Physics GRE Sample Test Solutions

DRAFT

The Physics GRE Solution Guide

Sample Test

http://groups.yahoo.com/group/physicsgre_v2

November 3, 2009

Author:David S. Latchman

http://groups.yahoo.com/group/physicsgre_v2

DRAFT

2

David S. Latchman ©2009

DRAFTPreface

This solution guide initially started out on the Yahoo Groups web site and was prettysuccessful at the time. Unfortunately, the group was lost and with it, much of the thehard work that was put into it. This is my attempt to recreate the solution guide andmake it more widely avaialble to everyone. If you see any errors, think certain thingscould be expressed more clearly, or would like to make suggestions, please feel free todo so.

David Latchman

Document Changes

05-11-2009 1. Added diagrams to GR0177 test questions 1-25

2. Revised solutions to GR0177 questions 1-25

04-15-2009 First Version

DRAFT

ii


DRAFTContents

Preface i

1 Classical Mechanics 1

1.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Newton’s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Work & Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Oscillatory Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5 Rotational Motion about a Fixed Axis . . . . . . . . . . . . . . . . . . . . 8

1.6 Dynamics of Systems of Particles . . . . . . . . . . . . . . . . . . . . . . . 10

1.7 Central Forces and Celestial Mechanics . . . . . . . . . . . . . . . . . . . 10

1.8 Three Dimensional Particle Dynamics . . . . . . . . . . . . . . . . . . . . 12

1.9 Fluid Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.10 Non-inertial Reference Frames . . . . . . . . . . . . . . . . . . . . . . . . 13

1.11 Hamiltonian and Lagrangian Formalism . . . . . . . . . . . . . . . . . . . 13

2 Electromagnetism 15

2.1 Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Currents and DC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Magnetic Fields in Free Space . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Lorentz Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.5 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.6 Maxwell’s Equations and their Applications . . . . . . . . . . . . . . . . . 20

2.7 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

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iv Contents2.8 AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.9 Magnetic and Electric Fields in Matter . . . . . . . . . . . . . . . . . . . . 20

2.10 Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.11 Energy in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.12 Energy in an Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.13 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.14 Current Destiny . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.15 Current Density of Moving Charges . . . . . . . . . . . . . . . . . . . . . 21

2.16 Resistance and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.17 Resistivity and Conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.18 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.19 Kirchoff’s Loop Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.20 Kirchoff’s Junction Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.21 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.22 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.23 Speed of Propagation of a Light Wave . . . . . . . . . . . . . . . . . . . . 23

2.24 Relationship between E and B Fields . . . . . . . . . . . . . . . . . . . . . 23

2.25 Energy Density of an EM wave . . . . . . . . . . . . . . . . . . . . . . . . 24

2.26 Poynting’s Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3 Optics & Wave Phonomena 25

3.1 Wave Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.4 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.5 Geometrical Optics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.6 Polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.7 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.8 Snell’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4 Thermodynamics & Statistical Mechanics 27

4.1 Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.2 Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . 27


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Contents v4.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.4 Ideal Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.5 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.6 Ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.7 Statistical Concepts and Calculation of Thermodynamic Properties . . . 28

4.8 Thermal Expansion & Heat Transfer . . . . . . . . . . . . . . . . . . . . . 28

4.9 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.10 Specific Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.11 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.12 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.13 Work done by Ideal Gas at Constant Temperature . . . . . . . . . . . . . 29

4.14 Heat Conduction Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.15 Ideal Gas Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.16 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equation . . . . . . . . 30

4.17 RMS Speed of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.18 Translational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.19 Internal Energy of a Monatomic gas . . . . . . . . . . . . . . . . . . . . . 30

4.20 Molar Specific Heat at Constant Volume . . . . . . . . . . . . . . . . . . . 31

4.21 Molar Specific Heat at Constant Pressure . . . . . . . . . . . . . . . . . . 31

4.22 Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.23 Adiabatic Expansion of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . 33

4.24 Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . 33

5 Quantum Mechanics 35

5.1 Fundamental Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.2 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

5.3 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.5 Wave Funtion Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.6 Elementary Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . 41

6 Atomic Physics 43

6.1 Properties of Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

©2009 David S. Latchman

DRAFT

vi Contents6.2 Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.3 Energy Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.4 Atomic Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

6.5 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.6 Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.7 Black Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.8 X-Rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

6.9 Atoms in Electric and Magnetic Fields . . . . . . . . . . . . . . . . . . . . 47

7 Special Relativity 51

7.1 Introductory Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.2 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.3 Length Contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

7.4 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.5 Energy and Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.6 Four-Vectors and Lorentz Transformation . . . . . . . . . . . . . . . . . . 53

7.7 Velocity Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.8 Relativistic Doppler Formula . . . . . . . . . . . . . . . . . . . . . . . . . 54

7.9 Lorentz Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7.10 Space-Time Interval . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8 Laboratory Methods 57

8.1 Data and Error Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

8.2 Instrumentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.3 Radiation Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.4 Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.5 Interaction of Charged Particles with Matter . . . . . . . . . . . . . . . . 60

8.6 Lasers and Optical Interferometers . . . . . . . . . . . . . . . . . . . . . . 60

8.7 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

8.8 Fundamental Applications of Probability and Statistics . . . . . . . . . . 60

9 Sample Test 61

9.1 Period of Pendulum on Moon . . . . . . . . . . . . . . . . . . . . . . . . . 61


DRAFT

Contents vii9.2 Work done by springs in series . . . . . . . . . . . . . . . . . . . . . . . . 62

9.3 Central Forces I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

9.4 Central Forces II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

9.5 Electric Potential I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

9.6 Electric Potential II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.7 Faraday’s Law and Electrostatics . . . . . . . . . . . . . . . . . . . . . . . 66

9.8 AC Circuits: RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

9.9 AC Circuits: Underdamped RLC Circuits . . . . . . . . . . . . . . . . . . 68

9.10 Bohr Model of Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . 70

9.11 Nuclear Sizes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

9.12 Ionization of Lithium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

9.13 Electron Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

9.14 Effects of Temperature on Speed of Sound . . . . . . . . . . . . . . . . . . 75

9.15 Polarized Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9.16 Electron in symmetric Potential Wells I . . . . . . . . . . . . . . . . . . . . 76

9.17 Electron in symmetric Potential Wells II . . . . . . . . . . . . . . . . . . . 77

9.18 Relativistic Collisions I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.19 Relativistic Collisions II . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

9.20 Thermodynamic Cycles I . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

9.21 Thermodynamic Cycles II . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

9.22 Distribution of Molecular Speeds . . . . . . . . . . . . . . . . . . . . . . . 79

9.23 Temperature Measurements . . . . . . . . . . . . . . . . . . . . . . . . . . 79

9.24 Counting Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

9.25 Thermal & Electrical Conductivity . . . . . . . . . . . . . . . . . . . . . . 80

9.26 Nonconservation of Parity in Weak Interactions . . . . . . . . . . . . . . 81

9.27 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

9.28 Lorentz Force Law I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

9.29 Lorentz Force Law II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

9.30 Nuclear Angular Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

9.31 Potential Step Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85


DRAFT

viii ContentsA Constants & Important Equations 87

A.1 Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A.2 Vector Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A.3 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

A.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89


DRAFTList of Tables

4.22.1Table of Molar Specific Heats . . . . . . . . . . . . . . . . . . . . . . . . . 32

9.4.1 Table of Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

A.1.1Something . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

DRAFT

x List of Tables


DRAFTList of Figures

9.5.1 Diagram of Uniformly Charged Circular Loop . . . . . . . . . . . . . . . 65

9.8.1 Schematic of Inductance-Resistance Circuit . . . . . . . . . . . . . . . . . 67

9.8.2 Potential Drop across Resistor in a Inductor-Resistance Circuit . . . . . . 68

9.9.1 LRC Oscillator Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

9.9.2 Forced Damped Harmonic Oscillations . . . . . . . . . . . . . . . . . . . 70

9.15.1Waves that are not plane-polarized . . . . . . . . . . . . . . . . . . . . . . 76

9.15.2φ = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

9.22.1Maxwell-Boltzmann Speed Distribution of Nobel Gases . . . . . . . . . . 79

9.27.1Hoop and S-shaped wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

9.28.1Charged particle moving parallel to a positively charged current carry-ing wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

9.31.1Wavefunction of particle through a potential step barrier . . . . . . . . . 85

DRAFT

xii List of Figures


DRAFTChapter 1Classical Mechanics

1.1 Kinematics

1.1.1 Linear Motion

Average Velocity

v =∆x∆t

=x2 − x1

t2 − t1(1.1.1)

Instantaneous Velocity

v = lim∆t→0

∆x∆t

=dxdt

= v(t) (1.1.2)

Kinematic Equations of Motion

The basic kinematic equations of motion under constant acceleration, a, are

v = v0 + at (1.1.3)

v2 = v20 + 2a (x − x0) (1.1.4)

x − x0 = v0t +12

at2 (1.1.5)

x − x0 =12

(v + v0) t (1.1.6)

1.1.2 Circular Motion

In the case of Uniform Circular Motion, for a particle to move in a circular path, aradial acceleration must be applied. This acceleration is known as the Centripetal

DRAFT

2 Classical MechanicsAcceleration

Centripetal Acceleration

a =v2

r(1.1.7)

Angular Velocity

ω =vr

(1.1.8)

We can write eq. (1.1.7) in terms of ω

a = ω2r (1.1.9)

Rotational Equations of Motion

The equations of motion under a constant angular acceleration, α, are

ω = ω0 + αt (1.1.10)

θ =ω + ω0

2t (1.1.11)

θ = ω0t +12αt2 (1.1.12)

ω2 = ω20 + 2αθ (1.1.13)

1.2 Newton’s Laws

1.2.1 Newton’s Laws of Motion

First Law A body continues in its state of rest or of uniform motion unless acted uponby an external unbalanced force.

Second Law The net force on a body is proportional to its rate of change of momentum.

F =dpdt

= ma (1.2.1)

Third Law When a particle A exerts a force on another particle B, B simultaneouslyexerts a force on A with the same magnitude in the opposite direction.

FAB = −FBA (1.2.2)


DRAFT

Work & Energy 31.2.2 Momentum

p = mv (1.2.3)

1.2.3 Impulse

∆p = J =w

Fdt = Favgdt (1.2.4)

1.3 Work & Energy

1.3.1 Kinetic Energy

K ≡12

mv2 (1.3.1)

1.3.2 The Work-Energy Theorem

The net Work done is given byWnet = K f − Ki (1.3.2)

1.3.3 Work done under a constant Force

The work done by a force can be expressed as

W = F∆x (1.3.3)

In three dimensions, this becomes

W = F · ∆r = F∆r cosθ (1.3.4)

For a non-constant force, we have

W =

x fw

xi

F(x)dx (1.3.5)

1.3.4 Potential Energy

The Potential Energy is

F(x) = −dU(x)

dx(1.3.6)

for conservative forces, the potential energy is

U(x) = U0 −

xw

x0

F(x′)dx′ (1.3.7)


DRAFT

4 Classical Mechanics1.3.5 Hooke’s Law

F = −kx (1.3.8)

where k is the spring constant.

1.3.6 Potential Energy of a Spring

U(x) =12

kx2 (1.3.9)

1.4 Oscillatory Motion

1.4.1 Equation for Simple Harmonic Motion

x(t) = A sin (ωt + δ) (1.4.1)

where the Amplitude, A, measures the displacement from equilibrium, the phase, δ, isthe angle by which the motion is shifted from equilibrium at t = 0.

