the current (in A) and V is either potential difference orelectromotive force (in V), depending on the componentbeing measured.Since power is the amount of energy changing form per.second, the amount of energy being given out each secondwill equal the power of the component giving out energy.You should be able to substitute in values for I and V fromother formulae (V=IR, Q=It) in order to relate power toresistance, conductance, charge and time, giving formu-lae like these:P = I2R

P = V 2

R

P = QVt

4.4.1 Questions

1. The potential difference across a 9W light bulb is240V. How much current is flowing through the lightbulb?2. How much energy is dissipated by a 10W componentin 1 hour?3. The potential difference across a top-notch kettle,which can hold up to 1 litre of water, is 240V, and thecurrent is 12.5 A. 4.2 kJ of energy is required to raise thetemperature of 1kg of water by 1°C. Assuming 100% ef-ficiency and that the temperature has to be raised 80°C(20°C to 100°C), how long does it take to boil 1 litre ofwater?4. How much energy is dissipated by a 100Ω resistor in10 seconds if 2A of current are flowing?5. The charge on an electron is−1.6 x 10−19 C. How longdoes it take for a mole (6 x 1023 particles) of electrons toflow through a 40W light bulb on a 240V ring main?Worked Solutions

4.5 Resistance and Conductance

Conductance is a measure of how well an artefact (suchas an electrical component, not a material, such as iron)carries an electric current. Resistance is a measure of howwell an artefact resists an electric current.Resistance is measured in Ohms (usually abbreviated us-ing the Greek letter Omega, Ω) and, in formulae, is rep-resented by the letter R. Conductance is measured inSiemens (usually abbreviated S) and, in formulae, is rep-resented by the letter G.Resistance and conductance are each other’s reciprocals,so:

R = 1G and G = 1

R

4.5.1 Ohm’s Law

Ohm’s Law states that the potential difference across anartefact constructed from Ohmic conductors (i.e. con-ductors that obey Ohm’s Law) is equal to the product ofthe current running through the component and the resis-tance of the component. As a formula:V = IRwhere V is potential difference (in V), I is current (in A)and R is resistance (in Ω).

In terms of Resistance

This formula can be rearranged to give a formula whichcan be used to calculate the resistance of an artefact:R = V

I

In terms of Conductance

Since conductance is the reciprocal of resistance, we candeduce a formula for conductance (G):1G = V

I

G = IV

The Relationship between Potential Difference andCurrent

From Ohm’s Law, we can see that potential differenceis directly proportional to current, provided resistance isconstant. This is because two variables (let us call them xand y) are considered directly proportional to one anotherif:y = kx

where k is any positive constant. Since we are assumingthat resistance is constant, R can equal k, so V=RI statesthat potential difference is directly proportional to cur-rent. As a result, if potential difference is plotted againstcurrent on a graph, it will give a straight line with a posi-tive gradient which passes through the origin. The gradi-ent will equal the resistance.

4.5.2 In Series Circuits

In a series circuit (for example, a row of resistors con-nected to each other), the resistances of the resistors addup to give the total resistance. Since conductance is thereciprocal of resistance, the reciprocals of the conduc-tances add up to give the reciprocal of the total conduc-tance. So:

14 CHAPTER 4. ELECTRICITY

ΣR = R1 +R2 + ...+Rn

Σ 1G = 1

G1+ 1

G2+ ...+ 1

Gn

4.5.3 In Parallel Circuits

In a parallel circuit, the conductances of the componentson each branch add up to give the total conductance. Sim-ilar to series circuits, the reciprocals of the total resis-tances of each branch add up to give the reciprocal of thetotal resistance of the circuit. So:ΣG = G1 +G2 + ...+Gn

Σ 1R = 1

R1+ 1

R2+ ...+ 1

Rn

When considering circuits which are a combination of se-ries and parallel circuits, consider each branch as a sepa-rate component, and work out its total resistance or con-ductance before finishing the process as normal.

4.5.4 Questions

1. The potential difference across a resistor is 4V, and thecurrent is 10A. What is the resistance of the resistor?2. What is the conductance of this resistor?3. A conductor has a conductance of 2S, and the potentialdifference across it is 0.5V. How much current is flowingthrough it?4. A graph is drawn of potential difference across anOhmic conductor, and current. For every 3cm across,the graph rises by 2cm. What is the conductance of theconductor?5. On another graph of potential difference and current,the graph curves so that the gradient increases as currentincreases. What can you say about the resistor?6. 3 resistors, wired in series, have resistances of 1kΩ,5kΩ and 500Ω each. What is the total resistance acrossall three resistors?7. 2 conductors, wired in parallel, have conductances of10S and 5S. What is the total resistance of both branchesof the parallel circuit?8. The circuit above is attached in series to 1 10Ω resistor.What is the total conductance of the circuit now?Worked Solutions

4.6 Internal Resistance

Batteries, just like other components in an electric circuit,have a resistance. This resistance is known as internal re-sistance. This means that applying Ohm’s law (V = IR) tocircuits is more complex than simply feeding the correctvalues for V, I or R into the formula.

The existence of internal resistance is indicated by mea-suring the potential difference across a battery. This isalways less than the EMF of the battery. This is becauseof the internal resistance of the battery. This idea givesus the following formula:

PD across battery = EMF of battery - voltage to be ac-counted forLet us replace these values with letters to give the simplerformula:Vₑₓ ₑᵣ ₐ = E - Vᵢ ₑᵣ ₐSince V = IR:Vₑₓ ₑᵣ ₐ = E - IRᵢ ₑᵣ ₐYou may also need to use the following formula to workout the external potential difference, if you are not givenit:Vₑₓ ₑᵣ ₐ = IΣRₑₓ ₑᵣ ₐYou should also remember the effects of using resistorsin both series and parallel circuits.

4.6.1 Questions

1. A 9V battery is short-circuited. The potential differ-ence across the battery is found to be 8V, and the currentis 5A. What is the internal resistance of the battery?2. What is the EMFof the battery in the following circuit?

3. What is the internal resistance of the battery in the

4.7. POTENTIAL DIVIDERS 15

following circuit?

Worked Solutions

4.7 Potential Dividers

Circuit symbols for a potential divider

A potential divider, or potentiometer, consists of a num-ber of resistors, and a voltmeter. The voltage read by thevoltmeter is determined by the ratio of the resistances oneither side of the point at which one end of the voltmeteris connected.To understand how a potential divider works, let us con-sider resistors in series. The resistances add up, so, in acircuit with two resistors:ΣR = R1 +R2

If we apply Ohm’s law, remembering that the current isconstant throughout a series circuit:∑

VI = V1

I + V2

I

Multiply by current (I):ΣV = V1 + V2

So, just as the resistances in series add up to the totalresistance, the potential differences add up to the total

potential difference. The ratios between the resistancesare equal to the ratios between the potential differences.In other words, we can calculate the potential differenceacross a resistor using the formula:Vresistor =

∑V × Rresistor∑

Rexternal

In many cases, you will be told to assume that the internalresistance of the power source is negligible, meaning thatyou can take the total potential difference as the EMF ofthe power source.A potential divider may work by combining a variableresistor such as an LDR or thermistor with a constantresistor, as in the diagram below. As the resistance ofthe variable resistor changes, the ratio between the re-sistances changes, so the potential difference across anygiven resistor changes.

Alternatively, a potential divider may be made of manyresistors. A 'wiper' may move across them, varying thenumber of resistors on either side of the wiper as it moves,as in the following diagram:

4.7.1 Questions

1. A 12 kΩ resistor and a 20 kΩ resistor are connected toa 9V battery. A voltmeter is connected across the 12kΩresistor. What is the reading on the voltmeter? (Assumenegligible internal resistance.)2. A potential divider consists of 100 5Ω resistors, with awiper which moves on one resistor for every 3.6° a handle

16 CHAPTER 4. ELECTRICITY

connected to it turns. The wiper is connected to a volt-meter, and the circuit is powered by a 120V power sourcewith negligible internal resistance. What is the reading onthe voltmeter when the handle turns 120°?3. A 9V battery with internal resistance 0.8Ω is con-nected to 3 resistors with conductances of 3, 2 and 1Siemens. A voltmeter is connected across the 3 and 2Siemens resistors. An ammeter is placed in the circuit,between the battery and the first terminal of the volt-meter, and reads 2A. What is the reading on the volt-meter?Worked Solutions

4.8 Sensors

A sensor is a device which converts a physical propertyinto an electrical property (such as resistance). A sensingsystem is a system (usually a circuit) which allows thiselectrical property, and so the physical property, to bemeasured.

4.8.1 Temperature Sensor

Use of a potential divider and thermistor to measure temperature

A common example of a sensing system is a tempera-ture sensor in a thermostat, which uses a thermistor. Inthe most common type of thermistor (an NTC), the resis-tance decreases as the temperature increases. This effectis achieved by making the thermistor out of a semicon-ductor. The thermistor is then used in a potential divider,as in the diagram on the right. In this diagram, the po-tential difference is divided between the resistor and thethermistor. As the temperature rises, the resistance of thethermistor decreases, so the potential difference across itdecreases. This means that potential difference across theresistor increases as temperature increases. This is whythe voltmeter is across the resistor, not the thermistor.

4.8.2 Properties

There are three main properties of sensing systems youneed to know about:

Sensitivity

This is the amount of change in voltage output per unitchange in input (the physical property). For example, inthe above sensing system, if the voltage on the voltmeterincreased by 10V as the temperature increased by 6.3°C:S = 10

6.3 ≈ 1.59 V/°C

Resolution

This is the smallest change in the physical property de-tectable by the sensing system. Sometimes, the limitingfactor is the number of decimal places the voltmeter candisplay. So if, for example, the voltmeter can display thevoltage to 2 decimal places, the smallest visible change involtage is 0.01V. We can then use the sensitivity of thesensor to calculate the resolution.S = 1.59 = 0.01

R

R = 0.011.59 ≈ 0.006 °C

Response Time

This is the time the sensing system takes to display achange in the physical property it is measuring. It is oftendifficult to measure.

4.8.3 Signal Amplification

Sometimes, a sensing system gives a difference in outputvoltage, but the sensitivity is far too low to be of any use.There are two solutions to this problem, which can beused together:

Amplification

An amplifier can be placed in the system, increasing thesignal. The main problem with this is that the signal can-not exceed the maximum voltage of the system, so valueswill be chopped off of the top and bottom of the signalbecause it is so high.

Wheatstone Bridge

This solution is far better, especially when used prior toamplification. Instead of using just one pair of resistors, asecond pair is used, and the potential difference betweenthe two pairs (which are connected in parallel) is mea-sured. This means that, if, at the sensing resistor (e.g.

4.9. RESISTIVITY AND CONDUCTIVITY 17

A wheatstone bridge, using a thermistor

thermistor / LDR) the resistance is at its maximum, a sig-nal of 0V is produced. This means that the extremes ofthe signal are not chopped off, making for a much bettersensor.

4.8.4 Questions

An LDR’s resistance decreases from a maximum resis-tance of 2kΩ to a minimum resistance of 0Ω as light in-tensity increases. It is used in a distance sensing systemwhich consists of a 9V power supply, a 1.6 kΩ resistor,the LDR and a multimeter which displays voltage to 2decimal places measuring the potential difference acrossone of the two resistors.1. Across which resistor should the multimeter be con-nected in order to ensure that, as the distance from thelight source to the sensor increases, the potential differ-ence recorded increases?2. In complete darkness, what voltage is recorded on themultimeter?3. When a light source moves 0.5m away from the sensor,the voltage on the multimeter increases by 2V. What isthe sensitivity of the sensing system when using this lightsource, in V m−1?4. When the same light source is placed 0m from the sen-sor, the potential difference is 0V. When the light sourceis 1m away, what voltage is displayed on the multimeter?5. What is the resolution of the sensing system?6. Draw a circuit diagram showing a similar sensing sys-tem to this, using a Wheatstone bridge and amplifier toimprove the sensitivity of the system.7. What is the maximum potential difference that canreach the amplifier using this new system (ignore the am-plification)?8. If this signal were to be amplified 3 times, would itexceed the maximum voltage of the system? What wouldthe limits on the signal be?Worked Solutions

4.9 Resistivity and Conductivity

Resistivity and conductivity are material properties: theyapply to all examples of a certain material anywhere.They are not the same as resistance and conductance,which are properties of individual artefacts. This meansthat resistivity and conductivity only apply to a given ob-ject. They describe how well a material resists or con-ducts an electric current.

4.9.1 Symbols and Units

Resistivity is usually represented by the Greek letter rho(ρ), and is measured in Ω m. Conductivity is usually rep-resented by the Greek letter sigma (σ), and is measuredin S m−1.

4.9.2 Formulae

The formula relating resistivity (ρ) to resistance (R),cross-sectional area (A) and length (L) is:ρ = RA

L

Conductivity is the reciprocal of resistivity, just as con-ductance (G) is the reciprocal of resistance. Hence:1σ =

1G×A

L = 1G × A

L = AGL

σ = GLA

You should be able to rearrange these two formulae to beable to work out resistance, conductance, cross-sectionalarea and length. For example, it all makes a lot moresense if we write the first formula in terms of ρ, A and L:R = ρL

A

From this, we can see that the resistance of a lump ofmaterial is higher if it has a higher resistivity, or if it islonger. Also, if it has a larger cross-sectional area, itsresistance is smaller.

4.9.3 Questions

1. The substation of the bar bus is made up of 2-inchesround copper bars 20ft long.what is the resistance of eachbar if resistivity is1.724×10?2. A pure copper wire has a radius of 0.5mm, a resistanceof 1 MΩ, and is 4680 km long. What is the resistivity ofcopper?3. Gold has a conductivity of 45 MS m−1. What is theresistance of a 0.01m across gold connector, 0.05m long?4. A strand of metal is stretched to twice its originallength. What is its new resistance? State your assump-tions.5. Which has the greater resistivity: a plank or a piece ofsawdust, made from the same wood?

18 CHAPTER 4. ELECTRICITY

Worked Solutions

4.10 Semiconductors

P

Si Si

Si

Si

SiSi

Si

Si

SiSiSi

SiSiSi

SiSi

SiSiSie-

Silicon, doped with phosphorous

A semiconductor has a conductivity between that of aconductor and an insulator. They are less conductive thanmetals, but differ from metals in that, as a semiconductorheats up, its conductivity rises. In metals, the oppositeeffect occurs.The reason for this is that, in a semiconductor, very fewatoms are ionised, and so very few electrons can move,creating an electric current. However, as the semicon-ductor heats up, the covalent bonds (atoms sharing elec-trons, causing the electrons to be relatively immobile)break down, freeing the electrons. As a result, a semicon-ductor’s conductivity rises at an increasing rate as temper-ature rises.Examples of semiconductors include silicon and germa-nium. A full list of semiconductor materials is availableat Wikipedia. At room temperature, silicon has a con-ductivity of about 435 μS m−1.Semiconductors are usually 'doped'. This means that ionsare added in small quantities, giving the semiconductora greater or lesser number of free electrons as required.This is controlled by the charge on the ions.

4.10.1 Questions

1. What is the resistivity of silicon, at room temperature?2. What sort of variable resistor would a semiconductorbe useful in?3. If positive ions are added to silicon (doping it), howdoes its conductivity change?Worked Solutions

4.10.2 See Also

• The book on Semiconductors.

Chapter 5

Material Structure

5.1 Stress, Strain & the YoungModulus

5.1.1 Stress

Stress is a measure of the internal force an object is expe-riencing per unit cross sectional area. Hence, the formulafor calculating stress is the same as the formula for calcu-lating pressure: butσ = F

A

where σ is stress (in Newtons per square metre or, equiva-lently, Pascals). F is force (in Newtons, commonly abbre-viated N), and A is the cross sectional area of the sample.

Tensile Strength

The (ultimate) tensile strength is the level of stress atwhich a material will fracture. Tensile strength is alsoknown as fracture stress. If a material fractures by 'crackpropagation' (i.e., it shatters), the material is brittle.

Yield Stress

On a stress strain graph beyond the yield point (or elas-tic limit) the material will no longer return to its originallength. This means it has become permanently deformed.Therefore the yield stress is the level of stress at which amaterial will deform permanently. This is also known asyield strength.

5.1.2 Strain

Stresses lead to strain (or deformation). Putting pressureon an object causes it to stretch. Strain is a measure ofhow much an object is being stretched. The formula forstrain is:ϵ = ∆l

l0= l−l0

l0= l

l0− 1 ,

where l0 is the original length of a bar being stretched, andl is its length after it has been stretched. Δl is the extensionof the bar, the difference between these two lengths.

5.1.3 Young’s Modulus

Young’s Modulus is a measure of the stiffness of a mate-rial. It states how much a material will stretch (i.e., howmuch strain it will undergo) as a result of a given amountof stress. The formula for calculating it is:E = σ

ϵ

The values for stress and strain must be taken at as lowa stress level as possible, provided a difference in thelength of the sample can bemeasured. Strain is unitless soYoung’s Modulus has the same units as stress, i.e. N/m²or Pa.

5.1.4 Stress-Strain Graphs

Stress–strain curve for low-carbon steel.

Stress (σ) can be graphed against strain (ε). The tough-ness of a material (i.e., how much it resists stress, in Jm−3) is equal to the area under the curve, between they-axis and the fracture point. Graphs such as the one onthe right show how stress affects a material. This imageshows the stress-strain graph for low-carbon steel. It hasthree main features:

19

20 CHAPTER 5. MATERIAL STRUCTURE

Elastic Region

In this region (between the origin and point 2), the ratioof stress to strain (Young’s modulus) is constant, meaningthat the material is obeying Hooke’s law, which states thata material is elastic (it will return to its original shape) ifforce is directly proportional to extension of the material

Hooke’s Law

Hooke’s law of elasticity is an approximation that statesthat the Force (load) is in direct proportion with the ex-tension of a material as long as this load does not exceedthe proportional limit. Materials for which Hooke’s lawis a useful approximation are known as linear-elastic

F = −kx

The relation is often denoted

F ∝ x

The work done to stretch a wire or the Elastic Poten-tial Energy is equal to the area of the triangle on a Ten-sion/Extension graph, but can also be expressed as

12kx

2

Plastic Region

In this region (between points 2 and 3), the rate at whichextension is increasing is going up, and the material haspassed the elastic limit. It will no longer return to its orig-inal shape. After point 1, the amount of stress decreasesdue to necking at one point in the specimen. If the stresswas recorded where the necking occurs we would observean upward curve and an increase in stress due to this re-duction in area(stress = Force / area, thus stress increasesduring necking). The material will now 'give' and extendmore under less force.

Fracture Point

At point 3, the material has been fractured.

Other Typical Graphs

In a brittle material, such as glass or ceramics, the stress-strain graph will have an extremely short elastic region,and then will fracture. There is no plastic region on thestress-strain graph of a brittle material.

5.1.5 Questions

1. 100N of force are exerted on a wire with cross-sectional area 0.50mm2. How much stress is beingexerted on the wire?

2. Another wire has a tensile strength of 70MPa, andbreaks under 100N of force. What is the cross-sectional area of the wire just before breaking?

3. What is the strain on a Twix bar (original length10cm) if it is now 12cm long?

4. What is this strain, expressed as a percentage?

5. 50N are applied to a wire with a radius of 1mm. Thewire was 0.7m long, but is now 0.75m long. Whatis the Young’s Modulus for the material the wire ismade of?

6. Glass, a brittle material, fractures at a strain of 0.004and a stress of 240 MPa. Sketch the stress-straingraph for glass.

7. (Extra nasty question which you won't ever get in anexam) What is the toughness of glass?

8. Wire has a tensile strength of 0.95Mpa, and breaksunder 25N of force. what is the cross-sectional areaof the wire before and after breaking?

Worked Solutions

5.2 Metals

Metals are constructed from positive ions in a sea of electrons.This explains many of their properties.

There are several physical properties of metals you needto know about:

5.3. POLYMERS 21

Electrical Conductivity

Metals consist of positive metal ions in a 'soup' or 'sea' offree (delocalized) electrons. This means that the electronsare free to move through the metal, conducting an electriccurrent.

Stiffness

The electrostatic forces of attraction between the nega-tively charged electrons and the positively charged ionsholds the ions together, making metals stiff. Stiffness isindicated by the Young modulus of the material, so a stiffmaterial deforms little for a given tensile stress.

Ductility

Since there are no permanent bonds between the ions,they can move about and slide past each other. Thismakes metals ductile.

Toughness

Metals are tough for the same reason as they are duc-tile: the positive ions can slide past each other while stillremaining together. So, instead of breaking apart, theychange shape, resulting in increased toughness. This ef-fect is called plasticity. When a tough material breaks theratio of 'energy used / new surface area created' is verylarge.

Elasticity

When a metal is stretched, it can return to its originalshape because the sea of electrons which bonds the ionstogether can be stretched as well.

Brittle

The opposite of tough: a material is likely to crack orshatter upon impact or force. It will snap cleanly due todefects and cracks.

Malleability

Metals are malleable because their atoms are arranged inflat planes that can slide past each other. Their bonds arenon-directional. Metals also contain dislocations whichmean that ions in the structure can be moved unilaterallyrather than as a whole layer, which takes less energy todo.

Transformation

Diffusive transformation: occur when the planes of atomsin the material move past each other due to the stresses onthe object. This transformation is permanent and cannotbe recovered from due to energy being absorbed by thestructureDiffusionless transformation: occurs where the bonds be-tween the atoms stretch, allowing the material to de-form elastically. An example would be rubber or a shapememory metal/alloy (often referred to as SMA) such asa nickel-titanium alloy. In the shape memory alloy thetransformation occurs via the change of phase of the in-ternal structure frommartensitic to deformedmartensitic,which allows the SMA to have a high percentage strain(up to 8% for some SMA’s in comparison to approxi-mately 0.5% for steel). If the material is then heatedabove a certain temperature the deformed martensite willform austenite, which returns to twinned martensite aftercooling.

5.2.1 Questions

1. Would you expect a metal to have more or less con-ductivity than a semiconductor? Why?2. How can the stress-strain graph for a metal be ex-plained in terms of ions in a sea of electrons?3. As a metal heats up, what happens to its conductivity?Why?Worked Solutions

5.3 Polymers

A simple polymer consists of a long chain of monomers(components of molecules) joined by covalent bonds. Apolymer usually consists of many of these bonds, tangledup. This is known as a bulk polymer.

5.3.1 Types

A bulk polymer may contain two types of regions. Incrystalline regions, the chains run parallel to each other,whereas in amorphous regions, they do not. Intermolecu-lar bonds are stronger in crystalline regions. A polycrys-talline polymer consists of multiple regions, in which thechains point in a different direction in each region.

5.3.2 Properties

Transparency

Polymers which are crystalline are usually opaque ortranslucent. As a polymer becomes less polycrystalline, it

22 CHAPTER 5. MATERIAL STRUCTURE

Polycrystalline glass

Amorphous rubber

becomes more transparent, whilst an amorphous polymeris usually transparent. [1]

Elasticity

In some polymers, such as polyethene, the chains arefolded up. When they are stretched, the chains unravel,stretching without breaking. When the stress ceases, theywill return to their original shape. If, however, the bondsbetween the molecules are broken, the material reachesits elastic limit and will not return to its original shape.

Stiffness

Polymer chains may be linked together, causing the poly-mer to become stiffer. An example is rubber, which,

when heated with sulfur, undergoes a process known asvulcanization. The chains in the rubber become joinedby sulfur atoms, making the rubber suitable for use in cartyres. A stiffer polymer, however, will usually be morebrittle.

Plasticity

When a polymer is stretched, the chains become paral-lel, and amorphous areas may become crystalline. Thiscauses an apparent change in colour, and a process knownas 'necking'. This is when the chains recede out of an areaof the substance, making it thinner, with fatter areas oneither side.

Conductivity

Polymers consist of covalent bonds, so the electrons arenot free to move according to potential difference. Thismeans that polymers are poor conductors.

Boiling Point

Polymers do not have boiling points. This is because, be-fore they reach a theoretical boiling point, polymers de-compose. Some polymers do not have melting points forthe same reason.

5.3.3 Questions

1. Different crystalline structures have different refrac-tive indexes. Why does this mean that a polycrystallinepolymer is translucent?2. What sort of polymer is a pane of perspex?3. What sort of polymer does the pane of perspex becomewhen shattered (but still in one piece)?4. What sort of polymer is a rubber on the end of a pencil?5. What happens to the translucency of an amorphouspolymer when it is put under stress?Worked Solutions

5.3.4 References[1] C. A. Heaton, The Chemical industry, page 113.

Chapter 6

Waves

6.1 What is a wave?

At this point in the course, it is easy to get bogged down inthe complex theories and equations surrounding 'waves’.However, a better understanding of waves can be gainedby going back to basics, and explaining what a wave is inthe first place.

6.1.1 Definitions

A wave, at its most basic level, is a disturbance by whichenergy is transferred because this disturbance is a store,of sorts, of potential energy. This begs the question “Howis this disturbance transferred across space?" In somecases, this is easy to answer, because some waves travelthrough a medium. The easiest example to think about isa water wave. One area moves up, pulling the next one upwith it, and pressure and gravity pull it back to its originalposition.

disp

lace

men

t ⟶

Wave

distance ⟶

λ

y

λ = wavelength y = amplitude

Features of a wave

However, some waves (electro-magnetic waves) do notappear to travel through a medium. Physicists have puz-zled over how light, which behaves like a wave in manysituations, travels for a long time. One theory was thatthere was a mysterious 'ether' which pervaded all ofspace, and so light was just like water waves, except thatthe water was invisible. This theory is widely regardedto be incorrect, but, since light is assumed to be a wave,what is it a disturbance in?Another explanation is that light is not a wave, but insteadis a stream of particles. This idea would explain away the

need for an 'ether' for light to travel through. This, too,has its problems, as it does not explain why light behavesas a wave.So, we are left with a paradox. Light behaves as both awave and a particle, but it can be shown not to be either.Quantum physics attempts to explain this paradox. How-ever, since light behaves as both a wave and a particle, wecan look at it as both, even if, when doing this, we knowthat we don't fully understand it yet.The image on the right shows a waveform. This plotsthe distance through the medium on the x-axis, and theamount of disturbance on the y-axis. The amount of dis-turbance is known as the amplitude. Wave amplitudestend to oscillate between two limits, as shown. The dis-tance in the medium between two 'peaks’ or 'troughs’(maxima and minima on the waveform) is known as thewavelength of the wave.

6.1.2 Types of Waves

Waves can be categorised according to the direction ofthe effect of the disturbance relative to the direction oftravel. A wave which causes disturbance in the directionof its travel is known as a longitudinal wave, whereas awave which causes disturbance perpendicular to the di-rection of its travel is known as a transverse wave.

6.1.3 Superposition

One feature of waves is that they superpose. That is tosay, when they are travelling in the same place in themedium at the same time, they both affect the mediumindependently. So, if two waves say “go up” to the samebit of medium at the same time, the medium will risetwice as much. In general, superposition means that theamplitudes of two waves at the same point at the sametime at the same polarisation add up.

6.1.4 Interference

Consider two identical waveforms being superposed oneach other. The resultant waveform will be like the two

23

24 CHAPTER 6. WAVES

other waveforms, except its amplitude at every point willbe twice as much. This is known as constructive inter-ference. Alternatively, if one waveform moves on by halfa wavelength, but the other does not, the resultant wave-form will have no amplitude, as the two waveforms willcancel each other out. This is known as destructive in-terference. Both these effects are shown in the diagrambelow:

These effects occur because the wavefronts are travellingthrough a medium, but electromagnetic radiation also be-haves like this, even though it does not travel through amedium.

6.1.5 Velocity, frequency and wavelength

You should remember the equation v = fλ from earlier inthis course, or from GCSE. v is the velocity at which thewave travels through the medium, in ms−1, f (or nu, ν) isthe frequency of the wave, in Hz (no. of wavelengths per.second), and λ is the wavelength, in m.This equation applies to electromagnetic waves, but youshould remember that there are different wavelengths ofelectromagnetic radiation, and that different colours ofvisible light have different wavelengths. You also need toknow the wavelengths of the different types of electro-magnetic radiation:

6.1.6 Questions

1. What is wave?2. Through what medium are sound waves propagated?3. What aspects of the behaviour of light make it looklike a wave?4. What aspects of the behaviour of light make it looklike a particle?5. Consider the diagram on the right. White light is par-tially reflected by the transparent material. Some of thelight, however, is refracted into the transparent materialand reflected back by the opaque material. The result istwo waves travelling in the same place at the same time

6.3. STANDING WAVES 25

at the same polarisation(the light is not a single beam).Why does, say, the red light disappear? (Variations onthis question are popular with examiners.)6. What is the wavelength of green light?7. The lowest frequency sound wave humans can hear hasa frequency of approximately 20Hz. Given that the speedof sound in air is 343ms−1, what is the wavelength of thelowest frequency human-audible sound?Worked Solutions

6.2 Phasors

Consider the image on the right. It shows a wave trav-elling through a medium. The moving blue dot repre-sents the displacement caused to themedium by the wave.It can be seen that, if we consider any one point in themedium, it goes through a repeating pattern, moving upand down, moving faster the nearer it is to the centre ofthe waveform. Its height is determined by the amplitudeof the wave at that point in time. This is determined by asine wave.

A phasor

Phasors are a method of describing waves which showtwo things: the displacement caused to the medium, and

the point in the repeating waveform which is being repre-sented. They consist of a circle. An arrow moves roundthe circle anticlockwise as the wave pattern passes. Forevery wavelength that goes past, the arrowmoves 360°, or2πc, starting from the right, as in trigonometry. The angleof the arrow from the right is known as the phase angle,and is usually denoted θ, and the radius of the circle isusually denoted a. The height of the point at the end ofthe arrow represents the displacement caused by the waveto the medium, and so the amplitude of the wave at thatpoint in time. The time taken to rotate 360° is known asthe periodic time, and is usually denoted T.Phase difference is the difference between the angles (θ)of two phasors, which represent two waves. It is nevermore than 180°, as, since the phasor is moving in a circle,the angle between two lines touching the circumferencewill always be less than or equal to 180°. It can also beexpressed in terms of λ, where λ is the total wavelength(effectively, 360°). You can use trigonometry to calculatethe displacement from the angle, and vice-versa, providedyou know the radius of the circle. The radius is equal tothe maximum amplitude of the wave.Phasors can be added up, just like vectors: tip-to-tail.So, for example, when two waves are superposed on eachother, the phasors at each point in the reference materialcan be added up to give a new displacement. This ex-plains both constructive and destructive interference aswell. In destructive interference, the phasors for eachwave are pointing in exactly opposite directions, and soadd up to nothing at all. In constructive interference, thephasors are pointing in the same direction, so the totaldisplacement is twice as much.

6.2.1 Questions

1. A sine wave with wavelength 0.1m travels through agiven point on the surface of the sea. A phasor arrowrepresenting the effect of this wave on this point rotates1000°. Howmany wavelengths have gone past in the timetaken for the phasor to rotate this much?2. A sine wave has a maximum amplitude of 500nm.What is its amplitude when the phasor has rotated 60°from its start position?3. Two waves have a phase difference of 45°. When thefirst wave is at its minimum amplitude of−0.3m, what isthe total amplitude of the superposed waveforms?Worked Solutions

6.3 Standing Waves

When two coherent waves - waves of equal frequencyand amplitude - travel in opposite directions through thesame area, an interesting superposition effect occurs, asis shown in the following animation:

26 CHAPTER 6. WAVES

Standing wave with nodes labelled in red

Some areas of the resultant waveform consistently havean amplitude of 0. These are known as nodes. At otherpoints (half-way between the nodes), the resultant wave-form varies from twice the amplitude of its constituentwaveforms in both directions. These points are known asantinodes. Everywhere in between the nodes and antin-odes varies to a lesser degree, depending on its position.This effect only occurs if the two waveforms have thesame amplitude and frequency. If the two waves havedifferent amplitudes, the resultant waveform is similar toa standing wave, except that it has no nodes, and 'moves’.Because of these conditions, standing waves usually onlyoccur when a waveform is reflected back on itself. Forexample, in a microwave oven, the microwaves are re-flected by the metal on the other side of the oven fromthe transmitter. This creates nodes and antinodes. Sincenothing cooks at the nodes, a turntable is necessary to en-sure that all of the food passes through the antinodes andgets cooked.

6.3.1 Frequencies

Consider a string, attached at either end, but allowed tomove freely in between. If you pluck it, you create a wave

Fundamental frequency, and the first 6 harmonics

which travels along the string in both directions, and isreflected at either end of the string. This process keepson happening, and so a standing wave pattern is created.The string goes up, and then down, as shown in the firstrow of the diagram on the right. If you imagine the toparc as the first half of a waveform, you can see that whenthe string is vibrating at the fundamental frequency, thestring is half as long as the wavelength: it is ½λ long.Therefore, λ=2LIf you were to pinch the string in the middle, and thenpluck it on one side, a different standing wave patternwould be generated. By plucking, you have created anantinode, and by pinching, you have created a node. Ifyou then let go of the string, the standing wave patternspreads, and the string length now equals the wavelength.This is known as the first harmonic.As you pinch the string in descending fractions (½, ⅓,¼, etc.), you generate successive harmonics, and the totallength of the string is equal to additional ½λ wavelengths.

6.3.2 Pipes

Consider a pipe which is open at one end, and closed atthe other. In pipes, waves are reflected at the end of thepipe, regardless of whether it is open or not. If you blowacross the end of the tube, you create a longitudinal wave,with the air as the medium. This wave travels down thetube, is reflected, travels back, is reflected again, and soon, creating a standing wave pattern.The closed end of the pipe must be a node; it is the equiv-alent of pinching a string. Similarly, the open endmust bean antinode; blowing across it is the equivalent of pluck-ing the string.Harmonics can be present in pipes, as well. This is how

6.4. YOUNG’S SLITS 27

Standing waves in pipes, showing nodes and antinodes

musical instruments work: an open hole in a wind instru-ment creates an antinode, changing the frequency of thesound, and so the pitch.TomDuncan states that the fundamental frequency IS thesame as the first harmonic (Adavanced Physics 5th edi-tion page 317)

6.3.3 Questions

1. The air in a 3m organ pipe is resonating at the fun-damental frequency. Organ pipes are effectively open atboth ends. What is the wavelength of the sound?2. A string is vibrating at the second harmonic frequency.Howmanywavelengths long is the standingwave created?3. Express, in terms of λ, the length of a pipe which isclosed at one end, where λ is the length of one wave atthe fundamental frequency.Worked Solutions

6.4 Young’s Slits

You should be familiar with the idea that, when lightpasses through a slit, it is diffracted (caused to spread outin arcs from the slit). The amount of diffraction increasesthe closer the slit width is to the wavelength of the light.Consider the animation on the right. Light from a lightsource is caused to pass through two slits. It is diffractedat both these slits, and so it spreads out in two sets of arcs.Now, apply superposition of waves to this situation. Atcertain points, the peaks (or troughs) of the waves willcoincide, creating constructive interference. If this oc-curs on a screen, then a bright 'fringe' will be visible. Onthe other hand, if destructive interference occurs (a peak

coincides with a trough), then no light will be visible atthat point on the screen.

6.4.1 Calculating the angles at whichfringes occur

If we wish to calculate the position of a bright fringe,we know that, at this point, the waves must be in phase.Alternatively, at a dark fringe, the waves must be in an-tiphase. If we let the wavelength equal λ, the angle of thebeams from the normal equal θ, and the distance betweenthe slits equal d, we can form two triangles, one for brightfringes, and another for dark fringes (the crosses labelled1 and 2 are the slits):

28 CHAPTER 6. WAVES

The length of the side labelled λ is known as the pathdifference. For bright fringes, from the geometry above,we know that:sin θ = λ

d

Therefore:λ = d sin θHowever, bright fringes do not only occur when the sidelabelled λ is equal to 1 wavelength: it can equal multiplewavelengths, so long as it is a whole wavelength. There-forenλ = d sin θ ,where n is any integer.Now consider the right-hand triangle, which applies todark fringes. We know that, in this case:sin θ = 0.5λ

d

0.5λ = d sin θWe can generalise this, too, for any dark fringe. However,if 0.5λ is multiplied by an even integer, then we will geta whole wavelength, which would result in a bright, not adark, fringe. So, n must be an odd integer in the following

formula:0.5nλ = d sin θnλ = 2d sin θ

6.4.2 Calculating the distances angles cor-respond to on the screen

At this point, we have to engage in some slightly dodgymaths. In the following diagram, p is path difference, Lis the distance from the slits to the screen and x is theperpendicular distance from a fringe to the normal:

Here, it is necessary to approximate the distance from theslits to the fringe as the perpendicular distance from theslits to the screen. This is acceptable, provided that θ issmall, which it will be, since bright fringes get dimmer asthey get further away from the point on the screen oppo-site the slits. Hence:sin θ = x

L

If we substitute this into the equation for the path differ-

6.6. FINDING THE DISTANCE OF A REMOTE OBJECT 29

ence p:p = d sin θ = dx

L

So, at bright fringes:nλ = dx

L , where n is an integer.And at dark fringes:nλ = 2dx

L , where n is an odd integer.

6.4.3 Diffraction Grating

A diffraction grating consists of a lot of slits with equalvalues of d. As with 2 slits, when nλ = d sin θ , peaksor troughs from all the slits coincide and you get a brightfringe. Things get a bit more complicated, as all the slitshave different positions at which they add up, but you onlyneed to know that diffraction gratings form light and darkfringes, and that the equations are the same as for 2 slitsfor these fringes.

6.4.4 Questions

1. A 2-slit experiment is set up in which the slits are 0.03m apart. A bright fringe is observed at an angle 10° fromthe normal. What sort of electromagnetic radiation wasbeing used?2. Light, with a wavelength of 500 nm, is shone through2 slits, which are 0.05 m apart. What are the angles to thenormal of the first three dark fringes?3. SomeX-rays, with wavelength 1 nm, are shone througha diffraction grating in which the slits are 50 μm apart. Ascreen is placed 1.5m from the grating. How far are thefirst three light fringes from the point at which the normalintercepts the screen?Worked Solutions

6.5 Diffraction

We have already seen why fringes are visible when lightpasses through multiple slits. However, this does not ex-plain why, when light is only passing through 1 slit, a pat-tern such as the one on the right is visible on the screen.The answer to this lies in phasors. We already know thatthe phasor arrows add up to give a resultant phasor. Byconsidering the phasor arrows from many paths whichlight takes through a slit, we can explain why light anddark fringes occur.At the normal line, where the brightest fringe is shown,all the phasor arrows are pointing in the same direction,and so add up to create the greatest amplitude: a brightfringe.At other fringes, we can use the same formulæ as fordiffraction gratings, as we are effectively treating the sin-

gle slit as a row of beams of light, coming from a row ofslits.Now consider the central beam of light. By trigonometry:sin θ = W

L ,where θ = beam angle (radians), W = beam width and L= distance from slit to screen. Since θ is small, we canapproximate sin θ as θ, so:θ ≈ W

L

and since λ = d sin θ:θ ≈ λ

d

6.5.1 Questions

1. What is the width of the central bright fringe on ascreen placed 5m from a single slit, where the slit is 0.01mwide and the wavelength is 500nm?And that’s all there is to it ... maybe.Worked Solutions

6.6 Finding the Distance of a Re-mote Object

In the final section (Section C) of the exam, you have to beable to write about howwaves are used to find the distanceof a remote object. I would recommend that you pick amethod, and then answer the following questions about it:1. What sort of wave does your system use? What is anapproximate wavelength of this wave?2. What sort of distance is it usually used to measure?What sort of length would you expect the distance to be?3. Why is measuring this distance useful to society?4. Draw a labelled diagram of your system.5. Explain how the system works, and what data is col-

30 CHAPTER 6. WAVES

lected.6. Explain how the distance to the object is calculatedusing the data collected.7. What limitations does your system have? (e.g. accu-racy, consistency)8. What percentage error would you expect these limita-tions to cause?9. How might these problems be solved?

