PHOTOELECTRIC EFFECTS

Caesium Lithium

Work function of caesium, W = 2.14 eV Work function of lithium, W = 2.50 eV

Threshold frequency, f0 = 5.16 x 1014 Hz Threshold frequency, f0 = 6.03 x 1014 Hz

Maximum wavelength to produce photoelectric Maximum wavelength to produce photoelectric
current, λ0 = 579 nm current, λ0 = 496 nm

Conclusion:
▪ The higher the work function, the shorter the maximum wavelength required to produce

photoelectric current.
▪ As the light intensity increases, the photoelectric current in the photocell also increases

Work Function, W = Minimum ENERGY required to release a photoelectron
Threshold frequency, f0 = Minimum FREQUENCY required to release a photoelectron

Photoelectric effect exists Derive Formula:

f0 E = W + Kmax

Graph of Kmax against f hf = W + Kmax

W Based on Kmax-f graph

Kmax = hf - W

y = mx + c

y-intercept, gradient of the graph,

c=W m=h

Work Function, W = Minimum ENERGY required to release a photoelectron
Threshold frequency, f0 = Minimum FREQUENCY required to release a photoelectron

Vs Photoelectric effect exists Derive Formula:

f0 E = W + Kmax
hf = W + eVs
Graph of Vs against f Based on Vs-f graph
eVs = hf - W
W/e
Vs = hf/e – W/e

y = mx + c

y-intercept, gradient of the graph,

c = W/e m = h/e

In general, there are three types of photoelectric cell:

PHOTO EMISSIVE CELL PHOTO CONDUCTIVE CELL PHOTO VOLTAIC CELL

▪ Vacuum or gas filled tube used in ▪ Solid state semiconductor device ▪ Uses P-N junction to conduct
electricity where electrons jump from
photoelectric effect experiment. that changes resistance when light valency band to conduction band
when the electrons received enough
▪ Work function metal cathode will shines on the surface energy from photons.

determine photoelectric current. ▪ Used as LDR or automatic switch ▪ Work function is not relevant. Energy
gaps to overcome is about 1.1 eV only.
▪ Used in the last few decades when semi ▪ Work function is not relevant
▪ Used in Solar Panel modules and
conductor was not discovered. ▪ Used in light industry as photo arrays to generate electricity

▪ Got superceded by photo resistors and detectors and measure light intensity

photo diodes They are only similar in one aspect that is it requires photon to free

up the electrons or excite electrons to conduct electricity.



▪ The Noor Complex Solar Power Plant located
in the Sahara Desert is the world’s largest
concentrated solar power plant.

▪ This station is expected to be
completed in 2020 and is
capable of producing 580 MW
capacity for use by 1 million
residents.

• Light detectors at the
automatic doors use infrared
beam and photocells as switches.

• When the light path is disturbed,
photoelectric current in the
photocell circuit will be
disconnected and the door will
remain open.



▪ The image sensor is a main component in
high resolution cameras. This component is
used to convert light into electrical signals
which can be processed to form digital images.

What Makes a Digital Camera a Camera? The Image Sensor is King
An image sensor is a device that allows the camera to convert photons – that is, light – into electrical
signals that can be interpreted by the device. The first digital cameras used charge-coupled devices,
facilitating movement of the electrical charge through the device so it could be modulated. They were invented
in 1969 at Bell Laboratories, an unexpected result of semiconductor research.

https://www.youtube.com/watch?v=oLrOnEmy_GA

How does the International Space Station work?

QUANTUM PHYSICS

FORMATIVE PRACTICE
CG-ANA-AF
SMK ABDUL RAHMAN TALIB, KUANTAN, PAHANG

E = W + Kmax W = hf0
hf = W + ½ mv2max
W = hc
The minimum energy required for a
photoelectron to be emitted from a metal λ0
surface.

The minimum frequency for a light photon to
produce photoelectric effect.

W = hf0

Derive Formula:

E = W + Kmax
hf = W + Kmax

Gradient = Based on Kmax-f graph
Planck’s constant,h
Kmax = hf - W
f0 x-intercept
y = mx + c
= Threshold frequency of the metal, fo
y-intercept, gradient of the graph,
W
c=W m=h
y-intercept
= Work function, W

A black body is an ideal body that is able to absorb all the electromagnetic
rays that fall on it.
A black body can also emit thermal radiation depending on its temperature

Quantum energy is a discrete packet of energy

h = 6.63 x 10-34 Js λ = 680 nm
c = 3.0 x 108 ms-1 E = hc/λ = 1.83 eV
λ = 680 nm
W = 2.28 eV Kmax???

E = hc/λ W = 2.28 eV
= (6.63 x 10-34 )(3.0 x 108)/ 680 nm
= 2.925 x 10-19 J
= 1.83 eV

▪ E<W
▪ Photoelectric effect does not occur because of

the photon energy of the red light is LOWER
than work function of sodium metal.

