PHYSICS
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SYLLABUS
PHYSICS (Code No. 042)
COURSE STRUCTURE, CLASS XII (2021 - 22)
TERM - II
Time : 2 Hours Max Marks : 35
No. of Periods
Units-V
Units-VI
Units-VII
Units-VIII
Units-IX
Electromagnetic Waves
Chapter-8:Electromagnetic Waves
Optics
Chapter-9: Ray Optics and Optical Instruments
Chapter-10: Wave Optics
Dual Nature of Radiation and Matter
Chapter-11: Dual Nature of Radiation and Matter
Atoms and Nuclei
Chapter-12: Atoms
Chapter-13: Nuclei
Electronic Devices
Chapter-14: Semiconductor Electronics: Materials, Devices and Simple Circuits
Total
THEORY
Unit V: Electromagnetic Waves 2 Periods
Chapter-8 : Electromagnetic Waves
Electromagnetic waves, their characteristics, their Transverse nature
(qualitative ideas only).
Electromagnetic spectrum (radio waves, microwaves, infrared, visible,
ultraviolet, X-rays, gamma rays) including elementary facts about their
uses.
Unit VI: Optics 18 Periods
Chapter-9 : Ray Optics and Optical Instruments
Ray Optics: Refraction of light, total internal reflection and its applications,
optical fibres, refraction at spherical surfaces, lenses, thin lens formula,
lensmaker's formula, magnification, power of a lens, combination of thin
lenses in contact, refraction of light through a prism.
Optical instruments: Microscopes and astronomical telescopes (reflecting
and refracting) and their magnifying powers.
Chapter–10: Wave Optics
Wave optics: Wavefront and Huygens principle, reflection and refraction
of plane wave at a plane surface using wavefronts. Proof of laws of
reflection and refraction using Huygens principle. Interference, Young's
double slit experiment and expression for fringe width, coherent sources
and sustained interference of light, diffraction due to a single slit, width
of central maximum
Marks
02
17
18
07
11
11
11
7
73
35
Unit VII : Dual Nature of Radiation and
Matter
7 Periods
Chapter–11: Dual Nature of Radiation and Matter
Dual nature of radiation, Photoelectric effect, Hertz and Lenard's
observations; Einstein's photoelectric equation-particle nature of light.
Experimental study of photoelectric effect Matter waves-wave nature of
particles, de-Broglie relation
Unit VIII: Atoms and Nuclei 11 Periods
Chapter–12: Atoms
Alpha-particle scattering experiment; Rutherford's model of atom; Bohr
model, energy levels, hydrogen spectrum.
Chapter–13: Nuclei
Composition and size of nucleus Nuclear force Mass-energy relation, mass
defect, nuclear fission, nuclear fusion.
Unit IX: Electronic Devices 11 Periods
Chapter–14: Semiconductor Electronics: Materials, Devices
and Simple Circuits
Energy bands in conductors, semiconductors and insulators (qualitative
ideas only) Semiconductor diode - I-V characteristics in forward and
reverse bias, diode as a rectifier; Special purpose p-n junction diodes: LED,
photodiode, solar cell.
PRACTICALS
The second term practical examination will be organised by schools as per the directions of CBSE and viva will be taken by both internal and external
observers. The record to be submitted by the students at the time of second term examination has to include a record of at least 4 Experiments and 3
Activities to be demonstrated by teacher.
Evaluation Scheme
Time Allowed : One and half hours Two experiments to be performed by students at time of examination
Practical record [experiments and activities]
Viva on experiments, and activities
Total
Max. Marks : 15
8 marks
2 marks
5 marks
15 marks
CONTENTS
Chapter 8
Electromagnetic Waves
Chapter 9
Ray Optics and Optical Instruments
17-39
Chapter 10
Wave Optics
40-62
Chapter 11
Dual Nature of Radiation and Matter
63-81
Chapter 12
Atoms
82-102
Chapter 13
Nuclei
103-114
Chapter 14
Semiconductor Electronics :
Materials, Devices and Simple Circuits 115-136
Practice Papers 1-3
137-156
1-16
*As per the Circular Issued by CBSE on July 05, 2021, Special Scheme of Assessment
for Board Examination Class XII for the Session 2021-22 is as follows :
Term II Examination/Year-end Examination :
•At the end of the second term, the Board would organize Term II or Year-end Examination based on the rationalized
syllabus of Term II only (i.e. approximately 50% of the entire syllabus).
•
This examination would be held around March-April 2022 at the examination centres fixed by the Board.
•The paper will be of 2 hours duration and have questions of different formats (case-based/ situation based, open
ended- short answer/ long answer type).
•In case the situation is not conducive for normal descriptive examination a 90 minute MCQ based exam will be
conducted at the end of the Term II also.
•
Marks of the Term II Examination would contribute to the final overall score.
To cope up with ongoing unpredictable pandemic situation, this book contains chapterwise objective as well as subjective
questions.
In Objective Section, each question carry 1 mark and in Subjective Section, each VSA carry 1 mark, SA I carry 2 marks,
SA II carry 3 marks and LA carry 5 marks.
*As per the CBSE Term-II 2021-2022 curriculum, for latest information visit www.cbse.gov.in
Electromagnetic
Waves
CHAPTER
8
Recap Notes
Electromagnetic (E.M.) waves : E.M.
waves are those waves in which there is a
sinusoidal variation of electric and magnetic
fields at right angle to each other as well
as at right angle to the direction of wave
propagation.
X For a plane progressive electromagnetic
wave propagating along + Z direction,
the electric and magnetic fields can be
written as
E = E0 sin (kz – wt)
B = B0 sin (kz – wt)
Y
where m and e are permeability and
permittivity of the medium respectively.
c
1
v=
=
µ 0µ r ε 0 ε r
µr εr
X
E
X
Z
X
X
X
B
In electromagnetic wave, the electric and
magnetic fields vary with space and time
and have the same frequency and are in
the same phase.
