The photon theory received further support from the discovery of the Compton effect. This was in
1923. Arthur Compton aimed an x-ray beam at a chunk of graphite. An amazing thing took place; the scattered photons had a longer wavelength than the incident photons!


The energy of the scattered photon was less than the incident photons. The change in wavelength is called the Compton shift . So what happened to cause this energy to be lost? Compton figured that the photon, acting like a particle, has a collision with an electron. Energy and momentum must be conserved, and the electron, initially at rest, gains energy. This represents the energy lost by the photon. Since the photon has lost energy, its wavelength must increase. This was great proof for the photon theory of light.
The Compton shift depends on the angle and the wavelength of the photon. It is given by this equation: h   
The quantity is called the Compton Wavelength.
To sum things up: h

In a collision, the photon acts like particle.

Energy and momentum conserved.
 The electron gets some of photon’s energy and momentum.

The photon loses energy, so its wavelength must increase (frequency gets smaller, lower
E).
1


Proof of photon theory
Wave Properties of Particles: In 1905 Albert Einstein showed that waves behaved like particles.
This was very disconcerting – how could a wave be both a particle and a wave? But what about the opposite thing? Could matter (which is what particles are) exhibit wave-like properties? In
1923 (seems like a busy year, don’t it) Victor de Broglie (1892-1987) suggested just that. And he got Einstein to support him. It seemed logical, if not symmetrical, to de Broglie that if waves had particle characteristics, then particles ought to demonstrate some wave characteristics. Fair is fair!
The wavelength of a massive particle, like a baseball, would have matter-waves of such ultra-short wavelength, that it would be impossible to detect. But the wavelength of a small, high-speed particle such as an electron ought to be long enough that it could be measured. In 1927 an
American physicist, Clinton Davisson (1892-1975) did just that. He found that a beam of fast electrons could be diffracted and refracted. This was substantial evidence for the wave-like behavior of matter.
All matter behaves as both a particle and a wave .
It seems odd that a photon could have momentum since it has zero mass. But according to special relativity, photons do indeed have momentum.
E
 hf
 hc

The energy of a photon is of course:
Its momentum can be expressed as: p
 h
 p

E c
 hc c

 h

The momentum for a photon is:
The photon’s wavelength can be expressed as a function of its momentum:
  h p
This equation is provided on the AP Physics Test equation sheet.
So we now have an equation for the momentum of a photon.
2

Wavelength of Particle: We can now develop an equation for the wavelength of a particle. The momentum of a particle is: p
 mv p
 h

The momentum of a photon is:
Set these two quantities equal to each other: mv
 h

Now solve for the wavelength:
  h mv
This gives us an expression for the wavelength of a particle.
  h mv
In addition to this, de Broglie postulated that the frequency of a particle obeys Planck’s energy relationship:
E
 hf
Thus the frequency of a particle is: f

E h
Example 2: Find the
Davisson-Germer Experiment:
Particles should have wave properties
Show diffraction patterns
3

1927 Davisson and Germer measured wavelength of electrons low energy electrons scattered by nickel target in vacuum scattered electrons exhibited minima and maxima at specific angles
Crystal structure of nickel acted as diffraction grating
Confirmed de Broglie’s theory

Calculate de Broglie wavelength for a 1.25 kg rock moving at 25.5 m/s.
  h mv
 



1.25
kg

34
 s
2
25.5
m s s


  
34 m
 
35 m
Pretty small wavelength

Calculate de Broglie wavelength for an electron moving at 1.50 x 10 6 m/s.
  h mv
 




31

34 kg
 s
2 x
6 s s

 m
 
0.485 10

9 m
 
9 m
This we can measure
Atomic Spectra (chap 28)
Emission spectrum
4

Get series of spectral lines
Characteristic of given element
Absorption spectrum
Continuous spectrum with missing lines
For given element, lines up with emission spectrum
Discovered Helium in sun’s atmosphere
Bohr Atom Model
Why did atoms produce only discrete spectral lines?
1913 -- Niels Bohr’s Hydrogen atom model
1.
Electron moves in circular orbits around proton
Electromagnetic force
2.
Only certain electron orbits are stable
Electron doesn’t emit energy in stable orbits
3.
Radiation is emitted when electron jumps from higher energy state to lower energy state
4.
Size of orbital function of electron’s orbital angular momentum. me
F v
9 a
0 r
Frequency of emitted photon (energy) hf

E i

E f
Energy of photon:
E p

E i

E f
4 a
0 a
0
5

Lowest E State

Ground State
When electrons get excited, make quantum leaps n
E (eV)
0.00
5
4
3
2
1
Unstable, fall back down to lower state
Give off photon as they do so
Value for hc: hc

1.24 10
3 eV nm
- 13.6 eV

A monotamic gas is illuminated with light of wavelength 400.0 nm. It absorbs some of the light and gives off visible light at both 400.0 nm and 600.0 nm. (a) draw an energy level diagram, (b) If the ground state of the atom has energy – 5.00 eV, what is the energy of the state to which the atoms were excited by the 400.0 nm light? (c) At which wavelengths outside the visible range do these atoms emit radiation after they are excited by the 400.0 nm light?
6

(a)
(b) E
 hf
E
E v
 c


E
 hc


1.24 10
3 eV nm

1

400.0

3.10
eV nm

E p

E i

E f
E p

5.00
eV

3.10
eV

1.90
eV
(c) E
E


1.24 10
3 eV nm

1

600.0

2.07
eV nm

E (eV)
600.0 nm

0.0031 10
3 eV

400.0 nm x
3 eV
0.00
- 5.00 eV
E

3.10
eV

2.07
eV

1.03
eV
E
 hc

  hc
E
 
1.24 10
3 eV nm
1.03
 
1 200 nm eV
 x
3 nm
De Broglie Electron Waves & H:
Electrons act as standing waves in Bohr’s orbits
7

Heisenberg Uncertainty Principle
1927 -- Werner Heisenberg
Impossible to know momentum and location of particle at same time x p
X

4 h

Product of momentum and position is related to Planck’s constant
If
 x is small,
 p will be very large and vice versa
Schrodinger’s cat
Schrodinger’s wave function
Particle behavior treated as a wave
Very successful

Ground state energy of an atom is – 10.0 eV. When the atoms in their ground state are illuminated with light, only wavelengths of 207 nm and 146 nm are absorbed by the atoms. (a)
Calculate the energies of the photons of light of the two absorption-spectrum wavelengths (b) draw a energy level diagram (c) Show by arrows on the energy level diagram all the possible transitions that would produce emission spectrum lines (d) What would be the wavelength of the emission line corresponding to the transition from the second excited state to the first excited state? (e) would the emission line be visible?
8

(a) E
 hc

E


1.24 10
3 eV nm
E
E



5.99
(b) eV
1.24 10
3 eV nm


146
1 nm
E p

E i

E f



207
1 nm


8.49
eV

0.00599 10
3 eV
E p
E p

10.0
eV

10.0
eV

5.99
eV

8.49
eV

4.01
eV

1.51
eV
E (eV)
0.00
- 6
- 7
- 8
- 9
(d) E p

E i

E f
- 10
- 0
- 1
- 2
- 3
- 4
- 5
E p

8.49
eV

5.99
eV

2.50
eV
E
 hc

  hc
E
 
1.24 10
3 eV nm
2.50
eV
 
496 nm
 x
3 nm
(e) visible, visible light is 400 nm – 700 nm
- 10.0 eV
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