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CPP-1
Batches - PHONON
Class - XI
MODERN PHYSICS
Sol.
E = hv =
C=
2.
Sol.
3.
hc
= PC

E
P
The collector plote in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is
put on and a saturation photocurrent is recorded. An electric field is switched on which has vertically downward
direction :
(A) The photocurrent will increase
(B*) The kinetic energy of the electrons will increase
(C) The stopping potential will decrease
(D) The threshold wavelength will increase

Electric force CE will increase the K.E. of emitted electron.
In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W.
The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches
its saturation value. Assuming that on the average one of very 106 photons is able to eject a photoelectron, find the
4.
2  10 6 e
A = 1.6  A]
hc
5
–6
hc
 /   × 10 × e = 1.6 µA
Find the maximum magnitude of the linear momentum of a photoelectron emitted when light of wavelength 400 nm falls
on a metal having work function 2.5 eV.
Ans. [
P 2  1.24 104 

eV = 0.6eV;
2m  4000 
2  9.1 10 31  0.6  1.6  10 19 = 4.2 × 10–25 kg. m/s]
 hc
  1242

– 2.5  eV
maximum K.E. =  – 2.5  = 

400

 

FI
Sol.
Ans. [
IT
photo current in the circuit.
Sol.
E
The energy of a photon of frequency  is E = hand the momentum of a photon of wavelength  is p = h/. From this
statement one may conclude that the wave velocity of light is equal to :
2
E
E
(A) 3 × 108 ms–1
(B*)
(C) E p
(D)  
p
 p
JE
1.
P2
= 0.605
2m
P=
5.
Sol.
–25
2  9.110–37  0.605 1.6 10–19 = 4.2 × 10 kg-m/s
If the frequency of light in a photoelectric experiment is doubled then stopping potential will :
(A) Be doubled
(B) Be halved
(C*) Become more than double
(D) Become less than double
e (v) = K.E.max = hv – Q
v' = 2u
Page 16
 hv –  
v = 

 e 
increament in v is more than double.
7.
Two separate monochromatic light beams A and B of the same intensity (energy per unit area per unit time) are falling
normally on a unit area of a metallic surface. Their wavelength are A and B respectively. Assuming that all the
incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam A to that from
B is :
 
(A*)  A 
 B 
 
(B)  B 
 A 
nA
nB
=
A
B
Sol.
2
[as intensity fallen on metal surface is same]
nA  A
=
nB  B
8.
 
(C)  A 
 B 
E
Sol.
When stopping potential is applied in an experiment on photoelectric effect, no photocurrent is observed :
This means that
(A) The emission of photoelectrons is stopped
(B*) The photoelectrons are emitted but are reabsorbed by the emitter metal
(C) The photoelectrons are accumulated near the collector plate
(D) The photoelectrons are dispersed from the sides of the apparatus.
Electron are emitted from emittor plats but not reaches on collector plats.
 B 
(D) 

 A 
2
JE
6.
Which one of the following graphs in figure shows the variation of photoelectric current (l) with voltage (V) between
the electrodes in a photoelectric cell ?
(B)
(C)
(D)
IT
(A*)
As potential is decreased, photoelectric current decreases and when it reaches upto stopping potential photo current
becomes zero, hence curve would be (A)
9.
STATEMENT-1 : Though light of a single frequency (monochromatic light) is incident on a metal, the energies of
emitted photoelectrons are different.
STATEMENT-2 : The energy f electrons just after they absorb photons incident on metal surface may be lost in
collision with other atoms in the metal before the electron is ejected out of the metal.
(A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
(C) Statement-1 is True, Statement-2 is False
(D) Statement-1 is False, Statement-2 is True
Maximum kinetic energy is achieved by only few electrons howeres rest of then lose some amount of energy in
collision with atoms.
Sol.
10.
Sol.
FI
Sol.
In the figure shown electromagnetic radiations of wavelength 200 nm are incident on the
metallic plate A. The photo electrons are accelerated by a potential difference 10V. these
electrons strike another metal plate B from which electromagnetic radiations are emitted.
The minimum wavelength of the emitted photons is 100nm. Find the work
function of the metal 'A'. use hc = 12400 eVÅ.


hc
hc

 10  eV = 3.8 eV]
Ans. [W = 
9
9

 200 10 e 100 10 e

(K.E.)max of electron after they are emitted from 1st surface :
Page 17
hc
–
1
(K.E.)m =
=
12400
–  = (6.2 – ) × v
2000
after accelerating from field.
K.E. = [(6.2 – ) + 10]eV =
12400
1000
FI
IT
JE
E
 6.2 – + 10 – 12.4 = 0