1.4.2 Period of Simple Harmonic Motion

T =2πω

(1.4.2)

1.4.3 Total Energy of an Oscillating System

Given thatx = A sin (ωt + δ) (1.4.3)

and that the Total Energy of a System is

E = KE + PE (1.4.4)

The Kinetic Energy is

KE =12

mv2

=12

mdxdt

=12

mA2ω2 cos2 (ωt + δ) (1.4.5)


DRAFT

Oscillatory Motion 5The Potential Energy is

U =12

kx2

=12

kA2 sin2 (ωt + δ) (1.4.6)

Adding eq. (1.4.5) and eq. (1.4.6) gives

E =12

kA2 (1.4.7)

1.4.4 Damped Harmonic Motion

Fd = −bv = −bdxdt

(1.4.8)

where b is the damping coefficient. The equation of motion for a damped oscillatingsystem becomes

− kx − bdxdt

= md2xdt2 (1.4.9)

Solving eq. (1.4.9) govesx = Ae−αt sin (ω′t + δ) (1.4.10)

We find that

α =b

2m(1.4.11)

ω′ =

√km−

b2

4m2

=

√ω2

0 −b2

4m2

=√ω2

0 − α2 (1.4.12)

1.4.5 Small Oscillations

The Energy of a system is

E = K + V(x) =12

mv(x)2 + V(x) (1.4.13)

We can solve for v(x),

v(x) =

√2m

(E − V(x)) (1.4.14)

where E ≥ V(x) Let the particle move in the potential valley, x1 ≤ x ≤ x2, the potentialcan be approximated by the Taylor Expansion

V(x) = V(xe) + (x − xe)[dV(x)

dx

]x=xe

+12

(x − xe)2

[d2V(x)

dx2

]x=xe

+ · · · (1.4.15)


DRAFT

6 Classical MechanicsAt the points of inflection, the derivative dV/dx is zero and d2V/dx2 is positive. Thismeans that the potential energy for small oscillations becomes

V(x) u V(xe) +12

k(x − xe)2 (1.4.16)

where

k ≡[d2V(x)

dx2

]x=xe

≥ 0 (1.4.17)

As V(xe) is constant, it has no consequences to physical motion and can be dropped.We see that eq. (1.4.16) is that of simple harmonic motion.

1.4.6 Coupled Harmonic Oscillators

Consider the case of a simple pendulum of length, `, and the mass of the bob is m1.For small displacements, the equation of motion is

θ + ω0θ = 0 (1.4.18)

We can express this in cartesian coordinates, x and y, where

x = ` cosθ ≈ ` (1.4.19)y = ` sinθ ≈ `θ (1.4.20)

eq. (1.4.18) becomesy + ω0y = 0 (1.4.21)

This is the equivalent to the mass-spring system where the spring constant is

k = mω20 =

mg`

(1.4.22)

This allows us to to create an equivalent three spring system to our coupled pendulumsystem. The equations of motion can be derived from the Lagrangian, where

L = T − V

=12

my21 +

12

my22 −

(12

ky21 +

12κ(y2 − y1

)2+

12

ky22

)=

12

m(y1

2 + y22)−

12

(k(y2

1 + y22

)+ κ

(y2 − y1

)2)

(1.4.23)

We can find the equations of motion of our system

ddt

(∂L∂yn

)=∂L∂yn

(1.4.24)

1Add figure with coupled pendulum-spring system


DRAFT

Oscillatory Motion 7The equations of motion are

my1 = −ky1 + κ(y2 − y1

)(1.4.25)

my2 = −ky2 + κ(y2 − y1

)(1.4.26)

We assume solutions for the equations of motion to be of the form

y1 = cos(ωt + δ1) y2 = B cos(ωt + δ2)y1 = −ωy1 y2 = −ωy2

(1.4.27)

Substituting the values for y1 and y2 into the equations of motion yields(k + κ −mω2

)y1 − κy2 = 0 (1.4.28)

−κy1 +(k + κ −mω2

)y2 = 0 (1.4.29)

We can get solutions from solving the determinant of the matrix∣∣∣∣∣(k + κ −mω2)−κ

−κ(k + κ −mω2)∣∣∣∣∣ = 0 (1.4.30)

Solving the determinant gives(mω2

)2− 2mω2 (k + κ) +

(k2 + 2kκ

)= 0 (1.4.31)

This yields

ω2 =

km

=g`

k + 2κm

=g`

+2κm

(1.4.32)

We can now determine exactly how the masses move with each mode by substitutingω2 into the equations of motion. Where

ω2 =km

We see that

k + κ −mω2 = κ (1.4.33)

Substituting this into the equation of motion yields

y1 = y2 (1.4.34)

We see that the masses move in phase with each other. You will also noticethe absense of the spring constant term, κ, for the connecting spring. As themasses are moving in step, the spring isn’t stretching or compressing and henceits absence in our result.

ω2 =k + κ

mWe see that

k + κ −mω2 = −κ (1.4.35)

Substituting this into the equation of motion yields

y1 = −y2 (1.4.36)

Here the masses move out of phase with each other. In this case we see thepresence of the spring constant, κ, which is expected as the spring playes a role.It is being stretched and compressed as our masses oscillate.


DRAFT

8 Classical Mechanics1.4.7 Doppler Effect

The Doppler Effect is the shift in frequency and wavelength of waves that results froma source moving with respect to the medium, a receiver moving with respect to themedium or a moving medium.

Moving Source If a source is moving towards an observer, then in one period, τ0, itmoves a distance of vsτ0 = vs/ f0. The wavelength is decreased by

λ′ = λ −vs

f0−

v − vs

f0(1.4.37)

The frequency change is

f ′ =vλ′

= f0

( vv − vs

)(1.4.38)

Moving Observer As the observer moves, he will measure the same wavelength, λ, asif at rest but will see the wave crests pass by more quickly. The observer measuresa modified wave speed.

v′ = v + |vr| (1.4.39)

The modified frequency becomes

f ′ =v′

λ= f0

(1 +

vr

v

)(1.4.40)

Moving Source and Moving Observer We can combine the above two equations

λ′ =v − vs

f0(1.4.41)

v′ = v − vr (1.4.42)

To give a modified frequency of

f ′ =v′

λ′=

(v − vr

v − vs

)f0 (1.4.43)

1.5 Rotational Motion about a Fixed Axis

1.5.1 Moment of Inertia

I =

∫R2dm (1.5.1)

1.5.2 Rotational Kinetic Energy

K =12

Iω2 (1.5.2)


DRAFT

Rotational Motion about a Fixed Axis 91.5.3 Parallel Axis Theorem

I = Icm + Md2 (1.5.3)

1.5.4 Torque

τ = r × F (1.5.4)τ = Iα (1.5.5)

where α is the angular acceleration.

1.5.5 Angular Momentum

L = Iω (1.5.6)

we can find the Torque

τ =dLdt

(1.5.7)

1.5.6 Kinetic Energy in Rolling

With respect to the point of contact, the motion of the wheel is a rotation about thepoint of contact. Thus

K = Krot =12

Icontactω2 (1.5.8)

Icontact can be found from the Parallel Axis Theorem.

Icontact = Icm + MR2 (1.5.9)

Substitute eq. (1.5.8) and we have

K =12

(Icm + MR2

)ω2

=12

Icmω2 +

12

mv2 (1.5.10)

The kinetic energy of an object rolling without slipping is the sum of hte kinetic energyof rotation about its center of mass and the kinetic energy of the linear motion of theobject.


DRAFT

10 Classical Mechanics1.6 Dynamics of Systems of Particles

1.6.1 Center of Mass of a System of Particles

Position Vector of a System of Particles

R =m1r1 + m2r2 + m3r3 + · · · + mNrN

M(1.6.1)

Velocity Vector of a System of Particles

V =dRdt

=m1v1 + m2v2 + m3v3 + · · · + mNvN

M(1.6.2)

Acceleration Vector of a System of Particles

A =dVdt

=m1a1 + m2a2 + m3a3 + · · · + mNaN

M(1.6.3)

1.7 Central Forces and Celestial Mechanics

1.7.1 Newton’s Law of Universal Gravitation

F = −(GMm

r2

)r (1.7.1)

1.7.2 Potential Energy of a Gravitational Force

U(r) = −GMm

r(1.7.2)

1.7.3 Escape Speed and Orbits

The energy of an orbiting body is

E = T + U

=12

mv2−

GMmr

(1.7.3)


DRAFT

Central Forces and Celestial Mechanics 11The escape speed becomes

E =12

mv2esc −

GMmRE

= 0 (1.7.4)

Solving for vesc we find

vesc =

√2GM

Re(1.7.5)

1.7.4 Kepler’s Laws

First Law The orbit of every planet is an ellipse with the sun at a focus.

Second Law A line joining a planet and the sun sweeps out equal areas during equalintervals of time.

Third Law The square of the orbital period of a planet is directly proportional to thecube of the semi-major axis of its orbit.

T2

R3 = C (1.7.6)

where C is a constant whose value is the same for all planets.

1.7.5 Types of Orbits

The Energy of an Orbiting Body is defined in eq. (1.7.3), we can classify orbits by theireccentricities.

Circular Orbit A circular orbit occurs when there is an eccentricity of 0 and the orbitalenergy is less than 0. Thus

12

v2−

GMr

= E < 0 (1.7.7)

The Orbital Velocity is

v =

√GM

r(1.7.8)

Elliptic Orbit An elliptic orbit occurs when the eccentricity is between 0 and 1 but thespecific energy is negative, so the object remains bound.

v =

√GM

(2r−

1a

)(1.7.9)

where a is the semi-major axis


DRAFT

12 Classical MechanicsParabolic Orbit A Parabolic Orbit occurs when the eccentricity is equal to 1 and the

orbital velocity is the escape velocity. This orbit is not bounded. Thus

12

v2−

GMr

= E = 0 (1.7.10)

The Orbital Velocity is

v = vesc =

√2GM

r(1.7.11)

Hyperbolic Orbit In the Hyperbolic Orbit, the eccentricity is greater than 1 with anorbital velocity in excess of the escape velocity. This orbit is also not bounded.

v∞ =

√GM

a(1.7.12)

1.7.6 Derivation of Vis-viva Equation

The total energy of a satellite is

E =12

mv2−

GMmr

(1.7.13)

For an elliptical or circular orbit, the specific energy is

E = −GMm

2a(1.7.14)

Equating we get

v2 = GM(2

r−

1a

)(1.7.15)

1.8 Three Dimensional Particle Dynamics

1.9 Fluid Dynamics

When an object is fully or partially immersed, the buoyant force is equal to the weightof fluid displaced.

1.9.1 Equation of Continuity

ρ1v1A1 = ρ2v2A2 (1.9.1)


DRAFT

Non-inertial Reference Frames 131.9.2 Bernoulli’s Equation

P +12ρv2 + ρgh = a constant (1.9.2)

1.10 Non-inertial Reference Frames

1.11 Hamiltonian and Lagrangian Formalism

1.11.1 Lagrange’s Function (L)

L = T − V (1.11.1)

where T is the Kinetic Energy and V is the Potential Energy in terms of GeneralizedCoordinates.

1.11.2 Equations of Motion(Euler-Lagrange Equation)

∂L∂q

=ddt

(∂L∂q

)(1.11.2)

1.11.3 Hamiltonian

H = T + V= pq − L(q, q) (1.11.3)

where

∂H∂p

= q (1.11.4)

∂H∂q

= −∂L∂x

= −p (1.11.5)


DRAFT

14 Classical Mechanics


DRAFTChapter 2Electromagnetism

2.1 Electrostatics

2.1.1 Coulomb’s Law

The force between two charged particles, q1 and q2 is defined by Coulomb’s Law.

F12 =1

4πε0

(q1q2

r212

)r12 (2.1.1)

where ε0 is the permitivitty of free space, where

ε0 = 8.85 × 10−12C2N.m2 (2.1.2)

2.1.2 Electric Field of a point charge

The electric field is defined by mesuring the magnitide and direction of an electricforce, F, acting on a test charge, q0.

E ≡Fq0

(2.1.3)

The Electric Field of a point charge, q is

E =1

4πε0

qr2 r (2.1.4)

In the case of multiple point charges, qi, the electric field becomes

E(r) =1

4πε0

n∑i=1

qi

r2i

ri (2.1.5)

DRAFT

16 ElectromagnetismElectric Fields and Continuous Charge Distributions

If a source is distributed continuously along a region of space, eq. (2.1.5) becomes

E(r) =1

4πε0

∫1r2 rdq (2.1.6)

If the charge was distributed along a line with linear charge density, λ,

λ =dqdx

(2.1.7)

The Electric Field of a line charge becomes

E(r) =1

4πε0

∫line

λr2 rdx (2.1.8)

In the case where the charge is distributed along a surface, the surface charge densityis, σ

σ =QA

=dqdA

(2.1.9)

The electric field along the surface becomes

E(r) =1

4πε0

∫Surface

σr2 rdA (2.1.10)

In the case where the charge is distributed throughout a volume, V, the volume chargedensity is

ρ =QV

=dqdV

(2.1.11)

The Electric Field is

E(r) =1

4πε0

∫Volume

ρ

r2 rdV (2.1.12)

2.1.3 Gauss’ Law

The electric field through a surface is

Φ =

∮surface S

dΦ =

∮surface S

E · dA (2.1.13)

The electric flux through a closed surface encloses a net charge.∮E · dA =

Qε0

(2.1.14)

where Q is the charge enclosed by our surface.