The amplitudes of electric and magnetic
fields in free space, in electromagnetic
waves are related by
E
E0 = cB0 or B0 = 0
c
The speed of electromagnetic wave in
free space is
E
1
= 3 × 108 m/s
c= 0 =
B0
µ0ε0
where m0 and e0 are the permeability and
permittivity of free space respectively.
X
The speed of electromagnetic wave in a
medium is
1
v=
µε
X
X
Properties of electromagnetic waves
– These waves do not carry any charge.
– These waves are not deflected by
electric and magnetic fields.
– They travel with the speed of light c
(= 3 ×× 108 m s–1) in vacuum.
– The frequency of electromagnetic wave
does not change when it goes from one
medium to another but its wavelength
changes.
– These waves are transverse in nature,
hence they can be polarised.
Production of electromagnetic waves
– Maxwell showed that an electric
charge oscillating harmonically with
frequency u produces electromagnetic
waves of the same frequency.
– An electric dipole is a basic source of
electromagnetic waves.
Energy density of electromagnetic
waves
– Electromagnetic waves carry energy
as they travel through space and this
energy is equally shared by electric field
and magnetic field of electromagnetic
wave.
– The energy density of the electric field
is
1
2
u E = ε 0 Erms
2
– The energy density of magnetic field is
uB =
2
1 Brms
2 µ0
CBSE Board Term-II Physics Class-12
2
– Average energy density of electromagnetic wave is
1
1 2
2
< u > = ε 0 Erms
+
B
2
2µ 0 rms
X
X
Intensity of electromagnetic wave
: It is defined as energy crossing per
unit area per unit time perpendicular
to the direction of propagation of
electromagnetic wave. The intensity of
electromagnetic wave is
B2
1
2
I = < u > c = ε 0 Erms
c + rms c
2
2µ 0
Momentum of electromagnetic wave
– An electromagnetic wave carries linear
momentum.
– Electromagnetic wave strikes the
surface at normal incidence and
transports a total energy U to the
surface in a time t , if the surface
absorbs all the incident energy, the
total momentum p transported to the
surface is
p=
Type
Radio
waves
U
(complete absorption)
c
Wavelength
range
> 0.1 m
Microwaves 0.1 m to 1 mm
Frequency
range (in Hz)
< 3 ×× 109
3 ××× 108 to
3 ×× 1011
3 ×× ×1011 to
4 ×× 1014
Infra-red
1 mm to 700 nm
Visible
light
700 nm to 400 nm 4 ×× 1014 to
8 ×× ×1014
– If the surface is a perfect reflector
and incidence is normal then the
momentum transported to the surface
is
2U
(complete reflection)
p=
c
Radiation pressure : It is defined as the
pressure exerted by the electromagnetic wave
on a surface.
X If I is the intensity of the incident
electromagnetic radiation, then the
radiation pressure for normal incidence
is
Pradiation =
I
(perfectly absorbing surface)
c
Pradiation =
2I
(perfectly reflecting surface)
c
Electromagnetic spectrum : The orderly
distribution of electromagnetic radiations
according to their wavelength or frequency
is known as electromagnetic spectrum.
Production
Detection
Rapid acceleration
and deceleration of
electrons in aerials
Klystron valve or
magnetron valve
Vibration of atoms
and molecules
Receiver’s aerials
Electrons in atoms
emit light when they
move from one energy
level to a lower
energy level
Inner shell electrons
in atoms moving from
one energy level to a
lower level
The eye, Photocells,
Photographic film
Point contact diodes
Thermopiles, Bolometer,
Infrared photographic
film
Ultraviolet 400 nm to 1 nm
8 ××× 1014 to
8 ×× 1016
X-rays
1 nm to 10–3 nm
1 ××× 1016 to
3 ×× 1021
X-ray tubes or inner
shell electrons
Photographic film,
Geiger tubes
Gamma
rays
< 10–3 nm
> 3 ××× 1021
Radioactive decay of
the nucleus
Photographic film, Ioniz
ation chamber
Photocells, Photographic
film
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. A plane electromagnetic wave travels in
vacuum along z-direction. If the frequency of
the wave is 40 MHz then its wavelength is
(a) 5 m
(b) 7.5 m
(c) 8.5 m
(d) 10 m
2. An electromagnetic wave of frequency
u = 3 MHz passes from vacuum into a dielectric
medium with permittivity e = 4. Then
(a) wavelength and frequency both become half.
(b) wavelength is doubled and frequency
remains unchanged.
(c) wavelength and frequency both remain
unchanged.
(d) wavelength is halved and frequency remains
unchanged.
3. The ratio of contributions made by the
electric field and magnetic field components to
the intensity of an electromagnetic wave is
(a) c : 1
(b) c2 : 1
(c) 1 : 1
(d) c :1
4. About 6% of the power of a 100 W light bulb
is converted to visible radiation. The average
intensity of visible radiation at a distance of
8 m is (Assume that the radiation is emitted
isotropically and neglect reflection).
(a) 3.5 × 10–3 W m–2
(b) 5.1 × 10–3 W m–2
(c) 7.4 × 10–3 W m–2
(d) 2.3 × 10–3 W m–2
5. Which of the following rays is not an
electromagnetic wave?
(a) X-rays
(b) g-rays
(c) b-rays
(d) Heat rays
6. A radio can tune to any station in 7.5 MHz
to 12 MHz band. The corresponding wavelengthband is
(a) 40 m to 25 m
(b) 30 m to 25 m
(c) 25 m to 10 m
(d) 10 m to 5 m
7. Which waves are used in sonography ?
(a) Microwaves
(b) Infrared rays
(c) Radio waves
(d) Ultrasonic waves
8. Light with an energy flux of 18 W cm–2 falls
on a non-reflecting surface at normal incidence.