6.6.1 Examples

Some example answers to these questions are given in thefollowing pages:

• Radar

• Sonar

Chapter 7

Quantum Physics

7.1 Light

We have already seen how light behaves like both a waveand a particle, yet can be proven not to be either. Thisidea is not limited to light, but we will start our brief lookat quantum physics with light, since it is easiest to under-stand.Quantum physics is the study of quanta. A quantum is,to quote Wiktionary, “The smallest possible, and there-fore indivisible, unit of a given quantity or quantifiablephenomenon”. The quantum of light is the photon. Weare not describing it as a particle or a wave, as such, butas a lump of energy which behaves like a particle and awave in some cases. We are saying that the photon is thesmallest part of light which could be measured, given per-fect equipment. A photon is, technically, an elementaryparticle. It is also the carrier of all electromagnetic radi-ation. However, its behaviour - quantum behaviour - iscompletely weird, so we call it a quantum.

7.1.1 Evidence for the Quantum Be-haviour of Light

Grainy image of a bowl of nuts. Much better images than thisone have been taken, clearly showing that light arrives in lumps.

Dim Photos

The easiest evidence to understand is dim photographs.When you take a photo with very little light, it appears'grainy', such as the image on the right. This means thatthe light is arriving at the camera in lumps. If light werea wave, we would expect the photograph to appear dim-mer, but uniformly so. In reality, we get clumps of lightdistributed randomly across the image, although the den-sity of the random lumps is higher on the more reflectivematerials (the nuts). This idea of randomness, accordingto rules, is essential to quantum physics.

The photoelectric effect

Photoelectric Effect

The second piece of evidence is more complex, but moreuseful since a rule can be derived from it. It can be shownexperimentally that, when light of an adequate frequencyfalls on a metallic surface, then the surface absorbs thelight and emits electrons. Hence, a current and voltage(between the surface and a positively charged terminalnearby) are produced, which can be measured.The amount of current produced varies randomly arounda certain point. This point changes depending on the fre-quency of the electromagnetic radiation. Furthermore, ifthe frequency of the radiation is not high enough, thenthere is no current at all! If light were a wave, we wouldexpect energy to build up gradually until an electron was

31

32 CHAPTER 7. QUANTUM PHYSICS

released, but instead, if the photons do not have enoughenergy, then nothing happens. This is evidence for theexistence of photons.

7.1.2 The Relationship between Energyand Frequency

The photoelectric effect allows us to derive an equationlinking the frequency of electromagnetic radiation to theenergy of each quantum (in this case, photons). This canbe achieved experimentally, by exposing the metallic sur-face to light of different colours, and hence different fre-quencies. We already know the frequencies of the differ-ent colours of light, and we can calculate the energy eachphoton carries into the surface, as this is the same as theenergy required to supply enough potential difference tocause the electron to move. The equation for the energyof the electron is derived as follows:First, equate two formulae for energy:P = E

t = IV

Rearrange to get:E = ItV

We also know that:Q = It

So, by substituting the previous equation into the equationfor energy:E = QV = e∆V ,where P = power, E = energy, t = time, I = current, V =potential difference, Q = charge, e = charge of 1 electron=−1.602 x 10−19 C, ΔV = potential difference producedbetween anode and cathode at a given frequency of radia-tion. This means that, given this potential difference, wecan calculate the energy released, and hence the energyof the quanta which caused this energy to be released.Plotting frequency (on the x-axis) against energy (on they-axis) gives us an approximate straight line, with a gradi-ent of 6.626 x 10−34. This number is known as Planck’sconstant, is measured in Js, and is usually denoted h.Therefore:E = hf

In other words, the energy carried by each quantum isproportional to the frequency of the quantum. The con-stant of proportionality is Planck’s constant.

7.1.3 Questions

1. How much energy does a photon with a frequency of50kHz carry?2. A photon carries 10−30J of energy. What is its fre-quency?3. How many photons of frequency 545 THz does a 20W

bulb give out each second?4. In one minute, a bulb gives out a million photons offrequency 600 THz. What is the power of the bulb?5. The photons in a beam of electromagnetic radiationcarry 2.5μJ of energy each. How long should the phasorsrepresenting this radiation take to rotate?Worked Solutions

7.2 Quantum Behaviour

So far, we have identified the fact that light travels inquanta, called photons, and that these photons carry anamount of energy which is proportional to their fre-quency. We also know that photons aren't waves or parti-cles in the traditional sense of either word. Instead, theyare lumps of energy. They don't behave the way we wouldexpect them to.

7.2.1 Many Paths

In fact, what photons do when they are travelling is to takeevery path possible. If a photon has to travel from point Ato point B it will travel in a straight line and loop the loopand go via Alpha Centauri and take every other possiblepath. This is the photon’s so-called 'quantum state'. It isspread out across all space.However, just because a photon could end up anywherein space does not mean that it has an equal probability ofending up in any given place. It is far more likely that aphoton from a torch I am carrying will end up hitting theground in front of me than it is that the same photon willhit me on the back of the head. But both are possible.Light can go round corners; just very rarely!

7.2.2 Calculating Probabilities

The probability of a photon ending up at any given pointin space relative to another point in space can be cal-culated by considering a selection of the paths the pho-ton takes to each point. The more paths considered, thegreater the accuracy of the calculation. Use the followingsteps when doing this:1. Define the light source.2. Work out the frequency of the photon.3. Define any objects which the light cannot pass through.4. Define the first point you wish to consider.5. Define a set of paths from the source to the point beingconsidered, the more, the better.6. Work out the time taken to traverse one of the paths.7. Work out how many phasor rotations this correspondsto.

7.2. QUANTUM BEHAVIOUR 33

8. Draw an arrow representing the final phasor arrow.9. Repeat steps 6-8 for each of the paths.10. Add all the phasor arrows together, tip-to-tail.11. Square the amplitude of this resultant phasor arrowto gain the intensity of the light at this point. It may helpto imagine a square rotating around, instead of an arrow.12. Repeat steps 4-11 for every point you wish to con-sider. The more points you consider, the more accurateyour probability distribution will be.13. Compare all the resultant intensities to gain aprobability distribution which describes the probabilitiesof a photon arriving at one point to another. For exam-ple, if the intensity of light at one point is two times theintensity of light at another, then it is twice as likely thata photon will arrive at the first point than the second.14. If all the points being considered were on a screen,the intensities show you the relative brightnesses of lightat each of the points.

7.2.3 Examples

If we now take this method and apply it to several situa-tions, we find that, in many cases, the results are similarto those obtained when treating light as a wave, with theexception that we can now reconcile this idea with theobservable 'lumpiness’ of light, and can acknowledge thefact that there is a certain probability that some light willnot behave according to some wave laws.

Travelling from A to B

Adding phasor arrows to gain resultant amplitude by consideringmultiple paths, showing variation in paths’ contribution to resul-tant amplitude.

This is the simplest example to consider. If we consider arange of paths going from point A to point B, and calcu-late the phasor directions at the end of the paths, we get aresultant phasor arrow which gives us some amplitude atpoint B. Since there are no obstructions, at any point this

far away from the source, we will get the same resultantamplitude.It is important to note that different paths contribute tothe resultant amplitude by different amounts. The pathscloser to the straight line between the two points are moreparallel to the resultant angle, whereas the paths furtheraway vary in direction more, and so tend to cancel eachother out. The conclusion: light travelling in straight linescontributes most to the resultant amplitude.

Young’s Slits

Here, we just need to consider two paths: one througheach slit. We can then calculate two phasor arrows, addthem together to gain a resultant phasor arrow, and squareits amplitude to gain the intensity of the light at the pointthe two paths went to. When calculated, these intensitiesgive a pattern of light and dark fringes, just as predictedby the wave theory.

The most contribution to amplitude comes from the centre of themirror.

Reflection

This situation is very similar to what happens when lighttravels in a 'straight line'. The only difference is that weconsider the paths which involve rebounding off an obsta-cle. The results are more or less the same, but the pathsfrom which they were obtained are different. This meansthat we can assume the same conclusions about these dif-ferent paths: that most of the resultant amplitude comesfrom the part of the mirror where the angle of incidenceequals the angle of reflection. In other words, the likeli-hood is that a photon will behave as if mirrors work ac-cording to wave theory.

Refraction

Different paths have different lengths, and so photonstake different amounts of time to traverse them (these are

34 CHAPTER 7. QUANTUM PHYSICS

Refraction works because the greatest contribution to amplitudeis caused by the paths with the shortest trip time.

known as trip times). In the diagram on the right, the pho-tons again traverse all possible paths. However, the pathswith the smallest difference between trip times have pha-sor arrows with the smallest difference in direction, so thepaths with the smallest trip times contribute most to theresultant amplitude. This shortest path is given by Snell’slaw. Yet again, quantum physics provides amore accuratepicture of something which has already been explained tosome degree.

Diffraction

Diffraction occurs when the photons are blocked fromtaking every other path. This occurs when light passesthrough a gap less than 1 wavelength wide. The result isthat, where the amplitudes would have roughly cancelledeach other out, they do not, and so light spreads out indirections it would not normally spread out in. This ex-plains diffraction, instead of just being able to observeand calculate what happens.

7.3 Electron Behaviour

So far, we have considered how quantum physics appliesto photons, the quanta of light. In reality, every otherparticle is also a quantum, but you only need to knowabout photons and electrons.The image on the right shows what happens when you fireelectrons through a pair of slits: it arrives in lumps, butyou get fringes due to superposition as well. The electronsare behaving as both waves and particles. Actually, theyare behaving as quanta. The equations describing quan-tum behaviour in electrons are similar to those describingit in photons.

Young’s slits, using electrons.

7.3.1 Frequency and Kinetic Energy

We know that, for photons:

7.3. ELECTRON BEHAVIOUR 35

f = Eh

In suggesting that electrons are quanta, we assume thatthey must have a frequency at which the phasors repre-senting them rotate. We also know that h is a constant;it does not change. So, when adapting the above equa-tion to apply to electrons, all we need to adapt is E. Inelectrons, this energy is their kinetic energy. If the elec-tron has some form of potential energy, this must first besubtracted from the kinetic energy, as this portion of theenergy does not affect frequency. So:f =

Ekinetic−Epotential

h

7.3.2 De Broglie Wavelength

If electrons exhibit some wavelike properties, they mustalso have a 'wavelength', known as the de Broglie wave-length, after its discoverer. This is necessary in order towork out a probability distribution for the position of anelectron, as this is the distance the electron travels foreach phasor arrow rotation. The de Broglie wavelengthλ is given by the equation:λ = h

p = hmv ,

where h = Planck’s constant, p = momentum, m = massof electron = 9.1 x 10−31kg and v = velocity of electron.

7.3.3 Potential Difference and Kinetic En-ergy

Potential difference is what causes electrons tomove. Youalready know how power is related to charge, voltage andtime:P = QV

t

Since power is the rate at which work is done:W = QV

We know that the charge on an electron equals −1.6 x10−19, and that work done is energy, so:Ekinetic = 1.6× 10−19 × V

Energy, in the SI system of units, is measured in Joules,but, sometimes it is measured in electronvolts, or eV. 1eV is the kinetic energy of 1 electron accelerated by 1Vof potential difference. So, 1 eV = 1.6 x 10−19 J.

7.3.4 Questions

1. An electron moves at 30,000 ms−1. What is its deBroglie wavelength?2. What is its frequency?3. What is its kinetic energy, in eV?4. Given that it is travelling out of an electron gun, whatwas the potential difference between the anode and the

Wave functions of an electron in different orbitals of a hydrogenatom. Brightness corresponds to the probability density functionfor the position of the electron.

cathode?5. An electron is accelerated by a potential difference of150V. What is its frequency?Worked Solutions

Chapter 8

Mechanics

8.1 Vectors

8.1.1 What is a vector?

Two types of physical quantity are scalars and vectors.Scalar quantities are simple: they are things like speed,distance, or time. They have a magnitude, but no direc-tion. A vector quantity consists of two parts: both a scalarand a direction. For example, the velocity of an object ismade up of both the speed of an object and the directionin which it is moving. Speed is a scalar; add a directionand it becomes velocity, a vector. Similarly, take a dis-tance and give it a direction and it becomes a displace-ment, such as '2 miles south-east'. Distance is a scalar,whereas displacement is a vector.Vectors and scalars are both useful. For example, if I runaround the room several times and end up back where Istarted, I may have covered a distance of 50m. My dis-placement is 0 - the null vector. The null vector is theonly vector which has no direction. If I want to calculatehow much work I have done, I should use the distance. IfI want to know where I am, I should use the displacement.As we shall see, the directional component of a vector canbe expressed in several different ways. '2miles south-east'is the same as saying '2 miles on a bearing of 135°', or '1.4miles east, −1.4 miles north'. The scalar component of avector is known as themodulus of a vector.

8.1.2 Vector Notation

You need to be able to understand the following algebraicrepresentations of vectors:

8.1.3 Vector Components

Sometimes, it is useful to express a vector in terms of twoother vectors. These two vectors are usually pointing upand right, and work similarly to the Cartesian co-ordinatesystem. So, for example, 'an acceleration of 3.4 ms−2

west' becomes 'a vertical acceleration of 0 ms−2 and anhorizontal acceleration of−3.4 ms−2 east. However, thisis a very simple example.

Calculating the components of a vector.

Consider the diagram on the right. The vector a consistsof a vertical component j and a horizontal component i.a has a modulus |a|. |i| and |j| can be calculated using |a|,the angle between i and a θ and some basic trigonometry.We know that:cos θ = |i|

|a| and sin θ = |j||a|

Hence:|i| = |a| cos θ and |j| = |a| sin θ.This will be given to you in the formula booklet in theexam.

8.1.4 Vector Addition

Adding vectors tip-to-tail.

You also need to know how to add vectors together. This

36

8.1. VECTORS 37

enables us to answer questions such as, “If I travel 5 milesnorth-west and then 6 miles east, where am I?", or “If Iaccelerate at 3 ms−2 in a northerly direction, and acceler-ate south-east at 1 ms−2, what is my total acceleration?".Vectors can be added 'tip-to-tail'; that is to say, the resul-tant vector is equal to 'travelling' the first vector, and thentravelling the second vector.This is shown in the diagram on the left. When vectorsa and b are added together, the resultant vector a + b isproduced, joining the tail of the first vector to the tip ofthe last, with the vectors joined together. In practise, theeasiest way to add two vectors together is to calculate (ifyou do not already know this) the vertical and horizontalcomponents, and add them all together to get two totalvertical and horizontal components. You can then usePythagoras’ theorem to calculate the modulus of the re-sultant vector, and some more basic trigonometry to cal-culate its direction.

Adding lots of vectors tip-to-tail.

In algebra:∑|a|n =

√(∑

|i|n)2 + (∑

|j|n)2 and θ =

arctan∑

|j|n∑|i|n ,

where a1 ... a are the vectors to be added together, i1... i are their horizontal components, j1 ... j are theirvertical components, and θ is the angle between the θ=0line and the resultant vector Σa , as shown in the diagramon the right.

8.1.5 Predicting Parabolas

If you use a diagram to represent vectors (making thelengths of the arrows proportional to the modulus of thevectors they represent, and the directions equal), you canpredict graphically the trajectory an object (such as a ball)will take. Use the following steps:

• Draw a vector to represent the velocity of the ball (inms−1). Since this is the number of metres travelledin 1 second, and each step of the process is 1 secondlong, this vector represents both the velocity and thedisplacement of the ball, i.e.

Illustration of how to estimate the parabola of a moving objectgraphically, taking into account the vectors of its starting velocity,and the change in velocity due to gravity.

∆v = ∆s∆t =

∆s1 = ∆s

• Copy this vector, and connect its tail to the tip of thefirst vector. This new vector represents the velocityand displacement that the ball would have had overthe next second, if gravity did not exist.

• Draw another vector to represent the change in ve-locity due to gravity (9.81 ms−2) on Earth. Thisshould be pointing downwards, and be to the samescale as the other vectors. This represents the factthat the velocity of the ball changes due to gravity(velocity is a vector, so both the speed and angle ofthe ball’s travel change).

• Add these vectors together, as shown above, to give anew vector. This new vector represents the velocityand displacement of the ball over the second second.

• Repeat this process until the ball hits something(draw the ground, if in doubt).

8.1.6 Questions

1. Which of the following are vectors?

• 20 cm

• 9.81 ms−2 towards the centre of the earth

38 CHAPTER 8. MECHANICS

• 5 km south-east

• 500 ms−1 on a bearing of 285.3°

2. A displacement vector a is the resultant vector of twoother vectors, 5 m north and 10 m south-east. What doesa equal, as a displacement and a bearing?3. If I travel at a velocity of 10 ms−1 on a bearing of030°, at what velocity am I travelling north and east?4. An alternative method of writing vectors is in a col-umn, as follows:a =

(xy

),

where x and y are the vertical and horizontal componentsof the vector respectively. Express |a| and the angle be-tween a and

(10

)in terms of x and y.

5. A more accurate method of modelling the trajectoryof a ball is to include air resistance as a constant force F.How would this be achieved?Worked Solutions

8.2 Graphs

There are two types of graphs of motion you need tobe able to use and understand: distance-time graphs andvelocity-time graphs.

8.2.1 Distance-time Graphs

An object travels at a constant rate for 6 seconds, stops for 5 sec-onds, returns to its original position in the next 7 seconds, travel-ling more slowly in the middle section of its return journey.

Adistance-time graph plots the distance of an object awayfrom a certain point, with time on the x-axis and distanceon the y-axis.There are several types of graphs of motionyou need to be able to use and understand: distance-timegraphs, position-time graphs, and velocity-time graphs.

8.2.2 Position-time Graphs or Displace-ment - Time Graphs

Distance-Time Graphs give you speed, but speed is nevernegative so you can only have a positive slope in adistance-time graph. Position-Time graphs show dis-placement, have direction, and from which you can cal-culate velocity. If we were to imagine the line on theposition-time graph to the right as a function f(t), givingan equation for s = f(t), we could differentiate this to gain:dsdt = f ′(t) ,where s is displacement, and t is time. By finding f'(t) atany given time t, we can find the rate at which distanceis increasing (or decreasing) with respect to t. This is thegradient of the line. A positive gradient means that dis-tance is increasing, and a negative gradient means thatdistance is decreasing. A gradient of 0 means that theobject is stationary. The velocity of the object is the rateof change of its displacement, which is the same as thegradient of the line on a distance-time graph. This is notnecessarily the same as the average velocity of the objectv:v = s

t

Here, t and s are the total changes in displacement andtime over a certain period - they do not tell you exactlywhat was happening at any given point in time.

8.2.3 Velocity-time Graphs

An object accelerates for 6 seconds, hits something, acceleratesfor 5 seconds and then decelerates to a stop in the remaining 6.5seconds of its journey.

A velocity-time graph plots the velocity of an object, rel-ative to a certain point, with time on the x-axis and ve-locity on the y-axis. We already know that velocity is thegradient (derivative) of the distance function. Since in-tegration is the inverse process to differentiation, if wehave a velocity-time graph and wish to know the distancetravelled between two points in time, we can find the areaunder the graph between those two points in time. In gen-eral:

8.3. KINEMATICS 39

If v = f(t)

s =∫ t2t1

f(t) dt

where v is velocity (in ms−1), t is time (in s), and s is thedistance travelled (in m) between two points in time t1and t2.Also, by differentiation, we know that the gradient (orderivative of v = f(t)) is equal to the acceleration of theobject at any given point in time (in ms−2) since:a = dv

dt

8.2.4 Questions

1. In the following distance-time graph, what is the veloc-ity 4 seconds after the beginning of the object’s journey?

2a)Describe this person’s movements.2b)What is the velocity at 12 seconds?3. In the following velocity-time graph, how fardoes the object travel between 7 and 9 seconds?

4. What is the object’s acceleration at 8 seconds?5. A car travels at 10ms−1 for 5 minutes in a straightline, and then returns to its original location over thenext 4 minutes, travelling at a constant velocity. Drawa distance-time graph showing the distance the car hastravelled from its original location.6. Draw the velocity-time graph for the above situation.The following question is more difficult than anything youwill be given, but have a go anyway:7. The velocity of a ball is related to the time since it wasthrown by the equation v = 30− 9.8t . How far has theball travelled after 2 seconds?Worked Solutions

8.3 Kinematics

Kinematics is the study of how objects move. One needsto understand a situation in which an object changesspeed, accelerating or decelerating, and travelling a cer-tain distance. There are four equations you need to beable to use which relate these quantities.

8.3.1 Variables

Before we can understand the kinematic equations, weneed to understand the variables involved. They are asfollows:

40 CHAPTER 8. MECHANICS

• t is the length of the interval of time being consid-ered, in seconds.

• v is the speed of the object at the end of the timeinterval, in ms−1.

• u is the speed of the object at the beginning of thetime interval, in ms−1.

• a is the acceleration of the object during the timeinterval, in ms−2. Has to be a constant.

• s is the displacement (distance traveled) of the objectduring the time interval, in meters.

8.3.2 Equations

The four equations are as follows:1. v = u+ at

2. s = u+v2 t

3. s = ut+ at2

2

4. v2 = u2 + 2as

8.3.3 Derivations

It is also useful to know where the above equations comefrom. We know that acceleration is equal to change inspeed per. unit time, so:a = v−u

t (*)at = v − u

v = u+ at (1)We also know that the average speed over the time inter-val is equal to displacement per. unit time, so:u+v2 = s

t

s = u+v2 t (2)

If we substitute the value of v from equation 1 into equa-tion 2, we get:s = u+(u+at)

2 t = 2u+at2 t = t(u+ at

2 ) = ut+ at2

2 (3)If we take the equation for acceleration (*), we can rear-range it to get:at = v − u

t = v−ua

If we substitute this equation for t into equation 2, weobtain:s = u+v

2v−ua = (v+u)(v−u)

2a = v2−u2

2a

2as = v2 − u2

v2 = u2 + 2as (4)

8.3.4 Questions

1. A person accelerates from a speed of 1 ms−1 to 1.7ms−1 in 25 seconds. How far has he travelled in this time?2. A car accelerates at a rate of 18 kmh−2 to a speed of60 kmh−1, travelling 1 km in the process. How fast wasthe car travelling before it travelled this distance?3. A goose in flight is travelling at 4 ms−1. It acceleratesat a rate of 1.5ms−2 for 7 seconds. What is its new speed?4. How far does an aeroplane travel if it accelerates from400 kmh−1 at a rate of 40 kmh−2 for 1 hour?Worked Solutions

8.4 Forces and Power

8.4.1 Forces

Forces are vectors. When solving problems involvingforces, it is best to think of them as lots of people pullingropes attached to an object. The forces are all pulling indifferent directions, with different magnitudes, but theeffect is in only one direction, with only one magnitude.So, you have to add the forces up as vectors.Forces cause things to happen. They cause an object toaccelerate in the same direction as the force. In otherwords, forces cause objects to move in a direction closerto the direction they are pulling in. If the object is alreadymoving, then theywill not cause it tomove in the directionof the force, as forces do not create velocities: they createaccelerations.If a force is acting on an object, it seems logical that theobject will not accelerate as much as a result of the forceif it has a lower mass. This gives rise to the equation:F = ma ,where F = force applied (in Newtons, denoted N), m =mass (in kg) and a = acceleration (in ms−2). If we rear-range the equation, it makes more sense:a = F

m

In other words, the acceleration in a given direction asthe result of a force is equal to the force applied per. unitmass in that direction.

8.4.2 Work Done

You should already know how to calculate some types ofenergy, for example:Kinetic Energy = mv2

2

Gravitational Potential Energy = mgh

The amount of energy converted by a force is equal to thework done, which is equal (as you already know) to the

8.4. FORCES AND POWER 41

force multiplied by the distance the object it is acting onmoves:Work Done = ∆E = F∆s

When answering questions about work done, you may begiven a force acting in a direction other than that of thedisplacement. In this case, you will have to find the dis-placement in the direction of the force, as shown in thesection on Vectors.

8.4.3 Power

Power is the rate of change of energy. It is the amount ofenergy converted per. unit time, and is measured in Js−1:P = ∆E

t ,where E = energy (in J) and t = time (in s). Since ΔE= work done, power is the rate at which work is done.Since:∆E = F∆s

∆Et = F ∆s

t

P = Fv ,where P = power (in Watts, denoted W), F = force and v= velocity.

8.4.4 Gravity

Gravity is something of a special case. The accelerationdue to gravity is denoted g, and is equal to 9.81359ms−2.It is uniform over small distances from the Earth. Theforce due to gravity is equal to mg, since F = ma. There-fore:a = F

m = mgm = g

Therefore, when things are dropped, they all fall at thesame acceleration, regardless of mass. Also, the acceler-ation due to gravity (in ms−2) is equal to the gravitationalfield strength (in Nkg−1).

8.4.5 Questions

1. I hit a ball of mass 5g with a cue on a billiards tablewith a force of 20N. If friction opposes me with a force of14.2N, what is the resultant acceleration of the ball awayfrom the cue?2. A 10g ball rolls down a 1.2m high slope, and leavesit with a velocity of 4ms−1. How much work is done byfriction?3. An electric train is powered on a 30kV power supply,where the current is 100A. The train is travelling at 90kmh−1. What is the net force exerted on it in a forwardsdirection?Worked Solutions

Chapter 9

A2

42

Chapter 10

Decay

10.1 Exponential Relationships

Graph of x = aebt, where b is positive.

Many things are governed by exponential relationships.The exponential relationships which we shall be dealingwith are of the following form:x = aebt ,where t is time, x is a variable, and a and b are constants.e is just a number, albeit a very special number. It is an ir-rational constant, like π. e is 2.71828182845904523536to 20 decimal places. However, it is far easier just to findthe e (or exp) button on your calculator.The inverse function of et is the natural logarithm, de-noted ln t:ln t = loge t

10.1.1 Growth and Decay

When b is positive, an exponential function increasesrapidly. This represents the growth of certain variablesvery well. When b is negative, an exponential functiondecreases, flattening out as it approaches the t axis. This

Graph of x = e-bt, showing several different values of b.

represents the decay of certain variables.

10.1.2 Exponential Relationships in theReal World

An exponential relationship occurs when the rate ofchange of a variable depends on the value of the vari-able itself. You should memorise this definition, as wellas understand it. Let us consider some examples:

A petri dish with bacteria growing on it.

43

44 CHAPTER 10. DECAY

Population Growth

Consider a Petri dish full of agar jelly (food for bacte-ria) with a few bacteria on it. These bacteria will repro-duce, and so, as time goes by, the number of bacteriaon the jelly will increase. However, each bacterium doesnot care about whether there are other bacteria aroundor not. It will continue making more bacteria at the samerate. Therefore, as the total number of bacteria increases,their rate of reproduction increases. This is an exponen-tial relationship with a positive value of b.Of course, this model is flawed since, in reality, the bac-teria will eventually have eaten all the agar jelly, and sothe relationship will stop being exponential.

Emptying Tank

If you fill a large tank with water, and make a hole in thebottom, at first, the water will flow out very fast. How-ever, as the tank empties, the pressure of the water willdecrease, and so the rate of flow will decrease. The rateof change of the amount of water in the tank depends onthe amount of water in the tank. This is an exponential re-lationship with a negative value of b - it is an exponentialdecay.

Cooling

A hot object cools down faster than a warm object. So,as an object cools, the rate at which temperature 'flows’out of it into its surroundings will decrease. Newtonexpressed this as an exponential relationship (known asNewton’s Law of Cooling):Tt = Tenv + (T0 − Tenv)e

−rt ,where T is the temperature at a time t, T0 is the temper-ature at t = 0, Tₑ ᵥ is the temperature of the environmentaround the cooling object, and r is a positive constant.Note that a here is equal to (T0 - Tₑ ᵥ) - but a is still aconstant since T0 and Tₑ ᵥ are both constants. The '-' signin front of the r shows us that this is an exponential decay -the temperature of the object is tending towards the tem-perature of the environment. The reason we add Tₑ ᵥ ismerely a result of the fact that we do not want the temper-ature to decay to 0 (in whatever unit of temperature wehappen to be using). Instead, we want it to decay towardsthe temperature of the environment.

10.1.3 Mathematical Derivation

We have already said that an exponential relationship oc-curs when the rate of change of a variable depends onthe value of the variable itself. If we translate this intoalgebra, we get the following:dxdt = ax , where a is a constant.

By separating the variables:dx = axdt

1xdx = adt∫

1xdx =

∫adt

lnx = at+ c (where c is the constant of integration)x = eat+c = eatec

If we let b = ec (b is a constant, since ec is a constant):x = beat

10.1.4 Questions

1. Simplify Newton’s Law of Cooling for the case whenI place a warm object in a large tank of water which is onthe point of freezing. Measure temperature in °C.2. What will the temperature of an object at 40 °C beafter 30 seconds? (Take r=10−3 s−1.)3. A body is found in a library (as per Agatha Christie) at8am. The temperature of the library is kept at a constanttemperature of 20 °C for 10 minutes. During these 10minutes, the body cools from 25 °C to 24 °C. The bodytemperature of a healthy human being is 36.8 °C. At whattime was the person murdered?4. Suppose for a moment that the number of pages onWikibooks p can be modelled as an exponential relation-ship. Let the number of pages required on average to at-tract an editor be a, and the average number of new pagescreated by an editor each year be z. Derive an equationexpressing p in terms of the time in years since Wiki-books was created t.5. Wikibooks was created inmid-2003. Howmany pagesshould there have been 6 years later? (Take a = 20, z =10 yr−1.)6. The actual number of pages inWikibooks in mid-2009was 35,148. What are the problems with this model?What problems may develop, say, by 2103?Worked Solutions

10.2 Capacitors

If you place two conducting plates near each other, withan insulator (known as a dielectric) in between, and youcharge one plate positively and the other negatively, therewill be a uniform electric field between them. Since:E = V

d ,as the distance between the two plates decreases, the en-ergy stored increases. This system is known as a capacitor- it has a capacitance for storing charge. The capacitanceC of a capacitor is:C = Q

V ,

10.2. CAPACITORS 45

A 10 μF capacitor

where Q is the charge stored by the capacitor, and V isthe potential difference between the plates. C is thereforethe amount of charge stored on the capacitor per unit po-tential difference. Capacitance is measured in farads (F).Just as 1 coulomb is a massive amount of charge, a 1Fcapacitor stores a lot of charge per. volt.Any capacitor, unless it is physically altered, has a con-stant capacitance. If it is left uncharged, Q = 0, and so thepotential difference across it is 0. If a DC power source isconnected to the capacitor, we create a voltage across thecapacitor, causing electrons to move around the circuit.This creates a charge on the capacitor equal to CV. Ifwe then disconnect the power source, the charge remainsthere since it has nowhere to go. The potential differenceacross the capacitor causes the charge to 'want' to crossthe dielectric, creating a spark. However, until the voltagebetween the plates reaches a certain level (the breakdownvoltage of the capacitor), it cannot do this. So, the chargeis stored.If charge is stored, it can also be released by reconnectingthe circuit. If we were to connect a wire of negligibleresistance to both ends of the capacitor, all the chargewould flow back to where it came from, and so the chargeon the capacitor would again, almost instantaneously, be0. If, however, we put a resistor (or another componentwith a resistance) in series with the capacitor, the flowof charge (current) is slowed, and so the charge on thecapacitor does not become 0 instantly. Instead, we canuse the charge to power a component, such as a cameraflash.

10.2.1 Exponential Decay

Current is the rate of flow of charge. However, current isgiven by the formula:I = V

R

But, in a capacitor, the voltage depends on the amount ofcharge left in the capacitor, and so the current is a functionof the charge left on the capacitor. The rate of change ofcharge depends on the value of the charge itself. And so,we should expect to find an exponential relationship:Q = Q0e

− tRC ,

where R is the resistance of the resistor in series with thecapacitor, Q is the charge on the capacitor at a time t andQ0 was the charge on the capacitor at t = 0. Since Q =IΔt:I∆t = I0∆te−

tRC

I = I0e− t

RC ,where I is the current flowing at a time t and I0 was theinitial current flowing at t = 0. Since V = IR:V = V0e

− tRC

The power being dissipated across the resitors in the cir-cuit is IV, so:P = I0V0e

− tRC e−

tRC = P0e

− 2tRC

10.2.2 Energy

The energy stored by a capacitor E is defined as:E =

∫ V

0Q dV

In other words, it is the area under a graph of chargeagainst potential difference. Charge is proportional to po-tential difference (Q = CV), so the area under the graphis that of a triangle with base V and height Q. You canshow this mathematically:

E =∫ V

0CV dV = C

[V 2

2

]V0= 1

2CV 2

Since Q = CV:E = 1

2QV

10.2.3 Circuits

The circuit symbol for a capacitor is . A simple circuitwith a capacitor in series with a resistor, an ideal amme-ter (no resistance), and in parallel with an ideal voltmeter(infinite resistance) looks like the following:

46 CHAPTER 10. DECAY

In the position shown, the capacitor is charging. If theswitch were put in the other position, the capacitor wouldbe discharging exponentially through the resistor. In thiscircuit, the capacitor charges instantly since there is noresistance to slow it down. In reality there will be inter-nal resistance in the battery, meaning that the capacitorcharges exponentially.If capacitors are placed in parallel, they act as one ca-pacitor with a capacitance equal to the total of all the ca-pacitances of all the individual capacitors. If capacitorsare placed in series, the distances between the plates ineach of them result in the capacitance of the imaginaryresultant capacitor ΣC being given by:1

ΣC = 1C1

+ 1C2

+ ...+ 1Cn

10.2.4 Questions

1. A 2 mF capacitor is connected to a 10V DC powersupply. Howmuch charge can be stored by the capacitor?2. What is the highest possible energy stored by this ca-pacitor?3. The capacitor is placed in series with a 5Ω resistorand charged to capacity. How long would it take for thecharge in the capacitor to be reduced to 1 mC?4. After this time has elapsed, how much energy is storedin the capacitor?5. What is the capacitance of the equivalent capacitor tothe following network of capacitors?

Worked Solutions

10.3 Radioactive Decay

10.3.1 Decay Constant

We can model radioactive decay by assuming that theprobability that any one nucleus out of N nuclei decaysin any one second is a constant λ. λ is known as the decayconstant, and is measured in s−1 (technically the same asHz, but it is a probability, not a frequency, so we use s−1).

10.3.2 Activity

As our N nuclei decay, the number of nuclei decreases.The activity of the N nuclei we have left is, on average,the probability that any one nucleus will decay per. unittime multiplied by the number of nuclei. If we have 200nuclei, and the decay constant is 0.5, we would expect,on average, 100 nuclei to decay in one second. This ratewould decreases as time goes by. This gives us the follow-ing formula for the activity A of a radioactive sample:A = −dN

dt = λN

Activity is always positive, and is measured in becquerels(Bq). It is easy to see that the rate of change of the num-ber of nuclei is -A = -λN.

10.3.3 Decay

The solution of the differential equation for activity givenabove is an exponential relationship:N = N0e

−λt ,where N is the number of nuclei present at a time t, andN0 is the number of nuclei present at time t=0. You candefine t=0 to be any point in time you like, provided youare consistent. Since A = λN and therefore A0 = λN0:A = A0e

−λt ,where A is the activity of the sample at a time t, and A0

is the activity at time t=0.

10.4. HALF-LIVES 47

10.3.4 Questions

1 mole = 6.02 x 1023 atoms1u = 1.66 x 10−27kg1. Americium-241 has a decay constant of 5.07 x10−11s−1. What is the activity of 1 mole of americium-241?2. How many g of lead-212 (λ = 18.2μs−1) are requiredto create an activity of 0.8 x 1018Bq?3. How long does it take for 2kg of lead-212 to decay to1.5kg of lead-212?4. Where does the missing 0.5kg go?5. Some americium-241 has an activity of 3kBq. Whatis its activity after 10 years?6. This model of radioactive decay is similar to takingsome dice, rolling them once per. second, and removingthe dice which roll a one or a two. What is the decayconstant of the dice?7. If you started out with 10 dice, how many dice wouldyou have left after 10s? What is the problem with thismodel of radioactive decay?Worked Solutions

10.4 Half-lives

The half life of something that is decaying exponentiallyis the time taken for the value of a decaying variable tohalve.

10.4.1 Half Life of a Radioisotope

The most common use of half-lives is in radioactive de-cay. The activity is given by the equation:At = A0e

−λt

At t=t½, A = ½A0, so:A0

2 = A0e−λt 1

2 = A0

eλt 1

2

2 = eλt 1

2

ln 2 = λt 12

Therefore:t 12= ln 2

λ

It is important to note that the half-life is completely unre-lated to the variable which is decaying. At the end of thehalf-life, all decaying variables will have halved. This alsomeans that you can start at any point in the decay, withany value of any decaying variable, and the time taken forthe value of that variable to halve from that time will bethe half-life.

10.4.2 Half-Life of a Capacitor

You can also use this formula for other forms of decaysimply by replacing the decay constant λ with the constantthat was in front of the t in the exponential relationship.So, for the charge on a capacitor, given by the relation-ship:Qt = Q0e

−tRC

So, substitute:λ = 1

RC

Therefore, the half-life of a capacitor is given by:t 12= RC ln 2

10.4.3 Time Constant of a Capacitor

However, when dealing with capacitors, it is more com-mon to use the time constant, commonly denoted τ,where:

τ = RC =t 12

ln 2

At t = τ:Qt = Q0e

−RCRC = Q0

e

So, the time constant of a capacitor can be defined asthe time taken for the charge, current or voltage from thecapacitor to decay to the reciprocal of e (36.8%) of theoriginal charge, current or voltage.

10.4.4 Questions

1. Radon-222 has a decay constant of 2.1μs−1. What isits half-life?2. Uranium-238 has a half-life of 4.5 billion years. Howlong will it take for a 5 gram sample of U-238 to decayto contain 1.25 grams of U-238?3. How long will it be until it contains 0.5 grams of U-238?4. Tritium, a radioisotope of Hydrogen, decays intoHelium-3. After 1 year, 94.5% is left. What is the half-life of tritium (H-3)?5. A large capacitor has capacitance 0.5F. It is placed inseries with a 5Ω resistor and contains 5C of charge. Whatis its time constant?6. How long will it take for the charge in the capacitor toreach 0.677C? ( 0.677 = 5

e2 )Worked Solutions

Chapter 11

Gravity

11.1 Force

Gravity is a force. Any object with mass exerts a gravita-tional force on any other object with mass. The attractiveforce of one object on another is proportional to the prod-uct of their masses. However, this force is also inverselyproportional to the distance between the objects squared.This means that, if two objects are twice as far away, theforces they exert on each other are four times smaller.Thus gravitational force exerted by a sphere of mass Mon another sphere of mass m is given by the followingformula:Fgrav = GMm

r2 ,where r is the distance between the spheres, and G is theGravitational constant. Experiments have shown that G =6.67 x 10−11 Nm2kg−2.