DRAFT

Electrostatics 172.1.4 Equivalence of Coulomb’s Law and Gauss’ Law

The total flux through a sphere is∮E · dA = E(4πr2) =

qε0

(2.1.15)

From the above, we see that the electric field is

E =q

4πε0r2 (2.1.16)

2.1.5 Electric Field due to a line of charge

Consider an infinite rod of constant charge density, λ. The flux through a Gaussiancylinder enclosing the line of charge is

Φ =

∫top surface

E · dA +

∫bottom surface

E · dA +

∫side surface

E · dA (2.1.17)

At the top and bottom surfaces, the electric field is perpendicular to the area vector, sofor the top and bottom surfaces,

E · dA = 0 (2.1.18)

At the side, the electric field is parallel to the area vector, thus

E · dA = EdA (2.1.19)

Thus the flux becomes,

Φ =

∫side sirface

E · dA = E∫

dA (2.1.20)

The area in this case is the surface area of the side of the cylinder, 2πrh.

Φ = 2πrhE (2.1.21)

Applying Gauss’ Law, we see that Φ = q/ε0. The electric field becomes

E =λ

2πε0r(2.1.22)

2.1.6 Electric Field in a Solid Non-Conducting Sphere

Within our non-conducting sphere or radius, R, we will assume that the total charge,Q is evenly distributed throughout the sphere’s volume. So the charge density of oursphere is

ρ =QV

=Q

43πR3

(2.1.23)


DRAFT

18 ElectromagnetismThe Electric Field due to a charge Q is

E =Q

4πε0r2 (2.1.24)

As the charge is evenly distributed throughout the sphere’s volume we can say thatthe charge density is

dq = ρdV (2.1.25)

where dV = 4πr2dr. We can use this to determine the field inside the sphere bysumming the effect of infinitesimally thin spherical shells

E =

∫ E

0dE =

∫ r

0

dq4πεr2

=ρ

ε0

∫ r

0dr

=Qr

43πε0R3

(2.1.26)

2.1.7 Electric Potential Energy

U(r) =1

4πε0qq0r (2.1.27)

2.1.8 Electric Potential of a Point Charge

The electrical potential is the potential energy per unit charge that is associated with astatic electrical field. It can be expressed thus

U(r) = qV(r) (2.1.28)

And we can see that

V(r) =1

4πε0

qr

(2.1.29)

A more proper definition that includes the electric field, E would be

V(r) = −

∫C

E · d` (2.1.30)

where C is any path, starting at a chosen point of zero potential to our desired point.

The difference between two potentials can be expressed such

V(b) − V(a) = −

∫ b

E · d` +

∫ a

E · d`

= −

∫ b

aE · d` (2.1.31)


DRAFT

Electrostatics 19This can be further expressed

V(b) − V(a) =

∫ b

a(∇V) · d` (2.1.32)

And we can show thatE = −∇V (2.1.33)

2.1.9 Electric Potential due to a line charge along axis

Let us consider a rod of length, `, with linear charge density, λ. The Electrical Potentialdue to a continuous distribution is

V =

∫dV =

14πε0

∫dqr

(2.1.34)

The charge density isdq = λdx (2.1.35)

Substituting this into the above equation, we get the electrical potential at some distancex along the rod’s axis, with the origin at the start of the rod.

dV =1

4πε0

dqx

=1

4πε0

λdxx

(2.1.36)

This becomesV =

λ4πε0

ln[x2

x1

](2.1.37)

where x1 and x2 are the distances from O, the end of the rod.

Now consider that we are some distance, y, from the axis of the rod of length, `. Weagain look at eq. (2.1.34), where r is the distance of the point P from the rod’s axis.

V =1

4πε0

∫dqr

=1

4πε0

∫ `

0

λdx(x2 + y2

) 12

=λ

4πε0ln

[x +

(x2 + y2

) 12]`

0

=λ

4πε0ln

[` +

(`2 + y2

) 12]− ln y

=λ

4πε0ln

` +(`2 + y2) 1

2

d

(2.1.38)


DRAFT

20 Electromagnetism2.2 Currents and DC Circuits

2

2.3 Magnetic Fields in Free Space

3

2.4 Lorentz Force

4

2.5 Induction

5

2.6 Maxwell’s Equations and their Applications

6

2.7 Electromagnetic Waves

7

2.8 AC Circuits

8

2.9 Magnetic and Electric Fields in Matter

9


DRAFT

Capacitance 212.10 Capacitance

Q = CV (2.10.1)

2.11 Energy in a Capacitor

U =Q2

2C

=CV2

2

=QV

2(2.11.1)

2.12 Energy in an Electric Field

u ≡U

volume=ε0E2

2(2.12.1)

2.13 Current

I ≡dQdt

(2.13.1)

2.14 Current Destiny

I =

∫A

J · dA (2.14.1)

2.15 Current Density of Moving Charges

J =IA

= neqvd (2.15.1)

2.16 Resistance and Ohm’s Law

R ≡VI

(2.16.1)


DRAFT

22 Electromagnetism2.17 Resistivity and Conductivity

R = ρLA

(2.17.1)

E = ρJ (2.17.2)

J = σE (2.17.3)

2.18 Power

P = VI (2.18.1)

2.19 Kirchoff’s Loop Rules

Write Here

2.20 Kirchoff’s Junction Rule

Write Here

2.21 RC Circuits

E − IR −QC

= 0 (2.21.1)

2.22 Maxwell’s Equations

2.22.1 Integral Form

Gauss’ Law for Electric Fieldsw

closed surface

E · dA =Qε0

(2.22.1)


DRAFT

Speed of Propagation of a Light Wave 23Gauss’ Law for Magnetic Fields

w

closed surface

B · dA = 0 (2.22.2)

Ampere’s Lawz

B · ds = µ0I + µ0ε0ddt

w

surface

E · dA (2.22.3)

Faraday’s Lawz

E · ds = −ddt

w

surface

B · dA (2.22.4)

2.22.2 Differential Form

Gauss’ Law for Electric Fields∇ · E =

ρ

ε0(2.22.5)

Gauss’ Law for Magnetism∇ · B = 0 (2.22.6)

Ampere’s Law

∇ × B = µ0J + µ0ε0∂E∂t

(2.22.7)

Faraday’s Law

∇ · E = −∂B∂t

(2.22.8)

2.23 Speed of Propagation of a Light Wave

c =1

√µ0ε0

(2.23.1)

In a material with dielectric constant, κ,

c√κ =

cn

(2.23.2)

where n is the refractive index.

2.24 Relationship between E and B Fields

E = cB (2.24.1)E · B = 0 (2.24.2)


DRAFT

24 Electromagnetism2.25 Energy Density of an EM wave

u =12

(B2

µ0+ ε0E2

)(2.25.1)

2.26 Poynting’s Vector

S =1µ0

E × B (2.26.1)


DRAFTChapter 3Optics & Wave Phonomena

3.1 Wave Properties

1

3.2 Superposition

2

3.3 Interference

3

3.4 Diffraction

4

3.5 Geometrical Optics

5

3.6 Polarization

6

DRAFT

26 Optics & Wave Phonomena3.7 Doppler Effect

7

3.8 Snell’s Law

3.8.1 Snell’s Law

n1 sinθ1 = n2 sinθ2 (3.8.1)

3.8.2 Critical Angle and Snell’s Law

The critical angle, θc, for the boundary seperating two optical media is the smallestangle of incidence, in the medium of greater index, for which light is totally refelected.

From eq. (3.8.1), θ1 = 90 and θ2 = θc and n2 > n1.

n1 sin 90 = n2sinθc

sinθc =n1

n2(3.8.2)


DRAFTChapter 4Thermodynamics & Statistical Mechanics

4.1 Laws of Thermodynamics

1

4.2 Thermodynamic Processes

2

4.3 Equations of State

3

4.4 Ideal Gases

4

4.5 Kinetic Theory

5

4.6 Ensembles

6

DRAFT

28 Thermodynamics & Statistical Mechanics4.7 Statistical Concepts and Calculation of Thermody-

namic Properties

7

4.8 Thermal Expansion & Heat Transfer

8

4.9 Heat Capacity

Q = C(T f − Ti

)(4.9.1)

where C is the Heat Capacity and T f and Ti are the final and initial temperaturesrespectively.

4.10 Specific Heat Capacity

Q = cm(T f − ti

)(4.10.1)

where c is the specific heat capacity and m is the mass.

4.11 Heat and Work

W =

∫ V f

Vi

PdV (4.11.1)

4.12 First Law of Thermodynamics

dEint = dQ − dW (4.12.1)

where dEint is the internal energy of the system, dQ is the Energy added to the systemand dW is the work done by the system.


DRAFT

Work done by Ideal Gas at Constant Temperature 294.12.1 Special Cases to the First Law of Thermodynamics

Adiabatic Process During an adiabatic process, the system is insulated such that thereis no heat transfer between the system and its environment. Thus dQ = 0, so

∆Eint = −W (4.12.2)

If work is done on the system, negative W, then there is an increase in its internalenergy. Conversely, if work is done by the system, positive W, there is a decreasein the internal energy of the system.

Constant Volume (Isochoric) Process If the volume is held constant, then the systemcan do no work, δW = 0, thus

∆Eint = Q (4.12.3)

If heat is added to the system, the temperature increases. Conversely, if heat isremoved from the system the temperature decreases.

Closed Cycle In this situation, after certain interchanges of heat and work, the systemcomes back to its initial state. So ∆Eint remains the same, thus

∆Q = ∆W (4.12.4)

The work done by the system is equal to the heat or energy put into it.

Free Expansion In this process, no work is done on or by the system. Thus ∆Q =∆W = 0,

∆Eint = 0 (4.12.5)

4.13 Work done by Ideal Gas at Constant Temperature

Starting with eq. (4.11.1), we substitute the Ideal gas Law, eq. (4.15.1), to get

W = nRT∫ V f

Vi

dVV

= nRT lnV f

Vi(4.13.1)

4.14 Heat Conduction Equation

The rate of heat transferred, H, is given by

H =Qt

= kATH − TC

L(4.14.1)

where k is the thermal conductivity.


DRAFT

30 Thermodynamics & Statistical Mechanics4.15 Ideal Gas Law

PV = nRT (4.15.1)

where

n = Number of molesP = PressureV = VolumeT = Temperature

and R is the Universal Gas Constant, such that

R ≈ 8.314 J/mol. K

We can rewrite the Ideal gas Law to say

PV = NkT (4.15.2)

where k is the Boltzmann’s Constant, such that

k =R

NA≈ 1.381 × 10−23 J/K

4.16 Stefan-Boltzmann’s FormulaStefan-Boltzmann’s Equa-tion

P(T) = σT4 (4.16.1)

4.17 RMS Speed of an Ideal Gas

vrms =

√3RTM

(4.17.1)

4.18 Translational Kinetic Energy

K =32

kT (4.18.1)

4.19 Internal Energy of a Monatomic gas

Eint =32

nRT (4.19.1)


DRAFT

Molar Specific Heat at Constant Volume 314.20 Molar Specific Heat at Constant Volume

Let us define, CV such that

Q = nCV∆T (4.20.1)

Substituting into the First Law of Thermodynamics, we have

∆Eint + W = nCV∆T (4.20.2)

At constant volume, W = 0, and we get

CV =1n

∆Eint

∆T(4.20.3)

Substituting eq. (4.19.1), we get

CV =32

R = 12.5 J/mol.K (4.20.4)

4.21 Molar Specific Heat at Constant Pressure

Starting with

Q = nCp∆T (4.21.1)

and

∆Eint = Q −W⇒ nCV∆T = nCp∆T + nR∆T

∴ CV = Cp − R (4.21.2)

4.22 Equipartition of Energy

CV =

(f2

)R = 4.16 f J/mol.K (4.22.1)

where f is the number of degrees of freedom.