If the surface has an area of 20 cm2, the average
force exerted on the surface during a 30 minute
time span is
(a) 2.1 × 10–6 N
(b) 1.2 × 10–6 N
6
(c) 1.2 × 10 N
(d) 2.1 × 106 N
9. T h e p a r t o f t h e s p e c t r u m o f t h e
electromagnetic radiation used to cook food is
(a) ultraviolet rays
(b) cosmic rays
(c) X-rays
(d) microwaves
10. A plane electromagnetic wave of frequency
25 MHz travels in free space along x-direction.
At a particular point in space and time, electric
field E = 6.3 V m–1. The magnitude of magnetic
field B at this point is
(a) 1.2 × 10–6 T
(b) 1.2 × 10–8 T
–6
(c) 2.1 × 10 T
(d) 2.1 × 10–8 T
11. A plane electromagnetic wave is incident on a
material surface. The wave delivers momentum
p and energy E. Then
(a) p ≠ 0, E ≠ 0
(b) p = 0, E = 0
(c) p = 0, E ≠ 0
(d) p ≠ 0, E = 0
12. Radiations of intensity 0.5 W m–2 are striking
a metal plate. The pressure on the plate is
(a) 0.166 × 10–8 N m–2 (b) 0.332 × 10–8 N m–2
(c) 0.111 × 10–8 N m–2 (d) 0.083 × 10–8 N m–2
13. One requires 11 eV of energy to dissociate
a carbon monoxide molecule into carbon and
oxygen atoms. The minimum frequency of the
appropriate electromagnetic radiation to achieve
the dissociation lies in
(a) visible region.
(b) infrared region.
(c) ultraviolet region. (d) microwave region.
14. The amplitude of the magnetic field of a
harmonic electromagnetic wave in vacuum is
B0 = 510 nT. The amplitude of the electric field
part of the wave is
(a) 120 N C–1
(b) 134 N C–1
–1
(c) 510 N C
(d) 153 N C–1
4
15. The decreasing order of wavelength of
infrared, microwave, ultraviolet and gamma
rays is
(a) microwave, infrared, ultraviolet, gamma rays
(b) infrared, microwave, ultraviolet, gamma rays
(c) gamma rays, ultraviolet, infrared, microwaves
(d) microwaves, gamma rays, infrared, ultraviolet
16. If µ0 be the permeability and k0 be the dielectric
constant of a medium, then its refractive index is
given by
1
1
(a)
(b)
(c)
µ0k0 (d) µ0k0
µ0k0
µ0k0
17. Which of the following statement is false for
the properties of electromagnetic waves?
(a) Both electric and magnetic field vectors
attain the maxima and minima at the same
place and same time.
(b) The energy in electromagnetic wave is
divided equally between electric and
magnetic field vectors.
(c) Both electric and magnetic field vectors are
parallel to each other and perpendicular to
the direction of propagation of wave
(d) These waves do not require any material
medium for propagation.
18. The source of electromagnetic waves can be
charge, when
(a) moving with a constant velocity
(b) moving in a circular orbit
(c) falling in an electric field
(d) both (b) and (c)
19. An electromagnetic wave radiates outwards
from a dipole antenna, with the amplitude of its
electric field vector E0. The electric field which
transports significant energy from the source
falls off as
1
1
1
(a) 3
(b) 2
(c)
(d) r
r
r
r
20. The electric field associated with an
electromagnetic wave in vacuum is given by
^
E = 40 cos(kz − 6 × 108t) i , where E , z and t are
in volt per meter, meter and second respectively.
The value of wave vector k is
(a) 2 m–1
(b) 0.5 m–1 (c) 6 m–1 (d) 3 m–1
21. A microwave and an ultrasonic sound wave
have the same wavelength. Their frequencies
are in the ratio (approximately)
(a) 102
(b) 104
(c) 106
(d) 108
CBSE Board Term-II Physics Class-12
22. The photon energy in units of eV for
electromagnetic waves of wavelength 2 cm is
(a) 2.5 × 10–19
(b) 5.2 × 1016
(c) 3.2 × 10–16
(d) 6.2 × 10–5
23. Which of the following electromagnetic
wave play an important role in maintaining the
earth’s warmth or average temperature through
the greenhouse effect?
(a) Visible rays
(b) Infrared waves
(c) Gamma rays
(d) Ultraviolet rays
24. The amplitude of an electromagnetic wave
in vacuum is doubled with no other changes
made to the wave. As a result of this doubling of
the amplitude, which of the following statement
is correct?
(a) The speed of wave propagation changes only
(b) The frequency of the wave changes only
(c) The wavelength of the wave changes only
(d) None of these.
25. The frequency of electromagnetic wave
which is best suitable to observe a particle of
radius 3 × 10–4 cm is of the order of
(a) 1015 Hz
(b) 1011 Hz
13
(c) 10 Hz
(d) 1012 Hz
26. Frequency of radiations arising from two
close energy levels in hydrogen, known as lamb
shift is 1057 MHz. This frequency falls in which
range of electromagnetic wave?