 = 3.6 eV
Page 18
CPP-2
Batches - PHONON
Class - XI
MODERN PHYSICS
=
where

E
Sol.
One milliwatt of light of wavelength  = 4560 Å is incident on a cesium metal surface. Calculate the electron current
liberated. Assume a quantum efficiency of  = 0.5 %. [work function for cesium = 1.89 eV].
Ans. [1.84 × 10–6 amp]
Electrons emitted per second
0.5
n
100
 hc 
n   = 1 × 10–3
 
n=
10–3  4560  10 –10
6.62 10 –34  3 108
= 23 × 1014
current =
0.5
×n×e
100
= 18.42 × 10–7 amp.
Sol.
The electric field at a point associated with a light wave is E = (100 V/m) sin [(3.0 × 1015 s–1)t] sin [(6.0 × 1015 s–1)t]. If
this light falls on a metal surface having a work function of 2.0 eV, what will be the maximum kinetic energy of the
photoelectrons ?
15


Ans. [  9  10 h  2  eV = 3.93 eV]
2

e



E = hv and v =
2
IT
2.
hence
= 1 + 2 = 3 × 1015 + 6 × 1015

 9 1015 
E =  2e  eV


Sol.
9 105
·h – 2 = 3.93 eV
2e
FI
(K.E.)m = E –  =
3.
JE
1.
Suppose the wavelength of the incident light is increased from 3000 Å to 3040 Å. Find the corresponding change in
the stopping potential. [Take the product hc = 12.4 × 10–7 eV m].
hc 
hc
.

Ans. [dVs =
× 107 = –5.5 × 10–2 volt]
e 2
228e
V =
Eincident
e
 hc hc 
=  – 
 1
2 
= 4.133 – 4.078
Page 19
= 5.5 × 10–2 V.
Calculate the number of photons emitted per second by a sodium vapour lamp consuming power 10 W. Assume that
60 % of the consumed energy is converted into light. Wavelength of sodium light = 590 nm.
Ans. [
Sol.
10 ×
60
hc
=n×
100

12400
× 1.6 × 10–19
5900

6= n ×

n = 1.77 × 1019
A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle
of incidence is 60º and the number of photons striking the mirror per second is 1.0 × 1019. Calculate the force exerted
by the light beam on the mirror. (h = 6.63 × 10–34 J.s.). Ans. [1.0 × 10–8 N]
Sol.
F=
E
5.
354 108
= 1.77 × 1019]
hc
P
t
h
n· ·2 cos 

=

11019  6.63 10 –34  2 
=
1
2
6630 10 –10
= 10–8 N
In an experiment on photoelectric effect, the emitter and the collector plates are
placed at a separation of 10 cm and are connected through an ammeter without
any cell. A magnetic field B exists parallel to the plates. The work function of the
emitter is 2.39eV and the light incident on it has wavelengths between 400 nm
and 600 nm. Find minimum value of B for which the current
registered by the ammeter is zero Neglect any effect of space charge.
Ans. [Bmin =
Sol.
IT
6.
JE
4.
 hc

10
2me 
 2.39 e  = 2.58 × 10 –5 T]

7
e
 4  10

hc
1
K.Emax = (400nm) – 2 – 39eV = meV2max
2
For no current registerd by ammeter
mV
eB
FI
Rmax = 10 cm =
Bmin =
meV
me
=
e(10cm) e(10cm)
2  hc

– 2.39eV 

mc  4nm

= 2.58 × 10–5 tex/a
7.
Sol.
A light ray of wavelength in the range of 450–700 nm emits descrete wavelengths of hydrogen spectrum of Balmer
series fall on a metal of work function 2eV. Find the maximum kinetic energy of one photo electron.
Ans. [Kmax = h –  = 0.55 eV]
1
=R

 1 1
 2 – 2
2 3 
Page 20
1
1 1
= 1.097 × 107  – 

4 9
9 – 4
= 1.097 × 107 

 30 
=
36
 10 –7 = 657 nm
5  1.097
And for n = 4 to n = 2 transition
=
1243
2.55
[for n = 4 E4 = –.55 eV, for n = 2 =3.4 eV]
Sol.
The graph is showing the photocurrent with the applied voltage of a photoelectric effect
experiment. Then :
(A*) A and B will have same intensity and B and C have same frequency.
(B) B and C will have same intensity and A and B have same frequency.
(C) A and B will have same frequency and B and C have same intensity.
(D) A and C will have same intensity and B and C have same frequency.
A 2 B will how same intensity becuase the saturation current are same.
B 2 C have same f requency have stopping potential is also same for both of them.
IP
A
B C
V
JE
8.
E
= 457 nm
Hence wavelength is ground 487 nm Hence energy 2.55 eV out of this 2eV is used as work function rest .55 eV is KE
of electrons.
Comprehension
A physicist wishes to eject electrons by shining light on a metal surface. The light source emits light of wavelength of
450 nm. The table lists the only available metals and their work functions :
Metal
W0(eV)
Barium
–
2.5
Lithium
–
2.3
Tantalum
–
4.2
Tungsten
–
4.5
a.
Which metal (s) can be used to produce electrons by the photoelectric effect from given source of light :
(A) Barium only
(B*) Barium or lithium
(C) Lithium, tantalum or tungsten
(D) Tungsten or tantalum
Energy of light
Sol.
IT
9.
hc 12400
=
= 2.755 eV

