Chapter-11
DUAL NATURE OF MATTER
AND
RADIATION
Work function (jo): The minimum energy
required for an electron to escape from the
surface of a metal
i.e. The energy required for free electrons
to escape from the metal surface
Work function depends on the properties of the
metal and nature of the surface.


Work function of alkali metals is
less as compared to other metals.
Work function of metals decreases
with increase in atomic number.


Cesium is the best photosensitive
metal as it has least Work
function.
Infrared
radiation
can
not
produce photoelectric effect on
any metal surface
The phenomenon of emission of electrons
from metal surfaces exposed to light energy of
suitable frequency is called
photoelectric effect.
Photoelectric current increases linearly
with increase in intensity of incident light.
I μA
Intensity (L)
i.e. the number of electrons emitted per second
is directly proportional to intensity of light.
Photoelectric current increases with
increase in +ve potential applied to the
anode.
I
μA
L2
L 2 > L1
Saturation Current
VS
L1
+
0
Potential of A (V)
When all the photoelectrons reach the plate
A, current becomes maximum and is known
as saturation current.
When the potential is decreased, the
current decreases but does not become
zero at zero potential.
i.e.even in the absence of accelerating
potential, a few photoelectrons manage to
reach the plate on their own due to their
K.E.
When –ve potential is applied to the plate A
with respect to cathode, photoelectric
current becomes zero at a particular value of
–ve potential called stopping potential or cutoff potential.
Intensity of incident light does not affect
the stopping potential.
If Kmax is the maximum kinetic energy of
emitted electron ,
Kmax=e Vo
I μA
Saturation Current
 2>  1
2
1
VS2
VS1 0
Potential of A (V)
+
The saturation current is same for
different frequencies of the incident lights
of same intensity.
Higher the frequency, higher the stopping
potential. i.e. VS α 
Variation of stopping potential with frequency of
incident radiation
VS
(V)
1
νo ν
1 o2
2
 Stopping potential 0 varies linearly with the
frequency of incident radiation
There exists a certain minimum cut-off
frequency ν0 for which the stopping potential is
zero.
Einstein’s photoelectric equation
Photon energy=work function+
Max.K.E
h  h o  K .Emax
h  h
 K .E
o
max
hc
hc

 K .E
max

o
hc
hc
1
2


mv

o
2
1
1
K .E
 hc 
max



o





Photon concept of electromagnetic radiation
 In interaction of radiation with matter, radiation
behaves as if it is made up of particles called
photons.
Each photon has energy E (=hν) and momentum
p (= h ν/c).
 All photons of light of a particular frequency ν,
or wavelength λ, photon energy is independent of
intensity of radiation.
 Photons are electrically neutral and are not
deflected by electric and magnetic fields.
 In a photon-particle collision , the total energy
and total momentum are conserved.
However, the number of photons may not be
conserved in a collision. The photon may be
absorbed or a new photon may be created.
Dual nature of matter
The wave associated with material particles in
motion is called matter wave or de Broglie wave.
h
h
 
p mv
Matter wave is neither electromagnetic nor
mechanical
Relation between de-Broglie wavelength and kinetic energy
of the particle
h
1

 
2mE
mE
h
1

 
2mqV
mqV
If electron is accelerated through a potential
difference of V volt
h
12.27