(a) Infrared rays
(b) X-rays
(c) g-rays
(d) Radio waves
27. X-rays, gamma rays and microwaves
travelling in vacuum have
(a) same wavelength but different velocities
(b) same frequency but different velocities
(c) same velocity but different wavelengths
(d) same velocity and same frequency
28. The electric field part of an electromagnetic
wave in vacuum is E = 3.1 N C−1 cos (1.8 rad m −1 ) y

8
−1  ^
+ (5.4 × 10 rad s ) t i . The wavelength of this

part of electromagnetic wave is
(a) 1.5 m
(b) 2 m
(c) 2.5 m (d) 3.5 m
29. The ratio of amplitude of magnetic field to the
amplitude of electric field for an electromagnetic
wave propagating in vacuum is equal to
(a) the speed of light in vacuum
(b) reciprocal of speed of light in vacuum
(c) the ratio of magnetic permeability to the
electric susceptibility of vacuum
(d) unity
5
Electromagnetic Waves
30. The electric field part of an electromagnetic
wave in a medium is represented by Ex = 0,
E y = 2 .5
N
rad 


−2 rad  
cos   2π × 106
 t −  π × 10
x ,

C
m
s  

Ez = 0. The wave is
(a) moving along x direction with frequency
106 Hz and wavelength 100 m
(b) moving along x direction with frequency
106 Hz and wavelength 200 m
(c) moving along –x direction with frequency
106 Hz and wavelength 200 m
(d) moving along y direction with frequency
2p × 106 Hz and wavelength 200 m
31. A plane electromagnetic wave of frequency
25 MHz travels in free space along the
x-direction. At a particular point in space and
^
time, E = 6.3 j V m −1 . At this point B is equal
to
^
^
(a) 8.33 × 10−8 k T
(b) 18.9 × 10−8 k T
(c) 2.1 × 10−8 k T
(d) 2.1 × 10−8 k T
^
^
34. An electromagnetic wave propagating along
north has its electric field vector upwards. Its
magnetic field vector point towards
(a) north
(b) east
(c) west
(d) downwards
35. The waves used by artificial satellites for
communication is
(a) microwaves
(b) infrared waves
(c) radio waves
(d) X-rays
36. Assume a bulb of efficiency 2.5% as a point
source. The peak values of electric and magnetic
fields produced by the radiation coming from a
100 W bulb at a distance of 3 m is respectively
(a) 2.5 V m–1, 3.6 × 10–8 T
(b) 4.2 V m–1, 2.8 × 10–8 T
(c) 4.08 V m–1, 1.36 × 10–8 T
(d) 3.6 V m–1, 4.2 × 10–8 T
37. An electromagnetic radiation has an energy
of 13.2 keV. Then the radiation belongs to the
region of
(a) visible light
(b) ultraviolet
(c) infrared
(d) X-ray
32. Which one of the following is the property
of a monochromatic, plane electromagnetic wave
in free space?
(a) Electric and magnetic fields have a phase
π
difference of .
2
(b) The energy contribution of both electric and
magnetic fields are equal.
(c) The direction of propagation is in the
direction of B × E .
(d) The pressure exerted by the wave is the
product of its speed and energy density.
38. The magnetic field of a beam emerging
from a filter facing a flood light as given by
B = 12 × 10–8 sin (1.20 × 107 z – 3.60 × 1015 t) T.
The average intensity of the beam is
(a) 1.71 W m–2
(b) 2.1 W m–2
–2
(c) 3.2 W m
(d) 2.9 W m–2
33. An electromagnetic wave is propagating
along x-axis.
At x = 1 m and t = 10 s, its electric
vector |E | = 6 V/ m then the magnitude of its
magnetic vector is
(a) 2 × 10–8 T
(b) 3 × 10–7 T
(c) 6 × 10–8 T
(d) 5 × 10–7 T
40. Radio waves diffract around buildings,
although light waves do not. The reason is that
radio waves
(a) travel with speed larger than c
(b) have much larger wavelength than light
(c) are not electromagnetic waves
(d) none of these
39. The electric field of a plane electromagnetic
wave varies with time of amplitude 2 Vm –1
propagating along z-axis. The average energy
density of the magnetic field is (in J m–3)
(a) 13.29 × 10–12
(b) 8.85 × 10–12
(c) 17.72 × 10–12
(d) 4.43 × 10–12
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 41 to 44.
Directions of Electromagnetic Waves
In an electromagnetic wave both the electric and
magnetic fields are perpendicular to the direction
of propagation, that is why electromagnetic
waves are transverse in nature. Electromagnetic
waves carry energy as they travel through
space and this energy is shared equally by the
electric and magnetic fields. Energy density of
an electromagnetic waves is the energy in unit
volume of the space through which the wave
travels.
CBSE Board Term-II Physics Class-12
6
41. The electromagnetic waves propagated
perpendicular to both E and B. The electromagnetic
waves travel in the direction of
(a) E ⋅ B (b) E × B
(c) B ⋅ E (d) B × E
(a) 500
(b) 100
250
500
(c)
(d)
3
3
Case III : Read the passage given below and
answer the following questions from 48 to 50.
42. Fundamental particle in an electromagnetic
wave is
(a) photon
(b) electron
(c) phonon
(d) proton
Oscillating Charge
A stationary charge produces only an electrostatic
field while a charge in uniform motion produces
a magnetic field, that does not change with time.
An oscillating charge is an example of accelerating
charge. It produces an oscillating magnetic field,
which in turn produces an oscillating electric
fields and so on. The oscillating electric and
magnetic fields regenerate each other as a wave
which propagates through space.
43. For a wave propagating in a medium, identify
the property that is independent of the others.
(a) velocity
(b) wavelength
(c) frequency
(d) all these depend on each other
44. The electric and magnetic fields of an
electromagnetic waves are
(a) in opposite phase and perpendicular to each
other
(b) in opposite phase and parallel to each other
(c) in phase and perpendicular to each other
(d) in phase and parallel to each other.
Case II : Read the passage given below and
answer the following questions from 45 to 47.
Momentum and Pressure of an Electromagnetic Wave
An electromagnetic wave transports linear
momentum as it travels through space. If
an electromagnetic wave transfers a total
energy U to a surface in time t, then total linear
U
momentum delivered to the surface is p = .
c
When an electromagnetic wave falls on a surface,
it exerts pressure on the surface. In 1903, the
American scientists Nichols and Hull succeeded
in measuring radiation pressures of visible light
where other had failed, by making a detailed
empirical analysis of the ubiquitous gas heating
and ballistic effects.
45. The pressure exerted by an electromagnetic
wave of intensity I(W m–2) on a non-reflecting
surface is (c is the velocity of light)
(a) Ic
(b) Ic2
(c) I/c
(d) I/c2
46. Light with an energy flux of 18 W/cm2 falls
on a non-reflecting surface at normal incidence.
