Chapter 39: Introduction to Quantum...

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Chapter 39:Introduction to Quantum Physics

You can’t see atoms with light, but you can with electrons because …

Blackbody Radiation and Planck’s Hypothesis

Thermal vibrations causes charged particles to accelerate, emitting radiation. Blackbody radiation depend only on temperature and not on the material.

Blackbody Radiation: Classical Treatment

/ 2L n

Consider a cubic cavity with length L on each side, inside which electromagnetic radiation due to thermal vibration of the walls are resonating in standing waves. Standing wave requires L to be exact multiples (n: integer) of the half-wavelength.

Allowed frequency is (for each axis) : / (2 )nc L

For a 3D standing wave with nx, ny, and nz, its frequency is

2 22 2 2 2 2

2 2( )4 4x y zc cn n n nL L

Each standing wave is an independent oscillator and, according to equipartition of energy principle, would have an average energy (per degree of freedom) of kT. We are therefore in position to estimate the energy spent in electromagnetic radiation due to thermal motion classically. We just need to find out how many oscillators are allowed in each frequency range!

22 24cd n dnL

Rayleigh-Jeans Law and the Ultraviolet Catastrophe

2 ck TI T

In the “phase space”, each non-negative (nx,ny,nz) represents an allowed point, which contains two allowed states for the two possible polarizations of light. In a shell of volume in phase space with radius N and thickness dN, the energy is

24 (1/ 8) (2)BdE k T n dn

3 48 BdE dk TL

Energy density (energy/volume)

Intensity (power/area)

Rayleigh-Jean Law

Actual Experimental Observations

1. The total power of the emitted radiation increases with temperature.

4P AeT2. The peak of the wavelength

distribution shifts to shorter wavelengths as the temperature increases.

3max 2.898 10 m KT

Stefan-Boltzmann

Planck’s Assumption: Quantization of Energy

nE nhf

Max Planck 346.626 10 J sh

Planck’s Model

3 48 BdE dk TL

Allowed oscillator modes unchanged from classical model. Average energy per oscillator is no longer kT!

3 4 8 PlanckdE dL

nh k T

nnh k T

nh k Th k T

/ ( )/ ( )

1 2/ ( )( ) 1

h k Tnh k T

h k Tn B

dZ h enh ed k T e

/ ( ) 1Bh k T

3 /( )5

8( 1)Bhc k T

dE hcL d e

Or, equivalently,

3 /( )3

8( 1)Bh k T

dE hL d c e

Planck’s Model

346.626 10 J sh

21hc k T

hcI Te

Planck’s Constant determined by fitting.

Planck’s Model (Prob. 49)

Total Energy Per Volume:5 43

3 /( ) 330

8 ( )815( )( 1)B

Bh k T

k TE h dL hcc e

Total Intensity:5 4

2 ( )15

Bk TI T

Stefan-Boltzmann Law

8 2 45.67 10 /W m K

multiply by c/4

Exercise Prob. 51

Derive this3

max 2.898 10 m KT

3 / ( )5

8( 1)Bhc k T

dE hcL d e

By differentiating

Infrared Radiation and the Ear Thermometer

normal

38 C 273 C 1.003237 C 273 C

4fever

normal

38 C 273 C37 C 273 C

The Photoelectric Effect

Stopping Potential

The Photoelectric Effectand Energy Conservation

0 0 0i s

max sK e V

The Photoelectric Effectand the Particle Theory of Light

Photoelectric effect should occur at any frequency

No electrons emitted for frequency below fc

Light intensity increases K of photoelectrons increases

Kmax independent of light intensity

No relationship between photoelectron energy and light frequency

Light frequency increases Kmax of photoelectrons increases

Photoelectrons need time to absorb incident radiation before escaping from

the metal

Electrons are emitted from the surface almost instantaneously even at low light

intensities.

Wave Theory Prediction Observation

Einstein’s Model for the Photoelectric Effect

EREK U T

0 0K hf

maxK hf

IDEA: A beam of light is made of particles called photons. Each photon transfers all of its energy (hf) to one electron of the metal

Explanations of Observations

Observation Explanation

No electrons emitted for frequency below fc

Photoelectrons created by absorbing single photon photon energy

Kmax independent of light intensity Kmax = hf – , no dependence on intensity

Light frequency increases Kmax of photoelectrons increases

Kmax linear in f

Electrons are emitted from the surface almost instantaneously even at low light

intensities.

Light energy is in packets; no time needed for electron to acquire energy to

escape metal

Work Function and Cutoff Frequency

c c hcf h

1240 eV nmhc

maxK hf

Application of the Photoelectric Effect:The Photomultiplier Tube

39.41 The Photoelectric Effect for Sodium

Use the graph to find (a) the work function of sodium, (b) the ratio h/e, and (c) the cutoff wavelength.

39.47 Photoelectric Effect

A light source emitting radiation at frequency 7.00x1014 Hz is incapable of ejecting photoelectrons from a certain metal. In an attempt to use this source to eject photoelectrons, the source is given a velocity toward the metal. (a) When the speed of the light source is equal to 0.280c, photoelectrons just begin to be ejected from the metal. What is the work function of the metal?

Application: Photoemission Spectroscopy

Powerful Light Source

Photoemission Spectroscopy

The Compton Effect

Arthur Holly Compton

Compton EquationAssume negligible kinetic energy for the electron initially. Conservation of energy and momentum.

2 2 4 2 2e e e

hc hcm c m c p c

ep p p

2 2 22 ep p p p p

22 2 4 2 2e e ecp cp m c m c p c

2 22 ( )e ep p m c p p p

2 (1 cos ) 2 ( )ep p m c p p

(1 cos )e

The Compton Wavelength and the Compton Shift Equation

0 1 cose

C 0.002 43 nme

The Nature of Electromagnetic Waves

Is light a wave or particle? Or both?

