chapter 39: introduction to quantum...
TRANSCRIPT
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Chapter 39:Introduction to Quantum Physics
You can’t see atoms with light, but you can with electrons because …
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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.
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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!
2
22 24cd n dnL
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Rayleigh-Jeans Law and the Ultraviolet Catastrophe
B4
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
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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
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Planck’s Assumption: Quantization of Energy
nE nhf
E hf
Max Planck 346.626 10 J sh
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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
/ ( )
/ ( )
B
B
nh k T
nnh k T
n
nh e
e
/ ( )
/ ( )
11
B
B
nh k Th k T
nZ e
e
/ ( )/ ( )
1 2/ ( )( ) 1
BB
B
h k Tnh k T
h k Tn B
dZ h enh ed k T e
/ ( ) 1Bh k T
he
3 /( )5
8( 1)Bhc k T
dE hcL d e
Or, equivalently,
3
3 /( )3
8( 1)Bh k T
dE hL d c e
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Planck’s Model
346.626 10 J sh
B
2
/5
21hc k T
hcI Te
Planck’s Constant determined by fitting.
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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
43 2
2 ( )15
Bk TI T
h c
Stefan-Boltzmann Law
8 2 45.67 10 /W m K
multiply by c/4
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Exercise Prob. 51
Derive this3
max 2.898 10 m KT
3 / ( )5
8( 1)Bhc k T
dE hcL d e
By differentiating
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Infrared Radiation and the Ear Thermometer
fever
normal
38 C 273 C 1.003237 C 273 C
TT
4fever
normal
38 C 273 C37 C 273 C
1.013
PP
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The Photoelectric Effect
Stopping Potential
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The Photoelectric Effectand Energy Conservation
0EK U
0 0 0i s
s i
K e V
e V K
max sK e V
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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
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Einstein’s Model for the Photoelectric Effect
EREK U T
max
max
0 0K hf
K 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
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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
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Work Function and Cutoff Frequency
/cc
c c hcf h
1240 eV nmhc
maxK hf
cf h
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Application of the Photoelectric Effect:The Photomultiplier Tube
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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.
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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?
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Application: Photoemission Spectroscopy
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Powerful Light Source
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Photoemission Spectroscopy
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The Compton Effect
Arthur Holly Compton
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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
hm c
hp
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The Compton Wavelength and the Compton Shift Equation
0 1 cose
hm c
C 0.002 43 nme
hm c
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The Nature of Electromagnetic Waves
Is light a wave or particle? Or both?
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The Wave Properties of Particles
E hf hpc c
h hp mu
Efh
Louis de BroglieA speculation!
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Principle of Complementarity
The wave and particle models of either matter or radiation complement each other
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The Davisson–Germer Experiment
Davisson and Germer
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Electron Diffraction
Low-Energy Electron Diffraction (LEED)~ 20 – 200 eV
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LEED: Surface Periodicity
Si(111) 7x7
Si(100) 2x1
NiSi2(111)
“Surface Reconstruction”
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Electron Diffraction
Reflective High-Energy Electron Diffraction (RHEED) ~ 5 – 50 keV
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Electron Diffraction
RHEED Intensity Oscillations!How many atomic layers have you grown so far?
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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.
34
1131 7
6.626 10 J s 7.27 10 m9.11 10 kg 1.00 10 m/se
hm u
(B) A rock of mass 50 g is thrown with a speed of 40 m/s. What is its de Broglie wavelength?
3434
3
6.626 10 J s 3.3 10 m50 10 kg 40 m/se
hm u
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The Electron Microscope
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The Electron Microscope
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A New Model: The Quantum Particle
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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
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A New Model: The Quantum Particle
1 2 1 22 cos cos2 2 2 2
k kky A x t x t
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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
gdvdk
g
ddvdk d k
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Phase and Group Speeds
22h f hf E
2
2h hk p
g
d dEvd k dp
221
2 2pE mum
2 1 2
2 2gdE d pv p udp dp m m
g
ddvdk d k
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The Double-Slit Experiment Revisited
sind m
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The Double-Slit Experiment Revisited
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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:
2xx p
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The Uncertainty Principle
/p h
2xx p
2E t
Time-Energy Uncertainty
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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
0
24
2
xkkikkxxk
)](exp[)(
)(exp
)(2)( 002
20
4 2 xxkixxx
xx
Integrate by parts to get
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Spreads In Real- and k-Space
For this “best case scenario”, 2
)( kx
Gaussian Distribution
202
( )1( ) exp42
x xf x
2xx
1
k x
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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
)(00
20
24
2
kxixkkikkxdkx
)](exp[)(
)(exp)(
2)0,( 002
204 2 xxki
xxx
xx
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
)2
exp(])(exp[)(4
)(exp2
)(),(2
002
0
24
2
mtkikxixkkikkxdkxtx
mk 2/22
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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
txtx
]2
exp[)(arg2
exp)}({
])/[()}0({
2exp20
2
2
200
2 mtkitxi
txmtkxx
xit
ieRarg
0 /w packetv k m demo
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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
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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
Hxm g
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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
34
31 3
4
1.055 10 J s2 2 9.11 10 kg 0.0000300 5.00 10 m/s
3.86 10 m 0.386 mm
x
xmfv
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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
6
8
1 8.0 10 Hz4 1.0 10 s
f