The pressure exerted on the surface is
(a) 2 N/m2
(b) 2 × 10–4 N/m2
2
(c) 6 N/m
(d) 6 × 10–4 N/m2
47 A point source of electromagnetic radiation
has an average power output of 1500 W. The
maximum value of electric field at a distance of
3 m from this source (in V m–1) is
x
E
E
B
B
O
y
z
B
E
B
Direction of
propagation
E
48 Magnetic field in a plane electromagnetic
^
wave is given by B = B0 sin(kx + ωt) j T .
Expression for corresponding electric field will
be (Where c is speed of light.)
^
(a) E = − B0c sin(kx + ωt) k V/m
^
(b) E = B0c sin(kx − ωt) k V/m
^
(c) E = B0 sin(kx + ωt) k
V/m
c
^
(d) E = B0c sin(kx + ωt) k V/m
49 The electric field component of a monochromatic
^
radiation is given by E = 2E0 i cos kz cos ωt. Its
magnetic field B is then given by
2E0
c
2E0
(c)
c
(a)
^
j cos kz cos ωt
^
j sin kz sin ωt
2E0 ^
j sin kz cos ωt
c
2E ^
(d) − 0 j sin kz sin ωt
c
(b)
50. A plane em wave of frequency 25 MHz
travels in a free space along x-direction. At a
^
particular point in space and time, E = (6.3 j ) V/m.
What is magnetic field at that time?
(a) 0.095 mT
(b) 0.124 mT
(c) 0.089 mT
(d) 0.021 mT
7
Electromagnetic Waves
Assertion & Reasoning Based MCQs
For question numbers 51-60, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
51. Assertion (A) : Radio waves cannot be
diffracted by the buildings.
Reason (R) : The wavelength of radio waves is
very small.
56. Assertion (A) : Velocity of light is constant
in all media.
Reason (R) : Light is an electromagnetic wave
which has constant velocity in all media.
52. Assertion : Only microwaves are used in radar.
Reason : Because microwaves have very small
wavelength.
57. Assertion (A) : X-rays in vacuum travel
faster than light waves in vacuum.
Reason (R) : The energy of X-rays photon is
less than that of light photon.
53. Assertion (A) : The electric field and
magnetic field have equal average values in
linearly polarised plane em wave.
Reason (R) : The electric energy and magnetic
energy have equal average values in linearly
polarised plane em wave.
58. Assertion (A) : Electromagnetic waves
exert pressure called radiation pressure.
Reason (R) : Electromagnetic waves carries
energy.
54. Assertion (A) : Light can travel in vacuum
whereas sound cannot do so.
Reason (R) : Light has an electromagnetic wave
nature whereas sound is mechanical wave.
59. Assertion (A) : Infrared waves sometimes
referred as heat waves.
Reason (R) : Infrared waves heat up the earth
surface.
55. Assertion (A) : The microwaves are better
carriers of signals than radio waves.
Reason (R) : The electromagnetic waves do not
required any material medium for propagation.
60. Assertion (A) : X-ray astronomy is possible
only from satellites orbiting the earth.
Reason (R) : Efficiency of X-rays telescope is
large as compared to any other telescope.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. To which part of the electromagnetic
spectrum does a wave of frequency 5 × 1019 Hz
belong?
2.
How are radio waves produced?
3. Welders wear special goggles or face masks
with glass windows to protect their eyes from
electromagnetic radiations. Name the radiations
and write the range of their frequency.
4. The small ozone layer on top of the stratosphere
is crucial for human survival. Why ?
5.
Write two uses of microwaves.
6.
If the Earth did not have atmosphere, would
its average surface temperature be higher or
lower than what it is now? Explain.
7. An e.m. wave exerts pressure on the surface
on which it is incident. Justify.
8. Do electromagnetic waves carry energy and
momentum?
9. How is the speed of em-waves in vacuum
determined by the electric and magnetic fields?
10. In which directions do the electric and magnetic
field vectors oscillate in an electromagnetic wave
propagating along the x-axis ?
CBSE Board Term-II Physics Class-12
8
Short Answer Type Questions (SA-I)
11. Illustrate by giving suitable examples, how
you can show that electromagnetic waves carry
both energy and momentum.
12. A plane electromagnetic wave travels in
vacuum along z-direction. What can you say
about the direction of electric and magnetic field
vectors?
13. An e.m. wave is travelling in a medium
^
with a velocity v = v i . Draw a sketch showing
the propagation of the e.m. wave, indicating the
direction of the oscillating electric and magnetic
fields.
16. The electric field intensity produced by the
radiations coming from 100 W bulb at a 3 m
distance is E. Find the electric field intensity
produced by the radiations coming from 50 W
bulb at the same distance.
17. The electric field in an electromagnetic wave
is given by E = (50 N C–1)sin w(t – x/c). Find the
energy contained in a cylinder of cross-section
10 cm2 and length 50 cm along the x-axis.
18. How are infrared waves produced? Why are
these referred as heat waves? Write their one
important use?
14. Name the physical quantity which remains
same for microwaves of wavelength 1 mm and
UV radiations of 1600 Å in vacuum.
19. Explain briefly how electromagnetic waves
are produced by an oscillating charge. How is
the frequency of the e.m. waves produced related
to that of the oscillating charge?
15. One requires 11 eV of energy to dissociate
a carbon monoxide molecule into carbon and
oxygen atoms. The minimum frequency of the
appropriate electromagnetic radiation to achieve
the dissociation lies in which region?
20. Identify the electromagnetic waves whose
wavelengths vary as
(a) 10–11 m < l < 10–14 m
(b) 10–4 m < l < 10–6 m
Write one use of each.
Short Answer Type Questions (SA-II)
21. Name the constituent radiation of electromagnetic spectrum which is used for
(i) aircraft navigation.
(ii) studying crystal structure.