11.1.1 Gravitational Force Inside an Ob-ject

Inside a roughly spherical object (such as the Earth), itcan be proved geometrically that the effects of the gravi-tational force resulting from all the mass outside a radiusat which an object is located can be ignored, since it allcancels itself out. So, the only mass we need to consideris the mass inside the radius at which the object is located.The density of an object ρ is given by the following equa-tion:ρ = M

V ,where M is mass, and V is volume. Therefore:M = ρV

If we substitute the volume of a sphere for V:M = 4

3πρr3

And if we substitute this mass into the formula for grav-itational force given above:

Fgrav =−Gm 4

3πρr3

r2 = − 43πGρmr

In other words, inside a sphere of uniformmass, the grav-itational force is directly proportional to the distance ofan object from the centre of the sphere. Incidentally, this

A lift acting under gravity in a lift shaft going through the centreof the Earth.

results in a simple harmonic oscillator such as the one onthe right. This means that a graph of gravitational forceagainst distance from the centre of a sphere with uniformdensity looks like this:

48

11.2. FIELD 49

11.1.2 Questions

1. Jupiter orbits the Sun at a radius of around 7.8 x1011m. The mass of Jupiter is 1.9 x 1027kg, and the massof the Sun is 2.0 x 1030kg. What is the gravitational forceacting on Jupiter? What is the gravitational force actingon the Sun?2. The force exerted by the Sun on an object at a certaindistance is 106N. The object travels half the distance tothe Sun. What is the force exerted by the Sun on theobject now?3. What is the magnitude of the gravitational force thattwo 1kg weights exert on each other when they are 5cmapart?4. The radius of the Earth is 6360km, and its mass is5.97 x 1024kg. What is the difference between the grav-itational force on 1kg at the top of your body, and on 1kgat your head, and 1kg at your feet? (Assume that you are2m tall.)Worked Solutions

11.2 Field

The gravitational field, or gravitational field strength is theforce exerted by gravity on an object per. unit mass of theobject:g =

Fgrav

m

As gravitational field strength is a measure of the forceexerted on each unit of mass, its unit is Nkg−1. If weconsider a planet, Body A, the gravitational field strength

experienced by another object, Body B, is given by:

g =−GMm

r2

m = −GMr2

This is the total force exerted on Body B divided by themass of Body B. Inside the planet, force is proportionalto the distance from the centre, so the field is also propor-tional to distance.

11.2.1 Acceleration

Force is given by:F = ma

This means that:g = ma

m = a

This shows that the gravitational field strength is also theacceleration due to gravity on any object. This accelera-tion is the same for any object, regardless of mass. Whenconsidering small heights above the Earth’s surface, suchas those in our day-to-day experiences, g remains roughlyconstant.

11.2.2 Field Lines

Gravitational field lines around the Earth.

The gravitational field can be represented using field lines.These run in the direction that a mass would be acceler-ated in initially. The object will not necessarily fall alongthe field lines, but the acceleration will always be in thedirection of the field lines. The closer the field lines aretogether, the denser the gravitational field.

11.2.3 Questions

G = 6.67 x 10−11 m3kg−1s−2

50 CHAPTER 11. GRAVITY

1. A 15kg object has a weight of 8000N. What is thegravitational field strength at this point?2. Draw a graph of gravitational field strength against dis-tance.3. What is the gravitational field strength of the Sun (mass2 x 1030kg) on the Earth (mass 6 x 1024kg, mean orbitalradius 15 x 1010m)?4. What is the difference in the acceleration due to gravityover a vertical distance d?5. How far would one have to travel upwards from theEarth’s surface to notice a 1Nkg−1 difference in gravita-tional field? (The Earth has a radius of 6400 km.)Worked Solutions

11.3 Potential Energy

If you throw a ball into the air, you give it kinetic en-ergy. The ball then slows down because of the effect ofthe Earth’s gravitational field on it. However, we knowthat energy cannot be created or destroyed. The kineticenergy you gave the ball is transformed into gravitationalpotential energy. The further away from the Earth youmanage to throw the ball, the greater the potential thereis for kinetic energy to be created on the way back down.However, for kinetic energy to be created, there must bean acceleration. If there is an acceleration, there must bea force.You should already know that energy is the same as thework done to move something a distance Δx:∆Egrav = F∆x

Work done is given by the force applied multiplied by thedistance moved in the direction of the force. To movean object against gravity, the force applied upwards mustequal the downwards force gravity exerts on the object,mg. So, if I move an object against gravity a distance Δx,the work done is given by:∆Egrav = mg∆x

It is usual to call this x the height, so you will often seeE ᵣₐᵥ=mgh. The deltas are important. They mean that itdoesn't matter which distance x I move the object across- I can decide the point at which gravitational potentialenergy is 0 in a way which makes the maths easy.The difficulty with this simple formula is that g does notremain the same over large distances:g = −GM

r2

So, over a distance Δr, x becomes r and so:Egrav = −GMmr

r2 = −GMmr

So, if you're dealing with gravitational potential energyover large distances, use this formula. If you're deal-ing with gravitational potential energy over short dis-

tances, such as with ramps on the Earth’s surface, whereg=9.81ms−2, use E ᵣₐᵥ = mgh.

11.3.1 Graphs

The area under a graph of gravitational force against distance isthe gravitational potential energy.

The gradient of a graph of gravitational potential energy againstdistance is the magnitude of the force.

We have just done something sneaky. You probablydidn't notice. Let’s see what happens when we integratethe gravitational force F with respect to r between r and∞:∫∞r

−GMmr2 dr =

[GMm

r

]∞r

= GMm∞ − GMm

r

Since dividing anything by infinity gets you practically 0:∫∞r

Fdr = −GMmr = Egrav

And therefore:dEgrav

dr = F

So, if you have a graph of gravitational potential energyagainst radius, the gradient of the graph is the gravita-tional force. If you have a graph of gravitational forceagainst radius, the area under the graph between any pointand the F-axis is the gravitational potential energy at thispoint. The area under the graph between any two pointsis the difference in gravitational potential energy betweenthem.

11.4. POTENTIAL 51

11.3.2 Questions

1. A ball rolls down a 3m-high smooth ramp. What speeddoes it have at the bottom?2. In an otherwise empty universe, two planets of mass1025 kg are 1012 m apart. What are their speeds whenthey collide?3. What is the least work a 2000kg car must do to driveup a 100m hill?4. How does the speed of a planet in an elliptical orbitchange as it nears its star?Worked Solutions

11.4 Potential

Gravitational potential is the gravitational potential en-ergy given to objects per unit mass:

Vgrav =Egrav

m =−GMm

r

m = −GMr

For short changes in distance Δh (Δh << r), gravitationalpotential energy is given by:Egrav ≈ mg∆h ,whereg = −GM

r2

So, for short changes of distance, gravitational potentialis equal to gΔh. Gravitational potential is measured inJkg−1.

11.4.1 Equipotentials

On a field diagram, lines can be drawn which, like con-tours on a map, show all the points which have the samegravitational potential. These lines are known as equipo-tentials. Equipotentials are always perpendicular to fieldlines, and get closer together as the field strength in-creases, and the density of field lines increases. Overshort distances, equipotentials are evenly spaced.

11.4.2 Summary of Gravity

You should now know (if you did the gravity section in theright order) about four attributes of gravitational fields:force, field strength, potential energy and potential. Thesecan be summarised by the following table:

11.4.3 Questions

G = 6.67 x 10−11 m3kg−1s−2

g = 9.81 ms−2

1. What is the gravitational potential at the Earth’s sur-face? (mass of Earth = 5.97 x 1024 kg,radius of Earth =6371 km)2. Taking the Earth’s surface as V ᵣₐᵥ = 0, what is thegravitational potential 2m above the ground?3. A 0.2kg firework reaches a gravitational potential rel-ative to the ground of 500Jkg−1. If the firework is 30%efficient, how much energy was expended to get there?4. Express gravitational potential in terms of gravitationalforce.5. Draw the equipotentials and field lines surrounding theEarth.Worked Solutions

Chapter 12

Mechanics

12.1 Simple Harmonic Motion

Simple harmonic motion occurs when the force on an ob-ject is proportional and in the opposite direction to thedisplacement of the object. Examples include masses onsprings and pendula, which 'bounce' back and forth re-peatedly. Mathematically, this can be written:F = −kx ,

T

A

t

x(t)

Graph of displacement against time in simple harmonic motion.

where F is force, x is displacement, and k is a positiveconstant. This is exactly the same as Hooke’s Law, whichstates that the force F on an object at the end of a springequals -kx, where k is the spring constant. Since F = ma,and acceleration is the second derivative of displacementwith respect to time t:md2x

dt2 = −kx

d2xdt2 = −kx

m

The solution of this second order differential equation is:x = A cosωt ,where A is the maximum displacement, and ω is the 'an-gular velocity' of the object. The derivation is given here,since it will seem very scary to those who haven't metcomplex numbers before. It should be noted that this so-lution, if given different starting conditions, becomes:x = A sinωt ,

12.1.1 Angular Velocity

Angular velocity in circular motion is the rate of changeof angle. It is measured in radians per. second. Since2π radians is equivalent to one complete rotation in timeperiod T:ω = 2π

T = 2πf

If we substitute this into the equation for displacement insimple harmonic motion:x = A cos 2πftThe reason the equation includes angular velocity is thatsimple harmonic motion is very similar to circular mo-tion. If you look at an object going round in a circleside-on, it looks exactly like simple harmonic motion. Wehave already noted that a mass on a spring undergoes sim-ple harmonic motion. The following diagram shows thesimilarity between circular motion and simple harmonicmotion:

52

12.2. ENERGY IN SIMPLE HARMONIC MOTION 53

12.1.2 Time Period

The time period of an oscillation is the time taken to re-peat the pattern of motion once. In general:T = 2π

ω

However, depending on the type of oscillation, the valueof ω changes. For a mass on a spring:

ω =√

km

For a pendulum:ω =

√gl ,

where g is the gravitational field strength, and l is thelength of the string. By substitution, we may gain thefollowing table:

12.1.3 Velocity and Acceleration

The displacement of a simple harmonic oscillator is:x = A cosωtVelocity is the rate of change of displacement, so:v = dx

dt = −Aω sinωtAcceleration is the rate of change of velocity, so:a = dv

dt = d2xdt2 = −Aω2cosωt = −ω2x

12.1.4 Questions

1. A 10N weight extends a spring by 5cm. Another10Nweight is added, and the spring extends another 5cm.What is the spring constant of the spring?2. The spring is taken into outer space, and is stretched10cm with the two weights attached. What is the timeperiod of its oscillation?3. What force is acting on the spring after 1 second? Inwhat direction?4. A pendulum oscillates with a frequency of 0.5 Hz.What is the length of the pendulum?5. The following graph shows the displacement of a sim-ple harmonic oscillator. Draw graphs of its velocity, mo-mentum, acceleration and the force acting on it.

6. A pendulum can only be modelled as a simple har-monic oscillator if the angle over which it oscillates issmall. Why is this?Worked Solutions

12.2 Energy in Simple HarmonicMotion

A mass oscillating on a spring in a gravity-free vacuumhas two sorts of energy: kinetic energy and elastic (po-tential) energy. Kinetic energy is given by:Ek = 1

2mv2

Elastic energy, or elastic potential energy, is given by:Ee =

12kx

2

So, the total energy stored by the oscillator is:ΣE = 1

2 (mv2 + kx2)

This total energy is constant. However, the proportions ofthis energy which are kinetic and elastic change over time,since v and x change with time. If we give a spring a dis-placement, it has no kinetic energy, and a certain amountof elastic energy. If we let it go, that elastic energy isall converted into kinetic energy, and so, when the massreaches its initial position, it has no elastic energy, and allthe elastic energy it did have has been converted into ki-netic energy. As the mass continues to travel, it is slowedby the spring, and so the kinetic energy is converted backinto elastic energy - the same amount of elastic energy asit started out. The nature of the energy oscillates backand forth, but the total energy is constant.

54 CHAPTER 12. MECHANICS

If the mass is oscillating vertically in a gravitational field,the situation gets more complicated since the spring nowhas gravitational potential energy, elastic potential energyand kinetic energy. However, it turns out (if you do themaths) that the total energy is still constant, although theequilibrium position will be lower.

12.2.1 Questions

1. A 10g mass causes a spring to extend 5cm. How muchenergy is stored by the spring?2. A 500g mass on a spring (k=100) is extended by 0.2m,and begins to oscillate in an otherwise empty universe.What is the maximum velocity which it reaches?3. Another 500g mass on another spring in another oth-erwise empty universe is extended by 0.5m, and beginsto oscillate. If it reaches a maximum velocity of 15ms−1,what is the spring constant of the spring?4. Draw graphs of the kinetic and elastic energies of amass on a spring (ignoring gravity).5. Use the trigonometric formulae for x and v to derive anequation for the total energy stored by an oscillating masson a spring, ignoring gravity and air resistance, which isconstant with respect to time.Worked Solutions

12.3 Damping

Previously, our mathematical models of simple harmonicmotion have assumed that the energy stored by a simpleharmonic oscillator is constant. In reality, of course, re-sistive forces slow an oscillator down, transferring its en-ergy to its surroundings. A pendulum will lose energy bymoving the air. In addition to this, the motion of a masson a spring will cause the spring to heat up, 'losing' theenergy. This process is known as damping.The principal effect of damping is to reduce the ampli-tude of an oscillation, not to change its frequency. So,the graph of the amplitude of a normal damped oscilla-tion might look like the following:

12.3.1 Critical Damping

Critical damping occurs when a system is designed toreturn an oscillator to its equilibrium position in theleast time possible. A critically damped oscillator, whendamped, ceases to oscillate, and returns to its equilibriumposition, where it stops moving. An example is the doorcloser. Normally, the door would swing back and forth,being damped by friction in the hinges, and air resistance.The door closer forces the door to stop swinging, and shutimmediately. When closed, the door is in its equilibriumposition.

12.3.2 Questions

1. Draw a graph of displacement for a critically dampedoscillation.2. How would you critically damp an oscillating pendu-lum?3. How would you damp an oscillating pendulum usingonly a weighted polystyrene block?4. What would the displacement graph look like for thisoscillation, before and after damping began?5. The graph above is an exponentially damped oscilla-tion. If the displacement of the undamped oscillation isgiven by sin ωt, what is an approximate equation for thedamped oscillation, in terms of a constant k which de-scribes the degree to which the oscillation is damped?Worked Solutions

12.5. CONSERVATION OF MOMENTUM 55

12.4 Resonance

Resonance occurs when an oscillating system is driven(made to oscillate from an outside source) at a frequencywhich is the same as its own natural frequency. All oscil-lating systems require some form of an elastic force anda mass e.g. a mass at the end of a spring. All oscillatorshave a natural frequency. If you have a mass on a spring,and give it an amplitude, it will resonate at a frequency:

f = 12π

√km

This frequency is independent of the amplitude you givethe oscillator to start with. It is the natural frequency ofthe oscillator. If you keep giving the oscillator amplitudeat this frequency, it will not change the frequency of theoscillation. But, you are still doing work. This energymust go somewhere. The only place it can go is into ad-ditional kinetic and gravitational potential energy in theoscillation. If you force an oscillation at its resonant fre-quency, you add significantly to its amplitude.Put simply, resonance occurs when the driving frequencyof an oscillation matches the natural frequency, givingrise to large amplitudes.If you were to force an oscillation at a range of frequen-cies, and measure the amplitude at each, the graph wouldlook something like the following:

There are many types of oscillators, and so practicallyeverything has a resonant frequency. This can be used,or can result in damage if the resonant frequency is notknown.

12.4.1 Reading

Instead of doing questions this time, read the followingarticles on Wikipedia about these different types:Tacoma Narrows BridgeResonance in Water Molecules (Microwave Ovens)“No Highway” - a novel with a plot that uses things sus-piciously similar to resonance.Millenium Bridge

12.5 Conservation of Momentum

Momentum is the product of the mass of an object andits velocity. It is usually denoted p:p = mv ,where m is mass, and v is velocity. The total momentumin a closed system is always conserved. This fact is use-ful, since it allows us to calculate velocities and masses incollisions.

12.5.1 Collisions

Let us consider a basic example: a ball of massM collideswith velocity u with a stationary ball of mass m. The sta-tionary ball has no momentum before the collision, andthe moving ball has momentum Mu. This must equal themomentum of both balls after the collision. If we let theirvelocities be v1 and v2:Mu = Mv1 +mv2

At this point, we would need to know one of the velocitiesafterwards in order to calculate the other.Alternatively, we could have one ball of mass M collidingwith another ball of mass m, with both balls moving inopposite directions with velocities u1 and u2 respectively.If we define the direction of motion of the ball with massM as the positive direction:Mu1 −mu2 = Mv1 +mv2

We do not need to worry about the signs on the right-hand side: they will take care of themselves. If one ofour velocities turns out to be negative, we know that it isin the opposite direction to u1.

12.5.2 Elasticity

Although momentum within a closed system is alwaysconserved, kinetic energy does not have to be. If kineticenergy is conserved in a collision, then it is known asa perfectly (or totally) elastic collision. If it is not con-served, then the collision is inelastic. If the colliding par-ticles stick together, then a totally inelastic collision has

56 CHAPTER 12. MECHANICS

The collisions in a Newton’s cradle are almost perfectly elastic.

occurred. This does not necessarily mean that the parti-cles have stopped. In a totally inelastic collision, the twoparticles become one, giving the equation:Mu1 +mu2 = (M +m)v

v = Mu1+mu2

M+m

12.5.3 Explosions

In an explosion, two particles which are stuck together areno longer stuck together, and so gain separate velocities:(M +m)u = Mv1 +mv2

12.5.4 Questions

1. A ball of mass 0.5kg collides with a stationary ball of0.6kg at a velocity of 3ms−1. If the stationary ball movesoff at a speed of 2ms−1, what is the new velocity of thefirst ball?2. Two balls are moving in opposite directions with ve-locities 5ms−1 and 10ms−1. They collide, and move offin opposite directions with new velocities of 7.5ms−1

each. If the mass of the first ball was 1.25kg, what isthe mass of the second ball?3. A totally elastic collision occurs between two balls ofequal mass. One of the balls is stationary. What happens?4. A particle explodes to become two particles withmasses 1kg and 2kg. The 1kg particle moves with veloc-ity 45ms−1. With what velocity does the other particlemove?5. A 3kg ball moving at 3ms−1 collides with a 5kg ballmoving at −5ms−1. The collision is perfectly elastic.What are the new velocities of the balls?6. A ball collides with a wall, and rebounds at the samevelocity. Why doesn't the wall move?Worked Solutions

12.6 Forces and Impulse in Colli-sions

You should already know that the force exerted on an ob-ject is proportional to its acceleration. The constant ofproportionality is known as the mass of the object:F = ma

In the case of a collision, for one of the particles in the col-lision, the acceleration is simply the difference betweenits velocity before the collision (u) and its velocity afterthe collision (v) per unit. time:F = m(v−u)

∆t = mv−mu∆t

So, force is the rate of change of momentum. The quan-tity on top is known as the impulse of the collision, mea-sured in Ns; Δ t is the length of time it took for the colli-sion to take place. So, the impulse I is given by:I = ∆p = mv −mu = F∆t

In a collision where a certain change in momentum (im-pulse) occurs, a force is exerted. If the collision time issmall, a larger force is exerted. If the collision time islong, a smaller force is exerted. If you have a graph offorce against time, impulse is the area under the graph,since:I =

∫F dt

The impulse on one particle in a simple collision is thenegative impulse on the other particle.

12.6.1 Questions

1. Escape velocity from the Earth is 11.2kms−1. Howmuch impulse must be exerted on a 47000kg payload toget it to travel away from the Earth?2. Two billiard balls, of mass 10g, collide. One is movingat 5ms−1, and the other at 2ms−1. After the collision,the first billiard ball is moving backwards at 4ms−1. Thecollision takes 1 ms. What force was exerted on this ball?3. What impulse and force were exerted on the secondball?4. A 60kg spacewalker uses a jet of gas to exert an im-pulse of 10Ns. How many times would he have to do thisto reach a speed of 1 ms−1 from stationary?5. A 5kg bowling ball collides with a stationary tennis ballof mass 0.1kg at 3ms−1, slowing to 2.5ms−1. It exerts aforce of 100N on the ball. How long did the collisiontake?Worked Solutions

12.8. CIRCULAR MOTION 57

12.7 Rockets, Hoses and MachineGuns

Apollo 15 launches itself to the moon by means of the change inmomentum of its fuel.

We have already seen that force is the rate of change ofmomentum. This applies to continuous flows of momen-tum as well as to collisions:F = dp

dt = v dmdt

If I have a machine gun, explosions give the bullets ofmass m momentum, causing them to move at a velocityv. This occurs several times each second - the momentumof the bullets is changing, and so there is a roughly contin-uous force acting on them. Momentum, of course, mustbe conserved. This results in a change in the momentumof the gun each time it fires a bullet. Overall, this resultsin a roughly continuous force on the gun which is equaland opposite to the force acting on the bullets.If I have a tank of water and a hose, with a pump, and Ipump the water out of the tank, a similar thing occurs -a force pushes me away from the direction of flow of thewater. This force is equal to the flow rate (in kgs−1) ofthe water multiplied by its velocity. Bear in mind that 1litre of water has a mass of about 1kg.Rockets work on this principle - they pump out fuel, caus-ing it to gain momentum. This results in a thrust on therocket. When designing propulsion systems for rockets,the aim is to give the fuel as high a velocity per. unit massas possible in order to make the system fuel-efficient, andto get a high enough change in momentum.

12.7.1 Questions

1. A machine gun fires 300 5g bullets per. minute at800ms−1. What force is exerted on the gun?2. 1 litre of water is pumped out of a tank in 5 secondsthrough a hose. If a 2N force is exerted on the tank, atwhat speed does the water leave the hose?3. If the hose were connected to the mains, what prob-lems would there be with the above formula?4. The thrust of the first stage of a Saturn V rocket is34 MN, using 131000kg of solid fuel in 168 seconds. Atwhat velocity does the fuel leave the tank?5. Escape velocity from the Earth is 11km−1. What isthe velocity of the rocket after the first stage is used up,if the total mass of the rocket is 3 x 106 kg? How doesthis compare to escape velocity?Worked Solutions

12.8 Circular Motion

Diagram showing the key variables involved in circular motion.

Very rarely, things move in circles. Some planets move inroughly circular orbits. A conker on a string might movearound my head in a circle. A car turning a corner might,briefly, move along the arc of a circle. The key thing tonote about circular motion is that there is no force pullingoutwards from the circle, and there is no force pulling themoving object tangential to the circle. Centrifugal forcedoes not exist. There is only one force acting in circularmotion, which is known as centripetal force. It alwaysacts towards the centre of the circle. The object does notfollow a circular path because two forces are balanced.Instead, the centripetal force accelerates the object with

58 CHAPTER 12. MECHANICS

a constant magnitude in an ever-changing direction. Theobject has a velocity, and will continue moving with thisvelocity unless acted on by the centripetal force, whichis perpetually adding velocity towards the centre of thecircle.If you were to subject a stationary object to the centripetalforce, it would simply fall. If you gave it a little bit of ve-locity, it would still fall, but it would not land directly be-neath its starting position. If you kept increasing the ve-locity and dropping it, there would come a point when itwould land infinitely far away - it would go into orbit. Therelationship between this 'magic' velocity and the magni-tude of the centripetal force is as follows:F = mv2

r ,where m is the mass of the object in circular motion, v isthe magnitude of its velocity, and r is the distance fromthe centre of the circle to the object. Since F=ma, thecentripetal acceleration is:a = v2

r

The centripetal force may manifest itself as many things:the tension in a string, friction, gravity or even an electricor magnetic field. In all these cases we can equate theequation for centripetal force with the equation for theforce it really is.

12.8.1 Angular Velocity

Velocity is the rate of change of displacement. Angularvelocity is the rate of change of angle, commonly denotedω and measured in radians per. second:ω = ∆θ

∆t

In circular motion:ω = 2π

T = 2πf ,where T is the time for one revolution and f is the fre-quency of rotation. However:v = 2πr

T

vr = 2π

T

Therefore, the relationship between velocity and angularvelocity is:ω = v

r

If we substitute this into the formula for centripetal ac-celeration:a = (ωr)2

r = ω2r2

r = ω2r

12.8.2 Questions

1. A tennis ball of mass 10g is attached to the end of a0.75m string and is swung in a circle around someone’shead at a frequency of 1.5Hz. What is the tension in thestring?

2. A planet orbits a star in a circle. Its year is 100 Earthyears, and the distance from the star to the planet is 70Gm from the star. What is the mass of the star?3. A 2000kg car turns a corner, which is the arc of acircle, at 20kmh−1. The centripetal force due to frictionis 1.5 times the weight of the car. What is the radius ofthe corner?4. Using the formulae for centripetal acceleration andgravitational field strength, and the definition of angularvelocity, derive an equation linking the orbital period ofa planet to the radius of its orbit.Worked Solutions

Chapter 13

Astrophysics

13.1 Radar and Triangulation

Radar and triangulation are two relatively easy meth-ods of measuring the distance to some celestial objects.Radar can also be used to measure the velocity of a ce-lestial object relative to us.

13.1.1 Radar

Essentially, radar is a system which uses a radio pulse tomeasure the distance to an object. The pulse is transmit-ted, reflected by the object, and then received at the siteof the transmitter. The time taken for all this to happenis measured. This can be used to determine the distanceto a planet or even the velocity of a spaceship.

Distance

The speed of electromagnetic waves (c) is constant in avacuum: 3 x 108 ms−1. If we fire a pulse of radio wavesto a planet within the Solar System, we know that:d = ct

where d is the distance to the object, and t is the timetaken for the pulse to travel there and back from the ob-ject. However, the pulse has to get both there and back,so:2d = ct

d = ct2

where d is the distance to the object, and t is the timetaken for the pulse to return.

Velocity

The velocity of an object can be found by firing two radarpulses at an object at different times. Two distances aremeasured. When asked to calculate the relative velocityof an object in this way, use the following method:1. Calculate the distance to the object at both times:d1 = c∆t1

2

d2 = c∆t22

2. Calculate the distance the object has travelled betweenthe two pulses. This is the difference between the twodistances previously calculated:∆d = d2 − d1

3. Calculate the time between the transmission (or recep-tion, but not both) of the two pulses:∆t = t2 − t1

4. Divide the distance calculated in step 2 by the time cal-culated in step 3 to find the average velocity of the objectbetween the transmission of the two pulses:v = ∆d

∆t

13.1.2 Triangulation

The distance from the solar system to a relatively nearby celestialobject can be found using triangulation.

We know that the Earth is, on average, about 150 Gmaway from the Sun. If we measure the angle between thevertical and the light from a nearby star 6 months apart(i.e. on opposite sides of the Sun), we can approximatethe distance from the Solar System to the star.Let r be the radius of the Earth’s orbit (assumed to beconstant for simplicity’s sake), a and b be the angles tothe star (from the horizontal) when the Earth is on eitherside of the Sun, and let d be the perpendicular distance

59

60 CHAPTER 13. ASTROPHYSICS

from the plane of the Earth’s orbit to the star, as shownin the diagram on the right. By simple trigonometry:2r = d

tan a + dtan b = d(tan a+tan b)

tan a tan b

Therefore:d = 2r tan a tan b

tan a+tan b

13.1.3 Questions

1. A radar pulse takes 8 minutes to travel to Venus andback. How far away is Venus at this time?2. Why can't a radar pulse be used to measure the dis-tance to the Sun?Regardless of (λ) wavelength, power density, or wave-front properties, the pulse would be absorbed with no re-flection possible. Distances to pure energy sources aregenerally measured in terms of received light intensity,shifts of the light spectrum, and radio interferometry.The RF spectrum; and Laser (light) spectrum can be usedto “listen” to radiation, but not bounce a pulse from an en-ergy source having no true angle of incidence. Just as anobservation, I will note that the sun can be seen on mostradars either sunrise or sunset, usually when the sun isjust above the horizon. But these receptions are unusablestrobes (interference) and not a result of receiving a radarpulse from the sun. Radar technicians also use the sun asa “known” exact position to align the system to true north(andmagnetic variations); this is called solar-boresightingand, again, only receives the radiation.3. Radar is used to measure the velocity of a spacecrafttravelling between the Earth and the Moon. Use the fol-lowing data to measure this velocity:4. The angles between the horizontal and a star are mea-sured at midnight on January 1 as 89.99980° and at mid-night on July 1 as 89.99982°. How far away is the star?5. Why can't triangulation be used to measure the dis-tance to another galaxy?Worked Solutions

13.2 Large Units

The distances in space are so large that we need some verylarge units to describe them with.

13.2.1 Light Years

A light year is the distance that light travels in one year.The velocity of light is constant (3 x 108ms−1), so 1 lightyear is:3× 108 × 365.24× 24× 60× 60 ≈ 9.46× 1015 mLight seconds, light minutes, light hours and light days are

less commonly used, but may be calculated in a similarfashion.

13.2.2 Astronomical Units

1 astronomical unit (denoted AU) is the mean averagedistance from the Earth to the Sun. This is approximately150 x 109m.

13.2.3 Parsecs

Degrees can be divided into minutes and seconds. Thereare 60 minutes in a degree, and 60 seconds in a minute.This means that 1 second is 1/3600 of a degree. A degreeis denoted °, a minute ' and a second ". The definitionof a parsec uses a simplified form of triangulation. It as-sumes that the perpendicular to the plane of the Earth’sorbit passes through the Sun and a celestial object. Aparsec is the distance from the Sun to this celestial objectif the angle between the perpendicular and the line con-necting the Earth to the celestial object is one arcsecond(1/3600 of a degree). This gives us the following right-angled triangle (the distance from the Earth to the Sun is1 AU):

tan 1′′ = tan(

13600

)= 4.85× 10−6 = 1parsec

Therefore, a parsec is 206,265 AU.

13.2.4 Questions

1. What is one parsec in m?2. Convert 3 light days into km.3. Convert 5.5 parsecs into light years.4. The difference in angle of a star on the perpendicularto the plane of the Earth’s orbit which passes through theSun when viewed from either side of the Earth’s orbit is0.1°. How far away is the star in parsecs?Worked Solutions

13.3 Orbits

When a planets orbit a star, theoretically, their orbit maybe circular. This case is dealt with under circular motion.

13.4. DOPPLER EFFECT 61

The total of the distances from any point on an ellipse to its fociis constant.

In reality, planets orbit in ellipses. An ellipse is a shapewhich has two foci (singular 'focus’). The total of the dis-tances from any point on an ellipse to its foci is constant.All orbits take an elliptical shape, with the sun as one ofthe foci. As the planet approaches its star, its speed in-creases. This is because gravitational potential energy isbeing converted into kinetic energy.A circle is an ellipse, in the special case when both fociare at the same point.

13.3.1 Kepler’s Third Law

Kepler’s third law states that:Mathematically, for orbital period T and semi-major axisR:T 2 ∝ R3

This result was derived for the special case of a circularorbit as the fourth circular motion problem. The semi-major axis is the distance from the centre of the ellipse(the midpoint of the foci) to either of the points on theedge of the ellipse closest to one of the foci.

13.3.2 Questions

1. The semi-major axis of an elliptical orbit can be ap-proximated reasonably accurately by the mean distanceof the planet for the Sun. How would you test, using thedata in the table above, that the inner planets of the SolarSystem obey Kepler’s Third Law?2. Perform this test. Does Kepler’s Third Law hold?3. If T2 α R3, express a constant C in terms of T and R.4. Io, one of Jupiter’s moons, has a mean orbital radiusof 421600km, and a year of 1.77 Earth days. What is the

value of C for Jupiter’s moons?5. Ganymede, another of Jupiter’s moons, has a meanorbital radius of 1070400km. How long is its year?Worked Solutions

13.4 Doppler Effect

The Doppler effect is a change in the frequency of a wavewhich occurs if one is in a different frame of referencefrom the emitter of the wave. Relative to us, we observesuch a change if an emitter of a wave is moving relativeto us.All waves travels in a medium. So, they have a velocityrelative to this medium v. They also have a velocity rela-tive to their source v and a velocity relative to the placewhere they are received vᵣ. The frequency at which theyare received f is related to the frequency of transmissionf0 by the formula:

f =(

v+vrv+vs

)f0

The Doppler effect can be used to measure the velocityat which a star is moving away from or towards us bycomparing the wavelength receive λ with the wavelengthwe would expect a star of that type to emit λ0. Sincethe speed of light c is constant regardless of referencemedium:c = fλ = f0λ0

Therefore:f = c

λ and f0 = cλ0

By substitution:cλ =

(v+vrv+vs

)cλ0

1λ =

(v+vrv+vs

)1λ0

λ = λ0(v+vs)v+vr

In this case, v is the speed of light, so v = c. Relative tous, we are stationary, so vᵣ = 0. So:λ = λ0(c+vs)

c

λλ0

= (c+vs)c = 1 + vs

c

If we call the change in wavelength due to Doppler shiftΔλ, we know that λ = λ0 + Δλ. Therefore:λ0+∆λ

λ0= 1 + ∆λ

λ0= 1 + vs

c

So, the important result you need to know is that:∆λλ0

= vsc = z

This value is known as the red-shift of a star, denotedz. If z is positive, the star is moving away from us - thewavelength is shifted up towards the 'red' end of the elec-tromagnetic spectrum. If z is negative, the star is movingtowards us. This is known as blue shift. Note that we have

62 CHAPTER 13. ASTROPHYSICS

Redshift of spectral lines in the optical spectrum of a superclusterof distant galaxies (right), as compared to that of the Sun (left).

assumed that v is much smaller than c. Otherwise, specialrelativity makes a significant difference to the formula.

13.4.1 Questions

1. M31 (the Andromeda galaxy) is approaching us atabout 120kms−1. What is its red-shift?2. Some light from M31 reaches us with a wavelength of590nm. What is its wavelength, relative to M31?3. Some light has a wavelength, relative to M31, of480nm. What is its wavelength, relative to us?4. A quasar emits electromagnetic radiation at a wave-length of 121.6nm. If, relative to us, this wavelength isred-shifted 0.2nm, what is the velocity of recession of thequasar?Worked Solutions

13.5 The Big Bang

Big Bang theory states that space-time began as a singlepoint, and that, as time passed, space itself expanded.

13.5.1 Hubble’s Law

Hubble’s Law describes the expansion of the universemathematically:v = H0d ,where v is the velocity of recession of a celestial object,and d is the distance to the object. H0 is the Hubble con-stant, where H0 = 70km s−1 Mpc−1. The '0' signifiesthat this is the Hubble constant now, not in the past or thefuture. This allows for the fact that the Hubble constantmight be changing, but very slowly.

13.5.2 The Age of the Universe

Imagine a galaxy which flies out from the big bang at thespeed of light (c). The distance it has travelled d is givenby:d = vt ,where t is the age of the universe, since the galaxy hasbeen travelling since the beginning. If we substitute inHubble’s Law for v, we get:d = H0dt

1 = H0t

t = 1H0

So, the reciprocal of the Hubble constant is the age of theuniverse - but be careful with the units.

13.5.3 More Doppler Effect

We have already seen that red-shift z is given by:z = ∆λ

λs= vs

c ,where Δλ is the amount by which radiation is red-shiftedfrom a celestial object, λs is the wavelength of the radi-ation relative to the celestial object, vs is the velocity ofrecession of the object, c is the speed of light, and v ismuch less than c. If λ is the wavelength of the radiationrelative to us:z = λ−λs

λs= λ

λs− 1

z + 1 = λλs

However, if it is actually space that is being stretched,then this is actually the ratio of the distances between usand the celestial object at two times: the time at which theradiation was emitted, and the time at which the radiationwas received. We can apply this to any distance betweenany two stars:

13.5. THE BIG BANG 63

Rnow

Rthen= z + 1

13.5.4 Evidence for the Big Bang

Red Shift

If we measure the red shift of celestial objects, we seethat most of them are moving away from us - the lightfrom them is red-shifted. This is not true of all celestialobjects - the Andromeda galaxy, for example, is blue-shifted; it is moving towards us due to the gravitationalattraction of the Milky Way. Some galaxies are partlyred-shifted, and partly blue-shifted. This is due to theirrotation - some parts of the galaxy are rotating towardsus, while others are rotating away from us. However, themajority of celestial objects are moving away from us. Ifwe extrapolate backwards, we find that the universe musthave started at a single point. However, we are assumingthat the universe has always expanded and that movementalone can cause redshift [1]. Red shift provides evidencefor a Big Bang, but does not prove it.

Cosmic Microwave Background Radiation

Models of the Big Bang show that, at the beginning of theuniverse, radiation of a relatively short wavelength wouldhave been produced. Now, this radiation, due to the ex-pansion of space, has been stretched - it has become mi-crowave radiation. Cosmic microwave background radi-ation fits in extremely well with Big Bang theory, and sois strong evidence for it.

13.5.5 Questions

1. What is the Hubble Constant in s−1?2. How old is the universe?3. What effect might gravity have had on this figure?4. Polaris is 132pc away. What is its velocity of recession,according to Hubble’s Law?Worked Solutions

13.5.6 Footnotes[1] Arp, Halton C.,Quasars, Redshifts and Controversies

(ISBN 0-521-36314-4)

Chapter 14

Thermodynamics

14.1 Heat and Energy

Matter is made of particles. These particles are constantlymoving. When we feel some matter, we feel somethingthat we call 'heat'. This is just our impression of how fastthe particles are moving. The higher the average speed ofthe particles, the hotter something is.Note that there is a technical difference between 'heat'and 'temperature'. Heat is energy in transit, also known aswork. Temperature is the internal energy of a substance.We feel heat, not temperature, for if the energy did notmove from an object, we would not be able to measure it.If we had some matter which was made of stationary par-ticles, then we would not be able to make the particlesmore stationary. The concept is meaningless. When mat-ter is in this state, it is at the coldest temperature possi-ble. We call this temperature 0°K. This corresponds to−273.15°C. If the temperature rises by 1°K, the temper-ature rises by 1°C. The only difference between the twoscales is what temperature is defined as 0 - in Kelvin, 0 isabsolute zero. In Celsius, 0 is the freezing point of water.In both scales, 1° is one hundredth of the difference intemperature between the freezing and boiling points ofwater.Some matter of temperature T consists of many particles.Their motion is essentially random - they are all movingat different speeds, and so they all have different kineticenergies. The temperature is related to the average energyper. particle E by the following approximate relationship:E ≈ kT ,where k is a constant known as the Boltzmann constant.k = 1.38 x 10−23JK−1. T must always be measured inKelvin.

14.1.1 Changes of State

Different substances change state at different tempera-tures. In other words, when the average energy per.particle reaches a certain level, the substance changesstate. The situation complicates itself since, in order tochange state, additional energy is required (or is givenout). When liquid water reaches its boiling point, it will

Dep

osi

tion

Sub

limati

on

Freezin

g

Melting

Recombination

Ionizatio

n

VaporizationCondensation

Plasma

Liquid

Solid

Enth

alp

y o

f sy

stem

Gas

The relationships between the states of matter.

stay at its boiling point until it has all changed into wa-ter vapour since the energy being taken in is being usedto change state, instead of to increase the temperatureof the water. The average energy per. particle requiredto change state can be approximated using the formulaabove, where T is the temperature at which the substancechanges state.

