PHOTOELECTRIC EFFECT

Table of Contents
Connections
Light Energy and Quantum Theory
Building on…..
Einstein’s Equation of Photoelectric Effect
Factors affecting Photoelectric Effect • X-Rays – Thermionic emmission
Applications of Photoelectric Effect • Electrostatics
Revision Exercise • Electromagnetic Spectrum – the
high energy radiations
Specific Objectives
Arriving at……
By the end of this topic, the learner should be able to:
a) perform and describe simple experiments to illustrate the • Explain Photoelectric emission
photoelectric effect • Einstein’s equation-derivation,
b) explain the factors affecting photoelectric emission and calculations
c) apply the equation E = h to calculate the energy of photons • Explain applications of
d) define threshold frequency, work function and the electron photoelectric effect
volt
e) explain photoelectric emission using Einstein equation(h = h 0 Looking forward to….
+ ½ mv2)
f) explain the applications of photoelectric effect • Radioactivity
g) solve numerical problems involving photoelectric emissions.

(15 Lessons)

Content
1. Photoelectric effect, photons, threshold frequency; work
function, Planck‟s constant, and electron-volt
2. Factors affecting photoelectric emission
3. Energy of Photons
4. Einstein’s equation h = h o + ½mv2
5. Applications of photoelectric effect:
- photo emissive,
- photo conductive,
- photovoltaic cells
6. Problems on photoelectric emissions
2 Modern Physics

PHOTOELECTRIC EFFECT
We have seen that a metal surface will, if provided with appropriate amount of heat energy, emit electrons.
Radiations of appropriate strengths can also produce a similar effect.
When an electromagnetic radiation of sufficient frequency is radiated on a metal surface, electrons are
emitted. These electrons are called photoelectrons and the phenomenon is known as photoelectric effect.
Experiments to demonstrate photoelectric effect

(a) Using Neutral Plates

When ultraviolet radiation is allowed to fall on metal plate A, the

galvanometer shows deflection. When the barrier is introduced so
that the radiation is cut off, the galvanometer shows no deflection.
When ultraviolet radiation energy falls on a metal surface, some
electrons absorb this energy and are dislodged from the surface. The
deflection of the galvanometer indicates that electrons are emitted
at plate A and attracted by the plate B, causing a current to flow.
The glass plate, however, cuts off the ultraviolet radiation.

(b) Using Charged Electroscope

For the positively charged electroscope, the leaf

divergence remains the same. However, for the
negatively charged electroscope, the leaf
divergence decreases.
When the zinc plate is irradiated with ultraviolet
radiation, electrons are emitted from its surface.
The photoelectrons emitted from the positively
charged zinc plate do not escape due to
attraction by the positive charge on the plate
and the leaf divergence therefore remains the
same.

However, the photoelectrons emitted from the

negatively charged zinc plate are repelled and
the electroscope becomes discharged as a result
of which the leaf divergence decreases.
If a sheet of ordinary glass (which absorbs ultraviolet radiation) is introduced between the negatively
charged zinc plate and the ultraviolet source, the leaf divergence remains same.

Light Energy and Quantum Theory

Max Planck in 1901 came up with the idea that light energy is propagated as small packets of energy. Each
packet is called a quantum of energy (plural: quanta). In light, these discrete amounts of energy are called
photons. According to Planck, the energy E possessed by one photon is given by;
 PHOTOELECTRIC EFFECT 3

E = hf, where, h is Planck’s constant equal to 6.63 × 10–34 Js and f the frequency of the radiation.
From the general wave equation;
c = fλ
c
Hence, f =
λ
c
Thus, E = h , where c is the velocity of the radiation in vacuum and λ the wavelength.
λ
Since the velocity c of the radiation and the Planck’s constant h are constants, a radiation of longer
wavelength λ has lower energy.

Example
Compare the energy contained in a photon of red light of wavelength 7.0 × 10 –7 m and violet light of
wavelength 4.0 × 10–7 m (Take Planck’s constant as 6.63 × 10–34 Js and c as 3.0 × 108 ms–1)
Solution
For red light;
c 6.63 × 10−34 × 3.0 × 108
E=h =
λ 7.0 × 10−7

= 2.84 × 10-19 J

For violet light;

6.63 × 10−34 × 3.0 × 108
E = 7.0 × 10−7

= 4.97 × 10-19 J
Hence, a photon of violet radiation contains more energy than that of red radiation.

