# Topic 1 - The Nature of Light ```PHY 102: Quantum Physics
Topic 1
The nature of light
Young’s Double Slit Experiment
•Demonstrates wave nature of light
•Each slit S1 and S2 acts as a separate source of coherent light (like the
loudspeakers for sound waves)
“real” intensity distribution for double slits…..
light behaving exactly like a wave………..
“ideal” pattern
Single slit pattern
“real” pattern
What kind of wave (what’s waving?)
•Key point: propagation of a light wave does NOT require a medium
•Light can propagate through empty space (vacuum)
Consider two of MAXWELL’S EQUATIONS OF
ELECTROMAGNETISM:
B
 E 
t
E
  B  0 J  0 0
t
(essentially Faraday’s law and Ampere’s law, respectively)
Electromagnetic Waves
•After some mathematical manipulation (see Y2
electromagnetism) we find that the electric and
magnetic fields obey the WAVE EQUATIONS:
 Ey
2
x
2
  0 0
 2 y( x, t )
x 2
 Ey
2
t 2
1  2 y( x, t )
 2
v
t 2
“standard” linear wave equation
 2 Bz
 2 Bz
  0 0
2
x
t 2
c
1
0 0
Electromagnetic Waves
Energy in Electromagnetic Waves
•For a wave, intensity = energy flow through unit area per
unit time = P/A
•In the case of an EM wave, intensity corresponds to the
c 0 2
I
E0
2
•Studies of the PHOTOELECTRIC EFFECT by Lennard
(~1900) gave results that could not be explained by the
classical wave picture of light………
The photoelectric effect
•In a metal, the flow of electric current is due to the flow of charged
particles called electrons
•The electrons are relatively free to move throughout the metal
(electron “gas” or “sea”), so most metals are good conductors of
electricity.
•By shining light on a metal, some of these electrons can be “knocked
out” of the metal to generate an electric current outside of the metal
•Although basic photoelectric effect can be explained with classical
wave model of light, the detailed results of such an experiment CAN
NOT!...
Electrons in metals
•Electrical current arises from the flow of charged
particles, usually electrons.
•In the “Drude model” of a metal, there are many free
electrons, which behave like an ideal gas of charged
particles.
•Because of this large number of free, charged particles,
metals are usually very good electrical conductors.
•If electrons in a metal behave as an ideal gas, we expect
them to have a spread of kinetic
energies…………………………………….
Maxwell-Boltzmann velocity distribution
•The probability that a molecule has a speed between v
and v + dv is f(v) dv where:
 m 
f (v )  

 2kT 
3/2
2 mv 2 / 2kT
4v e
Electrons in metals
•Electrons don’t normally spontaneously leak out of the
metal – there must be something that keeps them in place
Swimming pool analogy
WALL
WATER
Water is confined by a region of increased gravitational
potential energy (ie the wall!)
Electrons in metals
METAL
SURFACE

E=Emax
ELECTRON GAS
E=0
Electrons are confined by a region of increased
electrostatic potential energy (ie the metal surface)
The extra energy required to remove the most energetic
electrons from the metal is called the WORK
FUNCTION, , of the metal
Detection of emitted electrons
Vacuum
A
Detection of emitted electrons
Vacuum
electrons
CURRENT FLOW
A
ENERGY IN
Photoelectric Effect
Vacuum
electrons
CURRENT FLOW
A
MONOCHROMATIC LIGHT IN
VARIABLE VOLTAGE
The photoelectric effect
•Metallic anode and cathode
enclosed in evacuated glass
tube (like old-fashioned
electronic “valve”)
•Vacuum required so that
electrons can flow from
cathode to anode without
scattering from gas
molecules.
Potential divider
to vary anode-cathode
voltage Vac
•Monochromatic light is
shone on the cathode,
electrons flow from cathode
to anode, current detected
by galvanometer.
The photoelectric effect
•Pink lines: electric field
lines
•because electrons are
negatively charged, they
move in the opposite
direction to the field lines.
•When the cathode is
illuminated with light of
frequency above a certain
“threshold frequency” a
current is detected, ie
photoelectrons must be
ejected from the metal….
The photoelectric effect
•When the cathode is illuminated with light of frequency above a
certain “threshold frequency” a current is detected, ie photoelectrons
must be ejected from the metal….Electrons emitted
INSTANTANEOUSLY even for the lowest intensity of light
•If the frequency of the incident light is below the threshold
frequency, no electrons emitted, no matter how intense the light is.
•If light were behaving entirely like a classical EM wave, its power
would be proportional to intensity, but independent of frequency
(mechanical wave energy proportional to amplitude and frequency,
EM wave energy proportional to amplitude (intensity) only)
•Would expect intense light of any frequency to generate
photoelectrons, and also “time delay” for photoemission due to light
of low intensity…………..
The photoelectric effect: detailed results
current proportional to light intensity
battery polarity reversed to
impede flow of photoelectrons from
cathode to anode
The photoelectric effect: detailed results
battery polarity reversed to
impede flow of photoelectrons from
cathode to anode
Stopping potential proportional to
light frequency
Significance of the stopping potential
•If we assume the electrons in the
metal behave like a classical
ideal gas (?), then they won’t all
have the same velocity (Maxwell
velocity distribution)
•In other words, they won’t all
have the same kinetic energy
•The stopping potential is the
voltage required to stop the
MOST ENERGETIC electrons
from leaving the metal
Einstein’s postulate
A beam of light can be treated as a stream of particles (PHOTONS) with
zero rest mass
Each photon has energy:
E p  hf 
hc

