Share this:

Photoelectric effect Questions

1.  (a) Define the following:

  (i) Photoelectric effect

  (ii) threshold wavelength

  (b) The variation of frequency f with the maximum kinetic energy Ek of the emitted

electrons is shown on the graph below:

Using the graph above, determine

(i) the threshold frequency fo of the radiation  *RCH*


(ii) the value of the Planck’s constant h

(iii) the work function, Wo

(c) On the same graph in (b) above, draw a line to show the variation of frequency, f, with the

maximum kinetic energy, Ek, of the emitted electrons from a second metal which has a lower

work function that used in (b)

2.  Figure 8 below shows a mercury vapour lamp, which emits ultraviolet light held over a

  negatively charged electroscope:

Image From



Image From












 (i) What happens to the leaf after the lamp is switched on?

 (ii) Explain why it happens

 (iii) If the experiment is repeated with equally bright red light held the same distance from the  

plate in place of the mercury vapour lamp, what effect would this have on the leaf?

  Give a reason

 (iv) What does photoelectric effect suggest about the nature of light?

3.  Calculate the wavelength of Green light whose energy is 3.37 x 10-19 J .

(h = 6.63 x 10-34JS, C = 3.0 x 108m/s)

4.  a) Define the term work function

 b) Name one factor that determines the velocity of photoelectrons produced on a metal surface

when light shine on it

 c) In a photoelectric effect experiment, a certain surface was illuminated with radiations of

different wavelengths and stopping potential determined for each wavelength. The table

below shows the results obtained.


Stopping potential , Vs






Wave length, (x10-7m)






 i) On the grid provided plot a graph of stopping potential (Y –axis) against frequency  

 ii) From your graph determine:

a) The threshold frequency

b) The plank’s constant, h

(e = 1.6 x 10-19 Coulomb, C = 3.0 x 108 m/s)

5.  a) State the role of the Grid in a cathode ray tube

b) Explain why a magnetic field is used in the TV deflection system instead of an electric field  

 c) The time base of a CRO is 25ms/div while its gain is 2.5V/div. Use this information to

answer the questions that follow:

 i) Calculate the frequency of the signal  

 ii) What is the peak voltage of the signal  

6.  The graph below shows the relation between the stopping potential, Vs and the frequency

  of radiation when a certain surface is illuminated with light of different frequencies























  From the graph determine:-

 (i) The threshold frequency

  (ii) The value of plank’s constant (e = 1.6 x 10-19C)

(III) The work function of the material




7  a) State one reason why a C.R.O is a more accurate voltmeter than a moving coil voltmeter

(b)The diagram below represents a cathode ray oscilloscope (CRO)












 i) Name the parts labeled A and B

 ii) What are the functions of C and D?  

 iii) State how electrons are produced

8.  a) What is meant by the term photo electric effect  b) In an experiment using a photo cell, ultra violet light of varying frequency strikes a metal

surface. The maximum Kinetic energy (KE max) of the frequency F is measured. The graph

below shows how the maximum kinetic energy varies with frequency F

Image From













Image From
















 Use the graph to determine:-

 i) Threshold frequency F

 ii) The plank’s constant, h

 iii) Work function of the metal


9.  (a) The diagram fig 9 below shows a photo cell; connected in a circuit:-

fig. 9








(i) Complete the diagram by indicating the correct polarities in the gap for current to flow in

the circuit

(ii) State and explain the effect of using light of different wave lengths on the amount of

current flowing in the circuit given that the distance of the source of light remains the same

(b) Two fixed resistors one of 100 and the other of unknown resistance are connected in parallel.

The combination is placed in a circuit and current passing through the combination was  

  measured for various p.d. The graph in figure 10 below drawn to scale shows the results:-

Image From















(i) From the graph, calculate the total resistance of the combination  

(ii) Determine the value of the unknown resistance

(c) (i) Explain the cause of eddy currents and how they are minimized in a transformer  

(ii) A transformer with 4200 turns in the primary coil operates a 240V mains supply and gives an output of 8.0V. Determine the number of turns in the secondary coil (assuming it is 10% efficient)


Image From EcoleBooks.com10.  State one factor that affects photoelectric effect  

11.  a) i) What is photoelectric effect?

  ii) You are provided with the following; a photo cell; a source of UV light, a rheostat,

a source of e.m.f, a millimeter, a voltmeter and connecting wires. Draw a circuit

diagram to show how photoelectric effect may be demonstrated in the laboratory  

 b) In a photoelectric effect experiment, a certain surface was illuminated with radiation of

  different frequencies and stopping potential determined for each frequency. The following

results were obtained:

Frequency (f) (x 1014 Hz)






Stopping potential, (Vs), (V)










i) Plot a graph of stopping potential (Y-axis) against frequency  

ii) Determine plank’s constant, h and the work function of the surface given that

EVs = hf – hfo, where hfo = Qe = 1.6 x10-19 C  

c) A surface whose work function Q = 6.4 x 10-19 J is illuminated with light of frequency

3.0 x1015 Hz. Find the minimum K.E of the emitted photo electrons

(use value of h obtained in b(ii) above)  


Photoelectric effect Answers

1.  a)  i) – The emission of electrons from metal surface when radiation of unstable wave

length falls on it

ii) The maximum wavelength beyond which no photoelectric effect occurs

Image From  b)











i) f = 6.4 x 1014H

ii) EK = hf –W

h = gradient

= (6.2 – 22) x 10-19

(16 – 10) x 1014

= 6.667 x 10-34Js

iii) Wo= hfo

= 6.667 x 6.4 x 1014 x 10-34

= 4.267 x 10-19J

Image From EcoleBooks.com2.  (a) (i) X – Intercept = 4.5 x 1014Hz

  (ii) Slope = h – h = e x slope

Image From e

= e x 6.6 – 0)V

  (6-4.5) x 1014

Image From = 1.610-19 x 4 x 10-15

= 6.4 x 10-34Js

Image From EcoleBooks.comImage From (iii) W0 =hf0

= 6.4 x 10-34 Js x 4.5 x 1014s-1

Image From = 2.88 x 10-19J

Image From EcoleBooks.comImage From

(b) (i) The leaf falls  Collapses

Image From   (ii) The electrons are repelled causing the leaf potential to decrease

  (iii) NO effect on the leaf. Light emitted by red light doesn’t have enough energy to cause


Image From EcoleBooks.comImage From photoelectric effect.