W = hfo λ = 680 nm
= hc/λo E = hc/λ = 1.83 eV

2.28 x 1.6 x 10-19 = (6.63 x 10-34 )(3.0 x 108)/ λo Kmax???
λo = 5.4523 x 10-7 m
W = 2.28 eV
= 545.23 nm

De Broglie’s wavelength

λ =h K = mv2 h = 6.63 x 10-34
mv c = 3 x 108
me = 9.11 x 10-31
Squared both sides:
v2 = 2K λ = 590 nm
λ2 = h2 . m
m2v2

Substitute v:

λ2 = h2 . K = h2 .

m2v2 2m λ2

λ2 = h2 . K = . (6.63 x 10-34)2 .
m2 (2 K)
m 2 (9.11 x 10-31) (590 x 10-9)2

λ2 = h2 . = 6.93 x 10-25 J
m (2 K)

K = h2 .

2m λ2

λ = 555 nm E = hc/λ … (i)
P = 5.00 mW E = mc2 … (ii)
h = 6.63 x 10-34
c = 3 x 108 (i) =(ii) p = h/λ
hc/λ = mc2 = (6.63 x 10-34) / (555 x 10-9)
h/λ = mc = p = 1.1946 x 10-27 kg ms-1

P = nhf
(5.00 x 10-3) = nhc/λ

= n(6.63 x 10-34) (3 x 108)/ (555 x 10-9)
n =1.395 x 1016 photoelectrons

λ = 1.00 nm
h = 6.63 x 10-34
me = 9.11 x 10-31 kg

(a) Louis de Broglie hypothesized that particles such as electrons could have wave properties.
De Broglie wavelength, λe = h/p ; where p = momentum of the electron

(b) λe = h/p (c) p = mv

p = h/ λe 6.63 x 10-25 = (9.11 x 10-31) v

= (6.63 x 10-34) / (1.00 x 10-9) v = 7.2778 x 105 ms-1

= 6.63 x 10-25 kg ms-1

(c) K = ½ mv2
= ½ (9.11 x 10-31) (7.2778 x 105 )2
= 2.41 x 10-19 J

▪ The rays of light that enter the large cavity will

undergo repeated reflections on the inner walls of

the cavity.
▪ At each reflection, part of the rays are absorbed

by the inner walls of the cavity.
▪ Reflections continue to occur until all the rays are

absorbed and none of them can leave the cavity.
▪ Thus, the cavity acts like a black body

▪ As the temperature of the black body increases, the
intensity of the radiation emitted increases rapidly.

▪ The intensity of the violet-blue rays increases more
than the orange-yellow rays.

▪ Therefore, the black body is violet-blue at 9 000 K

Extra Info:

▪ Color of a black body from 800 K to 12200 K.

λ = 800 nm
P = 60 mW
h = 6.63 x 10-34
c = 3 x 108

p = h/λ Quantum communication is a field of applied
= (6.63 x 10-34) / (800 x 10-9) quantum physics closely related to quantum
= 8.2875 x 10-28 kg ms-1 information processing and quantum
teleportation
E = hc/λ
= (6.63 x 10-34)(3 x 108) / (800 x 10-9)
= 2.48625 x 10-19 J

λ = 800 nm
= 60 mW
P = 6.63 x 10-34
= 3 x 108
P = nhc/λ h
(60 x 10-3) = n (2.48625 x 10-19 ) c

n = 2.41327 x 1017 photons per second

From (a) p = 8.2875 x 10-28 kg ms-1
From (c) n = 2.41327 x 1017 photons per second

Total momentum = momentum of one photon x number of photons per second
= (8.2875 x 10-28 ) (2.41327 x 1017 )
= 1.9999 x 10-10 kg ms-2

8. Complete the table with information on the wavelength and photon
energy for several components of waves in the electromagnetic spectrum.

Wavelength Photon energy Region in the
500 nm electromagnetic
E = hc/λ
= (6.63 x 10-34)(3 x 108) / (500 x 10-9) spectrum
= 3.978 x 10-19 J = 2.486 eV
Visible light
(400 nm – 750 nm)

E = hc/λ 50 eV Ultraviolet
(50 x 1.6 x 10-19)= (6.63 x 10-34)(3 x 108) / λ 5.0 x 10-21 J Infrared

λ = 2.486 x 10-7 m
= 248.6 nm

E = hc/λ
(5 x 10-21) = (6.63 x 10-34)(3 x 108) / λ

λ = 3.978 x 10-5 m
= 39 780 nm
= 0.03978 μm

λ0 = 1110 nm W = hf0
h = 6.63 x 10-34 f0= (6.63 x 10-34) (2.702 x 1014)
c = 3 x 108
f0 = ? = 1.7914 x 10 -19 Hz
W =?

c = f0λ
f0 = c/λ0

= (3 x 108)/(1110 x 10-9)
= 2.702 x 1014 Hz

▪ At room temperature, the thermal energy is insufficient to release electrons in a

photocell or to activate the photocell.
▪ Energy of photons is low which lower than the material work function
▪ No photoelectrons emitted from the semiconductor material.

