Transcript e- e+ Metal e- eee- xp  12 Black Body Radiation: Light in a Box Consider a nearly enclosed container at uniform temperature: u() = energy/ volume /nm •Light.

e-
e+
Metal
e-
eee-
xp  12
Black Body Radiation: Light in a Box
Consider a nearly enclosed container at uniform temperature:
u() = energy/
volume /nm
•Light gets produced in hot
interior
•Bounces around randomly
inside before escaping
•Should be completely random
by the time it comes out
•Pringheim measures spectrum,
1899
The Electron Volt
•Quantum effects almost always involve individual particles
•These particles typically have charges like +e or –e
•The only practical way to push on them is with electric and
magnetic fields
U  qV
•Charge is e and voltage is measured in volts
•Define:
1 eV   e V  1.602 10 19 C   V   1.602 1019 J .
•This unit is commonly used in quantum mechanics
Attempting to Explain the Result
•Statistical Mechanics was a relatively new branch of physics
• It explained some things, like the kinetic theory of gasses
•Physicists tried to explain black body radiation in terms of this theory
•It said that the energy in black body could have any possible wavelength
•There are many, many ways to fit short wavelengths inside the box
•The amount of energy in any given wavelength could be any number from 0 to
infinity
•Therefore, there should be a lot more energy at short
8 k BT
u   
wavelengths then at long
4
•Prediction #1: There will be much more energy
at short wavelengths than long
•Prediction #2: The total amount of energy
will be infinity
• The “ultraviolet catastrophe”
Comparison Theory vs. Experiment:
Theory
Experiment
What went wrong?
•Not truly in thermal equilibrium?
•Possible state counting done wrong?
•The amount of energy in any given wavelength could
be any number from 0 to infinity
u   
8 k BT
4
Max Planck’s strategy (1900):
•Assume energy E must always be an integer multiple of
frequency f times a constant h
•E = nhf, where n = 0, 1, 2, …
•Perform all calculations with h finite
•Take limit h  0 at the end
Why this might help:
•Assume energy E must always be an integer multiple of
frequency f times a constant h
Experiment
•Notice the problem is at short
wavelength = high frequency
•Without this hypothesis,
energy can be small without
being zero
•Now add levels
•It can no longer have a little bit of energy
•For high frequency, it has almost no energy
Energy 
Theory
 Average
Average 
Energy 0 Energy
Planck’s Black Body Law
u   
8 hc
1
 5 exp  hc  kBT   1
Max Planck’s strategy (1900):
•Take limit h  0 at the end
•Except, it fit the curve with finite h!
h  6.626 1034 J  s
h  4.136 1015 eV  s
Planck Constant
E  hf
•Looks like light comes in chunks
with energy E = hf - PHOTONS
Photoelectric Effect: Hertz, 1887
e-
Metal
•Metal is hit by light
•Electrons pop off
•Must exceed minimum frequency
•Depends on the metal
•Brighter light, more electrons
•They start coming off immediately
•Even in low intensity
ee-
eEinstein, 1905
•It takes a minimum amount of energy to free an electron
•Light really comes in chunks of energy hf
•If hf < , the light cannot release any electrons from the metal
•If hf > , the light can liberate electrons
•The energy of each electron released will be K = hf – 
•Will the electron pass through a charged
plate that repels electrons?
•Must have enough energy
•Makes it if:
K  hf  
K  eV
Metal
Photoelectric Effect
e-
hf    eV
+
–
–
V
+
Vmax
Nobel Prize,
1921
f
Sample Problem
When ultraviolet light of wavelength 227 nm strikes calcium metal,
electrons are observed to come off with a kinetic energy of 2.57 eV.
1. What is the work function for calcium?
2. What is the longest wavelength that can free electrons from calcium?
3. If light of wavelength 312 nm were used instead, what would be the
energy of the emitted electrons?
K  hf  
3.00 108 m/s  1.32 1015 s1

f 
227 109 m

c
We need the frequency:
f c
15
15 1

4.136

10
eV

s
1.32

10
s    2.57 eV 
  hf  K 

 5.46 eV  2.57 eV  2.89 eV=
Continued . . .
Sample Problem continued
2. What is the longest wavelength that can free electrons from calcium?
3. If light of wavelength 312 nm were used instead, what would be the
energy of the emitted electrons?
K  hf  
  2.89 eV
f c
0  hf min  
•The lowest frequency comes when K = 0

2.89 eV
14 1
f min  

6.99

10
s
15
h
4.136 10 eV  s
•Now we get the wavelength:
c 3.00 108 m/s
7

4.29

10
m  429 nm
 
14 1
f 6.99 10 s
c 3.00 108 m/s
14 1
f  

9.61

10
s
9
•Need frequency for last part:

312 10 m
K  hf     4.136 1015 eV  s  9.611014 s 1   2.89 eV  1.08 eV
The de Broglie Relation
•We have a formula for the energy of a photon:
E  hf
•Now, steal a formula from special relativity:
E  cp
hf  cp
•Combine it with a formula from electromagnetic waves:
c f
hf  f  p
•And we get the de Broglie relation:
p h
•Photons should have momentum too
The Compton Effect
•By 1920’s X-rays were clearly light waves
•1922 Arthur Compton showed they carried momentum
Photon in
ee-

Atom
Photons carry energy
and momentum, just
like any other particle
e-
•Conservation of momentum and energy
implies a change in wavelength
   
h
1  cos  
mc
Waves and Electrons
Light is . . .
•Initially thought to be waves
•They do things waves do, like diffraction and interference
•Wavelength – frequency relationship
•Planck, Einstein, Compton showed us they behave like particles (photons)
•Energy and momentum comes in chunks
E  hf
•Wave-particle duality: somehow, they behave like both
Electrons are . . .
•They act like particles
•Energy, momentum, etc., come in chunks
•They also behave quantum mechanically
•Is it possible they have wave properties as well?
p h
The de Broglie Hypothesis
E  hf
•Two equations that relate the particle-like and
wave-like properties of light
p h
1924 – Louis de Broglie postulated that these
relationships apply to electrons as well
•Implied that it applies to other particles as well
•de Broglie was able to explain the spectrum of hydrogen using
this hypothesis
The Davisson-Germer Experiment
Same experiment as scattering X-rays, except
•Reflection probability from each layer greater
•Interference effects are weaker
•Momentum/wavelength is shifted inside the
material
•Equation for good scattering identical
e-
2d cos   m
 
d
Quantum effects are weird
•Electron must scatter off of all layers
What Objects are Waves?
•1928: Electrons have both wave and particle properties
•1900: Photons have both wave and particle properties
•1930: Atoms have both wave and particle properties
•1930: Molecules have both wave and particle properties
•Neutrons have both wave and particle properties
•Protons have both wave and particle properties
•Everything has both wave and particle properties
Dr. Carlson has a mass of 82 kg and leaves this room
at a velocity of about 1.3 m/s. What is his wavelength?
34
h
6.626

10
J s
36
h 

6.22

10
m


p mv  82 kg 1.3 m/s 
It’s a Particle, It’s a Wave, No It’s a . . .
•Interference only works with both slits open
•Every photon is going through both slits
•Sometimes, we say we have wave-particle
duality
•It acts sometimes like a particle,
sometimes like a wave
•It is a quantum object – something completely
new
•Consider the two slit experiment
•We can do it with photons or electrons, it doesn’t matter
•We can build a detector that counts individual photons or electrons
•We can put through particles one at a time
•We can count the number of photons on a screen
•Over time, we build up an interference pattern
Diffraction And Uncertainty
•The uncertainty in a quantity is how spread out the possible value is
T1  8.6
T1 score  79.2  8.6
•A plane wave approaches a small slit, width a
x  
•Initially it is very spread out in space
•But it has a very definite direction and
wavelength
•It therefore has a definite momentum p  0
•After it passes through the slit, it has a more definite position
•It now has a spread in angle

 min
a
•This creates an uncertainty in its momentum in this direction
p h
px   p 
xp  18 h

2a
a
x  14 a
The Uncertainty Relation
•Waves, in general, are not concentrated at a point
•They have some uncertainty x
•Unless they are infinitely spread out, they also typically contain more than
one wavelength
•The have some uncertainty in wavelength
•The have some uncertainty in momentum
•Hard mathematical theorem:
• Make a precise definition of the uncertainty in position
• Make a precise definition of the uncertainty in momentum
• There is a theoretical limit on the product of these:
xp 
h
4
Sample Problem
h
xp 
4
An experimenter determines the position of a proton to an accuracy of 10.0 nm.
1. What is the corresponding minimum uncertainty in the momentum?
2. As a consequence how far will the proton move (minimum) 1.00 ms later?
h
6.626 10 34 m 2 kg/s
27
p 


5.27

10
kg  m/s
8
4x
4 1.00  10 m 
•This corresponds to an uncertainty in the velocity of
p 5.27 1027 kg  m/s
v 
 3.15 m/s


27
m
1.673 10 kg
•This means the proton will move a minimum distance:
d  vt   3.15 m/s  1.00  10 3 s   3.15 mm
New Equations for Test 4
Reflection/
Refraction
i   r
Diffraction
Grating
m
sin  
d
m  0, 1, 2,
n1 sin 1  n2 sin 2
n2
sin  c 
n1
Light in
Materials

c
f v
k
n
c  3.00 108 m/s
Images
1 1 1
 
p q f
Quantum:
E  hf
K  hf  
p  h
 x  p  
h
4
Diffraction
Limit
 

a D
End of material
for Test 4