(iv) Light is a wave, it carries energy in small packets (protons).


3.  Calculate the wavelength of Green light whose energy is 3.37 x 10-19 J .

(h = 6.63 x 10-34JS, C = 3.0 x 108m/s)


 =
 = 3.37 X 10-19j  = 3.0 X10 8m/s

 6.63 X 10 -34 5.083X1014HZ

= 5.08 3X10 14HZ = 5.902 X 10 -7m

 =E


4.  a) This is the least radiation energy required to just dislodge an electron from a metal surface.

 b)The energy of the radiation. The higher the energy the higher the velocity of photo electrons

Frequency X 1014 Hz 7.959 7.43 6.88 6.10 5.49

 i) On the grid provided plot a graph of stopping potential (Y –axis) against frequency  

  Graph (diagram)

 ii) From your graph determine:

The threshold frequency  

o = 4.5 X 10 14 HZ  


 b) The plank’s constant, h

(e = 1.6 x 10-19 Coulomb, C = 3.0 x 108 m/s)

 eVs =hf –hfo gradient = 1.15 – 0.93  

Vs = h f – h fo (7.43 -6.98) X 10 14

= 0.22 X 10 -14

 gradient = h 0.55

e  = 0.4 X10-14

 = 0.4 X 10-14 X 1.602 X 10 -19 h = 6.408 X 10 -34js


5.  a) It controls the intensity of electron leaving the electron gun controlling the brightness

of the spot on the screen.

 b) The magnetic field deflection system make electrons span the whole screen unlike

the electric field deflection system.

c) i) Calculate the frequency of the signal  

T = 25ms/div X 2 div   f = 1

= 50 ms 50/1000

F = I

T   = 20HZ


ii) What is the peak voltage of the signal  

  • peak voltage = 21X 2.5v|d.v

=5 Volts


Image From EcoleBooks.com6.   (i) Graph is extrapolated to meet x-axis

Image From f0 = 7 x 1014Hz


 (ii) Gradient   = Vs


Image From = 1.75 – 0

12 – 7

Image From = 1.75 = 0.35 x 10-14

5 x 1014 = 3.5 x 10-15


h = Gradient x e

= 3.5 x 10-15 x 1.6 x 10-19

Image From EcoleBooks.comImage From  = 5.6 x 10-34Js


Image From W = hfo

  = 5.6 x 10-34 x 7 x 1014

Image From   = 3.92 x 10-19 J


7.   a) – Has infinite resistance/ does not take up any current

  – Sensitive/ does not require heating time

 b) i)   A – Grid

  B- Electron gun

  ii)  C – Vertical deflection of the beam

D- Horizontal deflection of the beam

iii) – By thermionic emission as heating the filament  


8.  a) Electrons being ejected from metal surfaces by use of electromagnetic waves

 b) i) X – intercept = 1.0 x 1015

  ii) From K.E = hf

Planks constant (h) = gradient of graph

= (8.2 – 0) x 10-19

(2.5 – 1.0) x 1015

= 8.2 x 10-19

1.5 x 1015

H = 5.5 x 10-34 JS


  iii) Work function, W0 = 5.5 x 10-34 x 1.0 x 10

= 5.5 x 10-19 J




Image From EcoleBooks.comImage From EcoleBooks.com9.  (a) (i)






Correct polarity

Image From No change in the amount of photo current. Change in wavelength/frequency of the

radiation does Image From not affect the amount of photo electrons produced. It is the number of  photo electrons that determines the photocurrent.


Image From EcoleBooks.comImage From (i) Total resistance = gradient

= 7.5 – 0 = 20

0.375 – 0

Image From

(ii) Combine d resistance = 100R

100 + R

Image From  100R = 20

100 + R

100R = 20R + 2000

Image From 80R = 2000 R = 25

(c) (i) Alternating magnetic flux in the coil induces current in the core of the same coil

Image From causing Image From EcoleBooks.comeddy currents.

– Eddy currents are minimized by lamination of the core

Image From

(ii) Vs = Ns = 8 = Ns Ns = 4200 x 8

  Xp Np 240 4200 240

10  – Intensity of the radiation

– Energy of the radiation

– Type of the metal

Ns = 140turns

11.  a)i) Emission of electrons from metal surface by electromagnetic radiation falling on

the surface

 b)  ii) M= u = 0.56 – 0 = 0.56 = 40 X 10 -15

e (6-4.6) x1014 1.4 X 1014


Image From h = 4.0 X 10-15 X 1.6 X 10 -19 = 6.4 X 10-34 Jl








 c)   hf = Q + K.E

6.4 X 10-34 X 3.0 X 1015 = 6.4 X 10 -19 + K.E

K.E = 19.2 X 10-19 – 6.4 X 10-19

= (19.2 – 6.4) X 10-19

= 12.8 X 10-19

= 1.28 X 10-18 J




Share this:

EcoleBooks | Photoelectric effect Questions And Answers (part 2)


Leave a Reply

Your email address will not be published. Required fields are marked *

Accept Our Privacy Terms.*