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General Physics (PHY 2140)

Lecture 27

Modern Physics

Quantum Physics

Blackbody radiation

Planks hypothesis

Chapter 27

http://www.physics.wayne.edu/~apetrov/PHY2140/


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Quantum Physics


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Introduction: Need for Quantum Physics

Problems remained from classical mechanics that relativitydidnt explain:

Blackbody Radiation The electromagnetic radiation emitted by a heated object

Photoelectric Effect Emission of electrons by an illuminated metal

Spectral Lines Emission of sharp spectral lines by gas atoms in an electric discharge tube


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Development of Quantum Physics

1900 to 1930

Development of ideas of quantum mechanics

Also called wave mechanics

Highly successful in explaining the behavior of atoms, molecules, and nuclei

Quantum Mechanics reduces to classical mechanics when applied

to macroscopic systemsInvolved a large number of physicists

Planck introduced basic ideas

Mathematical developments and interpretations involved such people asEinstein, Bohr, Schrdinger, de Broglie, Heisenberg, Born and Dirac


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26.1 Blackbody Radiation

An object at any temperature is known to emit

electromagnetic radiation

Sometimes called thermal radiation

Stefans Law describes the total power radiated

The spectrum of the radiation depends on the temperature andproperties of the object

4P AeT

emissivityStefans constant


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Blackbody

Blackbody is an idealized system that absorbs incident

radiation of all wavelengths

If it is heated to a certain temperature, it starts radiate

electromagnetic waves of all wavelengths

Cavity is a good real-life approximation to a blackbody


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Blackbody Radiation Graph

Experimental data fordistribution of energy inblackbody radiation

As the temperature increases,

the total amount of energyincreases

Shown by the area underthe curve

As the temperature increases,the peak of the distributionshifts to shorter wavelengths


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Wiens Displacement Law

The wavelength of the peak of the blackbody distribution

was found to follow Weins Displacement Law

max T = 0.2898 x 10-2

m K

maxis the wavelength at the curves peak

T is the absolute temperature of the object emitting the radiation


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The Ultraviolet Catastrophe

Classical theory did not matchthe experimental data

At long wavelengths, thematch is good

Rayleigh-Jeans law

At short wavelengths, classical

theory predicted infinite energyAt short wavelengths,experiment showed no energy

This contradiction is called theultraviolet catastrophe

4

2 ckTP


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Plancks Resolution

Planck hypothesized that the blackbody radiation was produced byresonators

Resonators were submicroscopic charged oscillators

The resonators could only have discrete energies

En= n h

n is called the quantum number

is the frequency of vibrationh is Plancks constant, h=6.626 x 10-34 J s

Key point is quantized energy states


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QUICK QUIZ

A photon (quantum of light) is reflected from a mirror. True or false:

(a) Because a photon has a zero mass, it does not exert a force on

the mirror.

(b) Although the photon has energy, it cannot transfer any energy to

the surface because it has zero mass.

(c) The photon carries momentum, and when it reflects off the mirror,it undergoes a change in momentum and exerts a force on

the mirror.

(d) Although the photon carries momentum, its change in momentum

is zero when it reflects from the mirror, so it cannot exert a

force on the mirror.

(a) False

(b) False

(c) True

(d) Falsep Ft


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27.2 Photoelectric Effect

When light is incident on certain metallic surfaces, electrons areemitted from the surface

This is called thephotoelectric effect

The emitted electrons are calledphotoelectrons

The effect was first discovered by HertzThe successful explanation of the effect was given by Einstein in1905

Received Nobel Prize in 1921 for paper on electromagnetic radiation, ofwhich the photoelectric effect was a part


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Photoelectric Effect Schematic

When light strikes E,

photoelectrons are emitted

Electrons collected at C and

passing through the ammeterare a current in the circuit

C is maintained at a positive

potential by the power supply


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Photoelectric Current/Voltage Graph

The current increases with

intensity, but reaches a

saturation level for largeVs

No current flows for voltagesless than or equal toVs, the

stopping potential

The stopping potential is

independent of the radiationintensity


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Features Not Explained by Classical

Physics/Wave Theory

No electrons are emitted if the incident light frequency is

below some cutoff frequencythat is characteristic of the

material being illuminated

The maximum kinetic energy of the photoelectrons is

independent of the light intensity

The maximum kinetic energy of the photoelectrons

increases with increasing light frequency

Electrons are emitted from the surface almost

instantaneously, even at low intensities


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Einsteins Explanation

A tiny packet of light energy, called aphoton, would be emittedwhen a quantized oscillator jumped from one energy level to thenext lower one