14.1.2 Activation Energy

Many things have an activation energy. In order for achemical reaction to start, for example, the average en-ergy per. particle must reach a certain level. However,most of the time, chemical reactions start at a lower aver-age energy per. particle than the activation energy. Thisis because there is always a chance that some particleshave the required activation energy, since the particlesare moving at random. If the reaction is exothermic (thismeans that it gives out heat, raising the average energyper. particle), then, once one reaction has happened,more of the particles have the activation energy, and sothe reaction accelerates until all the reagants are used up.

64

14.3. IDEAL GASES 65

The activation energy can be related to the temperatureof the substance using the formula E=kT.

14.1.3 Questions

1. Carbon dioxide sublimes at 195°K. Roughly what en-ergy per. particle does this correspond to?2. A certain chemical reaction requires particles withmass of the order 10−26kg to move, on average, at10ms−1. Roughly what temperature does this correspondto?3. The boiling point of water is 100°C. Roughly whatenergy per. particle does this correspond to?4. Thermionic emission from copper requires around5eV of energy per. particle. How hot will the wire beat this energy level?Worked Solutions

14.2 Specific Heat Capacity

It takes energy to heat things up, since heat is work. If weheat a moremassive thing up, it takes more work, becausewe have to give more particles, on average, an energy kT.Some substances require more work to heat up than oth-ers. This property is known as specific heat capacity. Thisgives us the formula:∆E = mc∆θ ,where ΔE is the work put in to heating something up (inJ), m is the mass of the thing we are heating up (in kg),c is the specific heat capacity (in Jkg−1K−1), and Δθ isthe difference in temperature due to the work done on thesubstance (in degrees Celsius or Kelvin).It should be noted that the specific heat capacity changesslightly with temperature, and more than slightly whenthe material changes state. A table of the specific heatcapacities of various substances is given below:

14.2.1 Questions

1. How much work would it take to heat 100kg of liquidwater from 20°C to 36.8°C?2. How much work would it take to heat a well-insulatedroom from 15°C to 21°C, if the room is a cube with sidelength 10m, and the density of the air is 1.2kgm−3?3. A 10kg block of iron at 80°C is placed in the roomabove once it has reached 21°C. If the iron cools by 40°C,what is the new temperature of the room?Worked Solutions

14.3 Ideal Gases

An animation showing the relationship between pressure and vol-ume when mass and temperature are held constant.

An animation demonstrating the relationship between volumeand temperature.

Real-world gases can bemodelled as ideal gases. An idealgas consists of lots of point particles moving at random,colliding with each other elastically. There are four sim-ple laws which apply to an ideal gas which you need toknow about:

14.3.1 Boyle’s Law

Boyle’s Law states that the pressure of an ideal gas is in-versely proportional to its volume, assuming that the massand temperature of the gas remain constant. If I com-press an ideal gas into half the space, the pressure on theoutsides of the container will double. So:p ∝ 1

V

14.3.2 Charles’ Law

Charles’ Law states that the volume of an ideal gas is pro-portional to its temperature:

66 CHAPTER 14. THERMODYNAMICS

V ∝ T

Tmust bemeasured in kelvin, where a rise of 1°K is equalto a rise 1°C, and 0°C = 273°K. If we double the temper-ature of a gas, the particles move around twice as much,and so the volume also doubles.

14.3.3 Amount Law

This law states that the pressure of an ideal gas is propor-tional to the amount of gas. If we have twice the numberof gas particles N, then twice the pressure is exerted onthe container they are in:p ∝ N

A mole is a number of particles. 1 mole = 6.02 x 1023particles. So, the pressure of a gas is also proportional tothe number of moles of gas present n:p ∝ n

14.3.4 Pressure Law

The pressure law states that the pressure of an ideal gas isproportional to its temperature. A gas at twice the tem-perature (in °K) exerts twice the pressure on the sides ofa container which it is in:p ∝ T

These laws can be put together into larger formulae link-ing p, V, T and N.

14.3.5 Questions

1. I heat some argon from 250K to 300K. If the pressureof the gas at 250K is 0.1 MPa, what is its pressure afterheating?2. The argon is in a 0.5m long cylindrical tank with radius10cm. What volume does it occupy?3. The argon is then squeezed with a piston so that inonly occupies 0.4m of the tank’s length. What is its newpressure?4. What is its new temperature?5. 25% of the argon is sucked out. What is its pressurenow?Worked Solutions

14.4 Kinetic Theory

One formula which sums up a lot of the kinetic theory ofan ideal gas is the following:pV = 1

3Nmc2 ,

where p is the pressure of the gas, V is its volume, N isthe number of molecules, m is the mass of each molecule,and c2 is the mean square speed of the molecules. If youknew the speeds of all the molecules, you could calculatethe mean square speed by squaring each speed, and thentaking the mean average of all the squared speeds.

14.4.1 Derivation

Some happy ideal gas particles

This formula can be derived from first principles by mod-elling the gas as a lot of particles colliding. The particleshave a momentum p = mc. If we put them in a box ofvolume V and length l, the change in momentum whenthey hit the side of the box is:∆p = m(c− (−c)) = 2mc

Every time the particle travels the length of the box (l)and back (another l), it hits the wall, so:c = 2l

t ,where t is the time between collisions. Therefore:t = 2l

c

Each collision exerts a force on the wall. Force is the rateof change of momentum, so:F = ∆p

∆t = 2mc2lc

= 2mc2

2l = mc2

l

However, we have got N particles all doing this, so thetotal force on the wall is given by:F = Nmc2

l

The molecules all have different velocities, so we have totaken an average - the mean square speed. This force isthe force in all three dimensions. The force in only onedimension is therefore:F = Nmc2

3l

Pressure, by definition, is:p = F

A = Nmc2

3Al

14.5. BOLTZMANN FACTOR 67

But area multiplied by length is volume, so:p = Nmc2

3V

Therefore:pV = 1

3Nmc2

14.4.2 Questions

1. Five molecules are moving at speeds of 1,5,6,8, and36ms−1. What is their mean square speed?2. What is the mass of one molecule of N2 (atomic mass14, 1u = 1.66 x 10−27kg)?3. Atmospheric pressure is 101,325 Pa. If one mole ofNitrogen takes up 2.3 m3 at about 10°C, what is the meansquare speed of the molecules in the air outside, assumingthat the atmosphere is 100% nitrogen (in reality, it is only78%)?4. What is the average speed of a nitrogenmolecule underthe above conditions?5. The particles in question 1 are duplicated 3000 times.If they have a completely unrealistic mass of 1g, what istheir pressure when they are crammed into a cube withside length 0.5m?Worked Solutions

14.5 Boltzmann Factor

Particles in a gas lose and gain energy at random due tocollisions with each other. On average, over a large num-ber of particles, the proportion of particles which haveat least a certain amount of energy ε is constant. This isknown as the Boltzmann factor. It is a value between 0and 1. The Boltzmann factor is given by the formula:nn0

= e−ϵkT ,

where n is the number of particles with kinetic energyabove an energy level ε, n0 is the total number of particlesin the gas, T is the temperature of the gas (in kelvin) andk is the Boltzmann constant (1.38 x 10−23 JK−1).This energy could be any sort of energy that a particle canhave - it could be gravitational potential energy, or kineticenergy, for example.

14.5.1 Derivation

In the atmosphere, particles are pulled downwards bygravity. They gain and lose gravitational potential energy(mgh) due to collisions with each other. First, let’s con-sider a small chunk of the atmosphere. It has horizon-tal cross-sectional area A, height dh, molecular density(the number of molecules per. unit volume) n and all themolecules have mass m. Let the number of particles in

the chunk be N.n = N

V = NA dh

Therefore:V = A dh (which makes sense, if you think about it)By definition:N = nV = nA dh

The total mass Σ m is the mass of one molecule (m) mul-tiplied by the number of molecules (N):Σm = mN = mnA dh

Then work out the weight of the chunk:W = gΣm = nmgA dh

The downwards pressure P is force per. unit area, so:P = W

A = nmgA dhA = nmg dh

Weknow that, as we go up in the atmosphere, the pressuredecreases. So, across our little chunk there is a differencein pressure dP given by:dP = −nmg dh (1) In other words, the pressure is de-creasing (-) and it is the result of the weight of this littlechunk of atmosphere.We also know that:PV = NkT

So:P = NkT

V

But:n = N

V

So, by substitution:P = nkT

So, for our little chunk:dP = kT dn (2)If we equate (1) and (2):dP = −nmg dh = kT dn

Rearrange to get:dndh = −nmg

kT

dhdn = −kT

nmg

Integrate between the limits n0 and n:h = −kT

mg

∫ n

n0

1n dn = −kT

mg [lnn]nn0=

−kTmg (lnn− lnn0) =

−kTmg ln n

n0

ln nn0

= −mghkT

nn0

= e−mgh

kT

Since we are dealing with gravitational potential energy,ε = mgh, so:nn0

= e−ϵkT

68 CHAPTER 14. THERMODYNAMICS

14.5.2 A Graph of this Function

This topic comes up in Q10 494 June 2010. The Valuesused for various things in that question arek = 1.4 × 10−23JK−1, g = 9.8, m = 4.9 ×10−26Kg, T = 290K

Shows how Energies are achieved with Height

14.5.3 Questions

1u = 1.66 x 10−27 kgg = 9.81 ms−2

1. A nitrogen molecule has a molecular mass of 28u.If the Earth’s atmosphere is 100% nitrous, with a tem-perature of 18°C, what proportion of nitrogen moleculesreach a height of 2km?2. What proportion of the molecules in a box of hydrogen(molecular mass 2u) at 0°C have a velocity greater than5ms−1?3. What is the temperature of the hydrogen if half of thehydrogen is moving at at least 10ms−1?4. Some ionised hydrogen (charge −1.6 x 10−19 C)isplaced in a uniform electric field. The potential differencebetween the two plates is 20V, and they are 1m apart.What proportion of the molecules are at least 0.5m fromthe positive plate (ignoring gravity) at 350°K?Worked Solutions

Chapter 15

Magnetic Fields

15.1 Flux

A coil of wire creates magnetic flux. The amount ofmagnetic flux created is dependant on three things: thenumber of coils in the wire, the amount of current flow-ing through the wire, and the permeance of the objectthrough which the flux is flowing. So:Φ = ΛNI ,where Φ is flux (in webers, denotedWb), Λ is permeance(in WbA−1) and I is current (in A). This is the total fluxinduced. NI is the number of “current-turns”. Perme-ance is related to permeability (a material property) bythe following equation:Λ = µA

L ,where μ is permeability, A is cross-sectional area, and Lis length. A permanent magnet is just like a coil, exceptthat a current does not need to be generated to maintainthe flux. Over smaller areas, we need to know the fluxdensity B. This is the amount of flux per. unit area:B = Φ

A

Therefore:Φ = AB

Lines of flux around a permanent magnet.

The flux around a coil of wire varies - ΛNI only gives thetotal flux, not the flux across a certain area. To show this,

we use lines of flux. These obey the following rules:1. Lines of flux go from the north pole of a permanentmagnet to the south pole.2. Lines of flux go clockwise about wires carrying currentaway from you.3. Lines of flux never touch, intersect, or cross.The direction of the flux is shown with an arrow. Flux is abit like electricity in that it must have a complete circuit.The lines of flux always take the route ofmost permeance.Iron has around 800 times as much permeability as air.So, flux goes through the iron, and not the air.

15.1.1 Questions

1. A circular steel core has a cross-sectional area of 9cm2, and a length of 0.5m. If the permeability of steel is875 μNA−2, what is the permeance of that core?2. A coil of insulated wire is wrapped 60 times aroundthe top of the core, and a 9A direct current is put throughthe coil. How much flux is induced?3. Assuming that all the flux goes through the core, whatis the flux density at any point in the core?4. Draw a diagram showing the lines of flux within thecore.Worked Solutions

15.2 Induction

A magnetic field creates a current in a wire movingthrough it. This process is known as induction.

15.2.1 Flux Linkage

A magnetic field going through a coil of wire has a prop-erty known as flux linkage. This is the product of the fluxΦ and the number of coils in the wire N.

69

70 CHAPTER 15. MAGNETIC FIELDS

15.2.2 Faraday’s Law

Electric current is only induced in a coil of wire if themagnetic field is moving relative to the coil. Faraday’sLaw gives the electromotive force (emf) ε produced in acoil by a magnetic field:ϵ = −N dΦ

dt

In other words, the emf (electric potential) induced in thecoil is proportional to the rate of change of flux linkage.In practice, this means that if the coil is stationary rel-ative to the magnetic field, no emf is induced. In orderto induce emf, either the coil or the magnetic field mustmove. Alternatively, we may change the number of coils,for example, by crushing the coil, or pressing a switchwhich added more coils into the circuit, or moving moreof the coils into the magnetic field.Faraday’s Law also works the other way. If we were tointegrate both sides and rearrange the formula in termsof Φ, we would find that the flux depends on the integralof the voltage - not on its rate of change. If we put anemf across a coil, it induces a magnetic field. The fluxdoes not depend on the rate of change of emf, but the emfdoes depend on the rate of change of the flux linkage.

15.2.3 Lenz’s Law

Lenz’s Law describes the direction of the current / emfinduced by a change inmagnetic flux. It states that currentinduced opposes the magnetic field. It does this by creat-ing its own magnetic field. This explains the minus signin Faraday’s Law. This also means that the flux inducedby a current (not a change in current) is proportional tothe current, since the flux is produced in response to thecurrent.So, a change in flux induces a current and a voltage whichis proportional to the rate of change of flux. This fits withOhm’s Law (V = IR). A current and a voltage in a coilinduce a flux which is proportional to the current and thevoltage.

15.2.4 Questions

1. What is the flux linkage of a 30cm coil of 0.5mm thickwire with a flux perpendicular to it of 10Wb?2. If the above coil is crushed steadily over a period of2s, what emf is maintained?3. The flux in a flux circuit varies according to the equa-tion Φ = sin ωt. What is the equation for the emf in-duced?4. Using a constant k, what is the equation for a currentwhich could induce the flux in the flux circuit above?5. Draw a graph of the flux, flux linkage, emf and currentas deduced in the previous two questions.

Worked Solutions

15.3 Force

Magnetic fields exert a force on a charge when the chargeis moving. If the charge is stationary, no force is exerted.This force is given by:−→F = q(−→v ×−→

B ) ,where q is the charge on the point charge, v is its velocityand B is themagnetic field strength. This involves a vectorcross product, which you don't need to know about for A-level. However, you do need to know a simplified versionof this. The magnitude of this force F is given by:F = Bqv sin θ ,where θ is the angle between the direction of motion ofthe point charge and the direction of the magnetic field.If the velocity and the magnetic field are in the same di-rection, the θ = 0, so sin θ = 0 and F = 0. If the velocityand the magnetic field are perpendicular to each other, θ= ½ π, so sin θ = 1. This means that, in the special casewhere velocity is perpendicular to the magnetic field:F = Bqv

If q is negative (for example, for an electron), the force isin the opposite direction.

15.3.1 Current

A current is just a flow ofmoving electrons, and so a mag-netic field will exert a force on a wire with a current flow-ing through it. The case you need to know about is whenthe magnetic field is perpendicular to the wire. In thiscase, the magnitude of the force on the wire is given by:F = BIl ,where I is current, and l is the length of the wire.

15.3.2 Direction

The direction of the force on either a point charge or on awire can be worked out using Fleming’s left-hand rule, asshown in the diagram on the right. The direction of thethumb is that of the force (or thrust), the direction of thefirst finger is that of the magnetic field, and the directionof the second finger is that of the current (or the motionof the point charge).On a 2D diagram, the direction of a magnetic field is rep-resented by one of two symbols, which resemble the pointand fletchings of an arrow pointing in the direction of themagnetic field. The symbol

⊙means that the field is

pointing towards you (just as the arrow would be, if youwere looking at the point). The symbol

⊗means that the

field is pointing away from you (just as the arrow would

15.4. TRANSFORMERS 71

B

I

F

Fleming’s left-hand rule

The point and fletchings of an arrow.

be, if you were looking at the fletching).

15.3.3 Questions

1. What force is exerted by a 1T magnetic field on anelectron (of charge−1.6 x 10−19C) moving at 5% of thespeed of light (3 x 108 ms−1)?2. What force is exerted by a 5mT magnetic field on a20cm wire with resistance 1μΩ attached to a 9V battery?3. The following diagram shows a positive charge movingthrough a magnetic field. Draw an arrow representing thedirection of the force on the charge.

4. The following diagram shows a wire in a magneticfield. Draw an arrow representing the direction of theforce on the wire.

Worked Solutions

15.4 Transformers

We have already seen that a change in flux induces an emfin a coil, given by Faraday’s Law:ϵ = −N dϕ

dt

We have also seen that a voltage in a coil induces a mag-netic flux inside the coil. If we were to connect two coilswith the same core, the flux, and the rate of change offlux, would be exactly the same inside both coils. Wewould have created a kind of flux circuit known as a trans-former. The ratio between the voltage at the primary coilV and the voltage at the secondary coil V would have tobe (since φ is constant):

72 CHAPTER 15. MAGNETIC FIELDS

PrimaryvoltageVP

Secondaryvoltage

VS

Primarycurrent

IP SecondarycurrentIS

PrimarywindingNP turns

NS turns

ΦMagneticFlux,

TransformerCore

Secondarywinding

+

+

−

−

An ideal step-down transformer showing magnetic flux in thecore.

Vp

Vs=

−Npdϕdt

−Nsdϕdt

=Np

Ns,

where N and N are the numbers of coils in the primaryand secondary coils respectively.In other words, we can change the voltage of some elec-tricity by varying the number of coils in each coil. Inorder for this to work, the current used must be an al-ternating current (AC). This means that the current andvoltage are constantly changing sinusoidally, and so thereis a sinusoidal change in flux. This means that an emf isinduced in the secondary coil. If the flux did not change(i.e. we were using direct current), then no emf would beinduced, and the transformer would be useless except asa magnet (since it would still have a flux circuit in it).

15.4.1 Ideal Transformers

An ideal transformer is one in which all the electrical en-ergy put into one coil comes out of the other coil. Anideal transformer does not exist, but, since it makes themaths easy, we like to pretend that it does. In this case,the power in must equal the power out:P = Pp = Ps = IpVp = IsVs ,where I and I are the currents in the primary and sec-ondary coils, respectively. So:Vp = P

Ipand Vs =

PIs

By substitution into the transformer equation for voltage:Np

Ns=

Vp

Vs=

PIpPIs

=1Ip1Is

= IsIp

So, in an ideal transformer, the ratio between the voltagesis equal to the ratio between the numbers of coils, but theratio between the currents is equal to the reciprocal of theratio between the numbers of coils.

15.4.2 Eddy Currents

In reality, the electrical energy is not all conserved - a lotof it is converted into heat by eddy currents. In a trans-former, the magnetic flux created by the primary coil in-duces a current in the core. This occurs in order to opposethe change that produced the magnetic flux (Lenz’s Law).The currents flowing in the core are called eddy currents.These currents produce heat, using up energy and so caus-ing inefficiency. One way of minimising the effects ofeddy currents is to make the core out of iron laminate.This is layers of iron separated by thin layers of an insu-lator such as varnish. The amplitude of the eddy currentsproduced is reduced as currents cannot flow through thelayers of insulator. (Note: OCR B question papers tendto have a question on eddy currents.)

15.4.3 Questions

1. A step-down transformer has 300 coils on one coil, and50 coils on the other. If 30 kV AC is put in, what voltagecomes out?2. A step-up transformer has 200 coils on one coil, and980 coils on the other. If 25 kV AC comes out, whatvoltage was put in?3. An ideal transformer transforms a 50A current intoa 1A current. It has 40 coils on the primary coil. Howmany coils are in the secondary coil?4. Transformers tend to vibrate. Why is this? What effectdoes this have on the efficiency of the transformer?5. Air does have some permeability. What effect doesthis have on the efficiency of the transformer? Why?Worked Solutions

15.5 Motors

Just as amovingmagnetic field induces current in conduc-tors, a changing current in a magnetic field induces mo-tion. When this motion is used to ensure that the currentkeeps changing relative to the magnetic field, the motionwill continue, and so we have an electric motor. Thereare several types of electric motor.

15.5.1 Simple DC Motor

When a coil is placed inside a stationary magnetic field,and a direct current is run through the coil, the coil triesto align itself with the field since it becomes an electro-magnet. This would be useless as a motor, since it wouldalways move to the same position when you turned it on,and then stop. If, however, we use a split-ring commuta-tor which changes the direction of the current every half-rotation, then the coil would try to align itself in the oppo-

15.5. MOTORS 73

Diagram of a simple DC motor, showing the repulsion betweenthe poles.

site direction every half-rotation. This means that, oncethe rotor starts to move, it continues to move. This is aDC electric motor. The permanent magnets can be re-placed with electromagnets as well. The main advantageof this type of motor is that the commutator works, re-gardless of the frequency of rotation.

Animation of a three-phase motor with an electromagnet as therotor.

15.5.2 Three-phase Motor

The three-phase power produced by a three-phase gener-ator may be used to power a motor. Each phase of poweris connected to one of three coils. This creates a magneticfield which rotates once for each cycle of the power. If apermanent magnet is placed in the middle, at any giventime, its north pole will be attracted to a south pole in oneof the coils, and will be repelled by a north pole in one ofthe coils. The converse would be true for its south pole.This means that the rotatingmagnetic field drags themag-

net around with it, causing the magnet to rotate with thesame frequency as the magnetic field. The disadvantageof this type of motor is that it goes at one frequency only- the frequency of the current.The permanent magnet can be replaced with a coil withdirect current in it. This creates a magnetic field, the ad-vantage being that there is no need for a permanent mag-net which is expensive and heavy. The main disadvan-tages are that electricity must be used to power the elec-tromagnet, and that a slip-ring commutator must be usedto prevent the coil getting tangled up and stopping themotor from running.

Animation of a squirrel-cage AC motor

Squirrel Cage Motor

A squirrel cagemotor works on a similar principle, exceptthat the rotor is no longer a permanent magnet. Instead,a series of metal rods run through the rotor, connected toeach other at either ends. The rods run perpendicular tothe rotating magnetic field. Once the rotor starts to rotate,an electric current is created in the rods - eddy currents.This creates a magnetic field which is perpendicular to therotating magnetic at all times. As the rotating magneticfield created by the stator rotates, it pulls the inducedmag-netic field around after it, causing the rotor to continue torotate.A squirrel cage motor relies on the fact that the two mag-netic fields are rotating at different rates. If they were not,then there would be no change in flux in the rotor, and sono eddy currents would be induced.

15.5.3 Questions

1. How could you adapt the simple DCmotor to use AC?2. Why does a three-phase motor have a constant angularvelocity?3. What is the difference between a split-ring and a slip-ring commutator?

74 CHAPTER 15. MAGNETIC FIELDS

4. How could the angular velocity of a three-phase motorbe increased?5. A squirrel-cage motor relies on eddy currents runningalong the rotor to function. However, if eddy currents runacross the rotor, then the force on the rotor is reduced.How may these eddy currents be reduced without reduc-ing the desired eddy currents?Worked Solutions

15.6 Generators

Fleming’s right-hand rule

We have seen that a change in flux induces an electriccurrent in a coil of wire. One way of changing the fluxis to move the magnet. Alternatively, we can move thecoil relative to the magnet. Generators work on this prin-ciple - a non-electrical source of energy is used to rotatesomething (known as the rotor), which induces an elec-tric current in either the rotor or the stator (the stationarypart of any electromagnetic machine). For a generator,the relationships between the directions of current, fieldand motion are given by Fleming’s right-hand rule (right).

15.6.1 Moving Coil

AC Generator

Attaching carbon brushes to a slip-ring commutator results in anAC generator.

If a coil of wire is placed in a magnetic field and rotated,an alternating (sinusoidal) current is induced. As it ro-tates, sometimes it is 'cutting' through lots of flux, and solots of current is induced. At other times, it is movingparallel to the flux, and so no flux is cut, and no currentis induced. In between, some current is induced. Thiscreates an alternating current.Either end of the coil can be connected to wires outside ofthe generator in order to use the current elsewhere. Thiswould be fine for the first few rotations, but after this, thewires would get tangled up and the generator would beuseless. To avoid this, we use a commutator. In an ACgenerator, this is a pair of rotating conducting 'slip rings’attached to either end of the coil. Carbon brushes bringthese into contact with the outside world.

DC Generator

Attaching wires to a split-ring commutator results in an DC gen-erator.

If we replace the slip-ring commutator in an AC genera-tor with a pair of brushes which the ends of the coil rotateinside, the generator creates direct current (DC) instead.Halfway through the rotation, the brushes come into con-tact with the other end of the coil, and so the AC changesdirection every half a rotation. This approximates to a di-rect current. This direct current is not perfect since it con-sists of a series of positive-voltage pulses. These pulsescan be smoothed out using a capacitor or a complex sys-tem of commutators.

15.6.2 Moving Magnet

Simple AC Generator

An alternative method of generating an alternating cur-rent is to rotate a permanent magnet in a gap betweentwo coils. This has the advantage of not requiring a com-mutator (the coil is the stator), but often a coil is lighterthan a magnet, and so it is more efficient to use a rotatingcoil.

15.6. GENERATORS 75

Three-Phase Generator

If we place three pairs of coils, evenly spaced, aroundthe rotating magnet, then three different alternating cur-rents, with three different phases, will be generated. Thisis a more efficient method of generating electricity, sincecurrent is always being generated. The sum of all threecurrents is zero, so three different cables must be used totransport the currents. Three-phase power is often usedin motors with three coils in the stator.

15.6.3 Questions

1. Draw diagrams of an alternating current, the 'directcurrent' produced by a DC generator, and this currentonce it has been smoothed with a capacitor.2. What is the phase difference (in radians) between thevoltages produced by a three-phase generator?3. According to Faraday’s law, what three things will in-crease the amplitude of the emf created by a generator?4. If an albatross touched two power cables carrying ACin phase, what would happen?5. What would happen if the two cables carried three-phase power?Worked Solutions

Chapter 16

Electric Fields

16.1 Force

Electric fields are caused by charge. This charge can beeither positive or negative. Like charges repel each other,and opposite charges attract each other. If we have twopoint charges of charge Q and q respectively, and they area distance r apart, the force on each of them is:Felectric =

kQqr2 = Qq

4πϵ0r2,

where k and ε0 are constants (k = 8.99 x 109 Nm2C−2,ε0= 8.85 x 10−12C2N−1m−2). This means that, twice as faraway from the point charge, the force on another chargedecreases by a factor of 4. Electric force around a pointcharge is very similar to gravitational force around a pointmass.A uniform electric field consists of two conducting plates.These plates are oppositely charged, and infinitely wide.Obviously, infinitely wide conducting plates do not exist,so uniform electric fields do not exist. However, fieldswhich approximate uniform electric fields do exist, pro-vided we look towards the middle of the plates, and theplates are not too far apart - at the ends, the formulae foruniform fields no longer apply.The force on a charge in a uniform electric field is givenby:Felectric =

qVd ,

where V is the potential difference between the twoplates, q is the charge of the point charge upon whichthe force is acting, and d is the distance between the twoplates. This force remains constant as the charge travelswithin the electric field.

16.1.1 Questions

e = 1.6 x 10−19C1. A positron (charge +e) is 1 μm from a lithium nucleus(charge +3e). What is the magnitude of the force actingon each of the particles? In what direction is it acting?2. An electron is 1mm from the positively charged plate ina uniform electric field. The potential difference betweenthe plates is 20V, and the plates are 10cm apart. Whatforce is acting on the electron? In what direction?

3. The acceleration due to gravity around a point massis constant, irrespective of the mass of the objects it isacting on. The acceleration due to electricity around apoint charge is not. Use Newton’s Second Law (F=ma)to explain this.4. An insulator contains charged particles, even thoughthe overall charge on the insulator is 0. Why is the insu-lator attracted by a nearby charge?5. Where in the charged conducting plates which create auniform electric field would you expect to find the chargelocated? Why?Worked Solutions

16.2 Field

Electric field E is the force per. unit charge caused by anelectric field:Eelectric =

Felectric

q

The unit of electric field is NC−1 or Vm−1.In general,the electric field is the rate of change of electric potential(voltage) with respect to distance:Eelectric = −dVelectric

dx

16.2.1 Special Cases

There are two different types of field which you need toknow about. Uniform fields occur between two plateswith opposite charges. Here, the electric field is simply:Eelectric =

Velectric

d

A charged sphere also has an electric field. To gain a for-mula for this, we divide the formula for force around acharged sphere by q, so:Eelectric =

Q4πϵ0r2

,

where ε0 = 8.85 x 10−12C2N−1m−2.The electric field around a point charge is called a radialfield. The field strength is highest at the centre and de-creases as the distance from the centre increases. This isreflected in the above formula, which shows that Eₑ ₑ ᵣᵢ

76

16.2. FIELD 77

is proportional to 1r2 .

16.2.2 Field Lines

We can represent electric field using field lines. These gofrom positive charge to negative charge. They are moreclosely packed together when the electric field is stronger.In a uniform field, they look like the following:

Around two oppositely charged spheres (known as adipole), they look like the following:

16.2.3 Questions

1. Two metal plates are connected to a 9V battery withnegligible internal resistance. If the plates are 10cmapart, what is the electric field at either of the plates?2. What is the electric field at the midpoint between theplates?3. The charge on an electron is −1.6 x 10−19 C. What isthe electric field 1μm from a hydrogen nucleus?4. What is the direction of this field?5. A 2C charge is placed 1m from a−1C charge. At whatpoint will the electric field be 0?Worked Solutions

78 CHAPTER 16. ELECTRIC FIELDS

16.3 Potential

16.3.1 Relationship to Electric PotentialEnergy

You will probably remember from AS (or even GCSE)that the energy U which flows along a wire is given by:U = V q ,where V is the potential difference between either end ofthe wire, and q is the amount of charge which flows. Asimple rearrangement shows that:V = U

q

This potential difference is the same thing as electricpotential. In a wire, the electric field is very simple.There are other electric fields, and in these fields as well,the electric potential is the electric potential energy per.unit charge. Electric potential energy between two pointcharges Q and q is given by:U = Qq

4πϵ0r

So, the electric potential at a distance r from any pointcharge Q (ignoring other charges) is:V = Q

4πϵ0r

16.3.2 Relationship to Electric FieldStrength

Electric potential is also the integral of electric fieldstrength. This is why it is often called potential differ-ence - it is an integral between two limits (two points inspace) with respect to distance. So, the potential differ-ence between two points a and b is:

Vab =∫ b

aE dx =

∫ b

aQ

4πϵ0x2 dx =[− Q

4πϵ0x

]ba

But, if we define b as infinity and a as r:

V =[− Q

4πϵ0x

]∞r

= − Q4πϵ0∞ −

(− Q

4πϵ0r

)= Q

4πϵ0r

So, the area under a graph of electric field strength againstdistance, between two points, is the potential differencebetween those two points.For a uniform electric field, E is constant, so:V =

∫∞r

E dx = Er

In other words, V is proportional to r. If we double thedistance between us and a point, the potential differencebetween us and that point will also double in a uniformelectric field.

16.3.3 Equipotentials

Equipotentials are a bit like contours on a map. Contoursare lines which join up all the points which have the sameheight. Equipotentials join up all the points which have

the same electric potential. They always run perpendic-ular to electric field lines. As the field lines get closertogether, the equipotentials get closer together.

16.3.4 Questions

ε0 = 8.85 x 10−12 Fm−1

1. Draw a diagram of a uniform electric field betweentwo plates, showing the field lines and the equipotentials.2. Do the same for the electric field around a point charge.3. The potential difference between two plates is 100V.What is the potential difference between a point halfwaybetween the plates and one of the plates?4. What is the electric potential at a point 0.2m from analpha particle (charge on an electron = −1.6 x 10−19C)?5. What is the electric potential energy of an electron atthe negative electrode of an electron gun if the potentialdifference between the electrodes is 10V?Worked Solutions

16.4 Potential Energy

Just as an object at a distance r from a sphere has grav-itational potential energy, a charge at a distance r fromanother charge has electrical potential energy εₑ ₑ . Thisis given by the formula:ϵelec = Velecq ,where Vₑ ₑ is the potential difference between the twocharges Q and q. In a uniform field, voltage is given by:Velec = Eelecd ,where d is distance, and Eₑ ₑ is electric field strength.Combining these two formulae, we get:ϵelec = qEelecd

For the field around a point charge, the situation is differ-ent. By the same method, we get:ϵelec =

−kQqr

If a charge loses electric potential energy, it must gainsome other sort of energy. You should also note that forceis the rate of change of energy with respect to distance,and that, therefore:ϵelec =

∫F dr

16.4.1 The Electronvolt

The electronvolt (eV) is a unit of energy equal to thecharge of a proton or a positron. Its definition is the ki-netic energy gained by an electron which has been accel-erated through a potential difference of 1V:1 eV = 1.6 x 10−19 J

16.4. POTENTIAL ENERGY 79

For example: If a proton has an energy of 5MeV then inJoules it will be = 5 x 106 x 1.6 x 10−19 = 8 x 10−13 J.Using eV is an advantage when high energy particles areinvolved as in case of particle accelerators.

16.4.2 Summary of Electric Fields

You should now know (if you did the electric fields sectionin the right order) about four attributes of electric fields:force, field strength, potential energy and potential. Thesecan be summarised by the following table:This table is very similar to that for gravitational fields.The only difference is that field strength and potentialare per. unit charge, instead of per. unit mass. Thismeans that field strength is not the same as acceleration.Remember that integrate means 'find the area under thegraph' and differentiate (the reverse process) means 'findthe gradient of the graph'.

16.4.3 Questions

k = 8.99 x 109 Nm2C−2

1. Convert 5 x 10−13 J to MeV.2. Convert 0.9 GeV to J.3. What is the potential energy of an electron at the neg-atively charged plate of a uniform electric field when thepotential difference between the two plates is 100V?4. What is the potential energy of a 2C charge 2cm froma 0.5C charge?5. What is represented by the gradient of a graph of elec-tric potential energy against distance from some charge?Worked Solutions

Chapter 17

Particle Physics

17.1 The Standard Model

The standard model of particle physics attempts to ex-plain everything in the universe in terms of fundamentalparticles. A fundamental particle is one which cannot bebroken down into anything else. These fundamental par-ticles are the building blocks of matter, and the thingswhich hold matter together.The standard model is usually represented by the follow-ing diagram:

The particles in the standard model can be put into twogroups: fermions and bosons. Fermions are the build-ing blocks of matter. They all obey the Pauli exclusionprinciple. Bosons are force-carriers. They carry the elec-tromagnetic, strong, and weak forces between fermions.

80

17.2. QUARKS 81

17.1.1 Bosons

There are four bosons in the right-hand column of thestandard model. The photon carries the electromagneticforce - photons are responsible for electromagnetic radia-tion, electric fields and magnetic fields. The gluon carriesthe strong nuclear force - they 'glue' quarks together tomake up larger non-fundamental particles. The W+, W-

and Z0 bosons carry the weak nuclear force. When onequark changes into another quark, it gives off one of thesebosons, which in turn decays into fermions.

17.1.2 Fermions

Fermions, in turn, can be put into two categories: quarksand leptons. Quarks make up, amongst other things, theprotons and neutrons in the nucleus. Leptons includeelectrons and neutrinos. The difference between quarksand leptons is that quarks interact with the strong nuclearforce, whereas leptons do not.

17.1.3 Generations

Fermions are also divided into three generations. The firstgeneration contains the fermions which we are made of- electrons, the up and down quarks, and the neutrino.The first generation particles have less mass than the sec-ond, and the second generation particles have less massthan their respective third generation particles. The sec-ond generation (the μ generation) contains two leptons:the muon and the muon-neutrino. It also contains thecharm and strange quarks. The third generation (the τgeneration) contains another two leptons: the tau and thetau-neutrino. Its quarks are the top and bottom quarks.

17.1.4 Antiparticles

Every fermion has its antiparticle. An antiparticle has thesame mass as a particle, but the opposite charge. So, thestandard model contains 12 quarks, 12 leptons, and thebosons (which are even more complex).

17.1.5 Questions

1. The third generation top quark was the last quark in theStandard Model to have its existence proven experimen-tally (in 1995). It is also the most massive of the quarks.Why was it so difficult to observe a top quark?2. What observable phenomena does the Standard Modelnot explain?3. How much more massive is an up quark than an elec-tron?4. How many fermions are there in the Standard Model?

5. The antiparticle of the electron (e-) is the positron.What is the charge and rest mass of a positron?Worked Solutions

17.2 Quarks

Quarks (pronounced like 'orcs’ with a 'qu' on the front)are a subset of the fermions - they make up part of mat-ter, most notably the nuclei of atoms. Quarks interactwith all four of the fundamental forces: gravity, electro-magnetism, and the weak and strong nuclear forces.

17.2.1 Generations

There are four quarks in each of the three generationsof fermions. The first contains the up quark ( u ), downquark ( d ), antiup quark ( u ) and antidown quark ( d). The second generation contains the charm quark ( c ),strange quark ( s ), anticharm quark ( c ) and antistrangequark ( s ). The third generation contains the top quark (t ), bottom quark ( b ), antitop quark ( t ) and antibottomquark ( b ).

17.2.2 Charge

The up, charm and top quarks have a charge of +⅔e, andso their respective antiparticles have a charge of -⅔e. Thedown, strange and bottom quarks have a charge of -⅓e,and so their respective antiparticles have a charge of +⅓e.

17.2.3 Hadrons

When quarks are combined, they form larger particles,which are not fundamental. These larger particles areknown as hadrons and are held together by the strong nu-clear force. There are two types of hadrons: baryons andmesons.

Baryons

Baryons are hadrons which are made up of three quarks.The two most common baryons are the proton and theneutron. Protons are made up of two up quarks and onedown quark, giving them a total charge of +1e. Neutronsare made up of one up quark and two down quarks, givingthem net charge of 0.

Mesons

Mesons are hadrons which are made up of a quark andan antiquark. For example, pions are made up of twofirst generation quarks - the π0 is made up of either anup quark and an antiup quark, or a down quark and an

82 CHAPTER 17. PARTICLE PHYSICS

antidown quark. The π+ is made up of an up quark and anantidown quark (total charge +1e), and The π- is made upof a down quark and an antiup quark (total charge −1e).

17.2.4 Questions

1. The Δ++ baryon is made up of up quarks. What is itstotal charge?2. The Δ- baryon has a total charge of −1e. Given thatit is made up of only one type of first generation quark,what is this quark?3. What is an antiproton made of? What is its charge?4. A K+meson is made of an up quark and an antistrangequark. What is its total charge?5. Lambda (Λ) baryons are made up of an up quark, adown quark, and another quark (not an antiquark). TheΛ0 is neutral, and contains a second generation quark.What is this quark?Worked Solutions

17.3 Bosons

Bosons are particles with an integer spin, such as 1, 2etc and mediate a specific force. All interactions can bedescribed by one of the four forces, gravity, electromag-netic, weak and strong which are caused by the release ofa corresponding boson.

17.3.1 Feynman Diagrams

One way of representing these interactions is the Feyn-man diagram. This is a graph with time on the verticalaxis, and space on the horizontal axis showing the pathsof particles through space and time as lines. So, a station-ary electron looks like this:

It is often useful to define our units of space and time insuch a way that, if something is travelling at the speed oflight, it makes a 45° angle. Bosons are virtual particles,so they are given wavy lines. So, a photon travelling at

the speed of light from A to B looks like the following:

Different particles can, of course, interact with eachother. These interactions must take place at a definitepoint in space-time. They can be represented by a cer-tain point on a Feynman diagram, with lines coming inand out of the point representing the velocities of parti-cles which take part in the interaction.

