3 Light As A Particle
Early in the 20th century a number of important experiment demonstrated
inadequacies in the classical physics view of light.
Each experiment provides evidence for 2 of the most basic quantum mechanic
equations for photons:
Planck’s Equations E  h  
De Broglie Equation  
h

 k
3.1 Black Body Radiation
Planck in 1900 studied the spectrum of EM energy emitted by a black body.
A black body is a cavity with a small aperture which has been coated black inside,
maintained at a constant temperature. It provides an accurate standard for radiation
intensity.
Lord Rayleigh studied black bodies classical and found the spectrum, I   
I  
1
4

However, experimental results yielded:
T3
T2
I  
T1
Where T3 > T2 > T1

Planck studied the problem and made a suggestion which lead to the precise fit of the
observatuibs. He suggested that the energy in the EM field takes only restricted
values: E  h
See handout:
3.2 Production Of X-Rays
If electrons are accelerated into a target with a high z, then following was observed:
I  

Observation Showed:
1 Unexpected min cut-off due to quantised energy of electrons
2 Sharp spectral lines, electrons have energy and excite the atoms, eventually
the electrons decay from upper to lower energy state E  h  E2  E1 .
Energy levels with high material are unique.
3.3 The Photoelectric Effect
The emission of an election when light strikes a surface (complement of x-ray
production) done by Leonard (1902). Experiment measure energies of emitted
electrons as a function of  and I(  ).
Vacuum, Light incidental on cathode
Cathode
Anode
A
Cathode and anode made of the same metal.
If V > 0 anode attracts electrons
If V < 0 anode repels electrons
However if the voltage to large the current will tend to 0
Handout
NB:







w is energy required to free electron from surface against electrostatic
attraction of positive metal ions.
As positive Voltage is decrease, current falls to 0 at V = V0. This is the
stopping potential.
V < V0 no current flows, no matter what the intensity.
When V is large and positive, the current saturates. All the electrons produced
are collected.
The value of V0 depends on the  of light. The larger  the more negative V0.
Value of V0 does not depend on I(  )
J max 2 I 2

J max 1 I 1
There is no time delay between illumination and current
Phenomena cannot be understood in basic classical theory. Classical theory would
state that V0  I   and not proportional to  . Also one would expect same current at
V  V0 if I(  ) was large enough.
Einstein in 1905 stated.
 Light energy arrives as “quanta” of energy. E  h
 No time delay because as long as V is large enough, one photon ejects one
electron.
 At V=V0 only most energetic electron reaches the anode. So maximum
1
Kinetic Energy must equal stopping force. MaxKE  me v 2 max  eV0
2
Since energy most overcome w energy balance is:
1
h  me v 2 max  w
2
h  eV0  w
Graph of eV0 vs  the gradient is h. and at V=0 eV0 = w
The Compton Effect: Photon Carrying Energy h And Momentum
h
HINTED

Compton (1928) studied scattering of X-rays ( 0  0.071nm ) by graphite. Since the
X-rays energy >> electron binding energy, the scattering was done essentially by free
electrons.
Classically one would expect scattering of  0 and possibly harmonics.
Compton saw scattering at 0 &  '  0 
From Relativity:
And Planck and Einstein:
E 2  m2c 4  p 2c 2
E  h
h 2 2  m 2 c 4  p 2 c 2
h 2 2  p 2 c 2
For m=0 photons:
h 2 2
p  2
c
h
p

2

0
X-Rays
Compared for all angles.
Graphite
'
0
 ' 
Compton Data Showed
h
1  cos   Hinted
me c
Compton Analysis
Assumed
X-rays were scattered from electrons as there binding energy was
much less then h
E  h
X-rays had
h
p

Compton then looked at Energy and Momentum Conservation
'
E  h '  p ' c

0
E  h 0  p 0 c
e
E  me c 4  pe c 2
2
pe
Energy Conservation
p0 c  me c 2  p' c  me c 4  pe c 2
Rest Mass
2
 p
0
 p'c  me c 2

2
2
 me c 2  p e c 2
2
2
Momentum Conservation
p0  p' pe
pe   p0  p' p0  p'
2
It can be shown that
2
2
pe   p0  p'  2 p0 p' 1  cos 
2
1
1
1
1  cos  


p ' p 0 me c
As p 
h

 '0 
NB
h
1  cos  
me c
Max  occurs at   180
Due to scattering from deeper, more strongly bonded electrons, mass needs to
be adjusted to mass of nucleus.