Extended Plancks idea of quantization to electromagneticradiation

The photons energy would be E = h

Each photon can give all its energy to an electron in the metal

The maximum kinetic energy of the liberated photoelectron is

KE = h

is called the work function of the metal


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Explanation of Classical Problems

The effect is not observed below a certain cutofffrequency since the photon energy must be greater thanor equal to the work function Without this, electrons are not emitted, regardless of the intensity

of the light

The maximum KE depends only on the frequency andthe work function, not on the intensity

The maximum KE increases with increasing frequency

The effect is instantaneous since there is a one-to-oneinteraction between the photon and the electron


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Verification of Einsteins Theory

Experimental observations of a

linear relationship between KE

and frequency confirm

Einsteins theory

The x-intercept is the cutoff

frequency

cf

h


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27.3 Application: Photocells

Photocells are an application of the photoelectric effect

When light of sufficiently high frequency falls on the cell,a current is produced

Examples Streetlights, garage door openers, elevators


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27.4 The Compton Effect

Compton directed a beam of x-rays toward a block of graphite

He found that the scattered x-rays had a slightly longer wavelengththat the incident x-rays

This means they also had less energy

The amount of energy reduction depended on the angle at which the

x-rays were scatteredThe change in wavelength is called the Compton shift


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Compton Scattering

Compton assumed the

photons acted like other

particles in collisions

Energy and momentum

were conserved

The shift in wavelength is)cos1(

cm

h

e

o

Compton wavelength


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Compton Scattering

The quantity h/mec is called the Compton wavelength

Compton wavelength = 0.00243 nm

Very small compared to visible light

The Compton shift depends on the scattering angle and not onthe wavelength

Experiments confirm the results of Compton scattering and

strongly support the photon concept


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Problem: Compton scattering

A beam of 0.68-nm photons undergoes Compton scattering from free

electrons. What are the energy and momentum of the photons that

emerge at a 45 angle with respect to the incident beam?


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QUICK QUIZ 1

An x-ray photon is scattered by an electron. The frequency of the

scattered photon relative to that of the incident photon (a)

increases, (b) decreases, or (c) remains the same.

(b). Some energy is transferred to the electron in the scattering

process. Therefore, the scattered photon must have less energy

(and hence, lower frequency) than the incident photon.


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QUICK QUIZ 2

A photon of energy E0 strikes a free electron, with the scattered photon

of energy Emoving in the direction opposite that of the incident

photon. In this Compton effect interaction, the resulting kinetic energy

of the electron is (a) E0 , (b) E, (c) E0 E , (d) E0 + E , (e) none of the

above.

(c). Conservation of energy requires the kinetic energy given to

the electron be equal to the difference between the energy of the

incident photon and that of the scattered photon.


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QUICK QUIZ 2

A photon of energy E0 strikes a free electron, with the scattered photon

of energy Emoving in the direction opposite that of the incident

photon. In this Compton effect interaction, the resulting kinetic energy

of the electron is (a) E0 , (b) E, (c) E0 E , (d) E0 + E , (e) none of the

above.

(c). Conservation of energy requires the kinetic energy given to

the electron be equal to the difference between the energy of the

incident photon and that of the scattered photon.


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27.8 Photons and Electromagnetic Waves

Light has a dual nature. It exhibits both wave and particlecharacteristics

Applies to all electromagnetic radiation

The photoelectric effect and Compton scattering offer evidence for

the particle nature of light When light and matter interact, light behaves as if it were composed of

particles

Interference and diffraction offer evidence of the wave nature of light


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28.9 Wave Properties of Particles

In 1924, Louis de Broglie postulated that because

photons have wave and particle characteristics, perhaps

all forms of matter have both properties

Furthermore, the frequency and wavelength of matter

waves can be determined

The de Broglie wavelength of a particle is

The frequency of matter waves is mv

h

h

E


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The Davisson-Germer Experiment

They scattered low-energy electrons from a nickel target

They followed this with extensive diffraction measurements fromvarious materials

The wavelength of the electrons calculated from the diffraction dataagreed with the expected de Broglie wavelength

This confirmed the wave nature of electrons

Other experimenters have confirmed the wave nature of otherparticles


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Review problem: the wavelength of a proton

Calculate the de Broglie wavelength for a proton (mp=1.67x10-27 kg )

moving with a speed of 1.00 x 107 m/s.