DRAFT

32 Thermodynamics & Statistical Mechanics

Degrees

ofFreedomPredicted

Molar

SpecificH

eats

Molecule

TranslationalR

otationalV

ibrationalTotal(f)

CV

CP

=C

V+

R

Monatom

ic3

00

332 R

52 RD

iatomic

32

25

52 R72 R

Polyatomic

(Linear)3

33n−

56

3R4R

Polyatomic

(Non-Linear)

33

3n−

66

3R4R

Table4.22.1:Table

ofMolar

SpecificH

eats


DRAFT

Adiabatic Expansion of an Ideal Gas 334.23 Adiabatic Expansion of an Ideal Gas

PVγ = a constant (4.23.1)

where γ = CPCV

.We can also write

TVγ−1 = a constant (4.23.2)

4.24 Second Law of Thermodynamics

Something.


DRAFT

34 Thermodynamics & Statistical Mechanics


DRAFTChapter 5Quantum Mechanics

5.1 Fundamental Concepts

1

5.2 Schrodinger Equation

Let us define Ψ to beΨ = Ae−iω(t− x

v ) (5.2.1)

Simplifying in terms of Energy, E, and momentum, p, we get

Ψ = Ae−i(Et−px)~ (5.2.2)

We obtain Schrodinger’s Equation from the Hamiltonian

H = T + V (5.2.3)

To determine E and p,

∂2Ψ

∂x2 = −p2

~2 Ψ (5.2.4)

∂Ψ∂t

=iE~

Ψ (5.2.5)

and

H =p2

2m+ V (5.2.6)

This becomes

EΨ = HΨ (5.2.7)

DRAFT

36 Quantum Mechanics

EΨ = −~

i∂Ψ∂t

p2Ψ = −~2∂2Ψ

∂x2

The Time Dependent Schrodinger’s Equation is

i~∂Ψ∂t

= −~2

2m∂2Ψ

∂x2 + V(x)Ψ (5.2.8)

The Time Independent Schrodinger’s Equation is

EΨ = −~2

2m∂2Ψ

∂x2 + V(x)Ψ (5.2.9)

5.2.1 Infinite Square Wells

Let us consider a particle trapped in an infinite potential well of size a, such that

V(x) =

0 for 0 < x < a∞ for |x| > a,

so that a nonvanishing force acts only at ±a/2. An energy, E, is assigned to the systemsuch that the kinetic energy of the particle is E. Classically, any motion is forbiddenoutside of the well because the infinite value of V exceeds any possible choice of E.

Recalling the Schrodinger Time Independent Equation, eq. (5.2.9), we substitute V(x)and in the region (−a/2, a/2), we get

−~2

2md2ψ

dx2 = Eψ (5.2.10)

This differential is of the formd2ψ

dx2 + k2ψ = 0 (5.2.11)

where

k =

√2mE~2 (5.2.12)

We recognize that possible solutions will be of the form

cos kx and sin kx

As the particle is confined in the region 0 < x < a, we say

ψ(x) =

A cos kx + B sin kx for 0 < x < a0 for |x| > a

We have known boundary conditions for our square well.

ψ(0) = ψ(a) = 0 (5.2.13)


DRAFT

Schrodinger Equation 37It shows that

⇒ A cos 0 + B sin 0 = 0∴ A = 0 (5.2.14)

We are now left with

B sin ka = 0ka = 0;π; 2π; 3π; · · ·

(5.2.15)

While mathematically, n can be zero, that would mean there would be no wave function,so we ignore this result and say

kn =nπa

for n = 1, 2, 3, · · ·

Substituting this result into eq. (5.2.12) gives

kn =nπa

=

√2mEn

~(5.2.16)

Solving for En gives

En =n2π2~2

2ma2 (5.2.17)

We cna now solve for B by normalizing the function∫ a

0|B|2 sin2 kxdx = |A|2

a2

= 1

So |A|2 =2a

(5.2.18)

So we can write the wave function as

ψn(x) =

√2a

sin(nπx

a

)(5.2.19)

5.2.2 Harmonic Oscillators

Classically, the harmonic oscillator has a potential energy of

V(x) =12

kx2 (5.2.20)

So the force experienced by this particle is

F = −dVdx

= −kx (5.2.21)


DRAFT

38 Quantum Mechanicswhere k is the spring constant. The equation of motion can be summed us as

md2xdt2 = −kx (5.2.22)

And the solution of this equation is

x(t) = A cos(ω0t + φ

)(5.2.23)

where the angular frequency, ω0 is

ω0 =

√km

(5.2.24)

The Quantum Mechanical description on the harmonic oscillator is based on the eigen-function solutions of the time-independent Schrodinger’s equation. By taking V(x)from eq. (5.2.20) we substitute into eq. (5.2.9) to get

d2ψ

dx2 =2m~2

(k2

x2− E

)ψ =

mk~2

(x2−

2Ek

)ψ

With some manipulation, we get

~√

mk

d2ψ

dx2 =

√mk~

x2−

2E~

√mk

ψThis step allows us to to keep some of constants out of the way, thus giving us

ξ2 =

√mk~

x2 (5.2.25)

and λ =2E~

√mk

=2E~ω0

(5.2.26)

This leads to the more compact

d2ψ

dξ2 =(ξ2− λ

)ψ (5.2.27)

where the eigenfunction ψ will be a function of ξ. λ assumes an eigenvalue anaglaousto E.

From eq. (5.2.25), we see that the maximum value can be determined to be

ξ2max =

√mk~

A2 (5.2.28)

Using the classical connection between A and E, allows us to say

ξ2max =

√mk~

2Ek

= λ (5.2.29)


DRAFT

Schrodinger Equation 39From eq. (5.2.27), we see that in a quantum mechanical oscillator, there are non-vanishing solutions in the forbidden regions, unlike in our classical case.

A solution to eq. (5.2.27) isψ(ξ) = e−ξ

2/2 (5.2.30)

where

dψdξ

= −ξe−ξ2/2

anddψ

dξ2 = ξ2e−xi2/2− e−ξ

2/2 =(ξ2− 1

)e−ξ

2/2

This gives is a special solution for λ where

λ0 = 1 (5.2.31)

Thus eq. (5.2.26) gives the energy eigenvalue to be

E0 =~ω0

2λ0 =

~ω0

2(5.2.32)

The eigenfunction e−ξ2/2 corresponds to a normalized stationary-state wave function

Ψ0(x, t) =

(mkπ2~2

) 18

e−√

mk x2/2~e−iE0t/~ (5.2.33)

This solution of eq. (5.2.27) produces the smallest possibel result of λ and E. Hence,Ψ0 and E0 represents the ground state of the oscillator. and the quantity ~ω0/2 is thezero-point energy of the system.

5.2.3 Finite Square Well

For the Finite Square Well, we have a potential region where

V(x) =

−V0 for −a ≤ x ≤ a0 for |x| > a

We have three regions

Region I: x < −a In this region, The potential, V = 0, so Schrodinger’s Equation be-comes

−~2

2md2ψ

dx2 = Eψ

⇒d2ψ

dx2 = κ2ψ

where κ =

√−2mE~


DRAFT

40 Quantum MechanicsThis gives us solutions that are

ψ(x) = A exp(−κx) + B exp(κx)

As x → ∞, the exp(−κx) term goes to ∞; it blows up and is not a physicallyrealizable function. So we can drop it to get

ψ(x) = Beκx for x < −a (5.2.34)

Region II: −a < x < a In this region, our potential is V(x) = V0. Substitutin this intothe Schrodinger’s Equation, eq. (5.2.9), gives

−~2

2md2ψ

dx2 − V0ψ = Eψ

ord2ψ

dx2 = −l2ψ

where l ≡

√2m (E + V0)~

(5.2.35)

We notice that E > −V0, making l real and positive. Thus our general solutionbecomes

ψ(x) = C sin(lx) + D cos(lx) for −a < x < a (5.2.36)

Region III: x > a Again this Region is similar to Region III, where the potential, V = 0.This leaves us with the general solution

ψ(x) = F exp(−κx) + G exp(κx)

As x→∞, the second term goes to infinity and we get

ψ(x) = Fe−κx for x > a (5.2.37)

This gives us

ψ(x) =

Beκx for x < aD cos(lx) for 0 < x < aFe−κx for x > a

(5.2.38)

5.2.4 Hydrogenic Atoms

c

5.3 Spin

3


DRAFT

Angular Momentum 415.4 Angular Momentum

4

5.5 Wave Funtion Symmetry

5

5.6 Elementary Perturbation Theory

6


DRAFT

42 Quantum Mechanics


DRAFTChapter 6Atomic Physics

6.1 Properties of Electrons

1

6.2 Bohr Model

To understand the Bohr Model of the Hydrogen atom, we will take advantage of ourknowlegde of the wavelike properties of matter. As we are building on a classicalmodel of the atom with a modern concept of matter, our derivation is considered to be‘semi-classical’. In this model we have an electron of mass, me, and charge, −e, orbitinga proton. The cetripetal force is equal to the Coulomb Force. Thus

14πε0

e2

r2 =mev2

r(6.2.1)

The Total Energy is the sum of the potential and kinetic energies, so

E = K + U =p2

2me− | f race24πε0r (6.2.2)

We can further reduce this equation by subsituting the value of momentum, which wefind to be

p2

2me=

12

mev2 =e2

8πε0r(6.2.3)

Substituting this into eq. (6.2.2), we get

E =e2

8πε0r−

e2

4πε0r= −

e2

8πε0r(6.2.4)

At this point our classical description must end. An accelerated charged particle, likeone moving in circular motion, radiates energy. So our atome here will radiate energy

DRAFT

44 Atomic Physicsand our electron will spiral into the nucleus and disappear. To solve this conundrum,Bohr made two assumptions.

1. The classical circular orbits are replaced by stationary states. These stationarystates take discreet values.

2. The energy of these stationary states are determined by their angular momentumwhich must take on quantized values of ~.

L = n~ (6.2.5)

We can find the angular momentum of a circular orbit.

L = m3vr (6.2.6)

From eq. (6.2.1) we find v and by substitution, we find L.

L = e√

m3r4πε0

(6.2.7)

Solving for r, gives

r =L2

mee2/4πε0(6.2.8)

We apply the condition from eq. (6.2.5)

rn =n2~2

mee2/4πε0= n2a0 (6.2.9)

where a0 is the Bohr radius.a0 = 0.53 × 10−10 m (6.2.10)

Having discreet values for the allowed radii means that we will also have discreetvalues for energy. Replacing our value of rn into eq. (6.2.4), we get

En = −me

2n2

(e2

4πε0~

)= −

13.6n2 eV (6.2.11)

6.3 Energy Quantization

3

6.4 Atomic Structure

4


DRAFT

Atomic Spectra 456.5 Atomic Spectra

6.5.1 Rydberg’s Equation

1λ

= RH

( 1n′2−

1n2

)(6.5.1)

where RH is the Rydberg constant.

For the Balmer Series, n′ = 2, which determines the optical wavelengths. For n′ = 3, weget the infrared or Paschen series. The fundamental n′ = 1 series falls in the ultravioletregion and is known as the Lyman series.

6.6 Selection Rules

6

6.7 Black Body Radiation

6.7.1 Plank Formula

u( f ,T) =8π~c3

f 3

eh f/kT − 1(6.7.1)

6.7.2 Stefan-Boltzmann Formula

P(T) = σT4 (6.7.2)

6.7.3 Wein’s Displacement Law

λmaxT = 2.9 × 10−3 m.K (6.7.3)

6.7.4 Classical and Quantum Aspects of the Plank Equation

Rayleigh’s Equation

u( f ,T) =8π f 2

c3 kT (6.7.4)


DRAFT

46 Atomic PhysicsWe can get this equation from Plank’s Equation, eq. (6.7.1). This equation is a classicalone and does not contain Plank’s constant in it. For this case we will look at thesituation where h f < kT. In this case, we make the approximation

ex' 1 + x (6.7.5)

Thus the demonimator in eq. (6.7.1) becomes

eh f/kT− 1 ' 1 +

h fkT− 1 =

h fkT

(6.7.6)

Thus eq. (6.7.1) takes the approximate form

u( f ,T) '8πhc3 f 3 kT

h f=

8π f 2

c3 kT (6.7.7)

As we can see this equation is devoid of Plank’s constant and thus independent ofquantum effects.