17.3.2 Photons

Photons are massless, have a spin of +1, neutral chargeand carry the electromagnetic force. They are 'given off'by one particle, causing it to change its velocity. Theyare then 'received' by another particle, causing it too tochange its velocity. This can be represented on a Feyn-man diagram in the following way:

e-

e-

e-

e-

17.3.3 W and Z Bosons

W and Z bosons carry the weak nuclear force betweenparticles. This occurs, for example, in β decay, which ac-tually takes place in two stages. First, a proton turns intoa neutron (or vice versa), emitting a W boson. Then, theW boson 'turns into' an electron / positron and an (anti-)neutrino. This is shown in the following Feynman dia-gram:

17.5. MILLIKAN’S EXPERIMENT 83

17.3.4 Gluons

Gluons carry the colour force between quarks, holdingthem together. Quarks have a property known as 'colour',as do gluons. The gluons carry colour between the quarks,mediating the colour force. The strong force is the resid-ual colour force that holds hadrons together. You proba-bly won't be asked about gluons in the exam.

17.3.5 Questions

1. A stationary light source emits single photons at regularintervals. Draw a Feynman diagram to represent this.2. Write two equations (including a W+ boson) whichdescribe positron emission.3. What is the charge on a W- boson?4. Read Richard Feynman’s excellent book, “QED - theStrange Theory of Light and Matter”, ISBN 978-0-140-12505-4.Worked Solutions

17.4 Leptons

Leptons are particles which interact with all the funda-mental forces except for the strong nuclear force. Thereare two types of leptons: electrons and neutrinos.

17.4.1 Electrons

Electrons are particles with a charge of −1.6 x 10−19C.They are responsible, amongst other things, for the wholeof chemistry since, as they occupy the quantum statesaround the nucleus. There are three types of electrons:the electron (e-), the muon (μ-), and the tauon(τ-), onefor each generation. These electrons have antiparticles,each with a charge of +1.6 x 10−19C: the positron (e+),the antimuon (μ+), and the antitauon (τ+), respectively.

17.4.2 Neutrinos

Neutrinos are chargeless, and almost massless. Loads ofthem travel around the universe and through you at speedsclose to the speed of light. The symbol for a neutrino isthe Greek letter nu (ν), with its generation (e, μ or τ) insubscript. If it is an antineutrino, the symbol has a barabove it. So, the symbol for a muon-antineutrino is νµ .

17.4.3 Lepton Number

All leptons have a lepton number of 1. All antileptonshave a lepton number of −1. In a nuclear reaction, thelepton number before the reaction must equal the leptonnumber after the reaction. This necessitates the existenceof neutrinos. When a nucleus gives off a beta particle(electron), the lepton number before the emission is 0.Without neutrinos, the lepton number after the emissionwould be 1, not 0. In reality, an electron-antineutrino isalso emitted, with a lepton number of−1, and so the totallepton number both sides of the reaction is 0.The situation is actually slightly more complicated, as thelepton number from each generation of particlesmust alsobe conserved. The lepton number from the beta particlecannot be balanced out by a tauon-antineutrino, since thisis from a different generation.

17.4.4 Questions

1. An electron is produced by a nuclear reaction, but anelectron-antineutrino is not produced. What other parti-cle is produced?2. Why do electrons not make up part of the nucleus?3. Why did it take until the 1950s to detect the first an-tineutrino?4. Complete the following equation for the emission of abeta particle from a nucleus:10n →1

1 p + ? + ?

5. Complete the following equation for the emission ofan antielectron from a nucleus:11p →1

0 n + ? + ?

6. Complete the following equation for the capture of anelectron by a nucleus:11p+ ? →1

0 n + ?

Worked Solutions

17.5 Millikan’s Experiment

Electrons have a finite charge, which is approximately1.6 x 10−19C. This was first proven by Robert Millikanin 1909. Millikan sprayed drops of oil which were then

84 CHAPTER 17. PARTICLE PHYSICS

charged (ionised) either by friction as they were sprayed,or with x-rays. They were then allowed to fall into a uni-form electric field.

Once in the uniform electric field, the strength of the fieldwas adjusted in order to keep an oil drop stationary. Thiswas done by hand, looking through a microscope. In astationary position, the gravitational force and the electricforce were balanced - there was no net force on the oildrop. So, at this point:qVd = mg

The electric field strength was adjusted by changing thevoltage between the two plates. The voltage at whichthe drops were stationary was measured. The charge oneach drop was then calculated. Millikan found that thesecharges were all multiples of 1.6 x 10−19C, thus show-ing that the charge of each drop was made up of smallercharges with a charge of 1.6 x 10−19C.

17.5.1 Questions

h = 6.63 x 10−34 Jsc = 3 x 108 ms−1

g = 9.81 ms−2

1. Rearrange the formula above in terms of q.2. The mass of an oil drop cannot be measured easily.Express the mass of an oil drop in terms of its radius rand its density ρ, and, by substitution, find a more usefulformula for q.3. An oil droplet of density 885kgm−3 and radius 1μmis held stationary in between two plates which are 10cmapart. At what potential differences between the plates isthis possible?4. If the X-rays used to ionise the oil are of wavelength1nm, how much energy do they give to the electrons?Why does this mean that the oil drops are ionised?5. In reality, the oil drops are moving when they enter theuniform electric field. How can this be compensated for?Worked Solutions

17.6 Pair Production and Annihi-lation

17.6.1 Pair Production

Sometimes, a photon turns into a particle and its antipar-ticle, for example, an electron and a positron. It could notturn into just an electron, since this would leave the leptonnumber unbalanced. The photon must have enough en-ergy to create the masses of the two particles. The energyrequired to create one of the particles is given by:E = mc2 ,where m is the mass of the particle, and c is the speed oflight (3 x 108 ms−1). However, two particles must be cre-ated. Since the two particles are each other’s antiparticle,they have identical masses. So, the total energy requiredis:E = 2mc2

17.6.2 Annihilation

When a particle meets its antiparticle, the two annihi-late each other to form 2 photons (due to conservationof momentum) with sum total energy equivalent to thetotal mass-energy of both particles.Sometimes, a pair of particles annihilates, but then thephoton produces another pair of particles. Also, a photoncould produce a pair of particles which then annihilateeach other.

17.6.3 Questions

h = 6.63 x 10−34 Js1. The mass of an electron is 9.11 x 10−31 kg. What isthe minimum amount of energy a photon must have tocreate an electron?2. A 1.1 MeV electron annihilates with a 1.1 MeVpositron. What is the total energy of the photon pro-duced?3. What is its frequency?4. What is its wavelength?5. A newly produced electron-positron pair are likelyto annihilate almost immediately. Under what circum-stances can this be avoided?Worked Solutions

17.7 Particle Accelerators

Modern experimental particle physics requires particlesto be accelerated to very high energies. This is accom-

17.8. CLOUD CHAMBERS AND MASS SPECTROMETERS 85

plished by passing them through an electric field multi-ple times, in a similar fashion to an electron gun. Typesof particle accelerator include linear accelerators and cy-clotrons.

17.7.1 Linear Accelerators

In a linear accelerator, particles pass through a series oftubes. At either end of each tube are electrodes. An al-ternating current is used. This means that, when particlespass an electrode to which they are being attracted, theelectrode switches charge, and starts to repel the parti-cle. The distances between electrodes increase as you goalong the accelerator, since, as the particles accelerate,they travel further per. oscillation of the current.

Aerial photo of the Stanford Linear Accelerator Center, with de-tector complex at the right (east) side

The original patent for a cyclotron.

17.7.2 Cyclotrons

A cyclotron is like a linear accelerator, except that, in-stead of using lots of different electrodes, it uses the sametwo over and over again. The particles move around in acircle due to a magnetic field. The radius of this circledepends on the velocity of the particles. The orbits of theparticles are enclosed by two semi-cylindrical electrodes.An alternating current is used to accelerate the particles.When the particles enter one half of the cyclotron, theyare pulled back to the other half. When they reach theother half, the current switches over, and they are pulledback to the first half. All the time, the magnetic fieldkeeps them moving in circles. As they gain energy fromthe electric field, the radii of their orbits increase, andtheir velocities increase, until the radius is as large as thecyclotron.

17.7.3 Questions

1. Use the formula for centripetal force to show that theradius of motion depends on the speed of the moving ob-ject.2. A cyclotron with a diameter of 1.5m is used to ac-celerate electrons (mass 9.11 x 10−31kg). The maximumforce exerted on an electron is 2.4 x 10−18N. What is themaximum velocity of the electrons?3. What are the problems involved in constructing a largecyclotron?4. Why don't particles stick to the electrodes when pass-ing through them?Worked Solutions

17.8 Cloud Chambers and MassSpectrometers

A cloud chamber without a magnetic field, so the particles movein straight lines.

17.8.1 Cloud Chambers

The magnitude of the magnetic force on a movingcharged particle is given by:F = qvB ,where B is themagnetic field strength, v is the speed of theparticle and q is the charge on the particle. This force isexerted in a direction perpendicular to both the magneticfield and the direction of motion. If a charged particleenters a uniform magnetic field which is perpendicular toits velocity, then it will move in a circle, since there willbe a force of constant magnitude acting on it in a direc-tion perpendicular to its motion. Using the equation forcentripetal force, we can derive a formula for the radiusof this circle:mv2

r = qvB

86 CHAPTER 17. PARTICLE PHYSICS

mvr = qB

r = mvqB = p

qB ,where p is the momentum of the particle, and m is themass of the particle. This equation makes sense. If theparticle has a higher momentum, then its circle of mo-tion will have a larger radius. A stronger magnetic fieldstrength, or a larger charge, will make the radius smaller.In a cloud chamber, particles enter a magnetic field, andalso a liquid which they ionise. This ionisation causesthe paths of the particles to become visible. When theparticle loses its charge, its track ceases. When the parti-cle loses momentum, the radius of the circle decreases,and so, particles spiral inwards. The direction of thisspiralling depends on the direction of the magnetic field.If the direction of the magnetic field causes a positivelycharged particle to spiral clockwise, then it will cause anegatively charged particle to spiral anticlockwise. Cloudchambers can, therefore, be used to identify particles bytheir charge and mass.

17.8.2 Mass Spectrometers

Mass spectrometers work on a similar principle. Parti-cles to be identified (such as nuclei) are accelerated usingan electric field. Then, a velocity selector is used to en-sure all the nuclei are at a known velocity - all the rest arediscarded. These nuclei enter a uniform magnetic fieldwhere they move in a circle. However, they are only al-lowed to move half a circle, since they are collected at thispoint, and the number of particles arriving at each pointis measured.

A velocity selector

Velocity Selector

In the velocity selector, both a uniform electric field anda uniform magnetic field act on the particle. The onlyway a particle can travel through the velocity selector ina straight line is if the electric force on it is equal andopposite to the magnetic force on it. If this is not the case,the particle’s path is bent, and so it does not get out of the

velocity selector into the rest of the mass spectrometer.If we equate these two forces, we get:qE = qvB ,where q is the charge on the particle, E is the strength ofthe uniform electric field, v is the velocity of the particle,and B is the strength of the uniform magnetic field. Thecharge may be eliminated from both sides:E = vB

Therefore:v = E

B

This means that, by adjusting the strengths of the electricand magnetic fields, we can choose the velocity at whichparticles emerge from the velocity selector.

Diagram of the mass spectrometer which went to the moon onApollo 16.

Finding Mass

The particles themmove at a speed v into another uniformmagnetic field. Here, as in the cloud chamber, the radiusof the circle in which the particle moves is given by:r = mv

qB = mEselector

qBBselector

If we know the charge on the particle (for example, weknowwhat element it is), we canmeasure the radius of thecircle, and find the mass of the particle (i.e. what isotopeit is, since neutrons have no charge) using the formula:m = qBr

v = qrBBselector

Eselector

If we do not know the charge, then we can find the massto charge ratio:mq = rBBselector

Eselector

17.8.3 Questions

Charge of electron = −1.6 x 10−19CMass of electron = 9.11 x 10−31kgu = 1.66 x 10−27kg

17.8. CLOUD CHAMBERS AND MASS SPECTROMETERS 87

1. An electron enters a cloud chamber, passing into a0.1T magnetic field. The initial curvature (the reciprocalof its radius) of its path is 100m−1. At what speed was itmoving when it entered the magnetic field?2. The electron spirals inwards in a clockwise direction,as show in the diagram on the right. What would the pathof a positron, moving with an identical speed, look like?3. Using a 2T magnetic field, what electric field strengthmust be used to get a velocity selector to select only par-ticles which are moving at 100ms−1?4. Some uranium (atomic number 92) ions (charge +3e)of various isotopes are put through the velocity selectordescribed in question 3. They then enter an 0.00002Tuniform magnetic field. What radius of circular motionwould uranium-235 have?Worked Solutions

Chapter 18

Nuclear Physics

18.1 Quantum Principles

There are two principles which you do not need to knowfor the exam, but may be helpful in understanding someof the concepts in the course.

18.1.1 Heisenberg Uncertainty Principle

The Heisenberg uncertainty principle states that the mo-mentum and position of an object are limited. Within acertain uncertainty, when we measure a quantum’s posi-tion, it does not have a definite momentum. When wemeasure its momentum, it ceases to have a definite posi-tion. If we try and measure both, the uncertainty in bothwill be limited. If we let the uncertainty in our knowledgeof momentum be Δp, and the uncertainty in our knowl-edge of position be Δx:∆x∆p = h

4π ,where h is Planck’s constant (6.63 x 10−34 Js). TheHeisenberg uncertainty principle explains what happenswhen electrons occupy energy levels - within these lev-els, they are limited to a certain range of momentumsand positions, but it is meaningless to say which exactmomentum and position they occupy. If we measure themomentum with no uncertainty, then the uncertainty inposition becomes infinite, and vice versa.

18.1.2 Pauli Exclusion Principle

The Pauli exclusion principle states that no two fermions(a set of particles including the electron) may occupy thesame quantum state as each other. In layman’s inaccurateterms, this means that, although two such particles can bein the same place as each other, if they are, they will bemoving at different velocities and so will shortly no longerbe in the same place as each other.This is why, for example, electrons appear to have 'shells’- there is only a limited number of quantum states that theelectrons can occupy, so some have to occupy a different'shell'. Also, without the Pauli exclusion principle, matterwould collapse in on itself - the attractive forces betweenparticles are greater than the repulsive forces. However,

the moment they try and do this, then they must be mov-ing at different velocities, and so no longer be collapsingin on each other.

18.2 Radioactive Emissions

Table of the number of neutrons and protons in the nuclei ofisotopes, showing their most common mode of decay.

'Radioactivity' is a catch-all term for several differentemissions from the nuclei of 'radioactive' atoms. Thereare three main types of radiation: alpha (α), beta (β) andgamma (γ). When radiation occurs, four things must beconserved:

• Mass

88

18.3. ENERGY LEVELS 89

• Charge

• Lepton Number

• Baryon Number

In formulae, mass and charge are shown next to the sym-bol of the particle. For example, a neutron with mass 1uand no charge is 1

0n . The charge on a nucleus is equal tothe number of protons in the nucleus (electrons can be ig-nored). The lepton and baryon numbers may be obtainedby counting the number of leptons and baryons on eitherside of the equation, remembering that antiparticles havenegative lepton and baryon numbers.

18.2.1 α Radiation

Unstable nuclei with a mass greater than 82u emit α ra-diation. This consists of an Helium nucleus ( 42He ). Thealpha particle simply splits off from the nucleus. Sincethe particle has no electrons, it has a charge of +2e. This,combined with its relatively large mass, means that it re-acts easily with other particles, ionising them, meaningthat it cannot penetrate more than a few centimetres ofair.

18.2.2 β Radiation

Unstable nuclei with a mass below 82u emit β radiation.There are two types of β radiation. β- radiation consistsof an electron ( 0

−1e ). This is produced by nuclei withmany more neutrons than protons. A neutron changesinto a proton, emitting an electron and an antineutrinoin order to balance the lepton number. β+ radiation con-sists of an positron ( 0

1e ). This is produced by nucleiwith roughly the same number of neutrons as protons. Aproton changes into a neutron, emitting a positron and aneutrino.β particles also ionise particles, but since they have lesscharge andmass, they do this less easily, and so they travelfurther (on average). Both α and β radiation result in thenucleus which emitted them being changed into anotherelement (Transmutation).

18.2.3 γ Radiation

The binding energies of nuclei are quantized - they canonly take on certain values. When an electron jumpsdown an energy level, this energy has to go somewhere- it takes the form of a γ photon. The structure of the nu-cleus is not changed by γ radiation. γ radiation is ionising,but only at the right frequency - the resonant frequency ofthe things it ionises. γ radiation travels very far, and onlya good thick layer of lead can stop it.

18.2.4 Questions

You will need a periodic table.1. Americium-241 is an α emitter. What element, andwhat isotope, is produced by this decay?2. Iodine-129 is a β- emitter. What element, and whatisotope, is produced by this decay?3. Gamma rays are used to kill microbes in food. Whydoesn't the food become radioactive?4. Plutonium-244 decays by emitting an α particle. Itdoes this twice, emits a β- particle, and then emits a fur-ther two α particles. The nucleus becomes a different el-ement each time. What element is produced at the end?5. Carbon-11 changes into Boron-11 by a radioactiveemission. What was emitted?6. Uranium-236 decays, following the equation:23692 U → 232

90 Th+ X

Identify the particle X in this equation.Worked Solutions

18.3 Energy Levels

As an electron approaches a nucleus from infinity, it be-comes 'bound' - it is attached to the nucleus, if you like.In this bound state, the electron occupies what is calledan energy level. A nucleus has a discrete number of en-ergy levels, and so electrons bound to a certain nucleuscan only take on certain potential energies. These ener-gies are negative by convention.The lowest (most negative) energy level is denoted n=1,the next lowest n=2, and so on. The values of these canbe found using formulae which you don't need to knowabout. Alternatively, theymay be determined experimen-tally.At random, electrons jump between energy levels. If theyjump to a lower energy level (more negative), they releaseenergy in the form of a photon. If they jump to a higherenergy level, theymust absorb a photon of the appropriateenergy. The energies of these photons can be calculatedusing the following formulae, which you should alreadyknow from AS:E = hf

c = λf ,where E is energy, h is Planck’s constant (6.63 x 10−34 Js), f is frequency, c is the speed of light, and λ is wave-length.The energy levels of different nuclei are different. Ev-idence for these energy levels comes from the emissionand absorption spectra of atoms. An emission spectrumcan be obtained by heating a sample of an element. This

90 CHAPTER 18. NUCLEAR PHYSICS

Energy levels in a hydrogen atom. The transition shown from then=3 level to the n=2 level gives rise to visible light of wavelength656 nm (red).

gives the electrons energy, so they jump up the energylevels. At random, they then jump down again, givingoff photons with measurable frequencies. The formu-lae above can be used to calculate the difference in en-ergy between the levels between which the electrons havejumped.An absorption spectrum can be found by passing lightthrough (for example) a gas, and observing the frequen-cies of light which are absorbed. These frequencies cor-respond to jumps between energy levels which electronshave undergone when they absorb the photons, gainingenergy.It should be noted that electrons do not always jump tothe next-door energy level - they can, in principle, jumpto any energy level. They cannot jump to an energy whichis not that of an energy level.

18.3.1 Questions

The following table gives the wavelengths of light givenoff when electrons change between the energy levels inhydrogen as described in the first row:1. Calculate the potential energy of an electron at leveln=2.2. Calculate the difference in potential energy betweenlevels n=2 and n=3.3. What is the potential energy of an electron at leveln=3?4. If an electron were to jump from n=7 to n=5, whatwould the wavelength of the photon given off be?Worked Solutions

18.4 Fission

Nuclear fission is the splitting of the nucleus of a massiveatom into smaller nuclei. This is used to produce energyin power stations and nuclear bombs.

18.4.1 Chain Reaction

In order to start nuclear fission, one nucleus must be madeto split apart. This is achieved by getting the nucleus toabsorb a slow-moving neutron. When the nucleus splits,it releases energy, two components, and possibly somemore neutrons. If at least one neutron is released, thena chain reaction occurs. This neutron goes on to makeanother nucleus unstable, which splits, and producesmoreneutrons, and so on.If this chain reaction is uncontrolled, a massive amount ofenergy is released very fast. This is an atomic explosion,which is used in nuclear bombs. In order to use nuclearfission in a power station, the number of neutrons releasedmust be controlled by inserting a substance such as boroninto the reactor, which absorbs the neutrons, preventingthem from going on to make more nuclei split.

18.4.2 Binding Energy

The reason nuclear fission produces energy is that thebinding energy of the original nucleus is greater than thebinding energy of the products of the fission reaction.This difference in binding energy is the amount of en-ergy released as photons (some of which are infra-red).This energy is used to heat up steam, pressurizing it, andenabling it to turn a turbine, producing electricity.

18.4.3 Neutron Moderator

Neutrons have to be moving slowly in order to cause anucleus to become unstable and split. If they are movingtoo fast, then they simply bounce off. A neutron modera-tor (such as graphite or heavy water) is used to slow themdown.

18.4.4 Questions

1. A neutron is fired at some Uranium-235. Barium-141and Krypton-92 are produced:10n+235

92 U →14156 Ba+92

36 Kr +N10n

How many neutrons are produced (i.e. what is the valueof N)?2. What proportion of the neutrons produced must beabsorbed in order to make the reaction stable?3. What would happen if too many neutrons were ab-sorbed?

18.6. BINDING ENERGY 91

4. Alternatively, Uranium-235 can split into Xenon-140,two neutrons and another element. What is this element?(You will need to use a periodic table.)Worked Solutions

18.5 Fusion

Nuclear fusion of deuterium and tritium.

Nuclear fusion is the joining together of atomic nuclei toform a larger nucleus, and possibly some other products,including energy. It occurs naturally in stars, where hy-drogen is fused together into larger isotopes of hydrogenand then into helium, releasing energy along the way.

18.5.1 Forces

Nuclei repel each other due to the electromagnetic force,since they have the same charge. However at a range ofbetween 1 and 3 femtometers the strong force causes nu-cleons to be attracted to one another, with the magnitudeof this force being far greater than that of electromagneticrepulsion. Therefore in order for two nuclei to fuse, theymust be sufficiently close enough together that the attrac-tive force between the baryons due to the strong nuclearforce is greater than the repulsive force due to the elec-tromagnetic force. If this is the case, then the two nucleiwill become a new, larger, nucleus.

18.5.2 Uses

Nuclear fusion was used by humans for the first time inthe hydrogen bomb, whereby a nuclear fission reactionwould occur, releasing enough thermal energy so that nu-clear fusion could occur, which would release free neu-trons allowing the nuclear fission reaction to be moreefficient, with more of the unstable isotope undergoingfission as well as a modest amount of energy being re-leased during the fusion process. At the time of writ-ing (2009), commercially viable fusion power has not yet

been achieved. However, research is under way, specif-ically at the Culham Centre for Fusion Energy, to bringa fusion reaction under control so that it can be used togenerate electricity. This would have the advantage ofminimal nuclear waste, since the main product would benon-radioactive helium, with some tritium, which has arelatively short 12-year half-life.

18.5.3 Binding Energy

The fusion of nuclei smaller than Iron-56 releases energy.This is because, if we were to take all the baryons of boththe nuclei apart, and then stick them all back together asone, we would do less work than would be required tostick them together as the two separate nuclei. The dif-ference in binding energy is the energy which is releasedby a fusion reaction. This energy might be given to the'real' particles which are given off, or to a 'virtual' particlesuch as a photon.

18.5.4 Questions

c = 3 x 108 ms−1

1. In the Sun, two tritium nuclei ( 31H ) are fused to pro-

duce helium-4 ( 42He ). What else is produced, apart fromenergy?2. In larger stars, carbon-12 ( 126 C ) is fused with protium( 11H ). What single nucleus does this produce?

3. In this reaction, 1.95MeV of energy is released. Whatdifference in binding energy does this correspond to?4. If all this energy was emitted as a photon, what wouldits frequency be?5. In order to contain a fusion reaction, electromagnetismmay be used. What other force could be used? Why isthis not being used for fusion reactors on Earth?Worked Solutions

18.6 Binding Energy

It takes energy to pull nuclei apart. The amount of work(energy) which must be done in order to pull all of theneutrons and protons in a nucleus infinitely far apart fromeach other is known as the binding energy of the nucleus.Practically, pulling them all apart far enough to stop theminteracting will do.If energy must be put in to a nucleus to break it apart,where does this energy go? The answer lies in the factthat if you add up the masses of all the protons and neu-trons in a nucleus individually, it is a little bit more thanthe actual mass of the nucleus. The binding energy put into break the nucleus apart has 'become' mass in the indi-vidual baryons. So, the binding energy of a nucleus can

92 CHAPTER 18. NUCLEAR PHYSICS

be calculated using the following formula:Eb = (nNmN + nZmZ −M)c2 ,where nN and nZ are the numbers of neutrons and protonsin the nucleus, mN andmZ are themasses of neutrons andprotons, M is the mass of the nucleus and c is the speedof light (3 x 108 ms−1).

18.6.1 The Unified Atomic Mass Unit

The unified atomic mass unit, denoted u, is roughly equalto the mass of one proton or neutron. 1 u = 1.660538782x 10−27 kg. They are useful since 1 mole of atoms witha mass of 1 u each will weigh exactly 1 gram. However,when dealing with binding energy, you must never useatomic mass units in this way. The mass defect is sosmall that using atomic mass units will result in a com-pletely wrong answer. If you want to use them with lotsof decimal places, then you will save writing in standardform.

18.6.2 Data

The following table gives the masses in kg and u of theproton and the neutron:

18.6.3 The Binding Energy Curve

Different nuclei have different binding energies. Theseare determined by the combination of protons and neu-trons in the nucleus. These are shown in the followinggraph:

U235

U238

Fe56O16

C12

He4

Li6Li7

He3H3

H2

H1

Number of nucleons in nucleus

Ave

rage

bin

ding

ene

rgy

per n

ucle

on (M

eV)

9

8

7

6

5

4

3

2

1

00 30 60 90 120 150 180 210 240 270

The position of Iron-56 at the top is important. If you taketwo nuclei completely apart, you do work. If you then putall the baryons back together again as one nucleus, youwill get energy back out. Sometimes, the energy you getback out will be more than the work you had to do to takethe nuclei apart. Overall, you release energy by fusing thenuclei together. This happens to nuclei which are smallerthan Iron-56. Nuclei which are larger than Iron-56 willgive out less energy when fused than you had to put in totake them apart into their constituent baryons in the firstplace. To the right of Iron-56, nuclear fusion, overall,requires energy.If you take only one nucleus apart you still do work. Ifyou stick its protons and neutrons back together, but thistime in two lumps, you will get energy out. Again, some-times this energy will be greater than the work you hadto do to take them apart in the first place. Nuclear fissionwill be releasing energy. This occurs when the nucleus islarger than Iron-56. If the energy released is less than theinitial work you put in, then nuclear fission, overall, re-quires energy. This happens when the nucleus is smallerthan Iron-56.This can be summarized in the following table:

18.6.4 Questions

1. Deuterium (an isotope of Hydrogen with an extra neu-tron) has a nuclear mass of 2.01355321270 u. What is its

18.7. RISKS, DOSES AND DOSE EQUIVALENTS 93

binding energy?2. Uranium-235 has a nuclear mass of 235.0439299 u. Itcontains 92 protons. What is its binding energy?3. How would you expect H-2 and U-235 to be used innuclear reactors? Why?Worked Solutions

18.7 Risks, Doses and Dose Equiv-alents

18.7.1 Risk

Radioactivity results in risk - this could be a risk of death,or a risk of developing cancer. In physics, risk is what youexpect, on average, to happen:risk = probability× consequenceSo, if there is a 1 in 500 chance that someone gets runover by a car when crossing the road, the risk involved inallowing 500 people to cross the road is one accident.

18.7.2 Absorbed Dose

Absorbed dose is measured in grays, commonly denotedGy. One gray is defined as one joule absorbed per kilo-gram. You may be expected to use the equation

E = hf =hc

λ

where

h is the Planck constantf is the frequency of the photonc is the speed of lightλ is the wavelength of the photon

or

E = ne

where

n is the number of particlese is the magnitude of the charge of one electron

to calculate absorbed dose in terms of numbers of par-ticles with a given frequency, wavelength or energy (ineV). If someone is exposed to a certain activity (parti-cles per second) over a period of time, the absorbed doseaccumulates.

18.7.3 Dose Equivalent

Hourly dose equivalent due to cosmic rays per. hour in Sv, acrossthe globe.

Absorbed dose does not give a full picture of the potentialharm radioactivity can do to you. Different types of ra-diation do different amounts of damage. Absorbed doseequivalent, measured in sieverts (denoted Sv) attempts tocompensate. To calculate the dose equivalent, multiplythe dose in grays by the quality factor of the particles ab-sorbed. These quality factors are given in the table below.

18.7.4 Questions

1. A mobile phone emits electromagnetic radiation. 1.2watts of power are absorbed per. kilogram. Assum-ing that the radiation is absorbed uniformly across a 5kghead, what dose of radiation would be delivered to thehead when making a 10-minute telephone call?2. What dose equivalent does this correspond to?3. How many nuclei are there in 1 mg of Americium-241?4. A ham sandwich becomes contaminated with 1 μg ofAmericium-241, and is eaten by an 80kg person. Thehalf-life of Americium-241 is 432 years. Given thatAmericium-241 gives off 5.638 MeV alpha particles,how long would it be before a dose equivalent of 6 Svis absorbed, making death certain?5. What assumptions have you made?Worked Solutions

Chapter 19

Appendices

19.1 Trigonometry

Trigonometry is the study of the ratios between sides ona right-angled triangle. Consider the following diagram:

A

B

C

ca

b

Adjacent

Opposite

Hypoten

use

Depending on the value of the angle BAC, the ratios be-tween the three sides of the triangle will vary. These threesides are generally given the names 'hypotenuse' (H), 'op-posite' (O) and 'adjacent' (A). The opposite side is op-posite the angle being considered. The hypotenuse is theside opposite the right angle, and the adjacent side is theremaining side, which is also next to the angle being con-sidered.The three basic trigonometric ratios, in relation to anacute angle θ, are the sine, cosine and tangent ratios, asfollows:sin θ = O

H

cos θ = AH

tan θ = OA

It is often helpful to remember them by using themnemonic 'SOH-CAH-TOA'.These ratios can be calculated using the appropriate but-tons on a scientific calculator. In addition, if you already

know the ratio between two sides, and wish to calculatethe angle from it, you can use the sin−1, cos−1 and tan−1

buttons to use the inverse function. These functions aretechnically known as arcsin, arccosin and arctan.

19.2 Logarithms

That formula involved logarithms. If ab = c, then:loga c = b

In other words, a logarithm is a way of asking the ques-tion, "a to the power of what is c?" a is known as the baseof the logarithm. log a is shorthand for log10a, and can becalculated using the 'log' button on a scientific calculator.Just as with indices, there are some laws of logarithms.At present, we are only concerned with this one:loga(cn) = n loga cIf we apply this to the formula for calculating the numberof possible values a pixel can take on (v = 2b), we canrearrange it to get a formula for b by taking the logarithmto the base 10 of both sides:log v = log(2b)log v = b log 2b = log v

log 2

We can then use a scientific calculator to calculate log 2,giving us the formula:b ≈ log v

0.3

19.3 Delta - 'difference in'

The Greek letter delta means 'difference in'. So, in thecase of current, it represents the difference in charge di-vided by the difference in time. If we write it as we wouldin calculus:I = dQ

dt

Without using calculus, we can only take an approxima-tion of I from values as close together as possible on eitherside of the point at which we wish to calculate the current.

94

19.5. DERIVATION OF EQUATIONS FOR SIMPLE HARMONIC MOTION 95

By using calculus, we can calculate the exact value of thecurrent at a given point. For example, if Q = 2t (i.e. 1coulomb every 2 seconds):I = dQ

dt = 1× 2t1−1 = 2t0 = 2 A

19.4 Sigma - 'sum of'

That’s the Greek letter Sigma. It means 'sum of', so Kir-choff’s First Law is stating that the sum of all the currentsgoing into a point equals the sum of all the currents leav-ing the point. So:Iᵢ 1 + Iᵢ 2 + Iᵢ 3 + ... + Iᵢ -₂ + Iᵢ -₁ + Iᵢ = Iₒᵤ 1 + Iₒᵤ 2 +Iₒᵤ 3 + ... + Iₒᵤ -₂ + Iₒᵤ -₁ + Iₒᵤ

19.5 Derivation of Equations forSimple Harmonic Motion

The following second-order differential equation de-scribes simple harmonic motion:d2xdt2 = −kx

m

d2xdt2 + kx

m = 0

Since we have a second derivative, we cannot separate thevariables, so let:x = ezt

Therefore:dxdt = zezt

d2xdt2 = z2ezt

By substitution:z2ezt + kezt

m = 0

ezt(z2 + k

m

)ezt is asymptotic at 0, so ezt cannot equal 0, and we cantherefore get away with dividing by ezt:z2 + k

m = 0

z2 = −km

z = ±√

−km = ±i

√km

Therefore:x = Peit

√km +Qe−it

√km ,

where P and Q are constants of integration. At this point,it is useful to clean things up a bit by letting:ω2 = k

m

x = Peiωt +Qe−iωt

It has been proven elsewhere (deMoivre’s Theorem) that,when n is a constant:eniθ = cosnθ + i sinnθ and e−niθ = cosnθ − i sinnθ

Therefore:x = P (cosωt+ i sinωt) +Q(cosωt− i sinωt)x = (P +Q) cosωt+ (P −Q)i sinωtLet: R = P +Q

S = P −Q

So, the general solution of the differential equation is:x = R cosωt+ Si sinωtThis describes what the simple harmonic oscillator willdo given any possible situation. However, we don't wantan equation which will cover anything and everything.We want to give our oscillator a starting position - let’ssay, at a position where x = A at t = 0:A = R cos (ω × 0) + Si sin (ω × 0)

A = R cos 0 + Si sin 0Therefore, R = A and S = 0.So, the specific solution is:x = A cosωt

Chapter 20

Worked Solutions

20.1 Lenses

1. A lens has a focal length of 10cm. What is itspower, in dioptres?Always use SI units, so 10cm = 0.1m.P = 1

f = 10.1 = 10 D

2. Light reflected off a cactus 1.5m from a 20D lensforms an image. How many metres is it from theother side of the lens?u must be negative, as per. the Cartesian covention, so u= −1.5.1v = 1

−1.5 + 20 = 583 D

v = 358 = 0.0517 m = 51.7 mm

3. A lens in an RGB projector causes an image tofocus on a large screen. What sort of lens is it? Is itspower positive or negative?The wavefronts are being caused to spread out (diverge)more by the lens. Hence, it is a diverging lens. Thewavefronts are losing curvature, so the lens has a nega-tive power.4. What is the focal length of a 100D lens?100 = 1

f

f = 1100 = 0.01 m = 10 mm

5. The film in a camera is 5mm from a lens whenautomatically focussed on someone’s face, 10m fromthe camera. What is the power of the lens?5mm = 0.005m

10.005 = 1

−10 + P

200 = P − 0.1

P = 200.1 DSince lenses are generally made to have nice round pow-ers, this was probably a 200D lens. However, this is inreal life, and this an hypothetical question with numberspicked out of thin air by me.6. The light from a candle is enlarged by a factorof 0.5 by a lens, and produces an image of a candle,0.05m high, on a wall. What is the height of the can-

dle?0.5 = 0.05

h1

2 = h1

0.05

h1 = 100 mm

20.2 Refraction

1. A ray of light is reflected from a mirror. Its angleto the normal when it reaches themirror is 70°. Whatis its angle of reflection?The angles of reflection on both sides of the normal arealways equal, so the angle of reflection is 70°.2. The speed of light in diamond is 1.24 x 108 m/s.What is its refractive index?n = v1

1.24×108

Assuming the speed of light in the reference medium is 3x 108:n = 3×108

1.24×108 = 31.24 ≈ 2.42

3. The refractive index of ice is 1.31. What is thespeed of light in ice?1.31 = 3×108

v2

v2 = 3×108

1.31 ≈ 2.29× 108 ms−1

4. A ray of light passes the boundary between air anda transparentmaterial. The angle of refraction is 20°,and the angle of incidence is 10°. What is the speedof light in this material? Why is it impossible for thismaterial to exist?3×108

v2= sin 10

sin 20 ≈ 0.508

v2 ≈ 3×108

0.508 ≈ 5.91× 108

This is greater than the speed of light in a vacuum (3 x108), and so is impossible.5. What is the critical angle of a beam of light leavinga transparent material with a refractive index of 2?sinC = 1

2

C = 30° (using the sin−1 button on a scientific calculator

96

20.5. DIGITISATION 97

- this means 'inverse sine', often denoted arcsine or asinfor short.)

20.3 Digital Storage

1. An image transmitted down a SVGA video cableis 800 pixels wide, and 600 pixels high. How manypixels are there in the image?800× 600 = 480000 pixels2. A grayscale image is encoded using 3 bits. Howmany possible values can each pixel have?23 = 83. The characters in a text document are numberedfrom 0 - 255. How many bits should each characterbe encoded with?There are 256 possible values.b = log 256

log 2 = 8

4. A page contains 30 lines of text, with an averageof 15 characters on each line. Each character is rep-resented by 4 bits. How many megabytes of uncom-pressed storage will a book consisting of 650 pageslike this fill on a computer’s hard disk?Total number of characters = 650 x 30 x 15 = 292 500Total amount of information = 292500 x 4 = 1 170 000bits1170000 bits = 1170000

8 bytes = 146250 bytes =14625010242 Mbytes ≈ 0.14Mbytes5. A 10cm wide square image is scanned into a com-puter. Each pixel is encoded using 3 channels (red,green and blue), and each channel can take on 256possible values. One pixel is 0.01 mm wide. Howmuch information does the scanned image contain?Express your answer using an appropriate unit.10cm = 0.1

0.01×10−3 pixels = 10000 pixelsTotal number of pixels = 10 0002 = 100 000 000 pixelsb = log 256

log 2 = 8 bits per. channel per. pixel = 8 ×3 bits per. pixel = 24 bits per. pixelTotal information = 24 x 100 000 000 = 2 400 000 000bits = 300 000 000 bytes300000000

10242 ≈ 286Mbytes (This is why we usually com-press images before storage, or at least use fewer bits per.pixel.)

20.4 Digital Processing

1. How could the above methods be applied to a dig-ital sound sample?By taking the median or mean of the sample and the val-

ues on either side of it. This would be unreliable, as soundis a wave, so the samples need to vary quite widely. How-ever, something similar could be used when comparingrepeating patterns in a waveform.2. Which of the above methods would be suitable forsmoothing sharp edges? Why?Mean smoothing - median smoothing would not blur theedges.3. Use median smoothing to remove noise from thefollowing image of a white cat in a snowstorm (theblack pixels have a value of 255):4. Why would mean sampling not be appropriate forsmoothing the image given in question 3?It would produce a really blurred mess, instead of an im-age, as the noise is too dense.5. Use mean smoothing to remove noise from the fol-lowing image of a black cat in a coal cellar:

20.5 Digitisation

1. Take samples for the signal below every 0.1ms, andthen produce a reconstructed signal. How does it dif-fer from the original?