A
2meV
V
o
1.The work function of a metal is the
minimum energy
required
1) to remove valence electron from a metal
atom
2) to remove any electron from a metal atom
3) to remove free electron from a metal
surface
4) to remove any electron from a metal
surface
Ans: To remove free electron from a metal
surface
2.The work functions of lithium and copper are
2.1 eV and 4.3 eV respectively. Out of these, the
one which is
suitable for the photoelectric cell
that works with the visible light is
1) Lithium
2) copper
3) both lithium and copper
4) neither lithium nor copper
Maximum energy of visible energy photon
is about 3.1 eV.
Hence visible light can produced photoelectric effect
on Lithium but not on Copper
3.A metallic surface has a threshold wave length of
5200Å. This surface is irradiated by monochromatic
light of wavelength 4500Å. Which of the following
statement is true?
1) the electrons are emitted from the surface with
energy between 0 to infinity
2) the electrons are emitted from the surface with
energy between 0 and finite maximum value
3) the electrons are emitted from the surface all
with certain finite energy
4) no electron is emitted from the surface
incident <o
Ans: 2)Photoemission takes place with
kinetic varying from zero to maximum
4.Blue light can cause photoelectric
emission from a metal but yellow light
cannot. If red light is incident on the metal
then,
1) photoelectric current will increase
2) rate of emission of photoelectrons will
decrease
3) no photoelectric emission will occur
4) energy of the photoelectrons will
increase
Threshold wavelength is less than that of
yellow light. But wavelength of red is
longer than that of yellow
So the red light cannot cause
photoelectric emission as R > Y > B
Ans: 3)
5. Light of frequency 1.5 times the
threshold frequency is incident on photo
sensitive material. If the frequency is
halved and intensity is doubled, the
photoelectric current becomes
1)quadrupled
3) halved
2) doubled
4) zero
Ans: 4) When frequency is halved, it becomes
0.750. No emission of light takes place (since 
< 0)
Ans: 4) zero
6. Which of the following radiation will
produce photoelectrons with highest
kinetic energy
1) X-rays
2) gamma rays
3) UV rays
4) visible light
Radiation with highest frequency will liberate
electrons with maximum kinetic energy
Ans: 2) Gamma rays
7. The work function of sodium and aluminium is
2eV and 4.2eV respectively. Then
1) Threshold of wavelength of sodium is less
than that of aluminium
2) Threshold of wavelength of sodium is more
than that of aluminium
3) Threshold of wavelength of sodium is equal
to that of aluminium
4) None of the above
1
work function
threshold wavlength
If sodium  alu minium  o(sodium)  o(alu minium)
Ans: 2) Threshold of wavelength of sodium is
more than that of aluminium
1
8. The graph between  and stopping potential
(V) for three metals is plotted as shown . Which of
the following statement is correct ?
1) Ratio of work functions = 1 : 2 : 4.
2) Ratio of work functions = 4 : 2 : 1.
3) tan is equal to Planck’s constant.
4) The violet colour light can eject
photoelectrons from metals 2 and 3
1


o
 : : 
1
2
3
1

1
:
1

:
2
 :  :   1: 2 : 4
1
2
Ans: 1
3
1

3
9.According to Einstein photoelectric equation
slope of plot of the stopping potential of the
emitted photo electron verses frequency
1Depends on the intensity of radiation and
metal used
2)Depends on the intensity of radiation
3)Depends on the metal used
4)Independent of metal used and incident
radiation
Ans: (4) always the value of slope is a
constant whose value is equal to plank’s
constant.
10. Relation between stopping potential Vo and
maximum velocity of photoelectron v is
1
1) Vo 
v
2
3) Vo  v
1
2)Vo  2
v
4)Vo  v
Vo K.E max and K.E max  vmax
 Vo  vmax
Ans: 3
2
2
11.Light of frequency 1.5 times the
threshold
frequency
produce
photoemission on a metal surface. If the
frequency is halved and intensity is
doubled,
the
photoelectric
current
becomes
1) quadrupled
3) halved
2) doubled
4) zero
When frequency is halved, it becomes 0.750.
Since  < 0 No emission of light takes place.
Ans: 4) zero
12. The photoelectrons emitted from a metal
surface
1) Have the same momentum
2) Have the same kinetic energy
3) Have speeds varying from zero up to a certain
maximum value
4) Are all at rest
Ans: photoelectron can have kinetic varying
from zero to certain maximum value.
Therefore photoelectrons can have speeds
varying from zero up to a certain maximum
value
13. The correct curve stopping potential verses
intensity of incident radiation
(X-axis Intensity Y-axis Stopping potential)
1
2
3
4
Stopping potential is independent of intensity.
Ans: (3)
14. The work function of a metal is 2eV . When
a radiation of frequency1.6x1015HZ.is
incident on it ,then
1) There will be photoemission
2)No photoemission
3)There will be photoemission if intensity is
very high
4)There will be photoemission if intensity is
very less
Energy of photon E  h  6.63x10 x1.6x10 J
34
34
15
15
6.63x10 x1.6x10