Einstein’s Equation of Photoelectric Effect

When a photon strikes an electron, all its energy is absorbed by the electron. Some of the absorbed energy is
used to dislodge the electron from the metal surface while the rest appears as the kinetic energy of the
emitted electron. The energy transformation during photoelectric emission is thus summed as follows;
energy needed to dislodge an maximum K. E. gained
Energy of photon = [ ]+[ ]
electron from metal surface by the electron

The minimum amount of energy needed to dislodge an electron from a metal surface is called the work
function (Wo) of the metal. This energy varies from metal to metal. The work function is expressed in a unit
called electron-volt (eV) or Joules (J).
1 eV = 1.6 × 10–19 J.
The minimum frequency required for photoelectric emission is known as the threshold frequency (fo) for the
metal.
The threshold wavelength λo, is the maximum wavelength beyond which no photoelectric emission will
occur.
The work function Wo is thus given by; Wo = hfo
c
This can be expressed as; Wo = h
λo
For any radiation of frequency f less than fo, hf will be lower than Wo and emission will not occur.
When the frequency of the radiation is fo, hfo = Wo (the work function), and emission occurs.
When the frequency of the radiation f > fo, hf > Wo and the excess energy in this case appears as the kinetic
energy of the emitted electron.
1
Thus, hf – Wo = 2mev2, where me is the mass of an electron and v the velocity of the electron emitted.
4 Modern Physics

1
hf = Wo + mev2
2
This is Einstein’s photoelectric equation.
Since Wo = hfo, this can also be written as;
1
hf = hfo + mev2
2
𝑐 1
Hence, hf = h + 2 m ev 2
𝜆𝑜

Example 1
The minimum frequency of light that will cause photoelectric emission from potassium surface is 5.37 × 10 14
Hz. When the surface is irradiated using a certain source, photoelectrons are emitted with a speed of 7.9 ×
105 ms–1. Calculate:
(a) the work function of potassium.
(b) the maximum kinetic energy of the photoelectron.
(c) the frequency of the source of irradiation.
(Take h = 6.63 × 10–34 Js, me = 9.11 × 10–31 kg)

Solution
(a) Work function, Wo = hfo
= 6.63 × 10–34 × 5.37 × 1014 J
= 3.56 × 10–19 J
1
(b) K.E.max = mev2max
2
1
= × 9.11 × 10-33 × (7.9 × 105 )2
2
= 2.84 × 10-19 J
1
(c) hf = hfo + mev2
2
= 3.56 × 10-19 + 2.84 × 10-19
= 6.4 × 10-19
6.4 × 10−19
∴f = × = 9.65 × 1014 Hz
6.63 × 10−34

Example 3
The threshold wavelength of a photoemissive surface is 0.45 µm. Calculate:
(a) its threshold frequency.
(b) the work function in eV.
(c) the maximum speed with which a photoelectron is emitted if the frequency of the radiation is 7.5 ×
1014 Hz (Take Planck’s constant h = 6.63 × 10–34 Js and me = 9.11 × 10–31 kg)

Solution
(a) λo = 0.45 µm = 4.5 × 10–7 m
c = foλo
𝑐 3.0 × 108
So, fo = 𝜆 = = 6.67 × 1014 Hz
𝑜 4.5 × 10−7
(b) Wo = hfo
= 6.63 × 10–34 × 6.67 × 1014 = 4.42 × 10–19 J
1 eV = 1.6 × 10-19 J
4.42 × 10−19
Wo = = 2.76 eV
1.6 × 10−19
 PHOTOELECTRIC EFFECT 5

1
(c) mev2max = hf - Wo
2
= 6.63 × 10–34 × 7.5 × 1014 - 4.42 × 10–19 J
= 4.97 × 10–19 J – 4.42 × 10-19 = 5.53 × 10-20
2 × 5.53 × 10−20
∴ v2max =
9.11 × 10−31
2 × 5.53 × 10−20
vmax =√ = 3.48 × 105 ms-1
9.11 × 10−31

Factors Affecting Photoelectric Effect

The ability of a radiation to eject photoelectrons from a metal surface is determined by three main factors,
namely:
(i) intensity of the radiation.
(ii) energy of the radiation.
(iii) type of metal.