450
2.755 eV > work function of Ba and Li.
Sol.
c.
Sol.
10.
Which option correctly identifies the metal that will produce the most energetic electrons and their energies ?
(A*) Lithium, 0.45 eV
(B) Tungston, 1.75 eV
(C) Lithium, 2.30 eV
(D) Tungston, 2.75 eV
Having minimum work function.
(K.E.)max = 2.75 – 2.3 = 0.45 eV
FI
b.
Suppose photoelectric experiment is done separately with these metals with light of wavelength 450 nm. The maximum
magnitude of stopping potential amongst all the metals is :
(A) 2.75 volt
(B) 4.5 volt
(C*) 0.45 volt
(D) 0.25 volt
Maximum stopping potential =
( K .E.) max
= 0.45 volt.
e
When the sun is directly overhead, the surface of the earth receives 1.4 × 103 W/m2 of sunlight. Assume that the light
is monochromatic with average wavelength 5000Å and that no light is absorbed in between the sun and the earth's
Page 21
surface. The distance between the sun and the earth is 1.5 × 1011 m.
(a) Calculate the number of the photons falling per second on each square meter of earth's surface directly below the
sun.
(b) How many photons are there in each cubic meter near the earth's surface at any instant ?
(c) How many photons does the sun emits per second ?
Sol.
n×

7  10 4  4(1.5  1011 ) 2
7 104
7 104
13
= 3.5 × 1021; (b)
=
1.2
×
10
;
(c)
= 9.9 × 1044]
hc
hc
hc 2
hc
= 1.4 × 103

n=
1.4  103  5000
= 3.52 × 1021
12400 1.6 10–19
Time spent by a photon within 1m height above each =
1
c
 total photon in 1m3 volume
=
(c)
n
= 1.2 × 1013
c
n = (n)1m2 × A
FI
IT
JE
= 3.5 × 1021 × 4× (1.5×1011)2 = 9.9 × 1044
E
Ans. [(a) N =
Page 22
CPP-3
Batches - PHONON
Class - XI
MODERN PHYSICS

hc
= E Å
m
=
(b)
E
Sol.
A beam of monochromatic light of wavelength  ejects photoelectrons from a cesium surface ( = 1.9 eV) These
photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of  for which (a)
hydrogen atoms may be ionised (b) hydrogen atoms may get excited from the ground state to the first excited state and
(c) the excited hydrogen atoms may emit visible light.
hc
hc
hc
Ans. [(a)  =
m = 80 nm; (b)  =
m = 102 nm; (c)  =
m = 89 nm]
(13.6  1.9)e
(10.2  1.9)e
(12.08  1.9)e
(a)
m = 13.6 + 1.9 eV
12400
= 800Å
15.5
 3 

Em =    13.6  1.9  eV
4
 

= 10.2 + 1.9 = 12.1 eV
=
12400
= 1020 Å
12.1
JE
1.
(c) To emit visible light, e should so to atleast 3rd shell.
1 1  8
E = 13.6    = × 13.6 = 12.08
1 9  9
IT

Energy of photon = (12.08 + 1.9) eV = 13.98 eV
=
hC
. When a monochromatic light of wavelength  < 0 is incident such that
0
the plate gains a total power P. If the efficiency of photoelectric emission is % and all the emitted photoelectrons are
captured by a hollow conducting sphere of radius R already charged to potential V, then neglecting any interaction
between plate and the sphere, expression of potential of the sphere at time t is :
The work function of a certain metal is
FI
2.
(A) V +
Sol.
12400
= 890 Å
13.98
100 pe t
4 0 RhC
(B*) V –
Power gain by plate (P) =