 6.63eV
19
1.6x10
 h  o
Ans: There will be photoemission
15. A photon of energy 8 eV is incident on
a metal surface of threshold frequency
1.6  1015 Hz. The K. E. of the photo
electrons emitted (in eV), (h = 6  10–34 Js)
1) 1.8
3) 2
2) 6
4) 1.2
34
6 10 1.6 10
work function  h o 

6eV
19
1.6 10

Emax = 8 – 6 = 2 eV
15
16. Maximum velocity of photoelectron from a
surface is 1.2 106 ms– 1. Assuming the specific
charge of electron to be 1.8 1011 Ckg– 1 ,the value
of stopping potential in volt will be
1) 4V
2)
3) 2V
4) 3 V
6V
1
2
mv max  eV0
2
2
6 2
1 v max 1 (1.2x10 )
 Vo 
 x
4
11
2 e
2 1.8x10
m
Ans: 1) 4 V
17. When a metal surface is illuminated with a
monochromatic radiation of wavelength λ the
stopping potential is 3V. When it is illuminated
with radiation of wavelength 2λ,stopping
potential reduces to V. Threshold wavelength
for the metal is
1) 8 λ
2) 6 λ
3) 4 λ
4) 4 λ/3
hc
hc
K.E =
λ λ
o
 3Ve = hc - hc .........(1)
λ λ
o
hc
hc
Ve=
...........(2)
2λ λ
o
  4
o
Ans:(3)
18. The kinetic energies of photoelectrons
emitted from a metal are K1 and K2 when it is
irradiated with lights of wavelength 1 and 2
respectively. The work function of the metal is
K1λ1  K 2 λ 2
1)
λ 2  λ1
K1λ1  K 2 λ 2
2)
λ 2  λ1
K1λ 2  K 2 λ1
3)
λ 2  λ1
K1λ 2  K 2 λ1
4)
λ 2  λ1
hc
 W + K1  h c  W λ1 + K1 λ1 ......................(1)
λ1
hc
 W + K 2  h c  W λ 2 + K 2 λ 2 .....................(2)
λ2
From (1) and (2)
W λ 2 + K 2 λ 2  W λ1 + K1 λ1
or W λ 2  W λ1  K1 λ1  K 2 λ 2
K1 λ1  K 2 λ 2
 W 
λ 2  λ1
Answer: (1)
19. When ultraviolet light is incident on a
photocell, its stopping potential is Vo and the
maximum momentum of electron is Po .When Xray is incident on the same cell
1) Both Vo and Po will increase.
2) Both Vo and Po will decrease
3) Vo increase and Po will decrease
4) Vo remain same and Po will increase.
Energy of X-ray photon is greater than ultraviolet
photon
Therefore Kinetic energy of emitted electron and
stopping potential w ill be more for ultraviolet light
In turn the momentum of the emitted electron
Ans: 1)Both Vo and Po will increase.
20. In an experiment with photoelectric effect,
the slope of stopping potential verses frequency
of incident light is found to be 4.130x10-15 Vs.
The value of plank’s constant from the graph is
1)6.591x10-34Js
2)3 6.650x10-34Js
3) 6.616x10-34Js
4) 6.631x10-34Js
Slope of Vo verses  = h/e
h =slope x e=4.130x10-15 x1.6x10-19
-34
6.616x10
=
Ans: (2)
21. When a monochromatic point source of light at
a distance of 0.2m from a photocell. The cut off
voltage and saturation current is 0.6V and 18mA. If
the same source is placed 0.6m away from the cell,
the cut off voltage and saturation current is
1) 0.2V and 2mA
3) 0.6V and 6mA
2) 1.8V and 6mA
4) 0.6V and 2mA
As the source is moved away ,the intensity of
light falling on the surface reduces according to
the relation
1
intensity α
2
distance
Therefore when the distance is increased by 3
times intensity reduces by 9 time and hence the
current reduces to 2mA
Frequency remains same and hence the cut off
voltage remains same (0.6V)
Ans:
4) 0.6V and 2mA