The high frequency elements of the signal have been lost.2. A signal is sampled for 5 seconds at a samplingrate of 20 kHz. How many samples were taken?

20× 103 =No. of samples

5

No. of samples = 20× 103 × 5 = 100000

3. Most sounds created by human speech except for'ss’ and 'ff' have a maximum frequency of 4 kHz.What is a suitable sampling rate for a low-qualitytelephone?4× 2 = 8 kHz4. Using a sampling rate of 20 kHz and 3 bits, sam-ple the following signal, and then produce a recon-structed signal. What is themaximum frequency thatcan be perfectly reconstructed using this samplingrate?First, calculate the length of each sample, by letting the

98 CHAPTER 20. WORKED SOLUTIONS

number of samples equal 1:20× 103 = 1

Length of one sampleLength of one sample = 1

20×103 = 0.00005 s =0.05 msThen, we can sample the waveform and create a recon-structed signal:

To calculate themaximum frequency that can be perfectlyreconstructed using this sampling rate (20 kHz):202 = 10 kHz

20.6 Signal Frequencies

1. What is the frequency of an X-ray (wavelength0.5nm)?X-rays are electromagnetic waves, so they travel at thespeed of light (3 x 108 ms−1).3 x 108 = f x 0.5 x 10−9

f = 3×108

0.5×10−9 = 600× 1015 Hz = 600 PHz2. A sound wave, with a frequency of 44 kHz, has awavelength of 7.7mm. What is the speed of sound?V = 44 x 103 x 7.7 x 10−3 = 338.8ms−1

Frequency spectrum from qs. 3 & 4

3. What is the fundamental frequency of the follow-ing signal?The big spike on the left is at approximately 750 Hz, sothis is the fundamental frequency.

4. Approximately how many harmonics does it con-tain?There are 14 other big spikes, plus a few other spikeswhich may be large enough to be harmonics.5. The three sine waves sin x°, 4sin(2x-50)° and0.5sin(3x+120)° are added together to form a signal.What are the frequencies of each of the waves? Whatis the signal’s fundamental frequency? Assume thatthe waves are travelling at the speed of light, and that60° = 1mmsin x° has a wavelength of 360°. Using this, we can calcu-late the wavelengths of the other two waves, since f(ax)stretches f(x) by the reciprocal of a on the x axis. Thefrequency of each wave is given by the formula:f = v

λ = 3×108

λ ×10−3

sin x° has the lowest frequency, so 50 GHz is the funda-mental frequency of the signal.

20.7 Bandwidth

1. A broadband internet connection has a bit rate of8Mbit s−1 when downloading information. What isthe minimum bandwidth required to carry this bitrate?2B = 8× 106

B = 8×106

2 = 4× 106 Hz = 4MHz2. The same connection has a bandwidth of 100kHz reserved for uploading information. What is themaximum bit rate that can be attained when upload-ing information using this connection?b = 2 x 100 x 103 = 200 x 103 bits / second = 24.4 kbytes/ second3. A lighthouse uses a flashing light and Morse Codeto communicate with a nearby shore. A 'dash' con-sists of the light being on for 2s. The light is left offfor 1s between dots and dashes. What is the band-width of the connection?One bit takes 3 seconds to transmit.B = 1

3 Hz ≈ 333 mHz4. The broadband connection in question two is usedto upload a 1Mbyte image to a website. How longdoes it take to do this?The bit rate is 200 x 103 bits / second.1 Mbyte = 1048576 bytes = 8388608 bits.

b = no. of bitsno. of seconds

200× 103 = 8388608

no. of secondsno. of seconds = 8388608

200×103 ≈ 42 s

20.10. VOLTAGE 99

20.8 Charge

1. How much charge do 1234 electrons carry?1234 x 1.6 x 10−19 = 1.9744 x 10−16 C = 197.44 aC(attocoulombs)2. How many electrons does is take to carry 5 C ofcharge?5 = 1.6 x 10−19 x nn = 5

1.6×10−19 = 31.25× 1018 electrons3. The total charge on 1 mole of electrons (6 x 1023particles) is equal to 1 faraday of charge. How manycoulombs of charge are equal to 1 faraday?1 faraday = 6 x 1023 x 1.6 x 10−19 coulombs = 96000coulombs =96kC

20.9 Current

1. 10 coulombs flow past a point in a wire in 1minute.How much current is flowing through the point?I = 10

60 ≈ 0.17 A2. How long does it take for a 2A current to carry 5C?2 = 5

t

t = 52 = 2.5 s

1R

1I 2I

2R

I

3. In the diagram on the right, I = 9A, and I1 = 4.5A.What is the current at I2?9 = 4.5 + I2I2 = 4.5 A

4. What would I equal if I1 = 10A and I2 = 15A?I = 10 + 15 = 25A5. In the diagram on the right, in 5 seconds, 5C ofcharged particles flow past I1, and 6.7C flow past I2.How long does it take for 10C to flow past I?10t = 5

5 + 6.75 = 2.34

t = 102.34 ≈ 4.27 s

20.10 Voltage

1. A battery has an EMF of 5V. What is the totalpotential difference across all the components in thecircuit?They are the same! The voltage decreases across the cir-cuit from the voltage at one end of the battery to the other.2. The voltages (relative to the voltage of the battery)on either side of a resistor are −6V and −5V. Whatis the potential difference across the resistor?−5 - −6 = −5 + 6 = 1V3. At a given point in a circuit, 5C of charge have10 kJ of potential energy. What is the voltage at thispoint?V = 10×103

5 = 2 kV4. Why do the electrons move to a point 1cm furtheralong the wire?The point further down the wire has a lower voltage (i.e.less energy per. coulomb) - the electrons will be at a lowerpotential energy at this point, and fall to it, just as objectsfall to lower potential energies due to gravity.

20.11 Power

1. The potential difference across a 9W light bulb is240V. How much current is flowing through the lightbulb?9 = 240II = 9

240 = 0.0375 A2. How much energy is dissipated by a 10W compo-nent in 1 hour?10 = E

602

E = 602 x 10 = 36000 J = 36 kJ3. The potential difference across a top-notch kettle,which can hold up to 1 litre of water, is 240V, and thecurrent is 12.5 A. 4.2 kJ of energy is required to raisethe temperature of 1kg of water by 1 C. Assuming100% efficiency and the temperature has to be raised80K (20C to 100C), how long does it take to boil 1 litreof water?

100 CHAPTER 20. WORKED SOLUTIONS

12.5× 240 = 3000 = 4.2×103×80t

t = 4.2×103×803000 = 112 s

In practice it may take longer than this as energy willbe lost heating the surrounding air, heating the materi-als used in the kettle and the water may be colder than20C.4. How much energy is dissipated by a 100Ω resistorin 10 seconds if 2A of current are flowing?E10 = 22 × 100 = 400

E = 4000 J = 4 kJ5. The charge on an electron is −1.6 x 10−19 C. Howlong does it take for amole (6 x 1023 particles) of elec-trons to flow through a 40W light bulb on a 240V ringmain?Total charge = 1.6 x 10−19 x 6 x 1023 = 96000 C40 = 96000×240

t = 23040000t

t = 2304000040 = 576000 s = 9600 min = 160 h =

6 days 16 hours

20.12 Resistance and Conductance

1. The potential difference across a resistor is 4V,and the current is 10A. What is the resistance of theresistor?R = V

I = 410 = 0.4 Ω

2. What is the conductance of this resistor?G = 1

R = 10.4 = 2.5 S

3. A conductor has a conductance of 2S, and the po-tential difference across it is 0.5V. Howmuch currentis flowing through it?I = GV = 2 ∗ 0.5 = 1A

I = 1A4. A graph is drawn of potential difference across anOhmic conductor, and current. For every 3cm across,the graph rises by 2cm. What is the conductance ofthe conductor?Gradient = 2

3 = ResistanceG = 1

23

= 32 = 1.5 S

5. On another graph of potential difference and cur-rent, the graph curves so that the gradient increasesas current increases. What can you say about the re-sistor?The resistor is not an Ohmic conductor.6. 3 resistors, wired in series, have resistances of1kΩ, 5kΩ and 500Ω each. What is the total resis-tance across all three resistors?ΣR = (1 x 103) + (5 x 103) + 500 = 6500Ω = 6.5kΩ

7. 2 conductors, wired in parallel, have conductancesof 10S and 5S. What is the total resistance of bothbranches of the parallel circuit?G = 10 + 5 = 15 S = 1

R

R = 115 ≈ 0.0667 Ω

8. The circuit above is attached in series to 1 10Ωresistor. What is the total conductance of the circuitnow?ΣR = 10 + 1

15 = 15115 Ω

G = 15151 ≈ 0.0993 S

20.13 Internal Resistance

1. A 9V battery is short-circuited. The potential dif-ference across the battery is found to be 8V, and thecurrent is 5A. What is the internal resistance of thebattery?8 = 9 - 5Rᵢ ₑᵣ ₐ5Rᵢ ₑᵣ ₐ = 1Rᵢ ₑᵣ ₐ = 0.2Ω

2. What is the EMF of the battery in the followingcircuit?∑

Rexternal =(

18+4 + 1

4

)−1

=(13

)−1= 3 Ω

Vₑₓ ₑᵣ ₐ = 5 x 3 = 15 V15 = E - (5 x 10)E = 50 + 15 = 65 V3. What is the internal resistance of the battery in thefollowing circuit?∑

Rexternal =(

12+4 + 1

3+6+1

)−1

=(

415

)−1= 3.75 Ω

Vₑₓ ₑᵣ ₐ = 7.5 x 3.75 = 28.125Which actually makes this question impossible becausethe current is too high for the resistance supplied to thecircuit.

20.15. SENSORS 101

20.14 Potential Dividers

1. A 12 kΩ resistor and a 20 kΩ resistor are con-nected to a 9V battery. A voltmeter is connectedacross the 12 kΩ resistor. What is the reading on thevoltmeter? (Assume negligible internal resistance.)V12kΩ = 9× 12

12+20 = 3.375V

2. A potential divider consists of 100 5Ω resistors,with a wiper which moves on one resistor for every3.6° a handle connected to it turns. The wiper is con-nected to a voltmeter, and the circuit is powered bya 120V power source with negligible internal resis-tance. What is the reading on the voltmeter when thehandle turns 120°?First, work out the number (n) of resistors between theterminals of the voltmeter:n = 120

3.6 = 33.3

The handle, then, has not turned enough to reach the 34thresistor, so the voltmeter is connected just after the 33rdresistor.V = 120× 33×5

100×5 = 39.6 V3. A 9V battery with internal resistance 0.8Ω is con-nected to 3 resistors with conductances of 3, 2 and 1Siemens. A voltmeter is connected across the 3 and 2Siemens resistors. An ammeter is placed in the cir-cuit, between the battery and the first terminal of thevoltmeter, and reads 2A. What is the reading on thevoltmeter?First, work out the resistances of the resistors:R = 1

G = 13 ,

12 , 1 Ω

Then, work out the external potential difference (i.e. ex-cluding the potential difference lost due to the battery’sinternal resistance):Vₑₓ ₑᵣ ₐ = E - IRᵢ ₑᵣ ₐ = 9 - (2 x 0.8) = 7.4V

V3Ω,2Ω = 7.4×13+

12

13+

12+1

≈ 3.37 V

20.15 Sensors

Note to the reader: I am not entirely happy with the an-swers given on this page. --Sjlegg (talk) 15:01, 19 De-cember 2007 (UTC)An LDR’s resistance decreases from a maximum re-sistance of 2kΩ to a minimum resistance of 0Ω aslight intensity increases. It is used in a distance sens-ing system which consists of a 9V power supply, a 1.6kΩ resistor, the LDR and a multimeter which dis-plays voltage to 2 decimal places measuring the po-tential difference across one of the two resistors.1. Across which resistor should the multimeter beconnected in order to ensure that, as the distancefrom the light source to the sensor increases, the po-tential difference recorded increases?As light intensity increases, distance decreases. So, asdistance increases, light intensity decreases. As light in-tensity decreases, the LDR’s resistance increases, as doesthe potential difference across it. So, as distance in-creases, the potential difference increases. Since we wantthe potential difference to change in the same directionas the distance, the multimeter must go across the LDR,not the resistor.2. In complete darkness, what voltage is recorded onthe multimeter?The voltage is split between the 1.6kΩ resistor and theLDR, which currently has a resistance of 2kΩ. Therefore,the potential difference across the LDR is:V = 2

2+1.6 × 9 = 5V

3. When a light source moves 0.5m away from thesensor, the voltage on themultimeter increases by 2V.What is the sensitivity of the sensing system when us-ing this light source, in V m−1?S = 2

0.5 = 4 Vm−1

4. When the same light source is placed 0m from thesensor, the potential difference is 0V. When the lightsource is 1m away, what voltage is displayed on themultimeter?Assuming a linear relationship between distance and po-tential difference, we know that:0.25 = m

V

Hence:V = m

0.25 = 10.25 = 4V

5. What is the resolution of the sensing system?The multimeter measures to 2 decimal places, so thesmallest measurable voltage is 0.01V.0.25 = m

V

Hence:m = 0.25× V = 0.25× 0.01 = 0.0025 m = 2.5 mm

102 CHAPTER 20. WORKED SOLUTIONS

6. Draw a circuit diagram showing a similar sensingsystem to this, using aWheatstone bridge and ampli-fier to improve the sensitivity of the system.

7. What is themaximumpotential difference that canreach the amplifier using this new system (ignore theamplification)?The maximum potential difference will occur when theLDR has no resistance. This will result in 9V on the left-hand side of the voltmeter, and 5V on the right-hand side.The difference is 4V.8. If this signal were to be amplified 3 times, wouldit exceed the maximum voltage of the system? Whatwould the limits on the signal be?4× 3 = 12V > 9V

The signal would be limited to the range −9V < V < 9V.

20.16 Resistivity and Conductivity

1. A material has a conductivity of 106 S m−1. Whatis its resistivity?ρ = 1

σ = 1106 = 10−6 Ω m = 1 µΩ m

2. A pure copper wire has a radius of 0.5mm, a re-sistance of 1 MΩ, and is 4680 km long. What is theresistivity of copper?

ρ = 106×π×(0.5×10−3)2

4680×103 ≈ 168 × 10−9 Ω m =168 nΩ m3. Gold has a conductivity of 45MSm−1. What is theresistance of a 0.01m across gold connector, 0.05mlong?

First, work out resistivity:ρ = 1

45×106 = 22.2× 10−9 ΩmThen, substitute everything possible into the resistivityformula:22.2× 10−9 = (π×(0.5×0.01)2)R

0.05 ≈ 1.57× 10−3R

R ≈ 22.2×10−9

1.57×10−3 ≈ 14.2× 10−6 Ω = 14.2 µΩ

4. A strand of metal is stretched to twice its origi-nal length. What is its new resistance? State yourassumptions.The material does not change, so resistivity is constant.Length doubles, and we know that volume must be con-stant.V = ALA = V/Lρ = RA

L =R(V

L )

L = RVL2

Rold = ρ1 × L2

V = ρL2

V

When we double L, we get:

Rnew = ρ(2L)2

V = 4ρL2

V = 4×Rold

We are assuming that ρ and V are constant.5. Which has the greater resistivity: a plank or apiece of sawdust, made from the same wood?Sawdust and a plank are artefacts, not materials. Hence,they do not have a resistivity. Even if they did, they aremade of the same thing, so they would have equal resis-tivity.

20.17 Semiconductors

1. What is the resistivity of silicon, at room temper-ature?ρ = 1

σ = 1435×10−6 ≈ 2300Ωm = 2.3kΩm

2. What sort of variable resistor would a semicon-ductor be useful in?A thermistor, as the resistance of a semiconductor de-creases as heat increases (but, assuming use of a poten-tial divider, the voltmeter would have to be on the otherresistor).3. If positive ions are added to silicon (doping it), howdoes its conductivity change?Positive ions mean more free electrons, and so greaterconductivity.

20.19. METALS 103

20.18 Stress, Strain & the YoungModulus

1. 10N of force are exerted on a wire with cross-sectional area 0.5mm2. How much stress is being ex-erted on the wire?0.5mm2 = 0.5 x (10−3)2m2 = 0.5 x 10−6m2

σ = 10N0.5×10−6m2 = 20 000 000 Pa = 20MPa

2. Another wire has a tensile strength of 70MPa,and breaks under 100N of force. What is the cross-sectional area of the wire just before breaking?70× 106 = 100

A

A = 10070×106 ≈ 1.43× 10−6 m2 = 1.43 µm2

3. What is the strain on a Twix bar (original length10cm) if it is now 12cm long?E = 12−10

10 = 210 = 0.2

4. What is this strain, expressed as a percentage?0.2 x 100 = 20%5. 50N are applied to a wire with a radius of 1mm.The wire was 0.7m long, but is now 0.75m long. Whatis the Young’s Modulus for the material the wire ismade of?

Y =( 50

π×(1×10−3)2)

( 0.75−0.70.7 )

≈ 160000000.0714 ≈ 224000000 Pa =

224MPa6. Glass, a brittle material, fractures at a strain of0.004 and a stress of 240 MPa. Sketch the stress-strain graph for glass.

7. (Extra nasty question which you won't ever get inan exam) What is the toughness of glass?Toughness equals the area under the above graph. Sinceglass is brittle, we can assume that the gradient of thegraph is constant, and, since the graph passes through theorigin, it is a triangle. So:

Atriangle =base × height

2 = Toughness =strain × stress

2 = 0.004×240×106

2 = 480000 Jm−3 =

480 kJm−3

20.19 Metals

1. Would you expect a metal to have more or less con-ductivity than a semiconductor? Why?A metal has more conductivity than a semiconductor be-cause a metal has more ions than a semiconductor, andhence more free (delocalized) electrons.2. How can the stress-strain graph for a metal be ex-plained in terms of ions in a sea of electrons?In the elastic region, the ions are held together by thecharge between them and the electrons, and they canmove apart when under stress. Then, once the ions are toofar apart, the bonds aren't strong enough to pull them backtogether, so the metal entends under the stress. Eventu-ally, the bonds cannot maintain the structural integrity ofthe metal any more, so the metal 'necks’ in one place, andthen the metal fractures.3. As a metal heats up, what happens to its conduc-tivity? Why?As a metal heats up, both ions and electrons vibrate more.This means that the collision rate between ions and elec-trons goes up, so it is harder for electrons to travel throughthe metal. So, as a metal heats up, its conductivity goesdown.

20.20 Polymers

1. Different crystalline structures have different re-fractive indexes. Why does this mean that a polycrys-talline polymer is translucent?The refractive index of each area is different, so the angleof refraction is different for each area. As a result, theimage is blurred.2. What sort of polymer is a pane of perspex?Crystalline3. What sort of polymer does the pane of perspexbecome when shattered (but still in one piece)?Polycrystalline4. What sort of polymer is a rubber on the end of apencil?Amorphous5. What happens to the translucency of an amor-phous polymer when it is put under stress?The polymer becomes more translucent, as it becomesmore polycrystalline as the chains are stretched andstraightened.

104 CHAPTER 20. WORKED SOLUTIONS

20.21 What is a wave?

1. Through what medium are sound waves propa-gated?Usually the air, but most material objects will carrysound.2. What aspects of the behaviour of light make it looklike a wave?Superposition and diffraction (and others).3. What aspects of the behaviour of light make it looklike a particle?Photoelectric effect4. Consider the diagram on the right. White light ispartially reflected by the transparent material. Someof the light, however, is refracted into the transparentmaterial and reflected back by the opaque material.The result is two waves travelling in the same placeat the same time at the same polarisation(the lightis not a single beam). Why does, say, the red lightdisappear?If the refractive index and width of the transparent ma-terial are correct, then the two waveforms leaving it willbe half a wavelength apart, as one will have travelled fur-ther. Destructive interference will cause the red light tobecome 'invisible'.5. What is the wavelength of green light?500nm6. The lowest frequency sound wave humans can hearhas a frequency of approximately 20Hz. Given thatthe speed of sound in air is 343ms−1, what is thewavelength of the lowest frequency human-audiblesound?λ = v

f = 34320 = 17.15m

20.22 Phasors

1. A sine wave with wavelength 0.1m travels througha given point on the surface of the sea. A phasor ar-row representing the effect of this wave on this pointrotates 1000°. Howmany wavelengths have gone pastin the time taken for the phasor to rotate this much?0.1m is irrelevant information.1000360 = 2.78λ

2. A sine wave has a maximum amplitude of 500nm.What is its amplitude when the phasor has rotated60° from its start position?a sin θ = 500× 10−9 × sin 60 ≈ 433nm3. Two waves have a phase difference of 45°. Whenthe first wave is at its minimum amplitude of−0.3m,what is the total amplitude of the superposed wave-forms?The first wave has an amplitude of−0.300m. The secondis at an angle of 270 - 45 = 225°.a sin θ = 0.3× sin 225 ≈ −0.212m−0.300 +−0.212 = −0.512m

20.23 Standing Waves

1. The air in a 3m organ pipe is resonating at the fun-damental frequency. Organ pipes are effectively openat both ends. What is the wavelength of the sound?3 = λ

2

so λ = 2 x 3 = 6m.2. A string is vibrating at the second harmonic fre-quency. How many wavelengths long is the standingwave created?(2 + 1)× λ

2 = 3λ2

3. Express, in terms of λ, the length of a pipe whichis closed at one end, where λ is the length of one waveat the fundamental frequency.¼λ (this is the distance between a node and an antinode)

20.24 Young’s Slits

1. A 2-slit experiment is set up in which the slits are0.03 m apart. A bright fringe is observed at an angle10° from the normal. What sort of electromagneticradiation was being used?λ = d sin θ = 0.03× sin 10 ≈ 5.21× 10−3 mSo, microwaves were being used.2. Light, with a wavelength of 500 nm, is shonethrough 2 slits, which are 0.05 m apart. What are the

20.27. ELECTRON BEHAVIOUR 105

angles to the normal of the first three dark fringes?n×500×10−9 = 0.05×sin θ , where n is an odd integer.θ = arcsin n×500×10−9

0.05 = arcsin (n× 500× 10−8)

Then, substitute in n = 1,3 and 5 to gain correspondingvalues of θ:3. Some X-rays, with wavelength 1 nm, are shonethrough a diffraction grating in which the slits are 50μm apart. A screen is placed 1.5 m from the grating.How far are the first three light fringes from the pointat which the normal intercepts the screen?n × 10−9 = 50×10−6×x

1.5 ≈ 33.3 × 10−6 × x , where nis an integer.x = n×10−9

33.3×10−6 = 30× 10−6 × n

Then, substitute in n = 1,2 and 3 to gain correspondingvalues of x:

20.25 Diffraction

1. What is the width of the central bright fringe on ascreen placed 5m from a single slit, where the slit is0.01m wide and the wavelength is 500nm?WL ≈ λ

d

W5 ≈ 500×10−9

0.01

W=250μm, but since this is only half the beam, the beamis 500μm wide.

20.26 Light

1. How much energy does a photon with a frequencyof 50kHz carry?E = hf = 6.626×50×103×10−34 = 3.313×10−29J2. A photon carries 10−30J of energy. What is itsfrequency?f = E

h = 10−30

6.626×10−34 ≈ 1509Hz3. How many photons of frequency 545 THz does a20W bulb give out each second?First calculate the amount of energy given out per. sec-ond:P = ∆E

t

∆E = Pt = 20× 1 = 20JThen, calculate the amount of energy carried by eachphoton:E = hf = 6.626×545×1012×10−34 ≈ 3.61×10−19JThen divide the former by the latter to give the numberof photons n:n = 20

3.61×10−19 ≈ 5.54× 1019 photons

4. In one minute, a bulb gives out a million photonsof frequency 600 THz. What is the power of the bulb?First calculate the energy carried by one photon:E = hf = 6.626×10−34×600×1012 ≈ 3.98×10−19JThen work out the energy carried by 1,000,000 photons:E = 106 × 3.98× 10−19 = 3.98× 10−13JThen work out the power of the bulb:P = ∆E

t = 3.98×10−13

60 = 6.63× 10−15W...maybe its a nanobulb.5. The photons in a beam of electromagnetic radia-tion carry 2.5μJ of energy each. How long should thephasors representing this radiation take to rotate?First calculate the frequency of each photon:f = E

h = 2.5×10−6

6.626×10−34 ≈ 3.77 × 1027Hz (That’s onenasty gamma ray.)Then calculate the time taken for one 'wavelength' to goby:f = 1

t

t = 1f = 1

3.77×1027 ≈ 2.65× 10−28s

20.27 Electron Behaviour

1. An electron moves at 30,000 ms−1. What is its deBroglie wavelength?λ = h

mv = 6.626×10−34

9.1×10−31×30,000 = 2.43× 10−8m2. What is its frequency?

f = Ekinetich =

12 9.1×10−31×30,0002

6.626×10−34 = 6.18× 1011 Hz3. What is its kinetic energy, in eV?From the top half of the fraction in the previous question:Ekinetic =

12 9.1×10−31×30,0002 = 4.10×10−22 J =

4.10×10−22

1.6×10−19 eV = 2.56meV4. Given that it is travelling out of an electron gun,what was the potential difference between the anodeand the cathode?2.56 mV – that’s why we use eV!5. An electron is accelerated by a potential differenceof 150 V. What is its frequency?Ekinetic = 1.6× 10−19 × 150 = 2.4× 10−17 Jf = Ekinetic

h = 2.4×10−17

6.626×10−34 Hz = 3.62× 1016 Hz

20.28 Vectors

1. Which of the following are vectors?

106 CHAPTER 20. WORKED SOLUTIONS

• 20 cm is scalar, not a vector - it does not have adirectional component.

• 9.81 ms−2 towards the centre of the earth is avector.

• 5 km south-east is a vector.

• 500 ms−1 on a bearing of 285.3° is a vector.

2. A displacement vector a is the resultant vector oftwo other vectors, 5 m north and 10 m south-east.What does a equal, as a displacement and a bearing?So, for the resultant vector:Horizontal displacement = 0 + 10√

2≈ 7.07 m

Vertical displacement = 5− 10√2≈ −2.07 m

This gives us a right-angled triangle. So:θ = arctan 2.07

7.07 ≈ 16.3 °, so the bearing equals 90° +16.3° = 106.3°.Resultant displacement =

√7.072 + (−2.07)2 ≈

7.4 m3. If I travel at a velocity of 10 ms−1 on a bearing of030°, at what velocity am I travelling north and east?Vertical velocity (North) = 10×cos 30 = 5

√3 ms−1 ≈

8.66 ms−1

Horizontal velocity (East) = 10× sin 30 = 5 ms−1

4. An alternative method of writing vectors is in acolumn, as follows:a =

(xy

),

where x and y are the vertical and horizontal compo-nents of the vector respectively. Express |a| and theangle between a and

(10

)in terms of x and y.

By Pythagoras’ theorem:(|a|)2 = x2 + y2

|a| =√x2 + y2

Let θ be the angle between a and(10

).

tan θ = yx

θ = arctan yx

This angle θ is known as the argument of a.5. A more accurate method of modelling the trajec-tory of a ball is to include air resistance as a constantforce F. How would this be achieved?Once the arrow representing the acceleration due to grav-ity has been added on, add an horizontal arrow pushingagainst the motion of the ball. Since F = ma, the magni-tude of this acceleration is F divided bym.Note that this model is still not perfect. In fact, F is notconstant - it depends on the horizontal component of theball’s velocity.

20.29 Graphs

1. In the following distance-time graph,what is the velocity 4 seconds afterthe beginning of the object’s journey?

At this point, the graph is a straight line. So:v = ds

dt = 10−06−0 = 5

3 ≈ 1.67ms−1

2. What is the velocity at 12 seconds?Approximately, there is a rough straight line at this point.The object is travelling towards home 2 ms−1 each sec-ond (look at the previous second, for example). So, itsvelocity is −2 ms−1.3. In the following velocity-time graph, howfar does the object travel between 7 and 9 seconds?

20.29. GRAPHS 107

Distance travelled is equal to the area under the graph.Between 7 and 9 seconds, this is the shaded area of thegraph. So, calculate the area of the triangle:A = bh

2 = 2×32 = 3m

4. What is the object’s acceleration at 8 seconds?There is a straight line between 7 and 9 seconds which wecan use to answer this question. The acceleration is equalto the gradient of the graph, so:a = 3

2 = 1.5ms−2

5. A car travels at 10ms−1 for 5 minutes in a straightline, and then returns to its original location over thenext 4minutes, travelling at a constant velocity. Drawa distance-time graph showing the distance the carhas travelled from its original location.10ms−1 = 600 metres / minute.

6. Draw the velocity-time graph for the above situa-tion.The velocity from 0-5s is 10ms−1. The velocity from 5-9sis:v = 3×103

4 = 750 metres/minute = 12.5ms−1

108 CHAPTER 20. WORKED SOLUTIONS

7. The velocity of a ball is related to the time since itwas thrown by the equation v = 30− 9.8t . How farhas the ball travelled after 2 seconds?s =

∫ t2t1

f(t) dt =∫ 2

030 − 9.8t dt = [30t − 4.9t2]20 =

30(2)−4.9(22)−30(0)+4.9(02) = 60−19.6 = 40.4 m

20.30 Kinematics

1. A person accelerates from a speed of 1 ms−1 to 1.7ms−1 in 25 seconds. How far has he travelled in thistime?u = 1, v = 1.7, t = 25s = u+v

2 t = 1+1.72 × 25 = 1.35× 25 = 33.75m

2. A car accelerates at a rate of 18 kmh−2 to a speedof 60 kmh−1, travelling 1 km in the process. How fastwas the car travelling before it travelled this distance?a = 18, v = 60, s = 1v2 = u2 + 2as602 = u2 + (2 x 18 x 1)3600 = u2 + 36u2 = 3564u = 59.7kmh−1 (to 3 significant figures)3. A goose in flight is travelling at 4 ms−1. It acceler-ates at a rate of 1.5 ms−2 for 7 seconds. What is itsnew speed?

u = 4, a = 1.5, t = 7v = u + at = 4 + (1.5 x 7) = 14.5 ms−1

4. How far does an aeroplane travel if it acceleratesfrom 400 kmh−1 at a rate of 40 kmh−2 for 1 hour?u = 400, a = 40, t = 1s = ut+ at2

2 = (400× 1) + 40×12

2 = 420km

20.31 Forces and Power

1. I hit a ball of mass 5g with a cue on a billiardstable with a force of 20N. If friction opposes me witha force of 14.2N, what is the resultant acceleration ofthe ball away from the cue?Resultant force = 20 - 14.2 = 5.8NF = ma5.8 = (5 x 10−3)aa = 1160ms−2

2. A 10g ball rolls down a 1.2m slope, and leaves itwith a velocity of 4ms−1. How much work is done byfriction?g.p.e. at top of slope = mgh = 10 x 10−3 x 9.81 x 1.2 =0.117Jk.e. at bottom of slope = ½mv2 = 0.5 x 10 x 10−3 x 42= 0.08JAssuming the remainder of the g.p.e. becomes heat /sound due to friction:work done by friction = 0.117 - 0.08 = 0.037J.3. An electric train is powered on a 30kV power sup-ply, where the current is 100A. The train is travellingat 90 kmh−1. What is the net force exerted on it in aforwards direction?P = IV = Fv90 kmh−1 = 0.025 km−1 = 25 ms−1

100 x 30 x 103 = 25FF = 120000N

20.32 Exponential Relationships

1. Simplify Newton’s Law of Cooling for the casewhen I place a warm object in a large tank of waterwhich is on the point of freezing. Measure tempera-ture in °C.Newton’s Law of Cooling states that Tt = Tenv + (T0 −Tenv)e

−rt

The freezing point of water is 0 °C, so, if we measure Tin °C, Tₑ ᵥ = 0:

20.33. CAPACITORS 109

Tt = 0 + (T0 − 0)e−rt

Tt = T0e−rt

2. What will the temperature of an object at 40 °C beafter 30 seconds? (Take r=10−3 s−1.)Tt = T0e

−rt = 40× e−10−3×30 = 38.8 °C3. A body is found in a library (as per. AgathaChristie) at 8am. The temperature of the library iskept at a constant temperature of 20 °C for 10 min-utes. During these 10 minutes, the body cools from25 °C to 24 °C. The body temperature of a healthyhuman being is 36.8 °C. At what time was the personmurdered?First, we must calculate r:24 = 20 + (25− 20)e−10r

4 = 5e−10r

e−10r = 0.8

−10r = ln 0.8r = ln 0.8

−10 = 0.0223 minute−1

Then, calculate t - this is the time between the murder and8am:25 = 20 + (36.8− 20)e−0.0223t

5 = 16.8e−0.0223t

e−0.0223t = 516.8

−0.0223t = ln 516.8

t =ln 5

16.8

−0.0223 = 54 minutesTherefore, the murder occurred at 7:06am.4. Suppose for amoment that the number of pages onWikibooks p can be modelled as an exponential rela-tionship. Let the number of pages required on aver-age to attract an editor be a, and the average numberof new pages created by an editor each year be z. De-rive an equation expressing p in terms of the time inyears since Wikibooks was created t.Let n be the number of editors.n = p

a

dpdt = nz = z p

a

dp = z padt∫

1pdp =

∫zadt

ln p = zta + c (where c is the constant of integration)

p = ezta +c = e

zta ec = ke

zta (where k is a constant - k =

ec)Theremust have been a first page, whichmarked the pointwhere t = 0, so:1 = ke

za×0 = ke0 = k

Therefore:

p = ezta

5. Wikibooks was created in mid-2003. How manypages should there have been 6 years later? (Take a= 20, z = 10 yr−1.)p = e

zta = e

10×620 = e3 = 20

6. The actual number of pages in Wikibooks in mid-2009 was 35,148. What are the problems with thismodel? What problems may develop, say, by 2103?There are two key problems with this model:

• We have estimated the values of the constants.These should have been determined statistically.

• We have assumed that the constants are constant. Inreality, as the amount of content on Wikibooks in-creases, more people think “Wikibooks already con-tains this content, so I am not going to add anything.”This means that both z and a change with time. Ourexponential model only applies over small periodsof time. Each of these small periods of time hasdifferent values for the constants.

In the future, such as 2103, the constants will havechanged so radically as to be useless. Question 5 showshowmuch they change over just 6 years - howmuchmoremust they change over a whole century!

20.33 Capacitors

1. A 2 mF capacitor is connected to a 10V DC powersupply. Howmuch charge can be stored by the capac-itor?Q = CV = 2× 10−3 × 10 = 0.02 CNote that this is the maximum charge - since capacitorscharge, as well as discharge, exponentially, wewould haveto leave the capacitor charging for an infinitely long periodof time to charge it to capacity.2. What is the highest possible energy stored by thiscapacitor?E = 1

2QV = 0.5× 0.02× 10 = 0.1 J3. The capacitor is placed in series with a 5Ω resistorand charged to capacity. How long would it take forthe charge in the capacitor to be reduced to 1 mC?Q = Q0e

− tRC

Q0 = Qet

RC

et

RC = Q0

Q

tRC = ln Q0

Q

t = RC ln Q0

Q = 5× 0.002× ln 0.020.001 = 0.0300 s

4. After this time has elapsed, how much energy isstored in the capacitor?

110 CHAPTER 20. WORKED SOLUTIONS

E = 12QV = 1

2Q0V0e− 2t

RC = 0.5 × 0.02 × 10 ×e−

2×0.035×0.002 = 2.5 mJ

5. What is the capacitance of the equivalent capacitorto the following network of capacitors?First, ignore the fact that the capacitances are in mF, notF - we will not be using any other units, so if we put inmF, we will get out mF.Then work out the equivalent capacitance of the secondrow:1

ΣC2= 1

1 + 12 + 1

3 = 116

ΣC2 = 611 mF

Then work out the equivalent capacitance of the thirdrow:1

ΣC3= 1

2 + 12 = 1

ΣC3 = 1 mFThe equivalent capacitance of the first row is easy, sinceit contains just 1 capacitor: 3mF. So, the total equivalentcapacitance is:ΣC = 3 + 6

11 + 1 = 5011 ≈ 4.55 mF

20.34 Radioactive Decay

1 mole = 6.02 x 1023 atoms1u = 1.66 x 10−27kg1. Americium-241 has a decay constant of 5.07x 10−11s−1. What is the activity of 1 mole ofamericium-241?A = λN = 5.07 × 10−11 × 6.02 × 1023 = 3.05 ×1013 BqThis is whywe only need very small samples for use in, forexample, smoke detectors. In fact, safety considerationsnecessitate a small sample - otherwise we would all havecancer!2. How many g of lead-212 (λ = 18.2μs−1) are re-quired to create an activity of 0.8 x 1018Bq?

N = Aλ = 0.8×1018

18.2×10−6 = 4.40 × 1022 nuclei = 4.40 ×1022 × 212× 1.66× 10−27 = 15.5 g3. How long does it take for 2kg of lead-212 to decayto 1.5kg of lead-212?Mass is proportional to the number of atoms, so:m = m0e

−λt

mm0

= e−λt

m0

m = eλt

λt = ln m0

m

t =ln m0

m

λ =ln 2

1.5

18.2×10−6 = 15807 s = 4.39 hours4. Where does the missing 0.5kg go?It becomes another isotope - in this case, mercury-208.Some also becomes alpha particles.5. Some americium-241 has an activity of 3kBq.What is its activity after 10 years?A = A0e

−λt = 3000 ×e−5.07×10−11×10×365.24×24×60×60 = 2.952 kBq6. Thismodel of radioactive decay is similar to takingsome dice, rolling them once per. second, and remov-ing the dice which roll a one or a two. What is thedecay constant of the dice?The decay constant is the probability of removing a die -λ = ⅓.7. If you started out with 10 dice, how many dicewould you have left after 10s? What is the problemwith this model of radioactive decay?N = N0e

−λt = 10× e−13×10 = 0.357 dice

Obviously, you can't have 0.357 dice. The problem withthis model of radioactive decay is that, once you have suf-ficiently few nuclei, the decay ceases to be continuous.As time passes, the pattern becomes relatively more ran-dom. The model also says that the number of nuclei willalways decrease. In reality, since there can only be an in-teger number of nuclei, there will eventually come a pointwhen there are no nuclei left.

20.35 Half-lives

1. Radon-222 has a decay constant of 2.1μs−1. Whatis its half-life?t 12= ln 2

2.1×10−6 = 330070 s = 3.82 days2. Uranium-238 has a half-life of 4.5 billion years.How long will it take for a 5 gram sample of U-238 todecay to contain 1.25 grams of U-238?2 half-lives, since 1.25 is a quarter of 5. 2 x 4.5 = 9 billionyears.3. How long will it be until it contains 0.5 grams ofU-238?