Write the frequency range for each.
22. Gamma rays and radio waves travel with
the same velocity in free space. Distinguish
between them in terms of their origin and the
main application.
23. Identify the part of the electromagnetic
spectrum used in (i) radar and (ii) eye surgery.
Write their frequency range.
24. (i) Which segment of electromagnetic
waves has highest frequency? How are these
waves produced? Give one use of these waves.
(ii) Which e.m. waves lie near the high frequency
end of visible part of e.m. spectrum? Give its one
use. In what way this component of light has
harmful effects on humans?
25. Name the parts of the electromagnetic
spectrum which is
(a) suitable for radar systems used in aircraft
navigation.
(b) used to treat muscular strain.
(c) used as a diagnostic tool in medicine.
Write in brief, how these waves can be produced.
26. Prove that the average energy density of the
oscillating electric field is equal to that of the
oscillating magnetic field.
27. Name the types of em radiations which
(i)are used in destroying cancer cells,
(ii)cause tanning of the skin and (iii) maintain
the earth’s warmth.
Write briefly a method of producing any one of
these waves.
28. (a) Arrange the following electromagnetic
waves in the descending order of their
wavelengths :
(i) Microwaves
(ii) Infra-red rays
(iii) Ultra-violet-radiation
(iv) Gamma rays
(b) Write one use each of any two of them.
29. (a) Optical and radio telescopes are built on
the ground but X-ray astronomy is possible only
from satellites orbiting the earth. Why?
9
Electromagnetic Waves
(b) The small ozone layer on top of the stratosphere
is crucial for human survival. Why?
30. Suppose that the electric field part of an
electromagnetic wave in vacuum is
E = (3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 108 rad/s)t]t
(a) What is the direction of propagation?
(b) What is the wavelength l?
(c) What is the frequency n?
31. (a) When the oscillating electric and
magnetic fields are along the x-and y-direction
respectively.
(i) point out the direction of propagation of
electromagnetic wave.
(ii) express the velocity of propagation in terms
of the amplitudes of the oscillating electric
and magnetic fields.
(b) How do you show that the e.m. wave carries
energy and momentum?
Long Answer Type Questions (LA)
32. Answer the following questions:
(a) Show, by giving a simple example, how e.m.
waves carry energy and momentum.
(b) How are microwaves produced? Why is
it necessary in microwave ovens to select the
frequency of microwaves to match the resonant
frequency of water molecules?
(c) Write two important uses of infrared
waves.
33. State clearly how a microwave oven works to
heat up a food item containing water molecules.
Why are microwaves found useful for the radar
systems in aircraft navigation?
34. The intensity of light ranges from about
1W
m2
for a candle to about 30 mW/m for a modest size
laser. A particular laser produces a power 4 W
in a beam 0.4 mm in diameter.
2
(a) What is its average intensity?
(b) Find the peak electric field of the laser light.
(c) Find the peak magnetic field of the laser
light.
(d) If the laser beam is aimed upward to levitate
a 20 mm diameter sphere, what is the
maximum mass of this sphere. Assume the
sphere is perfectly reflecting.
35. A p l a n e E M w a v e t r a v e l l i n g a l o n g
z-direction is described by E = E0 sin(kz − ωt)i
and B = B sin(kz − ωt) j . Show that
0
(a) The average energy density of the wave is
1
1 B02
ε0 E02 +
.
4
4 µ0
(b) The time averaged intensity of the wave is
1
given by I av = cε0 E02 .
2
given by uav =
OBJECTIVE TYPE QUESTIONS
1.
(b) : Wavelength, λ =
c 3 × 108 m s−1
=
= 7.5 m
υ 40 × 106 s−1
2. (d) : The frequency of electromagnetic wave remains
unchanged but the wavelength of electromagnetic wave
changes when it passes from one medium to another.
1
c=
µ0 ε0
1
ε0
∴
c∝
∴
c
ε
4
=
=
=2
v
ε0
1
and v ∝
1
ε
c
υλ
λ
λ
=
=
= 2 or λ ′ =
v
2
υλ ′ λ ′
3.
(c) : Intensity of electromagnetic wave, I = Uav c
1
In terms of electric field, Uav = ε0 E02
2
1 B02
In terms of magnetic field, Uav =
2 µ0
1
Now Uav electric field = ε0 E02
2
1
1
1 2
2
E0


= ε0 ( cB0 ) = ε0 ×
B0 ∵ B = c
2
2
µ0 ε0
0
1 B02
=
= Uav (due to magnetic field)
2 µ0
CBSE Board Term-II Physics Class-12
10
Thus the energy in electromagnetic wave is divided equally
between electric field vector and magnetic field vector. Therefore,
the ratio of contributions by the electric field and magnetic field
components to the intensity of electromagnetic wave is 1 : 1.
4. (c) : Here, power of bulb = 100 W
Power of visible light
As intensity, I =
area
100 × 6 / 100
=
= 7.4 × 10–3 W m–2
4 π (8)2
5.
6.
(c)
(a) : Here, u1 = 7.5 MHz, u2 = 12 MHz
∴ λ1 =
8
3 × 10
c
=
= 40 m
υ1 7.5 × 106
3 × 108
c
and λ2 =
=
= 25 m
υ2 12 × 106
7. (d)
8. (b) : The total energy falling on the surface is
U = 18 × 20 × (30 × 60) = 6.48 × 105 J
Therefore, the total momentum delivered (for complete
absorption is)
U 6.48 × 105
p= =
= 2.16 × 10–3 kg m s–1
c
3 × 108
The average force exerted on the surface is
p 2.16 × 10 −3
F= =
= 1.2 × 10–6 N
t 0.18 × 104
9. (d) : Microwaves are used to cook food. Microwave oven
is a domestic application of these waves.
E
6.3
10. (d) : As, B = =
= 2.1 × 10 −8 T
c 3 × 108
11. (a) : When plane electromagnetic wave is incident on
a material surface, the wave delivers some momentum and
energy to the surface and hence p ≠ 0 and E ≠ 0.