Quantum

At large frequencies, where h f > kT, quantum effects become apparent. We canestimate that

eh f/kT− 1 ' eh f/kT (6.7.8)

Thus eq. (6.7.1) becomes

u( f ,T) '8πhc3 f 3e−h f/kT (6.7.9)

6.8 X-Rays

6.8.1 Bragg Condition

2d sinθ = mλ (6.8.1)

for constructive interference off parallel planes of a crystal with lattics spacing, d.

6.8.2 The Compton Effect

The Compton Effect deals with the scattering of monochromatic X-Rays by atomictargets and the observation that the wavelength of the scattered X-ray is greater thanthe incident radiation. The photon energy is given by

E = hυ =hcλ

(6.8.2)


DRAFT

Atoms in Electric and Magnetic Fields 47The photon has an associated momentum

E = pc (6.8.3)

⇒ p =E

c=

hυc

=hλ

(6.8.4)

The Relativistic Energy for the electron is

E2 = p2c2 + m2e c4 (6.8.5)

wherep − p′ = P (6.8.6)

Squaring eq. (6.8.6) givesp2− 2p · p′ + p′2 = P2 (6.8.7)

Recall that E = pc and E ′ = cp′, we have

c2p2− 2c2p · p′ + c2p′2 = c2P2

E 2− 2E E ′ cosθ + E ′2 = E2

−m2e c4 (6.8.8)

Conservation of Energy leads to

E + mec2 = E ′ + E (6.8.9)

Solving

E − E ′ = E −mec2

E 2− 2E E ′ + E ′ = E2

− 2Emec2 + m2e c4 (6.8.10)

2E E ′ − 2E E ′ cosθ = 2Emec2− 2m2

e c4 (6.8.11)

Solving leads to

∆λ = λ′ − λ =h

mec(1 − cosθ) (6.8.12)

where λc = hmec

is the Compton Wavelength.

λc =h

mec= 2.427 × 10−12m (6.8.13)

6.9 Atoms in Electric and Magnetic Fields

6.9.1 The Cyclotron Frequency

A test charge, q, with velocity v enters a uniform magnetic field, B. The force acting onthe charge will be perpendicular to v such that

FB = qv × B (6.9.1)


DRAFT

48 Atomic Physicsor more simply FB = qvB. As this traces a circular path, from Newton’s Second Law,we see that

FB =mv2

R= qvB (6.9.2)

Solving for R, we getR =

mvqB

(6.9.3)

We also see that

f =qB

2πm(6.9.4)

The frequency is depends on the charge, q, the magnetic field strength, B and the massof the charged particle, m.

6.9.2 Zeeman Effect

The Zeeman effect was the splitting of spectral lines in a static magnetic field. This issimilar to the Stark Effect which was the splitting in the presence in a magnetic field.

In the Zeeman experiment, a sodium flame was placed in a magnetic field and itsspectrum observed. In the presence of the field, a spectral line of frequency, υ0 wassplit into three components, υ0 − δυ, υ0 and υ0 + δυ. A classical analysis of this effectallows for the identification of the basic parameters of the interacting system.

The application of a constant magnetic field, B, allows for a direction in space in whichthe electron motion can be referred. The motion of an electron can be attributed to asimple harmonic motion under a binding force −kr, where the frequency is

υ0 =1

2π

√k

me(6.9.5)

The magnetic field subjects the electron to an additional Lorentz Force, −ev × B. Thisproduces two different values for the angular velocity.

v = 2πrυ

The cetripetal force becomesmev2

r= 4π2υ2rme

Thus the certipetal force is

4π2υ2rme = 2πυreB + kr for clockwise motion

4π2υ2rme = −2πυreB + kr for counterclockwise motion

We use eq. (6.9.5), to emiminate k, to get

υ2−

eB2πme

υ − υ0 = 0 (Clockwise)

υ2 +eB

2πmeυ − υ0 = 0 (Counterclockwise)


DRAFT

Atoms in Electric and Magnetic Fields 49As we have assumed a small Lorentz force, we can say that the linear terms in υ aresmall comapred to υ0. Solving the above quadratic equations leads to

υ = υ0 +eB

4πmefor clockwise motion (6.9.6)

υ = υ0 −eB

4πmefor counterclockwise motion (6.9.7)

We note that the frequency shift is of the form

δυ =eB

4πme(6.9.8)

If we view the source along the direction of B, we will observe the light to have twopolarizations, a closckwise circular polarization of υ0 + δυ and a counterclosckwisecircular polarization of υ0 − δυ.

6.9.3 Franck-Hertz Experiment

The Franck-Hertz experiment, performed in 1914 by J. Franck and G. L. Hertz, mea-sured the colisional excitation of atoms. Their experiement studied the current ofelectrons in a tub of mercury vapour which revealed an abrupt change in the currentat certain critical values of the applied voltage.1 They interpreted this observation asevidence of a threshold for inelastic scattering in the colissions of electrons in mer-cury atoms.The bahavior of the current was an indication that electrons could losea discreet amount of energy and excite mercury atoms in their passage through themercury vapour. These observations constituted a direct and decisive confirmation ofthe existence os quantized energy levels in atoms.

1Put drawing of Franck-Hertz Setup


DRAFT

50 Atomic Physics


DRAFTChapter 7Special Relativity

7.1 Introductory Concepts

7.1.1 Postulates of Special Relativity

1. The laws of Physics are the same in all inertial frames.

2. The speed of light is the same in all inertial frames.

We can define

γ =1√

1 − u2

c2

(7.1.1)

7.2 Time Dilation

∆t = γ∆t′ (7.2.1)

where ∆t′ is the time measured at rest relative to the observer, ∆t is the time measuredin motion relative to the observer.

7.3 Length Contraction

L =L′

γ(7.3.1)

where L′ is the length of an object observed at rest relative to the observer and L is thelength of the object moving at a speed u relative to the observer.

DRAFT

52 Special Relativity7.4 Simultaneity

4

7.5 Energy and Momentum

7.5.1 Relativistic Momentum & Energy

In relativistic mechanics, to be conserved, momentum and energy are defined as

Relativistic Momentum

p = γmv (7.5.1)

Relativistic Energy

E = γmc2 (7.5.2)

7.5.2 Lorentz Transformations (Momentum & Energy)

p′x = γ(px − β

Ec

)(7.5.3)

p′y = py (7.5.4)

p′z = pz (7.5.5)E′

c= γ

(Ec− βpx

)(7.5.6)

7.5.3 Relativistic Kinetic Energy

K = E −mc2 (7.5.7)

= mc2

1√1 − v2

c2

− 1

(7.5.8)

= mc2 (γ − 1)

(7.5.9)


DRAFT

Four-Vectors and Lorentz Transformation 537.5.4 Relativistic Dynamics (Collisions)

∆P′x = γ(∆Px − β

∆Ec

)(7.5.10)

∆P′y = ∆Py (7.5.11)

∆P′z = ∆Pz (7.5.12)∆E′

c= γ

(∆Ec− β∆Px

)(7.5.13)

7.6 Four-Vectors and Lorentz Transformation

We can represent an event in S with the column matrix, s,

s =

xyz

ict

(7.6.1)

A different Lorents frame, S′, corresponds to another set of space time axes so that

s′ =

x′

y′

z′

ict′

(7.6.2)

The Lorentz Transformation is related by the matrixx′

y′

z′

ict′

=

γ 0 0 iγβ0 1 0 00 0 1 0−iγβ 0 0 γ

xyz

ict

(7.6.3)

We can express the equation in the form

s′ = L s (7.6.4)

The matrix L contains all the information needed to relate position four–vectors forany given event as observed in the two Lorentz frames S and S′. If we evaluate

sTs =[

x y z ict]

xyz

ict

= x2 + y2 + z2− c2t2 (7.6.5)

Similarly we can show that

s′Ts′ = x′2 + y′2 + z′2 − c2t′2 (7.6.6)


DRAFT

54 Special RelativityWe can take any collection of four physical quantities to be four vector provided thatthey transform to another Lorentz frame. Thus we have

b =

bx

by

bz

ibt

(7.6.7)

this can be transformed into a set of quantities of b′ in another frame S′ such that itsatisfies the transformation

b′ = L b (7.6.8)

Looking at the momentum-Energy four vector, we have

p =

px

py

pz

iE/c

(7.6.9)

Applying the same transformation rule, we have

p′ = L p (7.6.10)

We can also get a Lorentz-invariation relation between momentum and energy suchthat

p′Tp′ = pTp (7.6.11)

The resulting equality gives

p′2x + p′2y + p′2z −E′2

c2 = p2x + p2

y + p2z −

E2

c2 (7.6.12)

7.7 Velocity Addition

v′ =v − u1 − uv

c2

(7.7.1)

7.8 Relativistic Doppler Formula

υ = υ0

√c + uc − u

let r =

√c − uc + u

(7.8.1)

We have

υreceding = rυ0 red-shift (Source Receding) (7.8.2)

υapproaching =υ0

rblue-shift (Source Approaching) (7.8.3)


DRAFT

Lorentz Transformations 557.9 Lorentz Transformations

Given two reference frames S(x, y, z, t) and S′(x′, y′, z′, t′), where the S′-frame is movingin the x-direction, we have,

x′ = γ (x − ut) x = (x′ − ut′) (7.9.1)y′ = y y = y′ (7.9.2)z′ = y y′ = y (7.9.3)

t′ = γ(t −

uc2 x

)t = γ

(t′ +

uc2 x′

)(7.9.4)

7.10 Space-Time Interval

(∆S)2 = (∆x)2 +(∆y

)2+ (∆z)2

− c2 (∆t)2 (7.10.1)

Space-Time Intervals may be categorized into three types depending on their separa-tion. They are

Time-like Interval

c2∆t2 > ∆r2 (7.10.2)

∆S2 > 0 (7.10.3)

When two events are separated by a time-like interval, there is a cause-effectrelationship between the two events.

Light-like Interval

c2∆t2 = ∆r2 (7.10.4)

S2 = 0 (7.10.5)

Space-like Intervals

c2∆t2 < ∆r2 (7.10.6)∆S < 0 (7.10.7)


DRAFT

56 Special Relativity


DRAFTChapter 8Laboratory Methods

8.1 Data and Error Analysis

8.1.1 Addition and Subtraction

x = a + b − c (8.1.1)

The Error in x is

(δx)2 = (δa)2 + (δb)2 + (δc)2 (8.1.2)

8.1.2 Multiplication and Division

x =a × b

c(8.1.3)

The error in x is (δxx

)2

=(δaa

)2

+

(δbb

)2

+(δcc

)2

(8.1.4)

8.1.3 Exponent - (No Error in b)

x = ab (8.1.5)

The Error in x isδxx

= b(δaa

)(8.1.6)

DRAFT

58 Laboratory Methods8.1.4 Logarithms

Base e

x = ln a (8.1.7)

We find the error in x by taking the derivative on both sides, so

δx =d ln a

da· δa

=1a· δa

=δaa

(8.1.8)

Base 10

x = log10 a (8.1.9)

The Error in x can be derived as such

δx =d(log a)

daδa

=ln a

ln 10

daδa

=1

ln 10δaa

= 0.434δaa

(8.1.10)

8.1.5 Antilogs

Base e

x = ea (8.1.11)

We take the natural log on both sides.

ln x = a ln e = a (8.1.12)

Applaying the same general method, we see

d ln xdx

δx = δa

⇒δxx

= δa (8.1.13)


DRAFT

Instrumentation 59Base 10

x = 10a (8.1.14)

We follow the same general procedure as above to get

log x = a log 10log x

dxδx = δa

1ln 10

d ln adx

δx = δa

δxx

= ln 10δa (8.1.15)

8.2 Instrumentation

2

8.3 Radiation Detection

3

8.4 Counting Statistics

Let’s assume that for a particular experiment, we are making countung measurementsfor a radioactive source. In this experiment, we recored N counts in time T. Thecounting rate for this trial is R = N/T. This rate should be close to the average rate, R.The standard deviation or the uncertainty of our count is a simply called the

√N rule.