20.37. GRAVITATIONAL FIELD 111

First calculate the decay constant:λ = ln 2

t 12

= ln 24.5×109 = 1.54× 10−10 yr−1

0.5 = 5e−1.54×10−10t

0.1 = e−1.54×10−10t

ln 0.1 = −1.54× 10−10t

t = ln 0.1−1.54×10−10 = 14.9 Gyr

4. Tritium, a radioisotope of Hydrogen, decays intoHelium-3. After 1 year, 94.5% is left. What is thehalf-life of tritium (H-3)?0.945 = e−λ×1 (if λ is measured in yr−1)λ = − ln 0.945 = 0.0566 yr−1 = ln 2

t 12

t 12= ln 2

0.0566 = 12.3 yr5. A large capacitor has capacitance 0.5F. It is placedin series with a 5Ω resistor and contains 5C of charge.What is its time constant?τ = RC = 5× 0.5 = 2.5 s6. How long will it take for the charge in the capacitorto reach 0.677C? ( 0.677 = 5

e2 )2 x τ = 5s

20.36 Gravitational Force

1. Jupiter orbits the Sun at a radius of around 7.8 x1011m. The mass of Jupiter is 1.9 x 1027kg, and themass of the Sun is 2.0 x 1030kg. What is the gravita-tional force acting on Jupiter? What is the gravita-tional force acting on the Sun?Fgrav = −GMm

r2 = −6.67×10−11×2×1030×1.9×1027

(7.8×1011)2 =

−4.17× 1023 N2. The force exerted by the Sun on an object at a cer-tain distance is 106N. The object travels half the dis-tance to the Sun. What is the force exerted by the Sunon the object now?

1

( 12 )

2 = 4

So, the new force is 4 MN.3. How much gravitational force do two 1kg weights5cm apart exert on each other?Fgrav = −GMm

r2 = −6.67×10−11×1×10.052 = −2.67 ×

10−8 NIn other words, ordinary objects exert negligible gravita-tional force.4. The radius of the Earth is 6360km, and its massis 5.97 x 1024kg. What is the difference between thegravitational force on 1kg at the top of your body, andon 1kg at your head, and 1kg at your feet? (Assumethat you are 2m tall.)

∆Fgrav = GMm(

163600002 − 1

63600022

)= (1.55 ×

10−20)GMm = 1.55×10−20×6.67×10−11×5.97×1024 × 1 = 6.19 µNThis is why it is acceptable to approximate the accelera-tion due to gravity as constant over small distances.

20.37 Gravitational Field

G = 6.67 x 10−11 m3kg−1s−2

1. A 15kg object has a weight of 8000N. What is thegravitational field strength at this point?g =

Fgrav

m = 800015 = 533 Nkg−1

2. Draw a graph of gravitational field strengthagainst distance.

Since gravitational field strength is proportional to gravi-tational force, the two graphs look very similar.3. What is the gravitational field strength of the Sun(mass 2 x 1030kg) on the Earth (mass 6 x 1024kg,mean orbital radius 15 x 1010m)?g = GM

r2 = 6.67×10−11×2×1030

(15×1010)2 = 5.93 mNkg−1

4. What is the difference in the acceleration due togravity over a vertical distance d?

∆g = GM(r+d)2 − GM

r2 = GM(

1(r+d)2 − 1

r2

)=

GM(r2−(r+d)2)r2(r+d)2 = GM(r2−r2−2rd−d2)

r2(r+d)2 = −GMd(2r+d)r2(r+d)2

5. How far would one have to travel upwards fromthe Earth’s surface to notice a 1Nkg−1 difference ingravitational field? (The Earth has a radius of 6400

112 CHAPTER 20. WORKED SOLUTIONS

km.)1 = GMd(2r+d)

r2(r+d)2

GMd(2r + d) = r2(r + d)2

2GMrd+GMd2 = r2(r2+2rd+ d2) = r4+2r3d+r2d2

d2(r2 −GM) + dr(2r2 − 2GM) + r4 = 0

Using the quadratic formula:

d =2GMr−2r3±

√r2(2r2−2GM)2−4r4(r2−GM)

2(r2−GM) =

2GMr−2r3±√4r6−8GMr4+4G2M2r2−4r6+4GMr4

2(r2−GM)

= 2GMr−2r3±r√4r4−8GMr2+4G2M2−4r4+4GMr2

2(r2−GM)

Which is very horrible. If you plug in the numbers, youget:d = 4.598272×1021±4.853341552×1021

−7.1848×1014

Since we want a negative d (because of the minus sign weignored right at the beginning):d = 4.598272×1021+4.853341552×1021

−7.1848×1014 = −13155 kmAlthough, I have to confess, there was so much scope forerror that it is almost certain that this is the wrong answer.If you fancy wading through and checking, feel free! --Sjlegg (talk) 14:18, 23 April 2009 (UTC)

20.38 Gravitational Potential En-ergy

1. A ball rolls down a 3m-high smooth ramp. Whatspeed does it have at the bottom?mgh = 1

2mv2

gh = 12v

2

v =√2gh =

√2× 9.81× 3 = 7.67 ms−1

2. In an otherwise empty universe, two planets ofmass 1025 kg are 1012mapart. What are their speedswhen they collide?GMm0.5r = 1

2mv2

GM0.5r = 1

2v2

4GMr = v2

v = 2√

GMr = 2

√6.67×10−11×1025

1012 = 52 ms−1

(Not too sure about this one. Please check.)3. What is the least work a 2000kg car must do todrive up a 100m hill?mgh = 2000× 9.81× 100 = 1.962MJ4. How does the speed of a planet in an elliptical orbitchange as it nears its star?As it nears the star, it loses gravitational potential energy,

and so gains kinetic energy, so its speed increases.

20.39 Gravitational Potential

G = 6.67 x 10−11 m3kg−1s−2

g = 9.81 ms−2

1. What is the gravitational potential at the Earth’ssurface? (mass of Earth = 5.97 x 1024 kg,radius ofEarth = 6371 km)

Vgrav = −GMr = −6.67×10−11×5.97×1024MJkg−1

6371000 =

62.5MJkg−1

2. Taking the Earth’s surface as V ᵣₐᵥ = 0, what is thegravitational potential 2m above the ground?Vgrav ≈ g∆h = 9.81× 2 = 19.62 Jkg−1

3. A 0.2kg firework reaches a gravitational potentialrelative to the ground of 500Jkg−1. If the firework is30% efficient, how much energy was expended to getthere?Vgrav =

Egrav

m

Egrav = mVgrav = 0.2× 500 = 100 JHowever, this is only 30% of the energy expended, so:Eexpended = 100

0.3 ≈ 333 J4. Express gravitational potential in terms of gravi-tational force.Vgrav =

∫Fgrav dr

m

5. Draw the equipotentials and field lines surround-ing the Earth.

20.41. ENERGY IN SIMPLE HARMONIC MOTION 113

20.40 Simple Harmonic Motion

1. A 10N weight extends a spring by 5cm. Another10N weight is added, and the spring extends another5cm. What is the spring constant of the spring?∆F = k∆x

10 = 0.05k

So k = 100.05 = 200Nm−1

2. The spring is taken into outer space, and isstretched 10cm with the two weights attached. Whatis the time period of its oscillation?First calculate the mass of the two weights:20 = 9.81m

m = 2.04kg

T = 2π√

mk = 2π

√2.04200 = 0.634 s

3. What force is acting on the spring after 1 second?In what direction?We are starting the oscillation at t=0 with a displacement.This displacement is the amplitude of the oscillation, andwe need f(ωt) to be positive at t=0. So, we use an equationfor the displacement with a cosine in it. We have alreadyderived the acceleration in this case:a = −ω2 cosωtω = 2π

T = 2π6.34 = 0.99 rad s−1

So:a = −0.992 cos 0.99× t = −0.538 ms−2

The minus sign means that the acceleration is in the op-posite direction to the initial displacement.4. A pendulum oscillates with a frequency of 0.5 Hz.What is the length of the pendulum?ω = 2πf√

gl = 2× π × 0.5√9.81l = π

9.81l = π2

l = 9.81π2 = 0.994 m

5. The following graph shows the displacement of asimple harmonic oscillator. Draw graphs of its veloc-ity, momentum, acceleration and the force acting onit.x is a sine wave, so:x = A sinωtv = dx

dt = Aω cosωt

a = dvdt = d2x

dt2 = −Aω2 sinωtp = mv = mAω cosωtF = ma = −mAω2 sinωt

Since you haven't been given any details, the amplitudesof the waves don't matter. The phase differences, how-ever, do.

6. A pendulum can only be modelled as a simple har-monic oscillator if the angle over which it oscillates issmall. Why is this?Simple harmonic oscillators work because the force actsin the opposite direction to the displacement. As the pen-dulum moves away from the area immediately below thepeg it is hanging on, the force no longer acts in the oppo-site direction to the displacement.

20.41 Energy in Simple HarmonicMotion

1. A 10g mass causes a spring to extend 5cm. Howmuch energy is stored by the spring?k = ∆F

∆x = mg∆x = 0.01×9.81

0.05 = 1.962 Nm−1

E = 12kx

2 = 0.5× 1.962× 0.052 = 2.45 mJ2. A 500g mass on a spring (k=100) is extended by0.2m, and begins to oscillate in an otherwise emptyuniverse. What is the maximum velocity which itreaches?12mv2max = 1

2kx2max

v2max =kx2

max

m

vmax = xmax

√km = 0.2×

√1000.5 = 2.83 ms−1

3. Another 500g mass on another spring in anotherotherwise empty universe is extended by 0.5m, and

114 CHAPTER 20. WORKED SOLUTIONS

begins to oscillate. If it reaches a maximum velocityof 15ms−1, what is the spring constant of the spring?12mv2max = 1

2kx2max

k =mv2

max

x2max

= 0.5×152

0.52 = 450 Nm−1

4. Draw graphs of the kinetic and elastic energies ofa mass on a spring (ignoring gravity).Ee ∝ cos2 ωtEk ∝ sin2 ωt

5. Use the trigonometric formulae for x and v to de-rive an equation for the total energy stored by an os-cillating mass on a spring, ignoring gravity and airresistance, which is constant with respect to time.x = A cosωtv = −Aω sinωtSubstitute these into the equation for the total energy:ΣE = 1

2 (kx2 + mv2) = 1

2 (k(A cosωt)2 +m(−Aω sinωt)2) = 1

2 (kA2 cos2 ωt +

mA2ω2 sin2 ωt) = A2

2 (k cos2 ωt+mω2 sin2 ωt)We know that:

ω =√

km

Therefore:ω2 = k

m

By substitution:ΣE = A2

2 (k cos2 ωt + mkm sin2 ωt) = A2

2 (k cos2 ωt +k sin2 ωt) = kA2

2 (cos2 ωt+ sin2 ωt) = kA2

2

20.42 Damping

1. Draw a graph of displacement for a criticallydamped oscillation.

2. How would you critically damp an oscillating pen-dulum?Grab the weight, move it to its equilibrium position, andstop it moving.3. How would you damp an oscillating pendulum us-ing only a weighted polystyrene block?Put the block in the path of the pendulum, which willbounce off the weight, losing a bit of energy each oscilla-tion.4. What would the displacement graph look like forthis oscillation, before and after damping began?

20.43. CONSERVATION OF MOMENTUM 115

5. The graph above is an exponentially damped oscil-lation. If the displacement of the undamped oscilla-tion is given by sin ωt, what is an approximate equa-tion for the damped oscillation, in terms of a constantk which describes the degree to which the oscillationis damped?x = e−kωt sinωtIf k doubles, the e-kt will be squashed to half its size alongthe t-axis, so as k increases, the rate of damping increases.

20.43 Conservation of Momentum

1. A ball of mass 0.5kg collides with a stationary ballof 0.6kg at a velocity of 3ms−1. If the stationary ballmoves off at a speed of 2ms−1, what is the new velocityof the first ball?0.5× 3 = 0.5v + (0.6× 2)

1.5 = 0.5v + 1.2

0.5v = 0.3

v = 0.6 ms−1

2. Two balls are moving in opposite directions withvelocities 5ms−1 and 10ms−1. They collide, andmove off in opposite directions with new velocities of7.5ms−1 each. If the mass of the first ball was 1.25kg,what is the mass of the second ball?(5× 1.25) + (−10m) = (−7.5× 1.25) + (7.5m)

6.25− 10m = 7.5m− 9.375

17.5m = 15.625

m = 0.893 kg3. A totally elastic collision occurs between two ballsof equal mass. One of the balls is stationary. Whathappens?Since momentum must be conserved:mu = mv1 +mv2

u = v1 + v2 (1)Since kinetic energy must be conserved:mu2

2 = mv12

2 + mv22

2

u2 = v12 + v2

2

Substitute in the value of u from (1):(v1 + v2)

2 = v12 + v2

2

v12 + 2v1v2 + v2

2 = v12 + v2

2

2v1v2 = 0

Therefore, either v1 or v2 is 0. Using equation (1), if v1 iszero, then v2 = u, and vice versa. However, v1 cannot bethe same as u, as this would mean that the first ball hadto move through the second ball! So, the only physicalsolution is that the first ball stops, and the second ballcontinues moving with the first ball’s original velocity.4. A particle explodes to become two particles withmasses 1kg and 2kg. The 1kg particle moves withvelocity 45ms−1. With what velocity does the otherparticle move?0 = (1× 45) + 2v

2v = −45

v = −22.5 ms−1

i.e. in the opposite direction to the motion of the 1kgparticle.5. A 3kg ball moving at 3ms−1 collides with a 5kg ballmoving at −5ms−1. The collision is perfectly elastic.What are the new velocities of the balls?Since momentum must be conserved:(3× 3) + (5×−5) = 3v1 + 5v2 = −16 kgms−1 (1)Since kinetic energy must be conserved:3×32

2 + 5×(−5)2

2 = 3×v12

2 + 5×v22

2

3v12 + 5v2

2 = 152

From (1):v2 = −16−3v1

5

3v12 + 5

(−16−3v15

)2= 152

3v12 + 5

(9v1

2+96v1+25625

)= 152

3v12 + 9v1

2+96v1+2565 = 152

15v12 + 9v1

2 + 96v1 + 256 = 760

24v12 + 96v1 − 504 = 0

116 CHAPTER 20. WORKED SOLUTIONS

v12 + 4v1 − 21 = 0

(v1 + 7)(v1 − 3) = 0

So, v1 is either −7 or 3 ms−1.If v1 = −7ms−1, by (1):−21 + 5v2 = −16

5v2 = 5

v2 = 1 ms−1

If v1 = 3ms−1:9 + 5v2 = −16

5v2 = −25

v2 = −5 ms−1

This last solution is non-physical since it requires the ballsto move through each other. So, v1 = −7ms−1 and v2 =1ms−1

6. A ball collides with a wall, and rebounds at thesame velocity. Why doesn't the wall move?Let the mass of the wall be M, and the mass of the ballbe m:mu = Mvwall −mu

2mu = Mvwall

vwall =2muM

The mass of the wall is large. As M tends to infinity,therefore, v ₐ tends to 0.

20.44 Forces and Impulse in Colli-sions

1. Escape velocity from the Earth is 11.2km−1. Howmuch impulse must be exerted on a 47000kg payloadto get it to travel away from the Earth?I = m(v − u) = 47000(11200− 0) = 526.4MNs2. Two billiard balls, of mass 10g, collide. One ismoving at 5ms−1, and the other at 2ms−1. After thecollision, the first billiard ball is moving backwardsat 4ms−1. The collision takes 1 ms. What force wasexerted on this ball?I = m(v − u) = 0.01(5− (−4)) = 0.09 NsF = I

∆t =0.090.001 = 90 N

3. What impulse and force were exerted on the secondball?The impulse was −0.09Ns and the force was −90N.4. A 60kg spacewalker uses a jet of gas to exert animpulse of 10Ns. How many times would he have todo this to reach a speed of 1 ms−1 from stationary?∆v = I

m = 1060 = 1

6 ms−1

So, the spacewalker would have to do this 6 times to reacha speed of 1ms−1.5. A 5kg bowling ball collides with a stationary tennisball of mass 0.1kg at 3ms−1, slowing to 2.5ms−1. Itexerts a force of 100N on the ball. How long did thecollision take?F = m(v−u)

∆t

∆t = m(v−u)F = 5(3−2.5)

100 = 0.025 s

20.45 Rockets, Hoses and MachineGuns

1. A machine gun fires 300 5g bullets per. minute at800ms−1. What force is exerted on the gun?F = v dm

dt = 800× 300×0.00560 = 20 N

2. 1 litre of water is pumped out of a tank in 5 secondsthrough a hose. If a 2N force is exerted on the tank,at what speed does the water leave the hose?The flow rate is about 0.2 kg s−1.2 = 0.2v

v = 20.2 = 10 ms−1

3. If the hosewere connected to themains, what prob-lems would there be with the above formula?The force would not be exerted on the tank, but wouldinstead be referred back through the pipe - it is not gainingvelocity at the nozzle.4. The thrust of the first stage of a Saturn V rocket is34 MN, using 131000kg of solid fuel in 168 seconds.At what velocity does the fuel leave the tank?34× 106 = v 131000

168

v = 34×106×168131000 = 43600 ms−1

5. Escape velocity from the Earth is 11km−1. Whatis the velocity of the rocket after the first stage is usedup, if the total mass of the rocket is 3 x 106 kg? Howdoes this compare to escape velocity?ma = 34× 106 −mg

a = 34×106

3×106 − 9.81 = 1.52 ms−2

v = at = 168× 1.52 = 255.92 ms−1 ≈ 0.256 kms−1

This is 0.002% of escape velocity, hence the need for theother stages of the rocket.

20.46 Circular Motion

1. A tennis ball of mass 10g is attached to the end ofa 0.75m string and is swung in a circle around some-one’s head at a frequency of 1.5Hz. What is the ten-

20.47. RADAR AND TRIANGULATION 117

sion in the string?ω = 2πf = 2π × 1.5 = 3π rad s−1

F = T = mω2r = 0.01× (3π)2×0.75 = 0.0675π2 =0.666 N2. A planet orbits a star in a circle. Its year is 100days, and the distance from the star to the planet is70 Gm from the star. What is the mass of the star?100 days = 100 x 365.24 x 24 x 60 x 60 = 3155673600sf = 1

T = 13155673600 = 3.17× 10−10 Hz

ω = 2πf = 2π × 3.17× 10−10 = 1.99 nrad s−1

GMr2 = ω2r

M = ω2r3

G = (1.99×10−9)2×(70×109)3

6.67×10−11 = 2.04×1025 kg3. A 2000kg car turns a corner, which is the arc of acircle, at 20kmh−1. The centripetal force due to fric-tion is 1.5 times the weight of the car. What is theradius of the corner?20kmh−1 = 20000 / 3600 = 5.56ms−1

W = 2000× 9.81 = 19620 NFr = 1.5× 19620 = 29430 N29430 = mv2

r = 2000×5.562

r = 61728r

r = 6172829430 = 2.1 m

This is a bit unrealistic, I know...4. Using the formulae for centripetal accelerationand gravitational field strength, and the definition ofangular velocity, derive an equation linking the or-bital period of a planet to the radius of its orbit.ω2r = GMstar

r2

ω2r3 = GMstar

ω = 2πT

4π2r3

T 2 = GMstar

T 2 = 4π2r3

GMstar

So, orbital period squared is proportional to radius of or-bit cubed. Incidentally, this is Kepler’s Third Law in thespecial case of a circular orbit (a circle is a type of el-lipse).

20.47 Radar and Triangulation

1. A radar pulse takes 8 minutes to travel to Venusand back. How far away is Venus at this time?2d = ct = 3× 108 × 8× 60 = 1.44× 1011 md = 1.44× 1011 × 0.5 = 7.2× 1010 m2. Why can't a radar pulse be used to measure thedistance to the Sun?It would be impossible to pick up the reflected signal

due to all the other signals coming from the Sun. Also,the signal would almost certainly be absorbed anyway.<<Added>> Regardless of (λ) wavelength, power den-sity, or wavefront properties, the pulse would be absorbedwith no reflection possible. Distances to pure energysources are generally measured in terms of received lightintensity, shifts of the light spectrum, and radio interfer-ometry. The RF spectrum; and Laser (light) spectrumcan be used to “listen” to radiation, but not bounce a pulsefrom an energy source having no true angle of incidence.Just as an observation, I will note that the sun can be seenon most radars either sunrise or sunset, usually when thesun is just above the horizon. But these receptions areunusable strobes (interference) and not a result of receiv-ing a radar pulse from the sun. Radar technicians alsouse the sun as a “known” exact position to align the sys-tem to true north (and magnetic variations); this is calledsolar-boresighting and, again, only receives the radiation.3. Radar is used to measure the velocity of a space-craft travelling between the Earth and theMoon. Usethe following data to measure this velocity:First, calculate the distance of the spacecraft from theEarth at each time:d1 = c∆t1

2 = 3.0×108×(45.51213−45.31213)2 =

30, 000 km

d2 = c∆t22 = 3.0×108×(46.52785−46.32742)

2 =30, 064.5 kmNext, calculate the distance the spacecraft has travelledbetween the two pulses:∆d = d2 − d1 = 30, 064.5− 30, 000 = 64.5 kmNow, calculate the time elapsed between the transmissionof the two pulses:∆t = tβ − tα = 46.32742− 45.31213 = 1.01529 sFinally, divide the distance the spacecraft has travelledbetween the two pulses by the time between the trans-mission of the two pulses, to give the average velocity ofthe spacecraft in that interval of time:v = ∆d

∆t = 64.51.01529 ≈ 63.5 kms−1

4. The angles between the horizontal and a star aremeasured at midnight on January 1 as 89.99980° andat midnight on June 1 as 89.99982°. How far away isthe star?d = 2r tan a tan b

tan a+tan b = 2×150×109×tan 89.9998 tan 89.99982tan 89.9998+tan 89.99982 =

4.52× 1013 km5. Why can't triangulation be used to measure thedistance to another galaxy?The difference between the two angles becomes so tinythat we don't have good enough equipment to measure it.

118 CHAPTER 20. WORKED SOLUTIONS

20.48 Large Units

1. What is one parsec in m?206265 AU x 150 x 109 = 3.09 x 1016m2. Convert 3 light days into km.c = 3 x 105kms−1

3 x 3 x 105 x 24 x 60 x 60 = 7.78 x 1010km3. Convert 5.5 parsecs into light years.1 ly = 9.46 x 1015m5.5 pc = 5.5 x 3.09 x 1016m = 1.70 x 1017mDivide by 1 ly in metres to get a final answer of approxi-mately 18ly.4. The difference in angle of a star on the perpen-dicular to the plane of the Earth’s orbit which passesthrough the Sun when viewed from either side of theEarth’s orbit is 0.1°. How far away is the star in par-secs?The angle between the perpendicular and the line whichgoes through the Earth is 0.05° = 180”. So, the star is1/180 pc away.

20.49 Orbits

1. The semi-major axis of an elliptical orbit can beapproximated reasonably accurately by themean dis-tance of the planet for the Sun. How would you test,using the data in the table above, that the inner plan-ets of the Solar System obey Kepler’s Third Law?Divide T2 by R3 and for each planet and see if this valueis roughly constant.2. Perform this test. Does Kepler’s Third Law hold?So, Kepler’s Third Law does hold for the inner planets,using this rough approximation for the semi-major axis.3. If T2 α R3, express a constant C in terms of T andR.C = T 2

R3

This is the constant of proportionality. It should beroughly the same for all the planets around the Sun. Al-ternatively, you can use:C = R3

T 2

We will be using the former to answer the next two ques-tions, but you should be able to get the same answers usingthe latter.4. Io, one of Jupiter’s moons, has a mean orbitalradius of 421600km, and a year of 1.77 Earth days.What is the value of C for Jupiter’s moons?

C = T 2

R3 = (1.77×24×60×60)2

(421600000)3 = 3.12× 10−16 s2m−3

5. Ganymede, another of Jupiter’s moons, has amean orbital radius of 1070400km. How long is itsyear?3.12× 10−16 = T 2

10704000003

T =√3.12× 10−16 × 10704000003 = 618665 s =

7.16 daysWhich isn't that accurate, due to the approximations thatwe used.

20.50 Doppler Effect

1. M31 (the Andromeda galaxy) is approaching us atabout 120kms−1. What is its red-shift?z = vs

c = −120000300000000 = −0.4× 10−3

The minus sign is important! Andromeda is blue-shifted!2. Some light fromM31 reaches us with a wavelengthof 590nm. What is its wavelength, relative to M31?−0.0004 = ∆λ

λ0= λ−λ0

λ0= λ

λ0− 1 = 590×10−9

λ0− 1

0.9996 = 590×10−9

λ0

λ0 = 590×10−9

0.9996 = 590.23 nm3. Some light has a wavelength, relative to M31, of480nm. What is its wavelength, relative to us?0.9996 = λ

λ0= λ

480×10−9

λ = 0.9996× 480× 10−9 = 479.808 nm4. A quasar emits electromagnetic radiation at awavelength of 121.6nm. If, relative to us, this wave-length is red-shifted 0.2nm, what is the velocity of re-cession of the quasar?vs

c = ∆λλ0

vs3×108 = 0.2

121.6 = 0.00164

vs = 3× 108 × 0.00164 = 493 kms−1

However, this is about as high a velocity as we can use theclassical Doppler effect for.

20.51 The Big Bang

1. What is the Hubble Constant in s−1?H0 = 70 kms−1Mpc−1 = 0.07 ms−1pc−1

1 pc = 3.09 x 1016 mH0 = 0.07

3.09×1016 = 2.27× 10−18 s−1

2. How old is the universe?1H0

= 4.41× 1017 s = 14.0 bn years3. What effect might gravity have had on this figure?Gravity would slow the expansion of the universe, so the

20.54. IDEAL GASES 119

age of the universe would actually be lower than 14.0 bil-lion years.4. Polaris is 132pc away. What is its velocity of re-cession, according to Hubble’s Law?v = H0d = 70× 132× 10−6 = 9.24 ms−1

20.52 Heat and Energy

1. Carbon dioxide sublimes at 195°K. Roughly whatenergy per. particle does this correspond to?E ≈ kT = 1.38× 10−23 × 195 ≈ 10−21 J particle−1

2. A certain chemical reaction requires particleswith mass of the order 10−26 to move, on average, at10ms−1. Roughly what temperature does this corre-spond to?Ek = 1

2mv2 = 0.5× 10−26 × 102 = 5× 10−25

5× 10−25 = kT

T = 5×10−25

1.38×10−23 ≈ 10−2 °K3. The boiling point of water is 100°C. Roughly whatenergy per. particle does this correspond to?100°C = 373°KE ≈ kT = 1.38×10−23×373 ≈ 5×10−21 J particle−1

4. Thermionic emission from copper requires around5eV of energy per. particle. How hot will the wire beat this energy level?5eV = 5 x 1.6 x 10−19 = 8 x 10−19 J particle−1

T = 8×10−19

1.38×10−23 ≈ 6× 104 °K

20.53 Specific Heat Capacity

1. How much work would it take to heat 100kg ofliquid water from 20°C to 36.8°C?∆E = mc∆θ = 100× 4180× 16.8 = 7.022MJ2. How much work would it take to heat a well-insulated room from 15°C to 21°C, if the room is acube with side length 10m, and the density of the airis 1.2kgm−3?V = 103 = 1000 m3

ρ = mV

m = V ρ = 1000× 1.2 = 1200 kg∆E = mc∆θ = 1200× 1010× 6 = 7.272MJ3. A 10kg block of iron at 80°C is placed in the roomabove once it has reached 21°C. If the iron cools by40°C, what is the new temperature of the room?Energy must be conserved, so:10× 450× 40 = 1200× 1010×∆θ

∆θ = 1800001212000 = 0.149 °C

So, the new temperature of the room is 21.149°C.

20.54 Ideal Gases

1. I heat some argon from 250°K to 300°K. If thepressure of the gas at 250°K is 0.1 MPa, what is itspressure after heating?The temperature has increased by 20%, so the pressurewill also increase by 20% (using the pressure law). So,the new pressure is 0.1 MPa x 1.2 = 0.12 MPa.2. The argon is in a 0.5m long cylindrical tank withradius 10cm. What volume does it occupy?V = πr2l = π × 0.12 × 0.5 = 0.016 m3

3. The argon is then squeezed with a piston so that inonly occupies 0.4m of the tank’s length. What is itsnew pressure?The volume occupied decreases by 20%, which is equiv-alent to multiplying by 0.8. So, by Boyle’s Law, the pres-sure is multiplied by the reciprocal of 0.8 - 1.25. So, thepressure increases by 25% to 0.15 MPa.4. What is its new temperature?If we combine Charles’ Law with the pressure law, wefind that:T ∝ pV

Volume is multiplied by 0.8, but pressure is multiplied by1.25, so there is no change in temperature - it remains at300°K.5. 25% of the argon is sucked out. What is its pres-sure now?Using the amount law, if N decreases by 25%, the pres-sure must also decrease by 25% to 0.1125 MPa.

20.55 Kinetic Theory

1. Five molecules are moving at speeds of 1,5,6,8, and36ms−1. What is their mean square speed?c2 = 12+52+62+82+362

5 = 284.4 m2s−2

2. What is the mass of one molecule of N2 (atomicmass 14, 1u = 1.66 x 10−27kg)?2× 14× 1.66× 10−27 = 4.648× 10−26 kg3. Atmospheric pressure is 101,325 Pa. If onemole ofNitrogen takes up 2.3 m3 at about 10°C, what is themean square speed of themolecules in the air outside,assuming that the atmosphere is 100% nitrogen (inreality, it is only 78%)?pV = 1

3Nmc2

120 CHAPTER 20. WORKED SOLUTIONS

101235× 2.3 = 13 × 6.02× 1023 × 6.648× 10−22c2

c2 = 3×101235×2.36.02×1023×6.648×10−22 = 1745 m2s−2

4. What is the average speed of a nitrogen moleculeunder the above conditions?c =

√c2 =

√1745 = 41.8 ms−1

5. The particles in question 1 are duplicated 3000times. If they have a completely unrealistic mass of1g, what is their pressure when they are crammedinto a cube with side length 0.5m?p = Nmc2

3V = 5×3000×10−3×284.43×0.53 = 11376 Pa

20.56 Boltzmann Factor

1u = 1.66 x 10−27 kgg = 9.81 ms−2

1. A nitrogen molecule has a molecular mass of 28u.If the Earth’s atmosphere is 100% nitrous, with atemperature of 18°C, what proportion of nitrogenmolecules reach a height of 2km?ϵ = mgh = 28 × 1.66 × 10−27 × 9.81 × 2000 =9.12× 10−22 Jnn0

= e−9.12×10−22

1.38×10−23×291 = 0.797

So, 79.7% reach a height of at least 2km.2. What proportion of the molecules in a box of hy-drogen (molecular mass 2u) at 0°C have a velocitygreater than 5ms−1?ϵ = 1

2mv2 = 0.5 × 2 × 1.66 × 10−27 × 52 = 4.15 ×10−26 Jnn0

= e−4.15×10−26

1.38×10−23×273 = 0.99999

(Practically all of them.)3. What is the temperature of the hydrogen if half ofthe hydrogen is moving at at least 10ms−1?ϵ = 1

2mv2 = 0.5× 2× 1.66× 10−27 × 102 = 1.66×10−25 J0.5 = e

−ϵkT

ln 0.5 = − ln 2 = −ϵkT

T = ϵk ln 2 = 1.66×10−25

1.38×10−23×ln 2 = 0.0174 K(almost absolute zero)4. Some ionised hydrogen (charge −1.6 x 10−19 C)isplaced in a uniform electric field. The potential dif-ference between the two plates is 20V, and they are1m apart. What proportion of the molecules are atleast 0.5m from the positive plate (ignoring gravity)at 350°K?ϵ = 0.5× 20× 1.6× 10−19 = 1.6× 10−18 Jnn0

= e−1.6×10−18

1.38×10−23×350 ≈ 0

They would all fly onto the positive plate.

20.57 Magnetic Flux

1. A circular steel core has a cross-sectional area of 9cm2, and a length of 0.5m. If the permeability of steelis 875 μNA−2., what is the permeance of the core?

Λ = µAL = 875×10−6×9×10−4

0.5 = 1.58 µWbA−1

2. A coil of insulatedwire is wrapped 60 times aroundthe top of the core, and a 9A direct current is putthrough the coil. How much flux is induced?Φ = ΛNI = 1.58× 10−6 × 60× 9 = 0.851 mWb3. Assuming that all the flux goes through the core,what is the flux density at any point in the core?B = Φ

A = 0.851×10−3

9×10−4 = 0.945Wbm−2

4. Draw a diagram showing the lines of flux withinthe core.

20.58 Induction

1. What is the flux linkage of a 30cm coil of 0.5mmthick wire with a flux perpendicular to it of 10Wb?N = 30×10−2

0.5×10−3 = 600 coilsNΦ = 600× 10 = 6000Wb2. If the above coil is crushed steadily over a periodof 2s, what emf is maintained?| ϵ |= dNΦ

dt = 60002 = 3000 V

20.59. MAGNETIC FORCE 121

This principle of removing the flux linkage rapidly (al-though not usually by destroying one’s equipment) is usedto create very high voltages for short periods of time.3. The flux in a flux circuit varies according to theequation Φ = sin ωt. What is the equation for theemf induced?ϵ = −dNΦ

dt = −N ddt sinωt = −Nω cosωt

4. Using a constant k, what is the equation for a cur-rent which could induce the flux in the flux circuitabove?I = −k sinωt5. Draw a graph of the flux, flux linkage, emf andcurrent as deduced in the previous two questions.In order to keep the scales sane, I used a 2-coil (N=2) coilto plot the following graph:

Remember that the current here is inducing the flux (andflux linkage), which is in turn inducing the emf.

20.59 Magnetic Force

1. What force is exerted by a 1T magnetic field on anelectron (of charge −1.6 x 10−19C) moving at 5% ofthe speed of light (3 x 108 ms−1)?F = Bqv = 1 × −1.6 × 10−19 × 0.05 × 3 × 108 =2.4× 10−12 NThis may seem tiny, but remember that the accelerationof the electron will be far greater, as it has such a smallmass. The fact that the electron was moving at 0.05cmeans that we don't have to do any relativistic correc-

tions.2. What force is exerted by a 5mTmagnetic field on a20cm wire with resistance 1μΩ attached to a 9V bat-tery?V = IR

I = VR

F = BIl = BV lR = 5×10−3×9×0.2

10−6 = 9 kN3. The following diagram shows a positive chargemoving through amagnetic field. Draw an arrow rep-resenting the direction of the force on the charge.

Note that, if the charge had been negative, the forcewould be in the opposite direction (upwards).4. The following diagram shows a wire in a magneticfield. Draw an arrow representing the direction of theforce on the wire.

122 CHAPTER 20. WORKED SOLUTIONS

20.60 Transformers

1. A step-down transformer has 300 coils on one coil,and 50 coils on the other. If 30 kV AC is put in, whatvoltage comes out?30050 = 30000

Vs= 6

Vs =30000

6 = 5 kV2. A step-up transformer has 200 coils on one coil,and 980 coils on the other. If 25 kV AC comes out,what voltage was put in?200980 =

Vp

25000

Vp = 25000×200980 ≈ 5.1 kV

3. An ideal transformer transforms a 50A currentinto a 1A current. It has 40 coils on the primary coil.How many coils are in the secondary coil?501 = Ns

40 = 50

Ns = 40× 50 = 2000 coils4. Transformers tend to vibrate. Why is this? Whateffect does this have on the efficiency of the trans-former?There are wires with current in them going perpendicu-larly through a magnetic field. According to F = BIL, aforce is exerted on the wires. This means that some ofthe energy is converted into kinetic energy, reducing theefficiency of the transformer.5. Air does have some permeability. What effect doesthis have on the efficiency of the transformer? Why?Some of the flux passes through the air instead of the core,and so does not pass through the secondary coil, and is notused to create AC. Thismeans that some of the energy putin oscillates in and out of the magnetic field, reducing theefficiency of the transformer.

20.61 Motors

1. How could you adapt the simple DC motor to useAC?Connect the coil in the stator in series with the coil inthe rotor via a slip-ring commutator. This must be in thecorrect direction to ensure that the fields created by thetwo parts of the coil are constantly repelling each other,rather than constantly attracting each other.2. Why does a three-phase motor have a constant an-gular velocity?The vectorial sum of the three currents is constant, so thevectorial sum of the forces on the rotor due to each of thethree currents is constant.3. What is the difference between a split-ring and aslip-ring commutator?

A split-ring commutator makes the current change direc-tion every half-rotation, whereas a slip-ring commutatormerely maintains a connection between the moving rotorand the stationary stator.4. How could the angular velocity of a three-phasemotor be increased?Increase the frequency of the three-phase power.5. A squirrel-cage motor relies on eddy currents run-ning along the rotor to function. However, if eddycurrents run across the rotor, then the force on therotor is reduced. Howmay these eddy currents be re-duced without reducing the desired eddy currents?Laminate the plates which connect the rods to each otherso that currents are restricted from running in the wrongdirection.

20.62 Generators

1. Draw diagrams of an alternating current, the 'di-rect current' produced by a DC generator, and thiscurrent once it has been smoothed with a capacitor.

2. What is the phase difference (in radians) betweenthe voltages produced by a three-phase generator?2π3

c

3. According to Faraday’s law, what three things willincrease the amplitude of the emf created by a gen-erator?ϵ = −dNϕ

dt

So, more coils, a stronger magnetic field or a higher fre-quency of rotation will increase the amplitude of the emf.4. If an albatross touched two power cables carryingAC in phase, what would happen?Nothing - there is no potential difference between the twocables.5. What would happen if the two cables carried three-

20.64. ELECTRIC FIELD 123

phase power?Unfortunately, dead albatross - there is potential differ-ence between the two cables.

20.63 Electric Force

e = 1.6 x 10−19C1. A positron (charge +e) is 1 μm from a lithiumnucleus (charge +3e). What is the magnitude of theforce acting on each of the particles? In what direc-tion is it acting?

Felectric = kQqr2 = 8.99×109×1×3×(1.6×10−19)2

(10−6)2 =

6.90× 10−16 NThis may seem tiny, but the mass of an electron is alsotiny, so the acceleration is considerable.2. An electron is 1mm from the positively chargedplate in a uniform electric field. The potential dif-ference between the plates is 20V, and the plates are10cm apart. What force is acting on the electron? Inwhat direction?The 1mm is a red herring.Felectric =

qVd = 1.6×10−19×20

0.1 = 3.2× 10−17 N3. The acceleration due to gravity around a pointmass is constant, irrespective of the mass of the ob-jects it is acting on. The acceleration due to electricityaround a point charge is not. Use Newton’s SecondLaw (F=ma) to explain this.a = F

m

agrav = GMmmr2 = GM

r2

aelectric =kQqmr2

The m cancels for gravitational acceleration, but not forelectrical acceleration, since charge does not provide aresistance to a force.4. An insulator contains charged particles, eventhough the overall charge on the insulator is 0. Whyis the insulator attracted by a nearby charge?The electrons in the atoms can move within certainbounds. Being in an electric field means that they spendmore time closer to a positive charge, and so the parts ofthe insulator which are attracted to the charge end up be-ing closer to the charge than those which are repelled.This means that the attractive force on the insulator isgreater than the repulsive force. This causes the insulatoras a whole to be attracted to the charge if it is positive.If it is negative, the reverse happens, and the insulator isrepelled.5. Where in the charged conducting plates which cre-ate a uniform electric field would you expect to findthe charge located? Why?

The charges are free to move within a conductor. Theopposite charges in each plate are attracted to each otherand try to move as close to each other as possible. So,they end up on the inside edge of the plates.

20.64 Electric Field

1. Two metal plates are connected to a 9V batterywith negligible internal resistance. If the plates are10cm apart, what is the electric field at either of theplates?Eelectric =

Velectric

d = 90.1 = 90 Vm−1

2. What is the electric field at the midpoint betweenthe plates?The whole point of a uniform electric field is that the fielddoes not change anywhere between the plates - at themid-point, as anywhere, it is 90 NC−1.3. The charge on an electron is−1.6 x 10−19 C. Whatis the electric field 1μm from a hydrogen nucleus?