0.5
I
12. (a) : P = =
c 3 × 108
= 0.166 × 10–8 N m–2
13. (c) : Here, E = 11 eV = 11 × 1.6 × 10–19 J = hu
11 × 1.6 × 10 −19
\ υ=
h
=
11 × 1.6 × 10 −19
= 2.6 × 1015 Hz
6.62 × 10 −34
This frequency radiation belongs to ultraviolet region.
14. (d) : Here, B0 = 510 nT = 510 × 10–9 T
E0 = cB0 = 3 × 108 × 510 × 10–9 = 153 N C–1
15. (a)
16. (c) : Refractive index of medium is
c
1
µ = ,where c =
υ
µ 0 k0
and
∴
υ=
µ=
1
µ 0 k0 µ r k r
1 / µ 0 k0
1 / µ 0 k0 µ r k r
= µ r kr
Given µr = µ0 and kr = k0 then µ =
µ 0 k0
17. (c) : In an electromagnetic wave both electric and
magnetic vectors are perpendicular to each other as well as
perpendicular to the direction of propagation of wave.
18. (d) : An electromagnetic wave can be produced by
accelerated or oscillating charge.
In options (b) and (c), the charge is in accelerated state, hence
will be a source of electromagnetic waves.
19. (c) : From a dipole antenna, the electromagnetic waves
are radiated outwards. The amplitude of electric field vector
E0 which transports significant energy from the source falls
off intensity inversely as the distance r from the antenna
1
i.e., E0 ∝ .
r
20. (a) : Compare the given equation with
E = E0cos(kz – wt)
We get, w = 6 × 108 s–1
6 × 108 s −1
ω
= 2 m−1
Wave vector, k = =
c 3 × 108 m s −1
21. (c) : Frequency of microwaves, um ≈ 1011 Hz
Frequency of ultrasonic sound waves, uu ≈ 105 Hz
υ m 1011
∴
=
= 106
υ u 105
22. (d) : As E =
hc
λ
=
6.6 × 10 −34 × 3 × 108
= 9.9 ×10–24 J
2 × 10 −2
=
9.9 × 10 −24
eV = 6.2 × 10 −5 eV
1.6 × 10 −19
23. (b) : Infrared radiation plays an important role in
maintaining the earth’s warmth through greenhouse effect.
Incoming visible light when passes relatively easily through the
atmosphere is absorbed by the earth’s surface and radiated as
infrared (longer wavelength) radiation. This radiation is trapped
by greenhouse gases such as carbon dioxide and water vapour.
In this way an average temperature is maintained.
11
Electromagnetic Waves
24. (d) : Velocity of electromagnetic wave
1
c=
= 3 × 108 m s −1
µ0 ε0
It is independent of amplitude, frequency and wavelength of
electromagnetic wave.
25. (a) : Let l be the radius of the particle then
l = 3 × 10–4 × 10–2 m = 3 × 10–6 m
Frequency of electromagnetic wave, υ =
c 3 × 108
=
= 1014
λ 3 × 10 −6
Thus to observe the particle, the frequency of wave should be
more than 1014 i.e 1015 Hz or smaller value of wavelength.
26. (d) : This given frequency corresponds to the radio waves
i.e short wavelength or high frequency.
27. (c) : In vacuum X-rays, gamma rays and microwaves
travel with same velocity, i.e., with the velocity of light
c (= 3 × 108 m s–1) but have different wavelengths.
28. (d) : Given
^
E = 3.1 N C–1 cos [(1.8 radm−1 ) y + (5.4 × 108 rad s −1 ) t ] i ...(i)
Camparing (i) with the equation
E = E0 cos (ky + wt)...(ii)
We get, k = 1.8 rad m–1,
E0 = 3.1 N C–1, c = 3 × 108 m s–1, w = 5.4 × 108 rad s–1
2π 2 × 22
Now, λ =
=
= 3.5 m
k 1.8 × 7
29. (b) : The amplitudes of electric field and magnetic field
for an electromagnetic wave propagating in vacuum are
related as E0 = B0c
where c is the speed of light in vacuum.
B0 1
∴
=
E0 c
N 
rad  
rad 
30. (b) : E y = 2.5 cos  2π × 106  t −  π × 10 −2 
C 
m 
s 
Ez = 0, Ex = 0
The wave is moving in the positive direction of x.
This is in the form Ey = E0(wt – kx)
w = 2p × 106
2pu = 2p × 106 ⇒ u = 106 Hz
2π
2π
=k ⇒
= π × 10 −2
λ
λ
2π
⇒ λ=
= 2 × 102 = 200 m
−2
π × 10
^
31. (c) : Here, E = 6.3 j V m−1
The magnitude of B is
6.3 V m−1
E
B= =
= 2.1 × 10 −8 T
c 3 × 108 m s −1

x

E is along y-direction and the wave propagates along x-axis.
Therefore, B should be in a direction perpendicular to both
x and y-axis. Using vector algebra E × B should be along
x-direction.
^
^ ^
Since ( + j ) × ( + k ) = i , B is along z-direction.
^
Thus, B = 2.1 × 10 −8 k T
32. (b) : In electromagnetic wave, electric and magnetic
fields are in phase.
Electromagnetic wave carry energy as they travel through
space and this energy is shared equally by electric and
magnetic fields.
The direction of the propagation of electromagnetic wave is
the direction of E × B .
The pressure exerted by the wave is equal to its energy density.
33. (a) : Here, | E | = 6 V/m
The magnitude of magnetic vector is
|E|
6 V/ m
|B| =
=
= 2 × 10 −8 T
c
3 × 108 m/ s
34. (b)
35. (a) : In artificial satellite microwaves are used for
communication.