22. The maximum kinetic energy of
photoelectrons is 1eV and 3eV for incident
3
photon frequency  and 2  respectively.
Maximum kinetic energy of photoelectrons for
9
the incident frequency  4
1) 8eV
2) 6eV
3) 45eV
4) 3eV
Let Energy of light freuency   E1 ;
3
3
 Energy of light freuency   E1
2
2
9
9
 Energy of light freuency   E1
4
4
From Einstein 's equation
1  E1   

  E1  4 ,   3
3
3  E1   
2

9
 K.E  E1    6eV
4
Ans: 2) 6eV
23. A photosensitive metal is first incident with
the radiation of wavelength 400 nm and then with
radiations of wavelength 800 nm. The change in
the maximum kinetic energy of the photoelectron
is
1) 0.55 eV
3) 2.0 eV
2) 1.55 eV
4) 1.0 eV
1 1
E  hc   
 1 2 
 20x10
26
 1.56eV
Ans: (2)
1 
 1

7
 4x10 8x107 
24. Maximum velocity of photoelectrons emitted
by a photo emitter is 2X106 m/s. If e/m=1.8X1011
for electrons, the stopping potential of the
emitter is
1) 11.1V
3) 40V
2) 20V
4) 2V
1 2
mvmax  eV0
2
2
6 2
1 vmax 1 (2x10 )
 Vo 
 x

11.1V
11
2e
2 1.8x10
m
Ans: 2) 11.1V
25. Light photons of energies 1 eV and 2.5 eV
are successively incident on a metal surface of
work function 0.5 eV , then the ratio maximum
velocities of the emitted photo-electrons will be
1) 1 : 5
3) 1 : 3
2) 1 : 4
4) 1 : 2
26. When UV light of wavelength 100 nm is
incident on a silver surface of work function
4.7eV, a negative potential of 7.7V is required to
stop the photo-electrons from reaching the
collector plate. The potential which is required
to stop the photoelectrons when light of
wavelength 200 nm is incident on it will be
1) 1.5V
3) 4.5V
2) 3V
4) 6V
E
= 7.7 eV,
W=4.7eV
K1
Energy of photon: E = W + E
1
K1
= 4.7 + 7.7
= 12.4 eV
E
λ
2 = 1 = 100 nm = 1
E
λ
200 nm 2
1
2
E
E = 1  6.2eV
2
2
2nd case : Energy of photon : E  6.2 eV
2
E
 E -W  6.2 - 4.7  1.5 eV
K2
2
Hence the stopping potential is 1.5V
Ans:1
27. The ratio of linear momentum of an
electron and alpha particle when both are
accelerated through a potential difference of
100 volt
1) 1
2)
2me
ma
3)
me
ma
4)
me
2ma
p 
1

pe


p

mq & q  2qe
me qe

m q
Ans : 4)
me
2ma
me
2m
28. An X-ray tube operates at 10kV. The ratio of
X-ray wavelength to that of de-Broglie is
1)10:1
3) 1:100
2) 1:10
4) 100:1
de  Broglie wavelength d 
hc
Wavelength of X  ray x 
eV
d
2m
c
 10
x
eV
d : x  10 :1
Ans: 1)10:1
h
2meV
29. If the kinetic energy of the particle is
increased by 16 times ,the percentage change
in the de-Broglie wavelength is
1)25%
2)75%
3)60%
4)50%
1