Eelectric = Q4πϵ0r2

= 1.6×10−19

4π×8.85×10−12×(10−6)2 =

1440 Vm−1

4. What is the direction of this field?The hydrogen nucleus has a positive charge, so the electricfield goes away from the nucleus (by convention).5. A 2C charge is placed 1m from a −1C charge. Atwhat point will the electric field be 0?Define the distance r as shown:

24πϵ0(1+r)2 = 1

4πϵ0r2

2(1+r)2 = 1

r2

(1+r)2

2 = r2

(1 + r)2 = 2r2

1 + r = r√2

r(√2− 1) = 1

r = 1√2−1

=√2+1

(√2−1)(

√2+1)

=√2 + 1 ≈ 2.41 m

20.65 Electric Potential

ε0 = 8.85 x 10−12 Fm−1

1. Draw a diagram of a uniform electric field between

124 CHAPTER 20. WORKED SOLUTIONS

two plates, showing the field lines and the equipoten-tials.

2. Do the same for the electric field around a pointcharge.

3. The potential difference between two plates is100V. What is the potential difference between apoint halfway between the plates and one of theplates?Half of 100 is 50V.

4. What is the electric potential at a point 0.2m froman alpha particle (charge on an electron = −1.6 x10−19C)?The charge on an alpha particle is 3.2 x 10−19C.V = Q

4πϵ0r= 3.2×10−19

4π×8.85×10−12×0.2 = 1.44× 10−8 V5. What is the electric potential energy of an electronat the negative electrode of an electron gun if the po-tential difference between the electrodes is 10V?U = V q = 10 × −1.6 × 10−19 = −1.6 × 10−18 J =−10 eV

20.66 Electric Potential Energy

k = 8.99 x 109 Nm2C−2

1. Convert 5 x 10−13 J to MeV.5×10−13

1.6×10−19 = 3125000 eV = 3.125MeV2. Convert 0.9 GeV to J.0.9× 109 × 1.6× 10−19 = 1.44× 10−10 J3. What is the potential energy of an electron at thenegatively charged plate of a uniform electric fieldwhen the potential difference between the two platesis 100V?ϵelec = Velecq = 100×1.6×10−19 = 1.6×10−17 J =0.1 keV4. What is the potential energy of a 2C charge 2cmfrom a 0.5C charge?ϵelec =

kQqr = 8.99×109×2×0.5

0.02 = 4.5× 1011 JRemember, coulombs are big!5. What is represented by the gradient of a graph ofelectric potential energy against distance from somecharge?Force exerted on the charge with that energy.

20.67 The Standard Model

1. The third generation top quark was the last quarkin the Standard Model to have its existence provenexperimentally (in 1995). It is also the most massiveof the quarks. Why was it so difficult to observe a topquark?Creatingmassive particles requires a lot of energy. It tooka long time to develop a particle accelerator which accel-erated other particles to the energy required to create topquarks.2. What observable phenomena does the StandardModel not explain?Gravity, as well as mass, and the large number of con-

20.70. LEPTONS 125

stants in the Standard Model.3. How much more massive is an up quark than anelectron?2.4

0.511 = 4.7 times4. How many fermions are there in the StandardModel?12 + 12 = 24 (remember the antiparticles)5. The antiparticle of the electron (e-) is the positron.What is the charge and rest mass of a positron?Charge: +1.6 x 10−19CMass: 0.511 MeV/c2 = 9.11 x 10−31 kg

20.68 Quarks

1. The Δ++ baryon is made up of up quarks. What isits total charge?Baryons are made of three quarks. 3 x +⅔ = +2e2. The Δ- baryon has a total charge of −1e. Giventhat it is made up of only one type of first generationquark, what is this quark?Three identical quarks must have a total charge of −1e,so one of these quarks has a charge of -⅓e. This is a downquark.3. What is an antiproton made of? What is itscharge?uud , total charge −1e.4. A K+ meson is made of an up quark and an anti-strange quark. What is its total charge?⅔ + ⅓ = +1e5. Lambda (Λ) baryons are made up of an up quark,a down quark, and another quark (not an antiquark).The Λ0 is neutral, and contains a second generationquark. What is this quark?0 = +⅔ - ⅓ + qq = -⅓The second generation quark with a charge of -⅓e is thestrange quark.

20.69 Bosons

1. A stationary light source emits single photons atregular intervals. Draw a Feynman diagram to rep-resent this.

2. Write two equations (including aW+ boson) whichdescribe positron emission.p → n+W+

W+ → ν + e+

3. What is the charge on a W- boson?−1.6 x 10−19C, since it has to balance the charge on aproton.4. Read Richard Feynman’s excellent book, “QED -the Strange Theory of Light and Matter”, ISBN 978-0-140-12505-4.Don't look here. I can't read a book for you. Visit yourlibrary, bookshop, or whatever.

20.70 Leptons

1. An electron is produced by a nuclear reaction, butan electron-antineutrino is not produced. What otherparticle is produced?A positron, since this is the only other first-generationparticle with a lepton number of −1.2. Why do electrons not make up part of the nucleus?Electrons do not interact with the strong nuclear force.3. Why did it take until the 1950s to detect the firstantineutrino?They are tiny, chargeless, and almost massless.4. Complete the following equation for the emission

126 CHAPTER 20. WORKED SOLUTIONS

of a beta particle from a nucleus:10n →1

1 p + 0−1e

− + 00νe

5. Complete the following equation for the emissionof an antielectron from a nucleus:11p →1

0 n + 01e

+ + 00νe

6. Complete the following equation for the capture ofan electron by a nucleus:11p+

0−1e

− →10 n + 0

0νe

20.71 Millikan’s Experiment

h = 6.63 x 10−34 Jsc = 3 x 108 ms−1

g = 9.81 ms−2

1. Rearrange the formula above in terms of q.qVd = mg

qV = mgd

q = mgdV

2. Themass of an oil drop cannot bemeasured easily.Express the mass of an oil drop in terms of its radiusr and its density ρ, and, by substitution, find a moreuseful formula for q.ρ = m

Volumem = Volume× ρ

Assuming that the oil drop is spherical:Volume = 4

3πr3

m = 43ρπr

3

q = 4πρgdr3

3V

3. An oil droplet of density 885kgm−3 and radius1μm is held stationary in between two plates whichare 10cm apart. At what potential differences be-tween the plates is this possible?q must be a multiple of e, the charge on an electron:q = ne , where n is an integer.

ne = 4πρgdr3

3V

V = 4πρgdr3

3ne = 4π×885×9.81×0.1×(10−6)3

3×1.6×10−19×n = 25.7n

Then, to find some actual values, take n = 1,2,3 ... , so V= 25.7V, 12.8V, 8.6V ...4. If the X-rays used to ionise the oil are of wave-length 1nm, howmuch energy do they give to the elec-trons? Why does this mean that the oil drops areionised?c = fλ

f = cλ

E = hf = hcλ = 6.63×10−34×3×108

10−9 = 1.989 ×10−16 J = 1.24 keVThis provides the energy required for an electron to es-cape from its energy level, becoming unbound.5. In reality, the oil drops aremoving when they enterthe uniform electric field. How can this be compen-sated for?Either:

• Use a slightly stronger potential difference to slowthe oil drop down, and then reduce the potential dif-ference to keep the oil drop stationary.

• Measure the voltage required to keep the oil dropmoving at a constant (terminal) velocity, using moresophisticated equipment.

20.72 Pair Production and Annihi-lation

h = 6.63 x 10−34 Js1. The mass of an electron is 9.11 x 10-31 kg. What isthe minimum amount of energy a photon must haveto create an electron?E = 2mc2 = 2× 9.11× 10−31 × (3× 108)2 = 1.64×10−13 J = 1.02MeV2. A 1.1 MeV electron annihilates with a 1.1 MeVpositron. What is the total energy of the photon pro-duced?2.2 MeV = 3.52 x 10−13 J3. What is its frequency?f = E

h = 3.52×10−13

6.63×10−34 = 5.31× 1020 Hz4. What is its wavelength?λ = c

f = 3×108

5.31×1020 = 5.65× 10−13 m5. What classical physical conditions might cause anewly produced electron-positron pair to annihilatealmost immediately?Electrons and positrons have opposite charges, and sothey attract each other. When they are created, they musthave enough kinetic energy to escape each other’s attrac-tion. If this is not the case, they will annihilate.

20.73 Particle Accelerators

1. Use the formula for centripetal force to show thatthe radius ofmotion depends on the speed of themov-ing object.F = mv2

r

20.75. RADIOACTIVE EMISSIONS 127

r = mv2

F

So, if F is constant:r ∝ v2

2. A cyclotron with a diameter of 1.5m is used to ac-celerate electrons (mass 9.11 x 10−31kg). The max-imum force exerted on an electron is 2.4 x 10−18N.What is the maximum velocity of the electrons?F = mv2

r

v2 = Frm = 2.4×10−18×0.75

9.11×10−31 = 1.97× 1012 m2s−2

v = 1.41× 106 ms−1

3. What are the problems involved in constructing alarge cyclotron?Creating a uniform magnetic field over such a large area(for example, the land area enclosed by the LHC!) meansthat cyclotrons have been largely replaced in experimentalparticle physics by tubes, instead of flat cylinders.4. Why don't particles stick to the electrodes whenpassing through them?There are two reasons. Firstly, alternating current is sinu-soidal, so there is no charge on the electrodewhen the par-ticles are passing through them. Secondly, even if therewere charge on the electrode, the net force on the particlewould still be 0, since the force would be equal in everydirection.

20.74 Cloud Chambers and MassSpectrometers

Charge of electron = −1.6 x 10−19CMass of electron = 9.11 x 10−31kgu = 1.66 x 10−27kg1. An electron enters a cloud chamber, passing into a0.1T magnetic field. The initial curvature (the recip-rocal of its radius) of its path is 100m−1. At whatspeed was it moving when it entered the magneticfield?r = mv

qB

v = qBrm = 1.6×10−19×0.1×0.01

9.11×10−31 = 1.76 × 108 ms−1 =0.585c

This is too close to the speed of light to ignore specialrelativity, however we just did.2. The electron spirals inwards in a clockwise di-rection, as show in the diagram on the right. Whatwould the path of a positron, moving with an identi-cal speed, look like?

3. Using a 2T magnetic field, what electric fieldstrength must be used to get a velocity selector to se-lect only particles which are moving at 100ms−1?v = E

B

E = Bv = 2× 100 = 200 Vm−1

4. Some uranium (atomic number 92) ions (charge+3e) of various isotopes are put through the veloc-ity selector described in question 3. They then enteran 0.00002T uniform magnetic field. What radius ofcircular motion would uranium-235 have?mq = rBBselector

Eselector

235×1.66×10−27

3×1.6×10−19 = r × 2×0.00002200 = 0.0000002r

r = 235×1.66×10−27

0.0000002×3×1.6×10−19 = 4.06 m

20.75 Radioactive Emissions

You will need a periodic table.1. Americium-241 is an α emitter. What element,and what isotope, is produced by this decay?24195 Am → b

a? +42α

b = 241 - 4 = 237a = 95 - 2 = 93So, 23793 Np (Neptunium-237) is produced.2. Iodine-129 is a β- emitter. What element, and whatisotope, is produced by this decay?12953 I → b

a? +0−1β

− + 00ν

128 CHAPTER 20. WORKED SOLUTIONS

b = 129a = 53 + 1 = 54So, 12954 Xe (Xenon-129) is produced.3. Gamma rays are used to kill microbes in food.Why doesn't the food become radioactive?Gamma radiation only interacts with the electrons of theatom, not with the nucleus. Radioactivity is due to nuclearproperties, not chemical properties.4. Plutonium-244 decays by emitting an α particle. Itdoes this twice, emits a β- particle, and then emits afurther two α particles. The nucleus becomes a dif-ferent element each time. What element is producedat the end?24494 Pu → b

a? + 4 42α+ 0

−1β−

b = 244 - 16 = 228a = 94 - 8 + 1 = 87So, at the end of this decay chain, 22887 Fr (Francium-228)is produced.5. Carbon-11 changes into Boron-11 by a radioactiveemission. What was emitted?116 C → 11

5 B + ba?

b = 11 - 11 = 0a = 6 - 5 = 1So, a positron ( 01β+ ) was emitted. However, in order forthe lepton numbers to balance, a neutrino ( 00ν ) must alsobe emitted.6. Uranium-236 decays, following the equation:23692 U → 232

90 Th+ X

Identify the particle X in this equation.Solve these type of problems by applying the conserva-tion laws: e.g. the decay must balance electric charge,nucleon number, proton number, lepton number etc. Nu-cleon number doesn't balance (236 = 232 + ?), so X musthave a nucleon number of 4.The Proton number does not balance either (92 = 90 + ?)therefore X must have 2 Protons.Therefore it can be concluded that X is an alpha particle.

20.76 Energy Levels

The following table gives the wavelengths of lightgiven off when electrons change between the energylevels in hydrogen as described in the first row:1. Calculate the potential energy of an electron atlevel n=2.c = λf

3× 108 = 364.6× 10−9 × f

f = 8.23× 1014 Hz∆E = −hf = −6.63×10−34×8.23×1014 = −5.46×10−19 J = −3.41 eV2. Calculate the difference in potential energy be-tween levels n=2 and n=3.This time, let’s derive a general formula:f = ∆E

h

c = λ∆Eh

∆E = chλ = 3×108×6.63×10−34

656.3×10−9 = 3.03 × 10−19 J =1.89 eV3. What is the potential energy of an electron at leveln=3?−3.41 + 1.89 = −1.52 eV4. If an electron were to jump from n=7 to n=5, whatwould the wavelength of the photon given off be?

∆E = chλ5,2

− chλ7,2

= ch(

1λ5,2

− 1λ7,2

)= 3 × 108 ×

6.63 × 10−34(

1434.1×10−9 − 1

397×10−9

)= −4.28 ×

10−20 Jλ = −ch

∆E = 3×108×6.63×10−34

4.28×10−20 = 4.65 µm

20.77 Fission

1. A neutron is fired at some Uranium-235. Barium-141 and Krypton-92 are produced:10n+235

92 U →14156 Ba+92

36 Kr +N10n

How many neutrons are produced (i.e. what is the valueof N)?Consider the mass numbers:1 + 235 = 141 + 92 +N

N + 233 = 236

N = 3

2. What proportion of the neutrons produced mustbe absorbed in order to make the reaction stable?Two thirds, assuming that Uranium-235 always producesthree neutrons when split, which it doesn't.3. What would happen if too many neutrons wereabsorbed?The reaction would stop.4. Alternatively, Uranium-235 can split into Xenon-140, two neutrons and another element. What is thiselement? (You will need to use a periodic table.)Consider the mass numbers:1 + 235 = 140 +N + 2

N + 142 = 236

20.80. RISKS, DOSES AND DOSE EQUIVALENTS 129

N = 94

Then consider the atomic numbers:0 + 92 = 54 + n+ 2(0)

n = 38

So, the element is strontium-94:10n+235

92 U →14054 Xe+94

38 Sr + 210n

20.78 Fusion

c = 3 x 108 ms−1

1. In the Sun, two tritium nuclei ( 31H ) are fused

to produce helium-4 ( 42He ). What else is produced,apart from energy?231H → 4

2He + 220n

2. In larger stars, carbon-12 ( 126 C ) is fused with pro-tium ( 11H ). What single nucleus does this produce?126 C + 1

1H → 137 N

3. In this reaction, 1.95MeV of energy is released.What difference in binding energy, in J, does this cor-respond to?1.95MeV = 1.95 x 106 x 1.6 x 10−19 J = 3.12 x 10−13 J4. If all this energy was emitted as a photon, whatwould its frequency be?E = hf

f = Eh = 3.12×106

6.63×10−34 = 4.71× 1020 HzThis is a within the wavelength spectrum of a gamma-ray,so we can assume all energy is released as a gamma-rayphoton.5. In order to contain a fusion reaction, electromag-netismmay be used. What other force could be used?Why is this not being used for fusion reactors onEarth?Gravity, as in stars. However, the amount of mass re-quired to do this is far larger than the Earth, so we cannotdo this.

20.79 Binding Energy

1. Deuterium (an isotope of Hydrogen with an ex-tra neutron) has a nuclear mass of 2.01355321270 u.What is its binding energy?Total mass of neutron and proton is 1.00727647 +1.00866492 = 2.01594139 uMass defect = 2.01594139 - 2.01355321270 =0.002388178 u = 3.965662187 x 10−30 kgE = ∆mc2 = 3.57× 10−13 J = 2.23MeV

2. Uranium-235 has a nuclear mass of 235.0439299u. It contains 92 protons. What is its binding energy?235 - 92 = 143 neutronsTotal mass of baryons is (92 x 1.00727647) + (143 x1.00866492) = 236.9085188 uMass defect = 236.9085188 - 235.0439299 = 1.8645889u = 3.096222181 x 10−27 kgE = ∆mc2 = 2.79× 10−10 J = 1.74 GeV3. How would you expect H-2 and U-235 to be usedin nuclear reactors? Why?H-2 is used in experimental fusion reactors, since it has asmaller nucleus than Iron-56.U-235 is used in fission reactors, since it has amuch largernucleus than Iron-56.

20.80 Risks, Doses and DoseEquivalents

1. A mobile phone emits electromagnetic radiation.1.2 watts of power are absorbed per. kilogram.Assuming that the radiation is absorbed uniformlyacross a 5kg head, what dose of radiation would bedelivered to the head when making a 10-minute tele-phone call?1.2 x 5 = 6 grays are absorbed per. second. 10 minutes= 600 seconds.6 = dose

600

So, the total dose is 3600 Gy.Note to reader: There is currently something wrong withthis question, since 3600 Gy is enough to kill 600 people.Mobile phones don't do that. I'm working on it. --Sjlegg(talk) 14:30, 6 April 2009 (UTC) Suggestion : A mobilephone does not emit ionizing radiation as the kinetic en-ergy is to low. Therefore the gray unit cannot be used forthis exercise. The unit of measurement for absorption ofelectromagnetic energy by a body is the SAR.It measuresthe time rate of absorption of electromagnetic energy bya body. It is measured in W / kg.The SAR can be determined from the electric fieldstrength E in the body or from the rate of temperaturerise (Δt).2. What dose equivalent does this correspond to?Quality factor for photons is 1, so dose equivalent is 720Sv.3. How many nuclei are there in 1mg of Americium-241?The mass of 1 Americium-241 nucleus is 241u, whichcorresponds to 241 x 1.66 x 10−27 = 400 x 10−27kg.

10−3

400×10−27 = 2.5× 1021

130 CHAPTER 20. WORKED SOLUTIONS

4. An ham sandwich becomes contaminated with 1μg ofAmericium-241, and is eaten by an 80kg person.The half-life of Americium-241 is 432 years. Giventhat Americium-241 gives off 5.638 MeV alpha par-ticles, how long would it be before a dose equivalentof 6 Sv is absorbed, making death certain?First calculate the activity:A = λN

t 12= ln 2

λ

λ = ln 2t 12

A = N ln 2t 12

= 2.5×1021×ln 2432×365.24×24×60×60 = 1.27× 1011 Bq

Then calculate the power (energy per. second):P = 1.27×1011×5.638×106×1.6×10−19 = 0.115WThen calculate the dose per. second:0.11580 = 1.43 mGys−1

Multiply this by the quality factor for alpha particles (20)to get a dose equivalent per. second of 0.0287 Svs−1.

60.0287 = 209 s = 3 min 29 s5. What assumptions have you made?We assumed that the Americium was absorbed uniformlythroughout the body, and that the activity remained con-stant. The latter is acceptable since this period of time isrelatively short. The former is not acceptable.

Chapter 21

Text and image sources, contributors, andlicenses

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131

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21.1. TEXT 133

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• A-level Physics (Advancing Physics)/Rockets, Hoses and Machine Guns Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Rockets%2C_Hoses_and_Machine_Guns?oldid=2480003 Contributors: Sjlegg, Recent Runes, QuiteUnusual,Adrignola, Avicennasis and Anonymous: 3

• A-level Physics (Advancing Physics)/Circular Motion Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Circular_Motion?oldid=2756436 Contributors: Panic2k4, Sjlegg, QuiteUnusual, Adrignola and Anonymous: 5

• A-level Physics (Advancing Physics)/Radar and Triangulation Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radar_and_Triangulation?oldid=2965916 Contributors: Sjlegg, QuiteUnusual, Adrignola, Lemon123, Avicennasis and Anony-mous: 5

• A-level Physics (Advancing Physics)/Large Units Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Large_Units?oldid=2128732 Contributors: Sjlegg, Adrignola, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Orbits Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Orbits?oldid=2713285 Contributors: Panic2k4, Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Doppler Effect Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Doppler_Effect?oldid=2709829 Contributors: Sjlegg, Adrignola, Avicennasis, Robin van der Vliet and Anonymous: 1

• A-level Physics (Advancing Physics)/Big Bang Theory Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Big_Bang_Theory?oldid=2637551 Contributors: Thenub314, Sjlegg, Dwlegg, Adrignola, Avicennasis and Anonymous: 3

• A-level Physics (Advancing Physics)/Heat and Energy Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Heat_and_Energy?oldid=2348220 Contributors: Sjlegg, Adrignola, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Specific Heat Capacity Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Specific_Heat_Capacity?oldid=2970262 Contributors: Albmont, Sjlegg, Adrignola, Leitoxx and Anonymous: 1

• A-level Physics (Advancing Physics)/Ideal Gases Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Ideal_Gases?oldid=2554121 Contributors: Sjlegg, Recent Runes, Adrignola and Anonymous: 3

• A-level Physics (Advancing Physics)/Kinetic Theory Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Kinetic_Theory?oldid=2348232 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Boltzmann Factor Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Boltzmann_Factor?oldid=2611525 Contributors: Sjlegg, QuiteUnusual, NipplesMeCool, Avicennasis, Nerak99 and Anonymous: 2

• A-level Physics (Advancing Physics)/Flux Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Flux?oldid=2490270 Contributors: Panic2k4, Sjlegg, Adrignola, SMS and Anonymous: 4

• A-level Physics (Advancing Physics)/Induction Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Induction?oldid=2534375 Contributors: Sjlegg, Adrignola, TBDH144 and Anonymous: 3

• A-level Physics (Advancing Physics)/Force Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Force?oldid=2748136 Contributors: Sjlegg, Adrignola and Anonymous: 2

• A-level Physics (Advancing Physics)/Transformers Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Transformers?oldid=2586369Contributors: Jomegat, Sjlegg, CommonsDelinker, Adrignola, Knowledgebank007, Avicennasis and Anony-mous: 3

• A-level Physics (Advancing Physics)/Motors Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Motors?oldid=1924002 Contributors: Jomegat, Sjlegg, Recent Runes, Adrignola and Anonymous: 2

• A-level Physics (Advancing Physics)/Generators Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Generators?oldid=2525526 Contributors: Sjlegg, QuiteUnusual, Adrignola and Anonymous: 2

• A-level Physics (Advancing Physics)/Electric Force Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Force?oldid=2052899 Contributors: Sjlegg, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Electric Field Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Field?oldid=2472917 Contributors: Sjlegg, Adrignola, Knowledgebank007 and Anonymous: 1

• A-level Physics (Advancing Physics)/Electric Potential Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Potential?oldid=2052901 Contributors: Sjlegg, NipplesMeCool, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Electric Potential Energy Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Potential_Energy?oldid=2360053 Contributors: Sjlegg, Adrignola, Knowledgebank007, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/The Standard Model Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/The_Standard_Model?oldid=2600316 Contributors: Sjlegg, QuiteUnusual, Adrignola, Drumncars1996 and Anonymous: 1

• A-level Physics (Advancing Physics)/Quarks Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Quarks?oldid=1530086 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Bosons Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Bosons?oldid=2486132 Contributors: Sjlegg, Recent Runes, QuiteUnusual, Adrignola, Brutic and Anonymous: 2

• A-level Physics (Advancing Physics)/Leptons Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Leptons?oldid=2405285 Contributors: Sjlegg, QuiteUnusual, Adrignola, Avicennasis and Anonymous: 2

134 CHAPTER 21. TEXT AND IMAGE SOURCES, CONTRIBUTORS, AND LICENSES

• A-level Physics (Advancing Physics)/Millikan’s Experiment Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Millikan’s_Experiment?oldid=2052908 Contributors: Sjlegg, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Pair Production and Annihilation Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Pair_Production_and_Annihilation?oldid=2496874 Contributors: Panic2k4, Sjlegg, Adrignola and Anonymous: 6

• A-level Physics (Advancing Physics)/Particle Accelerators Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Particle_Accelerators?oldid=1530081 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Cloud Chambers and Mass Spectrometers Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Cloud_Chambers_and_Mass_Spectrometers?oldid=2482356 Contributors: Sjlegg, QuiteUnusual, Adrig-nola, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Quantum Principles Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Quantum_Principles?oldid=2096514 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Radioactive Emissions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radioactive_Emissions?oldid=2580415 Contributors: Sjlegg, Adrignola, SMS and Anonymous: 2

• A-level Physics (Advancing Physics)/Energy Levels Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Energy_Levels?oldid=2073560 Contributors: Sjlegg, Recent Runes and Adrignola

• A-level Physics (Advancing Physics)/Fission Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Fission?oldid=2093776 Contributors: Sjlegg, Adrignola, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Fusion Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Fusion?oldid=2486134 Contributors: Sjlegg, QuiteUnusual, Adrignola, Brutic and Anonymous: 1

• A-level Physics (Advancing Physics)/Binding Energy Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Binding_Energy?oldid=2966870 Contributors: Sjlegg, QuiteUnusual, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Risks, Doses and Dose Equivalents Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Risks%2C_Doses_and_Dose_Equivalents?oldid=2674204 Contributors: Sjlegg, Adrignola, Martinvl and Anony-mous: 2

• A-level Physics (Advancing Physics)/Vectors/Trigonometry Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Vectors/Trigonometry?oldid=1245765 Contributors: Sjlegg and Red4tribe

• A-level Physics (Advancing Physics)/Digital Storage/Logarithms Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Digital_Storage/Logarithms?oldid=1528468 Contributors: Sjlegg and Anonymous Dissident

• A-level Physics (Advancing Physics)/Current/delta Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Current/delta?oldid=2059596 Contributors: Retropunk, Sjlegg, Anonymous Dissident, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Current/sigma Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Current/sigma?oldid=1481260 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Simple Harmonic Motion/Mathematical Derivation Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Simple_Harmonic_Motion/Mathematical_Derivation?oldid=1538925 Contributors: Sjlegg andAdrignola

• A-level Physics (Advancing Physics)/Lenses/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Lenses/Worked_Solutions?oldid=1646564 Contributors: Sjlegg, Anonymous Dissident, Jolly Janner, Adrignola and Anonymous:1

• A-level Physics (Advancing Physics)/Refraction/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Refraction/Worked_Solutions?oldid=2100745 Contributors: Albmont, Sjlegg, Anonymous Dissident, Adrignolaand Anonymous: 1

• A-level Physics (Advancing Physics)/Digital Storage/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Digital_Storage/Worked_Solutions?oldid=2959494 Contributors: Xania, Sjlegg, Anonymous Dissident, Quite-Unusual, Adrignola and Anonymous: 5

• A-level Physics (Advancing Physics)/Digital Processing/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Digital_Processing/Worked_Solutions?oldid=1529799 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Digitisation/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Digitisation/Worked_Solutions?oldid=1529797 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Signal Frequencies/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Signal_Frequencies/Worked_Solutions?oldid=1529800 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Bandwidth/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Bandwidth/Worked_Solutions?oldid=2755209 Contributors: Sjlegg, Laleena, NipplesMeCool and Qaisjp

• A-level Physics (Advancing Physics)/Charge/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Charge/Worked_Solutions?oldid=2365142 Contributors: Sjlegg, Anonymous Dissident, Adrignola and Anonymous: 2

• A-level Physics (Advancing Physics)/Current/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Current/Worked_Solutions?oldid=1665979 Contributors: Sjlegg, QuiteUnusual, NipplesMeCool and Anonymous: 1

• A-level Physics (Advancing Physics)/Voltage/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Voltage/Worked_Solutions?oldid=2052925 Contributors: Sjlegg, Anonymous Dissident, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Power/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Power/Worked_Solutions?oldid=2027164 Contributors: Sjlegg, Anonymous Dissident, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Resistance and Conductance/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Resistance_and_Conductance/Worked_Solutions?oldid=2248963 Contributors: Sjlegg, Anonymous Dissi-dent, Adrignola and Anonymous: 1

21.1. TEXT 135

• A-level Physics (Advancing Physics)/Internal Resistance/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Internal_Resistance/Worked_Solutions?oldid=2474760 Contributors: Sjlegg, Anonymous Dissident, Computerjoe,Adrignola, Thefloogadooga and Anonymous: 2

• A-level Physics (Advancing Physics)/Potential Dividers/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Potential_Dividers/Worked_Solutions?oldid=2965089 Contributors: Sjlegg, Anonymous Dissident, Adrignola, Avi-cennasis, Rynbrgss and Anonymous: 1

• A-level Physics (Advancing Physics)/Sensors/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Sensors/Worked_Solutions?oldid=1534063 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Resistivity and Conductivity/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Resistivity_and_Conductivity/Worked_Solutions?oldid=2657853 Contributors: Sjlegg, Anonymous Dissi-dent, Adrignola and Anonymous: 4

• A-level Physics (Advancing Physics)/Semiconductors/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Semiconductors/Worked_Solutions?oldid=1529809 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/Stress, Strain & the Young Modulus/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Stress%2C_Strain_%26_the_Young_Modulus/Worked_Solutions?oldid=2022386 Contrib-utors: Sjlegg, Anonymous Dissident, Adrignola, Dared111~enwikibooks and Anonymous: 4

• A-level Physics (Advancing Physics)/Metals/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Metals/Worked_Solutions?oldid=1529817 Contributors: Sjlegg, Anonymous Dissident, CRB1 and Adrignola

• A-level Physics (Advancing Physics)/Polymers/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Polymers/Worked_Solutions?oldid=1529818 Contributors: Sjlegg, Anonymous Dissident and Adrignola

• A-level Physics (Advancing Physics)/What is a wave?/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/What_is_a_wave%3F/Worked_Solutions?oldid=1223676 Contributors: Sjlegg and Red4tribe

• A-level Physics (Advancing Physics)/Phasors/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Phasors/Worked_Solutions?oldid=1445909 Contributors: Sjlegg, QuiteUnusual and NipplesMeCool

• A-level Physics (Advancing Physics)/Standing Waves/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Standing_Waves/Worked_Solutions?oldid=1445911 Contributors: Sjlegg, QuiteUnusual and NipplesMeCool

• A-level Physics (Advancing Physics)/Young’s Slits/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Young’s_Slits/Worked_Solutions?oldid=2720811 Contributors: Sjlegg, Red4tribe, Syum90 and Anonymous: 2

• A-level Physics (Advancing Physics)/Diffraction/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Diffraction/Worked_Solutions?oldid=1225132 Contributors: Sjlegg and Red4tribe

• A-level Physics (Advancing Physics)/Light as a Quantum Phenomenon/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Light_as_a_Quantum_Phenomenon/Worked_Solutions?oldid=2965092 Contributors: Xania, Sj-legg, Red4tribe and Anonymous: 2

• A-level Physics (Advancing Physics)/Electron Behaviour as a Quantum Phenomenon/Worked Solutions Source:https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electron_Behaviour_as_a_Quantum_Phenomenon/Worked_Solutions?oldid=1717199 Contributors: Sjlegg, QuiteUnusual, NipplesMeCool, Sebastian Goll and Anonymous: 1

• A-level Physics (Advancing Physics)/Vectors/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Vectors/Worked_Solutions?oldid=1631296 Contributors: Sjlegg, QuiteUnusual, NipplesMeCool and Anonymous: 1

• A-level Physics (Advancing Physics)/Graphs/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Graphs/Worked_Solutions?oldid=1634695 Contributors: Sjlegg, QuiteUnusual and NipplesMeCool

• A-level Physics (Advancing Physics)/Kinematics/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Kinematics/Worked_Solutions?oldid=2618578 Contributors: Sjlegg, Red4tribe, Glaisher and Anonymous: 1

• A-level Physics (Advancing Physics)/Forces and Power/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Forces_and_Power/Worked_Solutions?oldid=2059705 Contributors: Sjlegg, QuiteUnusual, NipplesMeCool andAvicennasis

• A-level Physics (Advancing Physics)/Exponential Relationships/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Exponential_Relationships/Worked_Solutions?oldid=1891645 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Capacitors/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Capacitors/Worked_Solutions?oldid=2723056 Contributors: Sjlegg, Adrignola, Jordanosborn, Dan Milverton andAnonymous: 2

• A-level Physics (Advancing Physics)/Radioactive Decay/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radioactive_Decay/Worked_Solutions?oldid=2546694 Contributors: Sjlegg, Adrignola and Jordanosborn

• A-level Physics (Advancing Physics)/Half-lives/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Half-lives/Worked_Solutions?oldid=1529846 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Gravitational Forces/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Forces/Worked_Solutions?oldid=1529820 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Gravitational Fields/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Fields/Worked_Solutions?oldid=1529788 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Gravitational Potential Energy/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Potential_Energy/Worked_Solutions?oldid=1529848 Contributors: Sjlegg andAdrignola

• A-level Physics (Advancing Physics)/Gravitational Potential/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Potential/Worked_Solutions?oldid=1793793 Contributors: Sjlegg, Adrignola and Anony-mous: 1

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• A-level Physics (Advancing Physics)/Simple Harmonic Motion/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Simple_Harmonic_Motion/Worked_Solutions?oldid=1636868Contributors: Sjlegg, Adrignola andAnony-mous: 1

• A-level Physics (Advancing Physics)/Energy in Simple Harmonic Motion/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Energy_in_Simple_Harmonic_Motion/Worked_Solutions?oldid=1485189 Contributors: Sj-legg and Adrignola

• A-level Physics (Advancing Physics)/Damping/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Damping/Worked_Solutions?oldid=2719925 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Conservation of Momentum/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Conservation_of_Momentum/Worked_Solutions?oldid=2052897 Contributors: Sjlegg, Adrignola and Avi-cennasis

• A-level Physics (Advancing Physics)/Forces and Impulse in Collisions/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Forces_and_Impulse_in_Collisions/Worked_Solutions?oldid=2059704 Contributors: Sj-legg, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Rockets, Hoses and Machine Guns/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Rockets%2C_Hoses_and_Machine_Guns/Worked_Solutions?oldid=2539427 Contributors: Sj-legg, Adrignola, Avicennasis and Anonymous: 1

• A-level Physics (Advancing Physics)/Circular Motion/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Circular_Motion/Worked_Solutions?oldid=1529847 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Radar and Triangulation/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radar_and_Triangulation/Worked_Solutions?oldid=2471912 Contributors: Sjlegg, Adrignola, Lemon123and Anonymous: 3

• A-level Physics (Advancing Physics)/Large Units/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Large_Units/Worked_Solutions?oldid=2559625 Contributors: Sjlegg, Adrignola, Spittal and Anonymous: 1

• A-level Physics (Advancing Physics)/Orbits/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Orbits/Worked_Solutions?oldid=1794479 Contributors: Sjlegg, Adrignola and SMS

• A-level Physics (Advancing Physics)/Doppler Effect/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Doppler_Effect/Worked_Solutions?oldid=2472622 Contributors: Sjlegg, Adrignola, Avicennasis and Anonymous:1

• A-level Physics (Advancing Physics)/Big Bang Theory/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Big_Bang_Theory/Worked_Solutions?oldid=2836670 Contributors: Sjlegg, Adrignola and Reyk

• A-level Physics (Advancing Physics)/Heat and Energy/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Heat_and_Energy/Worked_Solutions?oldid=1704002 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Specific Heat Capacity/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Specific_Heat_Capacity/Worked_Solutions?oldid=1932814 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Ideal Gases/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Ideal_Gases/Worked_Solutions?oldid=1529904 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Kinetic Theory/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Kinetic_Theory/Worked_Solutions?oldid=2835507 Contributors: Sjlegg, Adrignola, Liam987 and Anonymous: 2

• A-level Physics (Advancing Physics)/Boltzmann Factor/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Boltzmann_Factor/Worked_Solutions?oldid=2059595 Contributors: Sjlegg, NipplesMeCool and Avicennasis

• A-level Physics (Advancing Physics)/Flux/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Flux/Worked_Solutions?oldid=1529852 Contributors: Sjlegg and Adrignola

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• A-level Physics (Advancing Physics)/Transformers/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Transformers/Worked_Solutions?oldid=2581018 Contributors: Sjlegg, Adrignola and Anonymous: 1

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• A-level Physics (Advancing Physics)/Generators/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Generators/Worked_Solutions?oldid=1529899 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Electric Force/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Force/Worked_Solutions?oldid=2093877 Contributors: Sjlegg, Indochinetn, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Electric Field/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Field/Worked_Solutions?oldid=1529854 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Electric Potential/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Electric_Potential/Worked_Solutions?oldid=2052902 Contributors: Sjlegg, Adrignola and Avicennasis

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• A-level Physics (Advancing Physics)/The Standard Model/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/The_Standard_Model/Worked_Solutions?oldid=2238507 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Quarks/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Quarks/Worked_Solutions?oldid=1530087 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Bosons/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Bosons/Worked_Solutions?oldid=1529783 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Leptons/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Leptons/Worked_Solutions?oldid=1834719 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Millikan’s Experiment/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Millikan’s_Experiment/Worked_Solutions?oldid=1529910 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Pair Production and Annihilation/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Pair_Production_and_Annihilation/Worked_Solutions?oldid=1530079 Contributors: Sjlegg andAdrignola

• A-level Physics (Advancing Physics)/Particle Accelerators/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Particle_Accelerators/Worked_Solutions?oldid=2052909 Contributors: Sjlegg, Adrignola, Avicennasis and Anony-mous: 1

• A-level Physics (Advancing Physics)/Cloud Chambers andMass Spectrometers/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Cloud_Chambers_and_Mass_Spectrometers/Worked_Solutions?oldid=2313732 Contribu-tors: Sjlegg, Adrignola and Anonymous: 2

• A-level Physics (Advancing Physics)/Radioactive Emissions/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radioactive_Emissions/Worked_Solutions?oldid=2349303 Contributors: Sjlegg, Adrignola, Icarntspel andSMS

• A-level Physics (Advancing Physics)/Energy Levels/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Energy_Levels/Worked_Solutions?oldid=2693812 Contributors: Sjlegg, Recent Runes, Adrignola and Anonymous:1

• A-level Physics (Advancing Physics)/Fission/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Fission/Worked_Solutions?oldid=2059701 Contributors: Sjlegg, Adrignola and Avicennasis

• A-level Physics (Advancing Physics)/Fusion/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Fusion/Worked_Solutions?oldid=2360995 Contributors: Sjlegg, Adrignola and Anonymous: 1

• A-level Physics (Advancing Physics)/Binding Energy/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Binding_Energy/Worked_Solutions?oldid=1529823 Contributors: Sjlegg and Adrignola

• A-level Physics (Advancing Physics)/Risks, Doses and Dose Equivalents/Worked Solutions Source: https://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Risks%2C_Doses_and_Dose_Equivalents/Worked_Solutions?oldid=2534018 Contributors: Sj-legg, Adrignola, Willywaller and Anonymous: 2

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