m = 5 x 10-10 kg p =h/λ
λ =h/p
v = 0.4 ms-1
= h / mv
a hole = 1 mm = (6.63 x 10-34) / (5 x 10-10)(0.4)
d sand = 0.07 mm = 3.315 x 10-24 m

▪ No
▪ The de Broglie wavelength of the sand is too short (10–24 m) compared to the size of the

hole (1 mm)
▪ If the size of the hole is further reduced to approximate the order of the de Broglie

wavelength, the sand will not be able to pass through it because the diameter of the sand

is 0.07 mm.

h = 6.63 x 10-34 Js hf = W + Kmax
λr = 700 nm
hc/ λ = W + Kmax
=
KmaxR = K hc/700nm = W + K ……(1)

= KmaxR = 2K 2hc/700nm = 2W + 2K ……(3)
W =?
f0 = ? (3) – (2) 2hc/700nm = 2W + 2K ……(3)

2hc/700nm - =W
W = 7.10357 x 10-20 J

(b) W = hc/λo h = 6.63 x 10-34 Js
λo = hc/W λ = 131 nm
= (6.63 x 10-34)(3 x 108) / (7.10357 x 10-20 )
λo = 2.8 x 10-6 m W = 7.10357 x 10-20 J

(c) hf = W + Kmax f0 = ?
hc/λ = W + Kmax Kmax = ?
λDB= ??

Kmax = hc/λ - W

= (6.63 x 10-34)(3 x 108) / (131 x 109) - 7.10357 x 10-20

= 1.447 x 10-18 J

λ = h/ √2Km
= (6.63 x 10-34)/ √2(1.447 x 10-18) (9.11 x 10-31)
= 4.0832 x 10-10 m



If the connection to the power supply is reversed, the potential
difference at the anode is set to negative and that will prevent the
arrival of the negatively charged photoelectrons.

If the potential divider, P is adjusted until the stopping potential, Vs
results in a zero milliammeter reading, then Vs is a measure of the
maximum kinetic energy, Kmax of the photoelectrons emitted, of
which Kmax = eVs.

a) Based on Einstein's Photoelectric Equation, derive an
equation that relates λ and Vs.

E = W + Kmax E = hf ; Kmax = EPE = eVs
E = hc
hc = W + eVs
λ λ

1/λ (x 106 m)
7.40
5.81
4.40
3.60
3.00
2.50

Vs(V) (i) To determine Planck’s constant

(8.0, 8.2) Vs = hc - W
eλ e
8
7 Draw a graph of Vs against (1/λ)
6
5 From Vs - 1/λ; Gradient = hc/e (8.0, 8.2)
4 (1.3 x 106, 0)
3 Gradient = . (8.2 – 0) .
2 (8.0 – 1.3) x 106
1
= 1.2239 x 10-6 Vm
(1.3, 0)
From Vs - 1/λ; Gradient = hc/e
12345678 Gradient = 1.2239 x 10-6 Vm

hc/e = 1.2239 x 10-6
h = (1.2239 x 10-6)(e)
c
= (1.222 x 10-6)(1.6 x 10-19)
(3.0 x 108)
= 6.527 x 10-34 Js

1/λ (106 m-1)

Vs(V) METHOD 1

8 (ii) To determine threshold wavelength, λ0
7 From Vs - 1/λ, x-intercept = 1/ λ0
6 1/λ0 = 1.3 x 106 m-1
5 λ0 = 7.692 x 10-7 m
4 λ0 = 769.2 nm
3
2 (iii) To determine work function, W
W = hc/λo
1 1 = x-intercept = 1.3 x 106 = (6.527 x 10-34 )(3 x 108)
λo 7.692 x 10-7
W = 2.546 x 10-19 J
12345678
1/λ (106 m-1)

λ =? hf = W + Kmax
h = 6.63 x 10-34
c = 3 x 108 hc = W + Kmax
Kmax = 10 eV = … J
W = 3.324 x 10-19 J λ

λ =? λ= hc .
h = 6.63 x 10-34
m = 9.1 x 10-31 kg W + Kmax
Kmax = 10 eV = … J
λ= (6.63 x 10-34) ( 3 x 108) .
K = mv2
(2.546 x 10-19) + (10 x 1.6 x 10-19)

λ = 1.072 x 10-7 m = 107.2 nm
v2 = 2K
m λ2 = h2 . λ2 = h2 .

m2v2 m (2 K)

λ2 = h2 . λ2 = (6.63 x 10-34)2 ..
m2 (2 K)
m (9.1 x 10-31) (2 x 10 x 1.6 x 10-19)

λ2 = h2 . λ2 = 1.5095 x 10-19 m
m (2 K)
λ = 3.885 x 10-10 m = 0.3885 nm

▪ Threshold wavelength is the maximum wavelength of photon required to release a

photoelectron
▪ From (b) the threshold wavelength, λ0 required for material X to emit photoelectron is 769.2

nm
▪ This threshold wavelength is longer than the wavelength of visible light (400 nm – 750 nm)
▪ Material X can be activated by radiation outside the wavelength of visible light and can

function in the dark.
▪ Material X can be used as a night vision device