E
2

1
E1

E2
E1
1

16 E1
4
1
2  1  0.251
4
change in wavelength  0.75 or 75%
Ans:
2)75%
30. The de-Broglie wavelength associated with
proton changes by 0.25% if its momentum
changes by p. The initial momentum is
1) 100 p
3) 401 p
2) p/400
4) p/100
h
       (1)
p
h
 0.25 
         (2)
1 
 
100 
p p

solving above equations p  401 p
Ans: 3) 401 p
31. The de-Broglie wavelength of electron in ground
state of hydrogen atom is
o
1) 0.5A
o
o
o
2)1.06A 3) 1.67A 4)3.33A
The circumference of ground state orbit is equal to
the wavelength of corresponding de-Broglie wave
i.e.
o
  2 ro (ro  0.53A)
o
 3.33A
o
Ans :4)3.33A
32. The maximum kinetic energy of
photoelectrons from the surface of silver of
threshold wavelength λo due to a radiation of
wavelength λ (λ< λo ) is given by
1) hc(o   )
hc
2)
(   )
1 1 
h
3)   
c   
   
4) hc 

  
o
0
o
o
K .E 
max
hc


hc

o
   
 hc 

  
o
o
Ans: (4)
33. Of the following the graph which represents the
variation of momentum (p) of a particle with the
wavelength () of matter wave is
p
p

p

p


h
1
   
p
p
Hence the plot is a rectangular hyperbola
Ans:3
34.If the velocity of a particle is reduced to
one-third then the percentage increase in its
de-Broglie wavelength is
1) 50%
2) 100%
3) 200%
4) 300%
h
 h 
and λ' =
= 3
= 3λ

mv
v

m 
3
Change in de - Broglie wavelength = 2λ
h
λ=
mv
2λ
Percentage change =
×100% = 200%
λ
Answer is (3)
35. The ratio of de-Broglie wavelength
associated with proton and  particle
accelerated through same potential
difference
1) 1
2)2
3)
1
4)
8
8
For a charged particle : λ =
For a proton : λ p 
h
2mq V
λα
=
8me
me
mq
1
me
1
For α - particle : λ α 
=
(4m) (2e)
λp

1
= 8
Answer is (3)
1
8me
36. If the momentum of two particles of
mass m and 2m are equal, then the ratio of
wavelength of the matter wave is
1) 1 : 2
3) 1 : 1
2) 2 : 1
4) 1 : 3
h
1 p 2
  
1
p
2 p1';
1 : 2 = 1 : 1
Ans: (3)
37. An X–ray photon has a wavelength of 0.02 Å.
Its momentum is
1) 3.3  10– 22 kg m/s
2) 6.6 10– 21 kg m/s
3) 6.6 10– 24 kg m/s
4) 1.65  1022 kg m/s
h
Momentum: p=
λ
-34
-34
6.6×10
6.6×10
=
=
-10
-12
0.02×10
2×10
-22
=3.3×10
Answer: (1)
38. If E is the energy, de-Broglie wavelength is
proportional to
1) E–1 for both photons and particles
2) E–1 for photons and E–1/2 particles
3) E–1/2 for both photons and particles
4) E–1/2 for photons and E–1 for particles
For photon: E =
hc
λ
1
λ
 λ  E 1
E
h
For a particle: λ=
2mE

Answer is
1
λ
E
(2)

1

λ E 2
39. For given kinetic energy which of the following
has the longest de-Broglie wavelength?
1) Electron
3)Neutron
2)Proton
4) alpha particle
h
1
λ=
or λ 
,
2mE
m
 >      (m < m < m < m )
e
p
n
Answer: (1)

e
p
n
α
40. Electrons used in an EM are accelerated by a
voltage of 25kV. If the voltage is increased to
100kV then the de-Broglie wavelength associated
with the electrons would
1) Increases to 2 times
2) increases to 4 times
3)decreases by 2 times
4)decreases by 4 times
λ
V
25
kV
1
1
1
2
λ
 


V
100 kV 2
λ
V
2
1
λ
1
λ 
2 2
Ans: 3