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PHY 102: Waves & Quanta
Topic 9
The photoelectric effect
John Cockburn (j.cockburn@... Room E15)
•EM radiation as waves
•Anomalous result: the photoelectric effect
•Photons
EM radiation as waves
•Interference (Young’s Slits)
•Single-slit diffraction
•X-ray diffraction…..
All show EM radiation behaving entirely as expected for classical
waves.
However, as was discovered at the beginning of the 20th century, the
behaviour of EM radiation is not always consistent with this…………
“real” intensity distribution for double slits…..
light behaving exactly like a wave………..
“ideal” pattern
Single slit pattern
“real” pattern
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 wave
model of light, the detailed results of such an experiment CAN NOT!...
The photoelectric effect
•Metallic anode and cathode
enclosed in evacuated glass
tube (like old-fashioned
“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
The work function
It requires a certain amount of energy to extract the electrons from the metal:
imagine the electrons as being trapped at the bottom of a well:
E=0

maximum electron energy
The work function of a metal is the least amount of energy required to remove
electrons from the metal, ie the energy required to remove the most energetic
electrons
The work function
Imagine we have incident light with energy Ep represented by length of
arrow)
Emax=Ep-
E=0


Ep
So, we expect the emitted photoelectrons to have a distribution of energies up
to a maximum of Ep - . So, going back to the stopping potential V0, we must
have:
eV0  E p  
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……..
Example Calculation 2
•In a photoelectric experiment, stopping potentials of 1V, 2 V and 3V
are measured for light of wavelengths 600nm, 400nm and 300nm
respectively. Use these data to determine the work function of the
cathode material and the value of Planck’s constant………….
Example Calculation 2