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Calculate the de Broglie wavelength for a proton (mp=1.67x10-27 kg ) moving with a

speed of 1.00 x 107 m/s.

Given:

v = 1.0 x 107m/s

Find:

p = ?

Given the velocity and a mass of the proton we can

compute its wavelength

p

p

h

m v

Or numerically,

34

14

31 7

6.63 10

3.97 10

1.67 10 1.00 10ps

J sm

kg m s


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QUICK QUIZ 3

A non-relativistic electron and a non-relativistic proton are movingand have the same de Broglie wavelength. Which of the

following are also the same for the two particles: (a) speed, (b)

kinetic energy, (c) momentum, (d) frequency?

(c). Two particles with the same de Broglie wavelength will have the same

momentump = mv. If the electron and proton have the same momentum, they

cannot have the same speed because of the difference in their masses. For the

same reason, remembering that KE = p2

/2m, they cannot have the same kineticenergy. Because the kinetic energy is the only type of energy an isolated particle can

have, and we have argued that the particles have different energies, Equation 27.15

tells us that the particles do not have the same frequency.


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QUICK QUIZ 2

A non-relativistic electron and a non-relativistic proton are movingand have the same de Broglie wavelength. Which of the

following are also the same for the two particles: (a) speed, (b)

kinetic energy, (c) momentum, (d) frequency?

(c). Two particles with the same de Broglie wavelength will have the same

momentump = mv. If the electron and proton have the same momentum, they

cannot have the same speed because of the difference in their masses. For the

same reason, remembering that KE = p2

/2m, they cannot have the same kineticenergy. Because the kinetic energy is the only type of energy an isolated particle can

have, and we have argued that the particles have different energies, Equation 27.15

tells us that the particles do not have the same frequency.


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The Electron Microscope

The electron microscope dependson the wave characteristics ofelectrons

Microscopes can only resolve details

that are slightly smaller than thewavelength of the radiation used toilluminate the object

The electrons can be accelerated tohigh energies and have smallwavelengths


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27.10 The Wave Function

In 1926 Schrdinger proposed a wave equation that

describes the manner in which matter waves change in

space and time

Schrdingers wave equation is a key element in

quantum mechanics

Schrdingers wave equation is generally solved for the

wave function,

i Ht


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The Wave Function

The wave function depends on the particles position and

the time

The value of ||2 at some location at a given time is

proportional to the probability of finding the particle at

that location at that time


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27.11 The Uncertainty Principle

When measurements are made, the experimenter isalways faced with experimental uncertainties in themeasurements

Classical mechanics offers no fundamental barrier toultimate refinements in measurements

Classical mechanics would allow for measurements witharbitrarily small uncertainties


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The Uncertainty Principle

Quantum mechanics predicts that a barrier to measurementswith ultimately small uncertainties does exist

In 1927 Heisenberg introduced the uncertainty principle

If a measurement of position of a particle is made with precisionxand a simultaneous measurement of linear momentum is made withprecision p, then the product of the two uncertainties can never besmaller than h/4


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The Uncertainty Principle

Mathematically,

It is physically impossible to measure simultaneously the

exact position and the exact linear momentum of aparticle

Another form of the principle deals with energy and time:

4

hpx x

4

htE


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Thought Experiment the Uncertainty

Principle

A thought experiment for viewing an electron with a powerfulmicroscope

In order to see the electron, at least one photon must bounce off it

During this interaction, momentum is transferred from the photon tothe electron

Therefore, the light that allows you to accurately locate the electronchanges the momentum of the electron


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Problem: macroscopic uncertainty

A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an

accuracy of 0.10%, what is the minimum uncertainty in its

position?


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A 50.0-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.10%,

what is the minimum uncertainty in its position?

Given:

v = 30 m/s

dv = 0.10%

m = 50.0 g

Find:

dx = ?

Notice that the ball is non-relativistic. Thus, p = mv,

and uncertainty in measuring momentum is

2 3 2

50.0 10 1.0 10 30 1.5 10

p m v m v v

kg m s kg m s

d

Thus, uncertainty relation implies

24

32

3

6.63 103.5 10

4 4 1.5 10

h J sx mp k g m s


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Problem: Macroscopic measurement

A 0.50-kg block rests on the icy surface of a frozen pond, which we

can assume to be frictionless. If the location of the block is measured

to a precision of 0.50 cm, what speed must the block acquire becauseof the measurement process?

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