Soσ =√

N (8.4.1)

Thus we can report our results as

Number of counts = N ±√

N (8.4.2)

We can find the count rate by dividing by T, so

R =NT±

√N

T(8.4.3)


DRAFT

60 Laboratory MethodsThe fractional uncertainty of our count is δN

N . We can relate this in terms of the countrate.

δRR

=δNTNT

=δNN

=

√N

N

=1N

(8.4.4)

We see that our uncertainty decreases as we take more counts, as to be expected.

8.5 Interaction of Charged Particles with Matter

5

8.6 Lasers and Optical Interferometers

6

8.7 Dimensional Analysis

Dimensional Analysis is used to understand physical situations involving a mis ofdifferent types of physical quantities. The dimensions of a physical quantity areassociated with combinations of mass, length, time, electric charge, and temperature,represented by symbols M, L, T, Q, and θ, respectively, each raised to rational powers.

8.8 Fundamental Applications of Probability and Statis-tics

8


DRAFTChapter 9Sample Test

9.1 Period of Pendulum on Moon

The period of the pendulum, T, is

T = 2π

√`g

(9.1.1)

where ` is the length of the pendulium string. The relationship between the weight ofan object on the Earth, We, and the Moon, Wm, is

Wm =We

6(9.1.2)

From eq. (9.1.2), we can determine the acceleration due to gravity on the Moon and onthe Earth; we use the same subscript notation as above.

gm =ge

6(9.1.3)

On Earth, the period of the pendulum, Te, is one second. From eq. (9.1.1), the equationfor the pendulum’s period on Earth is

Te = 2π

√`ge

= 1 s (9.1.4)

and similarly for the moon, the period becomes

Tm = 2π

√`

gm(9.1.5)

DRAFT

62 Sample TestSubstituting eq. (9.1.3) into eq. (9.1.5) gives

Tm = 2π

√`

gm

=√

6 Te =√

6 s

Answer: (D)

9.2 Work done by springs in series

Hooke’s Law tells us that the extension on a spring is proportional to the force applied.

F = −kx (9.2.1)

Springs in series follow the same rule for capacitors, see ??. The spring constants arerelated to each other by

k1 =13

k2 (9.2.2)

The springs are massless so we can assume that the weight is transmitted evenly alongboth springs, thus from Hooke’s Law the extension is

F1 = −k1x1 = F2 = −k2x2 (9.2.3)

where k1 and k2 are the spring constants for the springs S1 and S2 respectively. Thuswe see

k1

k2=

x2

x1=

13

(9.2.4)

The work done in stretching a spring or its potential energy is

W =12

kx2 (9.2.5)

Thus

W1

W2=

12

k1x21

12

k2x22

=k1

k2·

(x1

x2

)2

= 3 (9.2.6)

Answer: (D)


DRAFT

Central Forces I 639.3 Central Forces I

We are given a central force field where

V(r) = −kr

(9.3.1)

The Angular Momentum of an object is

L = r × p (9.3.2)

and the torque is definedτ = r × F = r × p (9.3.3)

From eqs. (9.3.2) and (9.3.3), we see that

τ =dLdt

(9.3.4)

We see that if τ = 0, then L is constant and therefore conserved. This can occur if r = 0,F = 0 or F ∝ r.

From 9.3.1, we can determine the force acting on the object since

F = −dVdr

=kr2 (9.3.5)

As our force is a central force, the force acts in the direction of our radius vector. Thusthe torque becomes

τ = r × F = rF cos 0= 0

We see that this means that our angular momentum is constant.

L = constant (9.3.6)

A constant angular momentum means that r and v remain unchanged. The totalmechanical energy is the sum of the kinetic and potential energies.

E = KE + PE

=12

mv2 +kr2 (9.3.7)

Both the kinetic and potential energies will remain constant and thus the total mechan-ical energy is also conserved.

Answer: (C)


DRAFT

64 Sample Test9.4 Central Forces II

The motion of particle is governed by its potential energy and for a conservative,central force the potential energy is

V(r) = −kr

(9.4.1)

we have shown in the above question that the angular momentum, L, is conserved.We can define three types of orbits given k and E.

Orbit k Total Energy

Ellipse k > 0 E < 0Parabola k > 0 E = 0Hyperbola k > 0 or k < 0 E > 0

Table 9.4.1: Table of Orbits

From, table 9.4.1, we expect the orbit to be elliptical; this eliminates answers (C), (D)and (E).

For an elliptical orbit, the total energy is

E = −k2a

(9.4.2)

where a is the length of the semimajor axis. In the case of a circular orbit of radius, r,eq. (9.4.2) becomes

E = −k2r

(9.4.3)

Recalling eq. (9.3.1), we see

E =12

V(r) = −K (9.4.4)

This is the minimum energy the system can have resulting in a circular orbit.

Answer: (A)


DRAFT

Electric Potential I 659.5 Electric Potential I

P2

P1

b

b

r2

r1

+z

Figure 9.5.1: Diagram of Uniformly Charged Circular Loop

The Electric Potential of a charged ring is given by1

V =1

4πε0

Q√

R2 + z2(9.5.1)

where R is the radius of our ring and x is the distance from the central axis of the ring.In our case, the radius of our ring is R = b.

The potential at P1, where z = b is

V1 =1

4πε0

Q√

b2 + b2=

14πε0

Q

b√

2(9.5.2)

The potential at P2, where z = 2b is

V2 =1

4πε0

Q√b2 + (2b)2

=1

4πε0

Q

b√

5(9.5.3)

Dividing eq. (9.5.3) by eq. (9.5.2) gives us

V2

V1=

√25

(9.5.4)

Answer: (D)1Add Derivation


DRAFT

66 Sample Test9.6 Electric Potential II

The potential energy, U(r), of a charge, q, placed in a potential, V(r), is[1]

U(r) = qV(r) (9.6.1)

The work done in moving our charge through this electrical field is

W = U2 −U1

= qV2 − qV1

= q (V2 − V1) (9.6.2)

Answer: (E)

9.7 Faraday’s Law and Electrostatics

We notice that our answers are in the form of differential equations and this leads usto think of the differential form of Maxwell’s equations[2]. The electrostatics form ofMaxwell’s Equations are[3]

Gauss’s Law∇ · E =

ρ

ε0(9.7.1)

Maxwell-Faraday Equation∇ × E = 0 (9.7.2)

Gauss’ Law for Magnetism∇ · B = 0 (9.7.3)

Ampere’s Law∇ × B = µ0J (9.7.4)

Comparing our answers, we notice that eq. (9.7.2) corresponds to Answer: (C) .

Answer: (C)

9.8 AC Circuits: RL Circuits

An inductor’s characteristics is opposite to that of a capacitor. While a capacitor storesenergy in the electric field, essentially a potential difference between its plates, aninductor stores energy in the magnetic field, which is produced by a current passingthrough the coil. Thus inductors oppose changes in currents while a capacitor opposeschanges in voltages. A fully discharged inductor will initially act as an open circuit


DRAFT

AC Circuits: RL Circuits 67with the maximum voltage, V, across its terminals. Over time, the current increasesand the potential difference across the inductor decreases exponentially to a minimum,essentially behaving as a short circuit. As we do not expect this circuit to oscillate, thisleaves us with choices (A) and (B). At t = 0, we expect the voltage across the resistorto be VR = 0 and increase exponentially. We choose (A).

V

I

LA

R

B

Figure 9.8.1: Schematic of Inductance-Resistance Circuit

We can see from the above schematic,

V = VL + VR (9.8.1)

where VL and VR are the voltages across the inductor and resistor respectively. Thiscan be written as a first order differential equation

V = LdIdt

+RL

I (9.8.2)

Dividing by L leavesVL

=dIdt

+RL

I (9.8.3)

The solution to eq. (9.8.3) leaves

I =

∫VL

exp(Rt

L

)dt + k

exp(Rt

L

)=

VR

+ k exp(−

RtL

)(9.8.4)

Multiplying eq. (9.8.4) by R gives us the voltage across the resistor

VR = V + kR exp(−

RtL

)(9.8.5)

at t = 0, VR = 0

0 = V + kR

∴ k = −VR

(9.8.6)


DRAFT

68 Sample TestSubstituting k into eq. (9.8.5) gives us

VR(t) = V[1 − exp

(−

RtL

)](9.8.7)

where τ = L/R is the time constant. Where τ = 2 s

0

1

2

3

4

5

6

7

0 5 10 15 20

Volt

age/

V

Time/s

V(x)

Figure 9.8.2: Potential Drop across Resistor in a Inductor-Resistance Circuit

Answer: (A)

9.9 AC Circuits: Underdamped RLC Circuits

When a harmonic oscillator is underdamed, it not only approaches zero much morequickly than a critically damped oscillator but it also oscillates about that zero. A quickexamination of our choices means we can eliminate all but choices (C) and (E). Thechoice we make takes some knowledge and analysis.


DRAFT

AC Circuits: Underdamped RLC Circuits 69

V

L

R

BC

A

Figure 9.9.1: LRC Oscillator Circuit

The voltages in the above circuit can be written

V(t) = VL + VR + VC

= LdI(t)

dt+ RI(t) +

1C

q(t) (9.9.1)

which can be written as a second order differential equation

Ld2q(t)

dt2 + Rdq(t)

dt+

1C

q(t) = V(t) (9.9.2)

or asd2q(t)

dt2 + γdq(t)

dt+ ω2

0q(t) = V(t) (9.9.3)

This can be solved by finding the solutions for nonhomogenoeus second order lineardifferential equations. For any driving force, we solve for the undriven case,

d2zdt2 + γ

dzdt

+ ω20 = 0 (9.9.4)

where for the underdamped case, the general solution is of the form

z(t) = A exp(−αt) sin(βt + δ) (9.9.5)

where

α = −γ

2(9.9.6)

β =

√4ω2

0 − γ2

2(9.9.7)

In the case of a step response,

V(t) =

1 t > 00 t < 0

(9.9.8)


DRAFT

70 Sample TestThe solution becomes

q(t) = 1 − exp(−

R2L

t) sin

√ω2

0 −

( R2L

)2

t + δ

sin δ

(9.9.9)

where the phase constant, δ, is

cos δ =R

2ω20L

(9.9.10)

where ω0 ≈ 3.162 kHz and γ = 5 ΩH−1

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 0.5 1 1.5 2

Volt

age/

V

Time/s

V(x)

Figure 9.9.2: Forced Damped Harmonic Oscillations

So in the case of our forced underdamped oscillator, we would expect the voltage toraise, overshoot a little and oscillate while slowly decaying. This resembles choice (C).

Answer: (C)

9.10 Bohr Model of Hydrogen Atom

Bohr’s theory of the atom proposed the existence of stationary states by blending newquantum mechanics ideas with old classical mechanics concepts.

Bohr’s model of the hydrogen atom starts as a system of two bodies bound together bythe Coulomb attraction. The charges and mass of one particle is −e and m and +Ze andM for the other. In the case of our hydrogen atom system, Z = 1, M = mp and m = me.


DRAFT

Bohr Model of Hydrogen Atom 71We will be taking into account the motion of both particles in our analysis. Normally,we expect the mass, M, to be stationary where M/m→∞ and as the proton-to-electronmass ratio is very large, we approximate to this limiting condition.

mp

me= 1836 (9.10.1)

This effect is detectable and should be retained as a small correction. As the effects canbe incorporated with little difficulty, we shall do so.2

We take the center of mass to be at th origin,3. We can reduce our two body systemto an equivalent one body description in terms of a single vector given by a relativecoordinate.

r = r1 − r2 (9.10.2)

As the center of mass is located at the origin

mr1 + Mr2 = 0 (9.10.3)

Solving eqs. (9.10.2) and (9.10.3) gives us

r1 =M

M + mr (9.10.4)

andr2 = −

mM + m

r (9.10.5)

Differentiating eqs. (9.10.4) and (9.10.5), gives us the corresponding velocities

v1 =M

M + mv (9.10.6)

andv2 = −

mM + m

v (9.10.7)

where the relative velocity is

v =drdt

(9.10.8)

The total energy can be found from eqs. (9.10.6) and (9.10.7)

K =12

mv21 +

12

Mv22

=12

mMM + m

v2 (9.10.9)

We can reduce this to the equivalent of a one body system where the reduced massfactor is

µ =mM

M + m(9.10.10)

2Put figure here3as seen in diagram


DRAFT

72 Sample TestEquation (9.10.9) becomes

K =12µv2 (9.10.11)

As the Coulomb Force is a central force, the total angular momentum of the systemwill be constant.