Intensity of Radiation
The intensity of a radiation is the rate of energy flow per unit area when the radiation is normal to the area.
E
Intensity =
At
E
I=
At
E
But = P, where P is power.
t

P
∴ Intensity =
A
It can be shown experimentally that intensity is inversely proportional to the square of distance r from
1
the source, i.e., I α
r2

The apparatus is set up as shown below. Intensity is varied

by changing the distance r of the photocell from the
source and noting the corresponding values of current.

Experimental results show that as distance r decreases, the

value of current increases. The photocurrent I is directly
proportional to the number of photoelectrons emitted per
second. Thus, the number of photoelectrons produced is
directly proportional to the intensity.

Energy of Radiation
The circuit shown below can be used to investigate the relationship between the frequency of the radiation
and the kinetic energy of the photoelectrons.

The frequency is varied by using different colour filters.

For each colour filter, the potential difference is varied by
moving the jockey J between X and Y until no current is
registered.
The battery is connected in such a way that it opposes
the ejection of the electrons by attracting the
photoelectrons back to the cathode. The voltmeter
6 Modern Physics

reading gives the stopping potential for a given frequency.

A graph of stopping potential VS against frequency is shown below.

The graph is a straight line. From Einstein’s photoelectric equation;

hf = hfo + mv2max
Work done by stopping potential is given by eVs.
1
By work energy theorem, eVS = mv2
2
Substituting in Einstein’s photoelectric equation;
hf = hfo + eVS
Therefore, eVs = hf – hfo
h(f−fo )
Thus, Vs =
e
But hfo is equal to work function Wo. Hence, the graph of Vs against f is straight line cutting the f-axis at fo.
h
The slope of the graph is and the Vs intercept is Wo. Both Planck’s constant h and the work function Wo can
e
therefore be calculated from the graph.

Type of Metal
Every metal surface has its own minimum frequency of radiation called threshold frequency (f o), below
which no photoemission takes place, no matter how intense the radiation is. For example, zinc metal has a
threshold frequency equal to 10.47 × 1014 Hz. Any radiation with a frequency lower than 10.47 × 1014 Hz
cannot therefore eject an electron from zinc plate.
The results on photoelectric effect can be summed up as follows:
(i) The rate of emission of photoelectrons is directly proportional to the intensity of the incident
radiation.
 PHOTOELECTRIC EFFECT 7

(ii) Each metal surface has its own minimum frequency for photoemission.
(iii) The emitted photoelectrons have kinetic energies ranging from zero to maximum value. Increasing
the frequency of incident radiation increases the kinetic energy of the photoelectrons.

Applications of Photoelectric Effect

Photoemissive Cell
The photoemissive cell has two electrodes, the anode and the cathode. The cathode is a curved
photosensitive plate. The emission surface of the cathode faces the anode.

When light falls on the cathode, photoelectrons are emitted.

These are attracted by the anode, causing a current to flow in a
given circuit.
Photoemissive cells are used in:
(i) counting vehicles or items on a conveyor belts in factories.
(ii) burglar alarms.
(iii) opening doors.

As the item moves along the conveyor belt, it blocks light from reaching the photocell, stopping the current
flow. After the item has passed, a current flows again and pulse is registered in the counter.
The photoemissive cell can also be used to reproduce sound from a film. An exciter lamp focuses light
through the sound track along the side of a moving film, onto a photocell.

The varying width of the track varies the intensity of the light passing to the cell. The cell accordingly creates
a varying current in line with current obtained from the microphone when the film was made. The current
develops a varying potential difference across the resistor R. This is amplified and the output converted into
sound by the loudspeaker.

Photovoltaic Cells
The cell consists of a copper disc which is oxidised on one face. On the free copper oxide surface, a film of
gold is deposited. This film is thin enough to allow light to pass through it.
8 Modern Physics

When light strikes the copper oxide surface,

electrons are knocked off. The copper oxide
acquires a negative potential and copper a
positive potential. A potential difference
therefore exists and a current flows through
a wire connecting the gold film and copper.
This can be shown by a galvanometer
included in the circuit as shown in (b).
However, this current is low in the order of
milliamperes, depending on the intensity of
the light source.