 pe t
4000 RhC
(C) V
(D)
 pet
40 RhC
hC

p
 = c
Charge captured by sphere q =

× e × t
100
 p
= 100  c × e × t
Page 23
Potential = V –
3.
q
 pe t
=V–
4  0 R
4000 RhC
The radiation force experienced by body exposed to radiation of intensity I, assuming
surface of body to be perfectly absorbing is :
2
(A) R I
c
IRH
(C)
2c
Radiation pressure =
2 R·H
[triangle of base 2R and height H]
2
Intensily
I
=
speed
C
force = P + A =
4.
E
effective area =
I
× RH
C
In the shown experimental setup to study photoelectric effect, two conducting electrodes are enclosed in an evacuated
glass-tube as shown. A parallel beam of monochromatic radiation, falls on photosensitive electrode. Assume that for
each photons incident, a photoelectron is ejected if its energy is greater than work function of electrode. Match the
statements in column I with corresponding graphs in column II :
Column-II
(p)
IT
Column-I
(A) Saturation photocurrent versus intensity of
radiation is represented by
JE
Sol.
RHI
c
IRH
(D*)
c
(B)
(B) Maximum kinetic energy of ejected photoelectrons (q)
versus frequency for electrodes of different work
function is represented by
FI
(C) Photo current versus applied voltage for different
intensity of radiation is represented by
(r)
(D) Photo current versus applied voltage at constant
intensity of radiation for electrodes of different work
function.
Ans. [(A) r, (B) s, (C) p (D) q]
Sol.
(A) Saturation current v/s intensity 
(s)
i
Intensity
Page 24
1
(K.E.)m
2
3
V
(B) (K.E.)max v/s frequency
i
I1
I2 I > I > I
1
2
3
I3
(C) Photocurrent v/s voltage.
I
1
(D) Photocurrent v/s voltage
2
3
E
V
5.
Sol.
6.
JE
N
In a discharge tube when 200 volt potential difference is applied 6.25 × 1018 electrons move from cathode to anode and
3.125 × 1018 singly charged positive ions move from anode to cathode in one second. Then the power of tube is :
(A) 100 watt
(B) 200 watt
(C*) 300 watt
(D) 400 watt
P = VI
= 200 × [6.25 × 1018 × 1.6 × 10–11 + 3.125 × 1019 × 1.6 × 10–19]
= 200 × [1 + 0.5] = 300 watt
Find the velocity of photoelectrons liberated by electromagnetic radiation of wavelength  = 18.0 nm from stationary
He+ ions in the ground state.

6
2  109 hc
.
 54.4e  = 2.2 × 10 m/s]
me  18

IT
Ans. [
Sol.
E=
12400
= 68.89eV
180
Energy required to free electrons
= 13.6 × 4
= 54.4
max.m energy of e–
= 68.89 – 54.4
= 14.49 eV

7.
A beam of white light is incident normally on a plane surface absorbing 70 % of the light and reflecting the rest. If the
incident beam carries 10 W of power, find the force exerted by it on the surface.
Ans. [
Sol.
1
m v2 = 14.49 × 1.6 ×10–19
2 e
v = 2.2 × 106 m/s
FI

13
× 10–8 = 4.3 × 10–8 N]
3
p
Force =
t
 hc 
Power = 10 =    × n
 
Page 25
and [(p)total]incident =

 p
10
=
t 
c
 incident
Force =
=
13 13
=
× 10–8
c
3
The graph between 1/ and stopping potential (V) of three metals having works
V
Metal 1 Metal 2
functions 1, 2 and 3 in an experiment of photo-electric effect is plotted as shown
in the figure. Which of the following statement(s) is/are correct ? [Here  is the
wavelength of the incident ray]
(A*) Ratio of work functions 1 : 2 : 3 = 1 : 2 : 4.
0.001 0.002 0.004
(B) Ratio of work functions 1 : 2 : 3 = 4 : 2 : 1.
(C*) tan  is directly proportional to hc/e, where h is Planck's constant and c is the speed of light.
(D) The violet colour light can eject photoelectrons from metals 2 and 3.
E = 0 + KEmax
E = 0 + eV0
if
v0 = 0

E = 0
hc
= 0




1

3 < 2 < 1
is
1 1 1
3 : 2 : 1=  :  : 
1
2
3
Here
hc
e
h
w
[Q v0 =   f –
Þ v0 =
e
c
 
h c 
  – ]
c  c
When a monochromatic source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the
saturation current are respectively 0.6 V and 18 mA. If the same source is placed 0.6 m away from the cell, then :
(A) The stopping potential will be 0.2 V
(B) The stopping potential will be 1.8 V
(C) The saturation current will be 6.0 mA
(D*) The saturation current will be 2.0 mA
 0.2 
I' = I0· 

 0.6 
2
FI
Sol.
tan  =
IT
= .001 : .002 : .003
=1:2 :4
9.
Metal 3
E
Sol.
p
10
10
= 0.7 ×
+ 2 × 0.3 ×
t
c
c
JE
8.
h
×h

=
18
= 2 MA
9
Stopping potantial will not change.
10.
Sol.
Photoelectric effect supports quantum nature of light because :
(A*) There is a minimum frequency below which no photoelectrons are emitted
(B*) The maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(C*) Even when the metal surface is faintly illuminated the photoelectrons (if  th) leave the surface immediately
(D) Electric charge of the photoelectrons is quantized
(a) reshord frequency
(b) K.E.max = hv –  (work function)
(c) if frequency (v) > v0 (tresherd frequency), then e emitted from metel surface.
Page 26