L = mv1r1 + Mv2r2

= m( MM + m

)2

vr + M( mM + m

)2

vr

= µvr (9.10.12)

The centripetal force of the system is equal to the Coulomb force, thus

F =mv2

1

r1=

Mv22

r2=µv2

r=

14πε0

Ze2

r2 (9.10.13)

The potential energy of the system comes from the Coulomb potential energy

V = −1

4πε0

Ze2

r(9.10.14)

The total energy of the system can be found by adding eqs. (9.10.11) and (9.10.14)

E = K + V

=12µv2−

14πε0

Ze2

r(9.10.15)

Substituting eq. (9.10.13) into eq. (9.10.15) gives

E = −12

Ze2

4πε0r(9.10.16)

We expect the total energy of the system to be negative as it is a bound system.

We have, so far, adhered to the principles of classical mechanics up to this point.Beyond this point, we must introduce quantum mechanical concepts. To producethe stationary states he was seeking, Bohr introduced the hypothesis that the angularmomentum is quantized.

L = n~ (9.10.17)

Equating this with eq. (9.10.12) and substitution into eq. (9.10.13) gives discrete valuesfor the orbital radius.

rn =4πε0

Ze2

n2~2

µ(9.10.18)

We can rewrite the above equation

rn =n2me

µZa0 (9.10.19)


DRAFT

Nuclear Sizes 73where

a0 =4πε0~2

e2me(9.10.20)

is the Bohr Radius.

As the orbital radii is discrete we expect the various orbital energies to also be discrete.Substitution of eq. (9.10.18) into eq. (9.10.16) gives

En = −12

Ze2

4πε0rn

= −Z2

n2

(e2

4πε0

)2 µ

2~2 (9.10.21)

or

En = −Z2

n2

µ

meE0 (9.10.22)

where

E0 =

(e2

4πε0

)2 me

2~2 = 13.6 eV (9.10.23)

We see that this analysis eliminates all but one answer.

Answer: (A)

9.11 Nuclear Sizes

We know from electron scattering experiments, the nucleus is roughly spherical anduniform density4. The Fermi model gives us an expression

r = r0A13 (9.11.1)

where r0 = 1.2 × 10−15 m and A is the mass number. In the case of hydrogen, we recallthe Bohr radius to be a0 = 0.0592 nm.

So in the case of hydrogen, A = 1,

r = r0 = 1.2 × 10−15 m (9.11.2)

Thus

r0

a0=

1.2 × 10−15

0.0592 × 10−9

= 2.02 × 10−5 (9.11.3)

Answer: (B)4Add diagram of nuclear and atomic sizes here


DRAFT

74 Sample Test9.12 Ionization of Lithium

The ionization energy of an electron is the energy to kick it off from its present state toinfinity. It can be expressed as

Eionization = E∞ − En

=Z2

n2

µ

meE0 (9.12.1)

where E0 = 13.6 eV and µ is the reduced mass where

µ

me=

MM + me

(9.12.2)

In the case of atoms, the above ratio is close to one and hence we can ignore it for thiscase.

Lithium has an atomic number, Z = 3 so its electron structure is5

1s2, 2s1 (9.12.3)

So the total ionization energy will be the total energy needed to completly remove eachelectron. This turns out to be

E =

[32

12 +32

12 +32

22

]13.6 eV

≈ 20 × 13.6 eV= 272.0 eV (9.12.4)

Answer: (C)

9.13 Electron Diffraction

We recall that in optics, one of the criteria for diffraction is a monochromatic wave. Weexpect the electron beam to also have wavelike effects. The de Broglie relations showthat the wavelength is inversely proportional to the momentum of a particle and thatthe frequency is directly proportional to the particle’s kinetic energy.

λ =hp

and E = h f

Thus, for electron we expect the beam to be monoenergetic.

Answer: (B)5Draw Lithium atom and its electrons


DRAFT

Effects of Temperature on Speed of Sound 759.14 Effects of Temperature on Speed of Sound

The speed of sound is determined by its Bulk Modulus and its density

v =

√Bρ

(9.14.1)

in the case of gases, the Bulk Modulus can be expressed

B = γP (9.14.2)

where γ is the adiabatic ratio and P is the pressure of the gas.

For an ideal gasPV = nRT (9.14.3)

Substituting eqs. (9.14.2) and (9.14.3) into eq. (9.14.1) gives us

v =

√nγRT

M(9.14.4)

So we see thatv ∝ T

12 (9.14.5)

Answer: (B)

9.15 Polarized Waves

Given the equations

y = y0 sin (ωt − kx)

z = z0 sin(ωt − kx − φ

)(9.15.1)

For our wave to be plane-polarized, the two waves, y and z must be in phase i.e. wheny is a maximum so too is z. So

sin (ωt − kx) = sin(ωt − kx − φ

)∴ φ = 0 (9.15.2)

A plane polarized wave will occur when φ = 0. We can also look at the waves belowand see that they are not in phase, except for Choice (E).


DRAFT

76 Sample Test

x

y

z y(x)z(x)

(a) φ =√

2

x

y

z y(x)z(x)

(b) φ = 3π/2

x

y

z y(x)z(x)

(c) φ = π/2

x

y

z y(x)z(x)

(d) φ = π/4

Figure 9.15.1: Waves that are not plane-polarized

x

y

z y(x)z(x)

Figure 9.15.2: φ = 0

Answer: (E)

9.16 Electron in symmetric Potential Wells I

As our potential is symmetric about the V-axis, then we will expect our wave functionto also be symmetric about the V-axis.

Answer: (E)


DRAFT

Electron in symmetric Potential Wells II 779.17 Electron in symmetric Potential Wells II

If the electrons do not interact, we can ignore Pauli’s Exclusion Principle. As a resultthey will not have spatially antisymmetric states but will have the same spatial wavefunctions.

Answer: (B)

9.18 Relativistic Collisions I

The Relativistic Momentum equation is

p =mv√

1 −(v

c

)2(9.18.1)

given that p = mc/2,

mc2

=mv√

1 −(v

c

)2

⇒c2

=v√

1 −(v

c

)2

∴ v =c√

5(9.18.2)

Answer: (D)

9.19 Relativistic Collisions II

Momentum is conserved. So in the horizontal direction,

p = 2p f cos 30 (9.19.1)

Solving this shows

p f =mc

2√

3(9.19.2)

Answer: (B)


DRAFT

78 Sample Test9.20 Thermodynamic Cycles I

We have a three stage cyclic process where

A(P1,V1,T1) , B(P2,V2,T2) , C(P3,V3,T3) (9.20.1)

Adiabatic Expansion, A −→ BP1Vγ

1 = P2Vγ2 (9.20.2)

Given that V2 = 2V1, we haveP2 = 2−γP1 (9.20.3)

andT2 = 21−γT1 (9.20.4)

Isochoric Expansion, B −→ C We have

P2

T2=

P3

T3(9.20.5)

where T3 = T1, we have

P3 = 2γ−1P2 (9.20.6)

=12

P1 (9.20.7)

This becomes

A(P1,V1,T1) , B(2−γP1, 2V1, 21−γT1) , C(P1

2, 2V1,T1) (9.20.8)

On a PV-graph, we see that this makes a clockwise cycle, indicating that positive workis done by the gas on the environment.

Answer: (A)

9.21 Thermodynamic Cycles II

We recall Calusius’s Therorem ∮dQT

= 0 (9.21.1)

for a reversible cycle, the change in entropy is zero.

Answer: (C)


DRAFT

Distribution of Molecular Speeds 799.22 Distribution of Molecular Speeds

The distribution of speeds of molecules follows the Maxwell-Boltzmann distribution,which has the form

f (v) = 4π[ M2πRT

] 32

v2 exp[−

Mv2

2RT

](9.22.1)

where R is the gas constant and M is the molar mass of the gas. The speed distributionfor noble gases at T = 298.15 K looks like

0

0.0005

0.001

0.0015

0.002

0.0025

0.003

0.0035

0.004

0.0045

0 500 1000 1500 2000

Max

wel

lSpe

edD

istr

ibut

ion,

f(v)

Molecular Speed, m/s

HeliumNeon

ArgonXenon

Figure 9.22.1: Maxwell-Boltzmann Speed Distribution of Nobel Gases

Answer: (D)

9.23 Temperature Measurements

All the thermometers won’t be able to survive that high a temperature except for theoptical pyrometer. Of course, a little knowledge always helps.

Optical Pyrometer Optical pyrometers work by using the human eye to match thebrightness of a hot object to a calibrated lamp filament inside the instrument.

Carbon Resistor These thermometers are typically used for very low temperaturesand not high ones. One of their main advantages is their sensitivity, their resis-tance increases exponentially to decreasing temperature and are not affected bymagnetic fields.


DRAFT

80 Sample TestGas-Bulb Thermometer May also be known as the constant volume gas thermometer.

Doubtful the glass bulb will survive such high temperatures.

Mercury Thermometer The boiling point of mercury is about 360 °C. This thermome-ter will be vaporized before you even had a chance to think about getting atemperature reading.

Thermocouple Thermocouples are made by joining two different metals together andproduces a voltage that is related to the temperature difference. They are typicallyused in industry to measure high temperatures, usually in the order ∼ 1800 °C.The metals would most likely start melting above these temperatures.

Even if we knew nothing about any of the above thermometers, we could have stilltake a stab at it. We should probably guess that at that high a temperature we won’twant to make physical contact with what we are measuring. The only one that can dothis is the optical pyrometer.

Answer: (A)

9.24 Counting Statistics

NOT FINISHED

Answer: (D)

9.25 Thermal & Electrical Conductivity

A metal is a lattice of atoms, each with a shell of electrons. This forms a positive ioniclattice where the outer electrons are free to dissociate from the parent atoms and movefreely through the lattic as a ‘sea’ of electrons. When a potential difference is appliedacross the metal, the electrons drift from one end of the conductor to the other underthe influence of the electric field. It is this free moving electron ‘sea’ that makes a metalan electrical conductor.

These free moving electrons are also efficient at transferring thermal energy for thesame reason. Thermal and electrical conductivity in metals are closely related to eachother as outlined in the Wiedemann-Franz Law.

κσ

= LT (9.25.1)

where the Lorenz number, L = 2.44 × 10−8 WΩK−1 and κ and σ are the thermal andelectrical conductivities respectively. This corelation does not apply to non-metals dueto the increased role of phonon carriers.

Answer: (E)


DRAFT

Nonconservation of Parity in Weak Interactions 819.26 Nonconservation of Parity in Weak Interactions

Of the four interations, electromagnetism, strong, weak and gravity, parity is conservedin all except for the weak interaction. To examine violations of these interactions wemust look at the helicity of our particles and see whether they are “left-handed” or“right-handed”.

A particle is said to be “right-handed” if the direction of its spin is the same as thedirection as its motion. It is “left-handed” if the directions of spin and motion areopposite to each other. Thus the helicity of a particle is the projection of the spin vectoronto the momentum vector where left is negative and right is positive.6

h ≡S · p∣∣∣S · p∣∣∣ (9.26.1)

Particles are not typically characterized as being “left-handed” or “right-handed”. Forexample, an electron could have both its spin and momentum pointing in the samedirection to the right and hence be classified as “right-handed”. But from the referenceframe of someone travelling faster than the speed of the electron, would see the electrontravelling to the left and hence conclude the electron is “left-handed”.

Neutrinos, on the other hand, travel very close to the speed of light and it would be verydifficult to accelerate to a point where one would be able to change the “handedness”of the neutrino. Thus, we say that the neutrino has an intrinsic parity, all of thembeing left-handed. Anti-neutrinos on the other hand are all right-handed. This causesweak interactions, neutrino emitting ones in particular, to violate the conservation ofparity law.7

For the pion decay,π+→ µ+ + υµ+ (9.26.2)

It is very difficult to detect and measure the helicity of the neutrino directly but we canmeasure it indirectly through the above decay and hence demonstrate nonconservationof parity.