Such cells are used in light meters, e.g.,

exposure meters in photography. (c) shows
the symbol for the photovoltaic cell.

Photoconductive cells

It consists of a grid like metal structure

having a thin layer of a semiconductor material called cadmium sulphide. It is enclosed in a casing covered
with a transparent window to allow light to fall on it.

Light energy reduces the resistance of the cell from 10

MΩ in the darkness to 1 kΩ in bright light. Photons of
light set the electrons on the semiconductor free, causing
an increase in conduction. Photoconductive cells are used
in fire alarms and also in exposure meters of cameras.
Other photoelectric devices are the solar cell and the
photodiode

Review Exercises
Where necessary, take: planck’s constant, h = 6.63 × 10–34 Js
mass of electron, me = 9.11 × 10–31 kg, charge on electron, e = 1.6 × 10–19 C
speed of light in a vacuum, c = 3.0 × 108 ms–1
1. Define the following:
(a) Photoelectric effect.
(b) Work function.
 PHOTOELECTRIC EFFECT 9

(c) Threshold frequency.

2. A clean zinc plate was charged and then placed on the cap of a positively charged electroscope as in

the figure below.

(a) State the charge on the plate before it was placed on the cap of the electroscope.
(b) What would happen to the leaf of the electroscope if ultraviolet light was made to fall on the
zinc plate?
(c) Why is the zinc plate cleaned?
3. Two metals X and Y are found to emit electrons when irradiated in turn with a light beam A, while only
metal X emits electrons when irradiated with a second light beam B.
(a) What are the possible distinctions between:
(i) light A and light B?
(ii) metal X and metal Y?
(b) What would be the effect of using on metal X:
(i) a light beam of the same colour but greater intensity.
(ii) light of shorter wavelength.
4. The minimum frequency of light which will cause photoelectric emission from a metal surface is 5.0 ×
1014 Hz. If the surface is illuminated by light of frequency 6.5 × 1014 Hz, calculate:
(a) the work function of the metal surface.
(b) the maximum K.E. (in eV) of the electrons emitted.
(c) the maximum speed of the electrons.
5. The work function of a clean metal surface is 4.5 eV. Calculate:
(a) the minimum frequency of radiation that will cause the emission of electrons from the surface.
(b) the maximum energy of the electrons emitted when the surface is illuminated with a radiation
of frequency 1.2 × 1015 Hz.

6. A sodium surface is illuminated by light of wavelength 3.0 × 10–7 m. The work function for sodium
metal is 2.46 eV. Find:
(a) the kinetic energy of the ejected photoelectrons.
(b) the cut-off frequency for sodium.
7. The table below shows the stopping potential and the corresponding frequencies for a certain
photocell:

Plot a graph of stopping potential Vs (y-axis) against frequency and from the graph, determine:
10 Modern Physics

(a) the threshold wavelength.

(b) Planck’s constant.
(c) the work function of the metal.
8. Sodium has work function of 2.3 eV. Calculate:
(a) its threshold frequency.
(b) the maximum velocity of the photoelectron produced when its surface is illuminated by light of
wavelength 5.0 × 10–7 m.
(c) the stopping potential of this energy.
9. A metal surface has work function of 4.92 eV.
(a) Find the maximum wavelength of radiation that will cause the emission of photoelectrons from
the metal.
(b) Calculate the maximum kinetic energy in joules of the electrons emitted when the metal is
irradiated with ultraviolet light of frequency 1.8 × 1015 Hz.
(c) What is the stopping potential of this metal?
10. (a) What is meant by threshold wavelength?
(b) The maximum kinetic energy of the photoelectrons emitted from a metal is 1.8 × 10 –19 J when
the frequency of the incident radiation is 8.5 × 1014 Hz. Calculate the maximum wavelength of
the radiation that can just emit an electron from the metal.
11. When light of wavelength 1.0 µm is radiated onto a metal, it ejects an electron with a velocity of 3.0
× 105 ms–1. Calculate the:
(a) work function of the metal.
(b) threshold frequency of the metal.
14. (a) Describe a simple experiment which can be used to explain Quantum Theory.
(b) One would suffer from skin burn when a photon of energy of about 3.0 eV is radiated to our
bodies. Find the wavelength of this radiation.

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