The Wave Properties of Particles

E hf hpc c

h hp mu

Louis de BroglieA speculation!

Principle of Complementarity

The wave and particle models of either matter or radiation complement each other

The Davisson–Germer Experiment

Davisson and Germer

Electron Diffraction

Low-Energy Electron Diffraction (LEED)~ 20 – 200 eV

LEED: Surface Periodicity

Si(111) 7x7

Si(100) 2x1

NiSi2(111)

“Surface Reconstruction”

Reflective High-Energy Electron Diffraction (RHEED) ~ 5 – 50 keV

RHEED Intensity Oscillations!How many atomic layers have you grown so far?

Example 39.5: Wavelengths for Microscopic and Macroscopic Objects

(A) Calculate the de Broglie wavelength for an electron (me = 9.11 1031 kg) moving at 1.00 107 m/s.

1131 7

6.626 10 J s 7.27 10 m9.11 10 kg 1.00 10 m/se

(B) A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength?

6.626 10 J s 3.3 10 m50 10 kg 40 m/se

The Electron Microscope

A New Model: The Quantum Particle

1 1 1 2 2 2cos and cosy A k x t y A k x t

1 2 1 1 2 2cos cosy y y A k x t A k x t

cos cos 2cos /2 cos /2a b a b a b

1 1 2 2 1 1 2 2

1 2 1 2

2 cos cos2 2

2 cos cos2 2 2 2

k x t k x t k x t k x ty A

k kky A x t x t

A New Model: The Quantum Particle

1 2 1 22 cos cos2 2 2 2

k kky A x t x t

Phase and Group Speeds

phasevk

/2coefficient of time variable coefficient of space variable /2g

tvx k k

cosy A kx t

1 2 1 22 cos cos2 2 2 2

k kky A x t x t

ddvdk d k

Phase and Group Speeds

22h f hf E

2h hk p

d dEvd k dp

2 2pE mum

2 2gdE d pv p udp dp m m

ddvdk d k

The Double-Slit Experiment Revisited

sind m

The Double-Slit Experiment Revisited

The Uncertainty Principle

If a measurement of the position of a particle is made with uncertainty x and a simultaneous

measurement of its x component of momentum is made with uncertainty px, the product of the two

uncertainties can never be smaller than /2:

The Uncertainty Principle

Time-Energy Uncertainty

Wave Packets Out Of Plane Waves

For a wave packet with wave vectors confined to a region k about a point in reciprocal space, the spatial spread of the wave packet is of the order

1|| kr

To show this we construct a specific wave packet at t=0 in one dimension, using Gaussian distribution

The Fourier transform of this wave packet also has the Gaussian form:

])(exp[)(4

)(exp2

)()( 002

xkkikkxxk

)](exp[)(

)(2)( 002

4 2 xxkixxx

Integrate by parts to get

Spreads In Real- and k-Space

For this “best case scenario”, 2

Gaussian Distribution

( )1( ) exp42

x xf x

Time Evolution of a Gaussian 1D Wave Packet

At t=0, a Gaussian wave packet centered at x0 is expressed as

)exp(])(exp[)(4

)(exp2

kxixkkikkxdkx

)](exp[)(

)(exp)(

2)0,( 002

204 2 xxki

Since each k component is an eigenstate of the free electron Hamiltonian, with the eigenvalue , the time dependence of the mixed state as specified by the above initial boundary condition can be written down, in the absence of external field, as

exp(])(exp[)(4

)(exp2

)(),(2

mtkikxixkkikkxdkxtx

mk 2/22

Motion Of 1D Wave Packets

Carrying out the integration in k-space, we get

Once we constructed a wave packet, the motion of the wave packet under the influence of external disturbances can be regarded as how electrons would react. This allows us to think of electrons with somewhat defined r and k coordinates. The dynamics of electrons (between collisions) can then be predicted to follow the time dependence of states in both of these coordinates.

)](exp[)}({

])/[(exp)}({

2),( 002

2004 2 xxki

txmtkxx

exp[)(arg2

exp)}({

])/[()}0({

2exp20

2 mtkitxi

txmtkxx

ieRarg

0 /w packetv k m demo

Electrons and Holes

Group Material Electron me Hole mh

IVSi (300K) 1.08 0.56

Ge 0.55 0.37

III-VGaAs 0.067 0.45

InSb 0.013 0.6

II-VIZnO 0.29 1.21

ZnSe 0.17 1.44

Prob. 48

A woman on a ladder (H: initial height) drops small pellets toward a point target on the floor. Show that, according to the uncertainty principle, the average miss distance must be at least

1/41/22 2f

Example 39.6:Locating an Electron

The speed of an electron is measured to be 5.00 103 m/s to an accuracy of 0.003 00%. Find the minimum uncertainty in determining the position of this electron.

x x xp m v mfv

1.055 10 J s2 2 9.11 10 kg 0.0000300 5.00 10 m/s

3.86 10 m 0.386 mm

Example 39.6:Locating an Electron

Atoms have quantized energy levels similar to those of Planck’s oscillators, although the energy levels of an atom are usually not evenly spaced. When an atom makes a transition between states separated in energy by E, energy is emitted in the form of a photon of frequency f = E/h. Although an excited atom can radiate at any time from t = 0 to t = , the average time interval after excitation during which an atom radiates is called the lifetime . If = 1.0 108 s, use the uncertainty principle to compute the line width f produced by this finite lifetime.

EE hf E h f fh

1 1 /2 1 12 2 4 4

hfh t h t t

1 8.0 10 Hz4 1.0 10 s

Chapter 39: Introduction to Quantum...

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