If the pion is at rest and has spin-0, the anti-muon and neutrino will come out in oppositedirections.8 In the figure below, the anti-muons are observed with their z-componentof angular momentum given by mµ = −1

2 . Angular momentum conversation thenimplies mυ = +1

2 for the neutrino.

It is very difficult if not impossible to detect neutrinos in a typical laboratory settingbut we can detect muons and measure their helicity.

Choice (A) The Q-value is the kinetic Energy released in the decay of the particle atrest. Parity deals with mirror symmetry violations and not energy.

6Draw Helicity Diagrams7Add section explaining parity8Draw Diagram Here


DRAFT

82 Sample TestChoice (B) We measure for violations of parity conservation by measuring the longi-

tudinal polarization of the anti-muon. We choose this answer.

Choice (C) The pion has spin-0 and is stationary. So it won’t be polarized. Measuringthis gives us no information on our decay products.

Choice (D) The angular correlation would be difficult as neutrinos are difficult todetect.

Choice (E) Parity deals with spatial assymetry and has nothing to do with time. Wecan eliminate this choice.

Answer: (B)

9.27 Moment of Inertia

The moment of inertia is

I =

∫r2 dm (9.27.1)

In the case of a hoop about its center axis,

I = MR2 (9.27.2)

From eq. (9.27.1), we see that the moment of inertia deals with how the mass isdistributed along its axis.9 We see that

A is equivalent to A

Figure 9.27.1: Hoop and S-shaped wire

Thus, see fig. 9.27.1, the moment of inertia of our S-shaped wire can be found from ahoop with its axis or rotation at its radius. This can be calculated by using the ParallelAxis Theorem

I = ICM + Md2 (9.27.3)

where d2 is the distance from the center of mass. This becomes

I = MR2 + MR2 = 2MR2 (9.27.4)

Answer: (E)9The moment of inertia of a 1 kg mass at a distance 1 m from the axis of rotation is the same as a hoop

with the same mass rotating about its central axis.


DRAFT

Lorentz Force Law I 839.28 Lorentz Force Law I

I+ + + + + + + + + + + + + + + + +

Bu

FE

FB

Figure 9.28.1: Charged particle moving parallel to a positively charged current carryingwire

The force on the charged particle is determined by the Lorentz Force Equation

F = e [E + u × B] (9.28.1)

where FE = eE and FB = e(u×B). For our charged particle to travel parallel to our wire,FE = FB.10

E =λ`

2πε0r(9.28.2)

and the magnetic field can be determined from Ampere’s Law∮B · ds = µ0Ienclosed (9.28.3)

In the case of our wire

B =µ0I2πr

(9.28.4)

Plugging eqs. (9.28.2) and (9.28.4) into eq. (9.28.1) gives

F = e[λ`

2πε0r+ u

µ0I2πr

]= 0 (9.28.5)

For the particle to be undeflected, FE + FB = 0

FE + FB = 0λ`

2πε0r− u

µ0I2πr

= 0 (9.28.6)

Now we can go about eliminating choices.

10Add derivation in a section


DRAFT

84 Sample TestDoubling charge per unit length We see from eq. (9.28.6), halving the current, I, and

doubling the linear charge density, λ, will not allow the particle to continueundeflected.

2λ`2πε0r

− uµ0I/22πr

= 2FE −FB

2, 0 (9.28.7)

Doubling the charge on the particle We see from, eq. (9.28.6) that the charge on theparticle, e, has no effect on the particle’s trajectory. We would be left with

λ`2πε0r

− uµ0I/22πr

= FE −FB

2, 0 (9.28.8)

Doubling both the charge per unit length on the wire and the charge on the particleAs shown above, the particle’s charge has no effect on the trajectory. This leavesus with the charge per unit length, λ and as we have seen before, this will changethe particle’s trajectory, see eq. (9.28.7).

Doubling the speed of the particle If we double the particle’s speed we will get

λ`2πε0r

− (2u)µ0I/22πr

=λ`

2πε0r− u

µ0I2πr

∴ FE = FB

This is our answer

Introducing an additional magnetic field parallel to the wire Recalling eq. (9.28.1), theforce due to the magnetic field is a cross product between the velocity and thefield. A charged particle moving in the same direction as the field will experienceno magnetic force.

FB = e [u × B]= uB sin 0= 0 (9.28.9)

Answer: (D)

9.29 Lorentz Force Law II

As we can see from eq. (9.28.5), the forces due to the electric and magnetic fields areequal.

F = e[λ`

2πε0r+ u

µ0I2πr

]= 0 (9.29.1)


DRAFT

Nuclear Angular Moment 85If we move our charged particle a distance 2r from the wire with a speed nu, eq. (9.28.5)becomes

e[

λ`2πε0(2r)

+ uµ0I

2π(2r)r

]= e

(12

) [λ`

2πε0r+ nu

µ0I2πr

]= e

(12

)[FE − nFB]

= 0 (9.29.2)

Thus

[FE − nFB] = FE − FB = 0⇒ n = 1 (9.29.3)

The speed of the particle is u.

Answer: (C)

9.30 Nuclear Angular Moment

NOT FINISHED

Answer: (B)

9.31 Potential Step Barrier

xO

Ψ(x)

VE

Region I Region II

Figure 9.31.1: Wavefunction of particle through a potential step barrier

The point, x = 0, divides the region into two regions, Region I, where classical motionis allowed and, Region II, where classical motion is forbidden. The barrier potential is

V(x) =

0 for x < 0V for x > 0

(9.31.1)


DRAFT

86 Sample TestFor, x < 0, the eigenfunction satisfies

d2ψ

dx2 = −k21ψ (9.31.2)

wherek2

1 =2m~2 E (9.31.3)

The general form of the eigenfunction is

ψI = Aeik1x + Be−ik1x (9.31.4)

For, x > 0, the eigenfunction satisfies

d2ψ

dx2 = k22ψ (9.31.5)

wherek2

2 =2m~2

(V − E) (9.31.6)

The general form of the eigenfunction becomes

ψII = Ce−k2x + Dek2x (9.31.7)

As x → ∞, the ek2x term blows up. So to allow eq. (9.31.7) to make any physical sensewe set D = 0, thus

ψII = Ce−k2x = Ce−αx (9.31.8)

We can continue solving for A, B and C but for the purposes of the question we seefrom eq. (9.31.8) that α is both real and positive.

Answer: (C)


DRAFTAppendix AConstants & Important Equations

A.1 Constants

Constant Symbol ValueSpeed of light in a vacuum c 2.99 × 108 m/sGravitational Constant G 6.67 × 10−11 m3/kg.s2

Rest Mass of the electron me 9.11 × 10−31 kgAvogadro’s Number NA 6.02 × 1023 mol-1

Universal Gas Constant R 8.31 J/mol.KBoltzmann’s Constant k 1.38 × 10−23 J/KElectron charge e 1.60 × 10−9CPermitivitty of Free Space ε0 8.85 × 10−12 C2/N.m2

Permeability of Free Space µ0 4π × 10−7 T.m/AAthmospheric Pressure 1 atm 1.0 × 105 M/m2

Bohr Radius a0 0.529 × 10−10 m

Table A.1.1: Something

A.2 Vector Identities

A.2.1 Triple Products

A · (B × C) = B · (C ×A) = C · (A × B) (A.2.1)A × (B × C) = B (A · C) − C (A · B) (A.2.2)

DRAFT

88 Constants & Important EquationsA.2.2 Product Rules

∇(

f g)

= f(∇g

)+ g

(∇ f

)(A.2.3)

∇ (A · B) = A × (∇ × B) + B × (∇ ×A) + (A · ∇) B + (B · ∇) A (A.2.4)∇ ·

(f A

)= f (∇ ·A) + A ·

(∇ f

)(A.2.5)

∇ · (A × B) = B · (∇ ×A) −A · (∇ × B) (A.2.6)∇ ×

(f A

)= f (∇ ×A) −A ×

(∇ f

)(A.2.7)

∇ × (A × B) = (B · ∇) A − (A · ∇) B + A (∇ · B) − B (∇ ·A) (A.2.8)

A.2.3 Second Derivatives

∇ · (∇ ×A) = 0 (A.2.9)∇ ×

(∇ f

)= 0 (A.2.10)

∇ × (∇ ×A) = ∇ (∇ ·A) − ∇2A (A.2.11)

A.3 Commutators

A.3.1 Lie-algebra Relations

[A,A] = 0 (A.3.1)[A,B] = −[B,A] (A.3.2)[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0 (A.3.3)

A.3.2 Canonical Commutator

[x, p] = i~ (A.3.4)

A.3.3 Kronecker Delta Function

δmn =

0 if m , n;1 if m = n;

For a wave function ∫ψm(x)∗ψn(x)dx = δmn (A.3.5)


DRAFT

Linear Algebra 89A.4 Linear Algebra

A.4.1 Vectors

Vector Addition

The sum of two vectors is another vector

|α〉 + |β〉 = |γ〉 (A.4.1)

Commutative|α〉 + |β〉 = |β〉 + |α〉 (A.4.2)

Associative|α〉 +

(|β〉 + |γ〉

)=

(|α〉 + |β〉

)+ |γ〉 (A.4.3)

Zero Vector|α〉 + |0〉 = |α〉 (A.4.4)

Inverse Vector|α〉 + | − α〉 = |0〉 (A.4.5)


DRAFT

90 Constants & Important Equations


DRAFTBibliography

[1] Stephen Gasiorowicz Paul M. Fishbane and Stephen T. Thornton. Physics for Scien-tists and Engineers with Modern Physics, chapter 24.2, page 687. Prentice Hall, thirdedition, 2005.

[2] Wikipedia. Maxwell’s equations — wikipedia, the free encyclopedia, 2009. [Online;accessed 21-April-2009].

[3] David J. Griffiths. Introduction to Electrodyanmics, chapter 5.3.4, page 232. PrenticeHall, third edition, 1999.

DRAFTIndex

RLC CircuitsSample Test Q09, 68

RL CircuitsSample Test Q08, 66

Angular Momentum, see Rotational Mo-tion

Bohr ModelHydrogen Model, 43

Celestial Mechanics, 10Circular Orbits, 11Escape Speed, 10Kepler’s Laws, 11Newton’s Law of Gravitation, 10Orbits, 11Potential Energy, 10

Central ForcesSample Test Q03, 63Sample Test Q04, 64

Circular Orbits, see Celestial MechanicsCommutators, 88

Canonical Commutators, 88Kronecker Delta Function, 88Lie-algebra Relations, 88

Compton Effect, 46Counting Statistics, 59

Doppler Effect, 8

Electric PotentialSample Test Q05, 65Work

Sample Test Q06, 66

Faraday’s LawSample Test Q07, 66

Franck-Hertz Experiment, 49

Gravitation, see Celestial Mechanics

Kepler’s Laws, see Celestial MechanicsKronecker Delta Function, 88

Linear Algebra, 89Vectors, 89

Maxwell’s EquationsSample Test Q07, 66

Moment of Inertia, see Rotational Motion

Newton’s Law of Gravitation, see CelestialMechanics

OscillationsUnderdamped

Sample Test Q09, 68Oscillatory Motion, 4

Coupled Harmonic Oscillators, 6Damped Motion, 5Kinetic Energy, 4Potential Energy, 5Simple Harmonic Motion Equation, 4Small Oscillations, 5Total Energy, 4

Parallel Axis Theorem, see Rotational Mo-tion

PendulumSimple

Sample Test Q01, 61

Rolling Kinetic Energy, see Rotational Mo-tion

Rotational Kinetic Energy, see RotationalMotion

DRAFT

Index 93Rotational Motion, 8

Angular Momentum, 9Moment of Inertia, 8Parallel Axis Theorem, 9Rolling Kinetic Energy, 9Rotational Kinetic Energy, 8Torque, 9

SpringsWork

Sample Test Q02, 62Subject, 30System of Particles, 10

Torque, see Rotational Motion

Vector Identities, 87Product Rules, 88Second Derivatives, 88Triple Products, 87


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