where h is a constant (Planck’s constant, h ≈ 6.63 x 10-34 Js)
f, λ, c, are frequency, wavelength and velocity of light (in vacuum) respectively.
Light intensity is proportional to PHOTON FLUX (no of photons passing
through unit area per second)
Results of photoelectric effect: the photon model
•When the cathode is illuminated with light of frequency above a
certain “threshold frequency” a current is detected, ie photoelectrons
must be ejected from the metal….Electrons emitted
INSTANTANEOUSLY even for the lowest intensity of light
•Photons behave like “bullets” with energy hf, which can “hit” the
electrons. If f is high enough, a single photon can instantaneously
remove an electron from the metal

•Photo current is proportional to light intensity (if f is above threshold
frequency).
•Light intensity = number of photons hitting surface per second
•Current = number of electrons emitted from surface per second. 
Results of photoelectric effect: the photon model
•So, in terms of the photon model, the stopping potential is related to
the photon frequency, Planck’s constant, and the work function of
the metal:
eV0  hf  
By measuring V0 as a function of f, we can measure Planck’s constant and the
work function of the metal……………………………….
Example Calculation 1
•In a photoelectric experiment a reverse potential of 2V is required to
stop the flow of current for light of a certain frequency. Calculate a)
the maximum kinetic energy and b) the maximum velocity of the
emitted photoelectrons……..
The Compton Effect
Photons: summary so far
•Einstein postulated the existence of a particle called a photon, to
explain detailed results of photoelectric experiment.
E p  hf 
hc

•Photon has zero rest mass, travels at speed of light
•Explains “instantaneous” emission of electrons in photoelectric
effect, frequency dependence.
•Further confirmation of the photon picture provided by the
COMPTON EFFECT (1922-23)…………………
•Einstein’s Special Theory of Relativity, 1905:
•The laws of physics are the same in every inertial frame of
reference (in which Newton’s first law is valid)
•The speed of light in a vacuum is the same in all inertial frames
of reference, and is independent of the motion of the source.
(corollary: the velocity of light can’t be exceeded)
Many important consequences: length contraction, time dilation
effects at high speeds, mass/energy equivalence……
Momentum and kinetic energy at high speeds…………….
Relativistic expressions for energy and momentum
E
mc 2
2
v
1 2
c
p
mv
2
v
1 2
c
(energy of stationary
particle = mc2)
Relativistic expressions for energy and momentum
E
mc
2
mv
p
2
2
v
1 2
c
v
1 2
c
E  mc
2
  pc 
2 2
2
E  mc
2
  pc 
2 2
2
consequently, particle with zero rest mass (eg photon) has momentum p
given by:

E hf h
p 

c
c 
The Compton Effect
•x-rays scattered from target containing very loosely bound electrons
•Wavelength of scattered x-rays found to be different from that of incident
X-rays AND to depend on detection angle :
h
1 cos  
   
me c
The Compton Effect
Compton Effect explained by photon model:
Treat Compton scattering as a 2-particle collision between photon and
initially stationary electron, obeying conservation laws for energy and
momentum:
Before
photon
electron
pi
photon
After

pe
pf
Compton Scattering: Conservation of momentum
Vector triangle:
pf
pe

pi
Consider magnitudes of vectors pi, pf and pe:
p  p  p  2 p f pi cos 
2
e
2
f
2
i
Compton Scattering: Conservation of energy
We are dealing with velocities at, or close to, speed of light so need to use
relativistic expressions:
Initial energy:
Final energy
pi c  me c
Ee  pi c  p f c  mec
p f c  Ee
E  mec
2
e
2
2
2
  p c  p c  p c  m c 
2
general expression
2
e
2
i
f
e
2
Compton Scattering: Conservation of energy
m c   p c  p c  p c  m c 
2
2
2
e
2
e
i
f
2
e
Divide both sides by c2 (be careful!………)
mec   pe   pi  p f  mec 
2
2
2
mec   pe   pi  p f   mec   2pi  p f mec
2
2
2
2
pe   pi  p f   2pi  p f mec
2
2

Compton Scattering: Conservation of energy
&amp; momentum
pe   pi  p f   2pi  p f mec
2
2
Energy
pe 
2

2
f

momentum

 p  p  2 pi p f  2 pi  p f me c
2
i
p  p  p  2 p f pi cos 
2
e
2
f
2
i
Compton Scattering: Conservation of energy
&amp; momentum
p  p  2 pi p f  2 pi  p f mec
2
i
=
2
f
p  p  2 p f pi cos 
2
f
2
i
 2 pi p f  2 pi  p f me c  2 pi p f cos 

p  p m c
1 
  cos 
i
f
pi p f
e
Compton Scattering: Conservation of energy
&amp; momentum
p  p m c

1
 cos 
i
f
e
pi p f

mc mc

 1 cos 
pf
pi
1
1
1
 
(1  cos  )
p f pi mc
h
   
(1  cos  )
mc

Compton Scattering: Summary
The observed experimental result:
h
   
(1  cos  )
mc
Is entirely explained by the photon-electron scattering model. Further proof
of the validity of the photon concept.
•maximum wavelength shift for  = 180&deg;, Δλ=2h/mc
•h/mc is known as the COMPTON WAVELENGTH of the electron.
```