36. (c) : Here intensity, I =
=
power
area
100 × 2.5
2.5
=
W m−2
2
4 π (3) × 100 36π
Half of this intensity belongs to electric field and half of that
to magnetic field.
I 1
∴
= ε0 E02 c
2 4
2.5
2×
2I
36π
or E0 =
=
= 4.08 V m–1
1
ε0 c
8
3
10
×
×
4 π × 9 × 109
∴
B0 =
E0
4.08
= 1.36 × 10–8 T
=
c 3 × 108
37. (d) : Given : E = 13.2 keV
λ(in Å) =
12400
hc
=
= 0.939 Å ≈ 1 Å
E (in eV) 13.2 × 103
X-rays cover wavelengths ranging from about 10 –8 m
(10 nm) to 10–13 m (10–4 nm).
An electromagnetic radiation of energy 13.2 keV belongs to
X-ray region of electromagnetic spectrum.
CBSE Board Term-II Physics Class-12
12
38. (a) : Here, B = 12 × 10–8 sin
(1.20 × 107 z –3.6 × 1015 t) T
Comparing it with, B = B0 sin (kz – wt),
we have B0 = 12 × 10–8 T
Iav =
1 B02 c
2 µ0
39. (b) : Amplitude of electric field and magnetic field are
related by the relation
E0
=c
B0
Average energy density of the magnetic field is
1 B02
uB =
4 µ0
1
= ε0 E02
4
1
= × 8.854 × 10 −12 × (2)2
4

∵
E 
B0 = 0 
c
1 

∵ c=

µ 0 ε0 
= 8.854 × 10–12 J m–3
≈ 8.85 × 10–12 J m–3
40. (b) : The wavelength of radiowaves being much larger
than light, has a size comparable to those of buildings, hence
diffract from them.
41. (b) : Electromagnetic waves propagate in the direction
of E × B.
42. (a) : Photon is the fundamental particle in an
electromagnetic wave.
43. (c) : Frequency u remains unchanged when a wave
propagates from one medium to another. Both wavelength
and velocity get changed.
44. (c) : The electric and magnetic fields of an electromagnetic
wave are in phase and perpendicular to each other.
45. (c) : Pressure exerted by an electromagnetic radiation,
I
P=
c
18 W/cm2
Energy flux
46. (d) : Prad =
=
Speed of light 3 × 108 m/s
=
18 × 104 W/m2
= 6 × 10 −4 N/m2
3 × 108 m/s
Vav =
1
e E2 × c
2 0 0
⇒ E0 =
1 (12 × 10 −8 )2 × 3 × 108
= ×
2
4 π × 10 −7
= 1.71 W m–2
1 E02
=
4 µ 0 c2
47. (b) : Intensity of EM wave is given by I =
P
4 πR2
P
1500
=
2
2
2πR ε0 c
2 × 3.14(3) × 8.85 × 10 −12 × 3 × 108
= 10, 000 = 100 V m−1
48. (d) : Given : B = B0 sin( kx + ωt )j T
The relation between electric and magnetic field is,
E
c = or E = cB
B
The electric field component is perpendicular to the direction
of propagation and the direction of magnetic field. Therefore,
the electric field component along z-axis is obtained as
^
E = cB0 sin (kx + wt) k V/m
49. (c) :
dE
dB
=−
dz
dt
dE
dB
= –2E0k sinkz coswt = −
dz
dt
dB = + 2 E0k sinkz coswtdt
B = +2E0k sin kz
E0 ω
= =c
B0 k
∫ cos ωt dt
= +2E0
k
sinkz sinwt
ω
2E
2E0
sin kz sinωt ∴ B = 0 sin kz sinωt j
c
c
E is along y-direction and the wave propagates along x-axis.
\ B should be in a direction perpendicular to both x-and
y-axis.
50. (d) : Here, E = 6.3j ; c = 3 × 108 m/s
B=
The magnitude of B is
E
6.3
Bz = =
= 2.1 × 10 −8 T = 0.021 µT
c 3 × 108
51. (d) : For wave to suffer diffraction, the wavelength
should be of the order of size of the obstacle. The wavelength
of radio waves (short radio waves) is order of the size of the
building and the other obstacles coming in their path and
hence they easily get diffracted.
52. (a) : In a radar, a beam signal is needed in particular
direction which is possible if wavelength of wave is very small.
Since the wavelength of microwaves is a few millimeter, hence
they are used in radar.
53. (b) : In case of a linearly polarised plane electromagnetic
wave, the average values of electric field and magnetic field
are equal and average values of electric energy and magnetic
energy are also equal.
13
Electromagnetic Waves
54. (a) : Light being electromagnetic wave do not require
any material medium for its propagation. Hence light can
travel in vacuum. On the other hand sound is a mechanical
wave and requires a material medium for its propagation.
Hence sound cannot travel in vacuum.
4.
The small ozone layer on the top of the atmosphere
is crucial for human survival because it absorbs harmful
ultraviolet radiations present in sunlight and prevents it from
reaching the earth’s surface. These radiations can penetrate
our skin and can cause harmful diseases like skin cancer etc.
55. (b) : Microwaves are the electromagnetic waves of
wavelength of the order of a few millimetres, which is less
than those of T.V. signals. On account of smaller wavelength,
the microwaves can be transmitted as beam signals in a
particular direction and are much better than radiowaves
because microwaves do not spread or bend around the
corners of any obstacle coming in their way. Therefore
microwaves are better carriers of signals than radiowaves.
5.
Uses of microwaves :
(i) In long distance communication
(ii) In radar systems used in aircraft navigation
56. (d) : Velocity of light has different values in different
media.
It depends on the refractive index of the medium. Related
by formula
velocity in vacuum
vmedium =
refractive index of medium
7.
An e.m. wave carries momentum with itself and given by
Energy of wave (U )
p=
Speed of the wave (c)
57. (d) : All electromagnetic waves including X-rays travels
with same velocity in vacuum. The energy of X-rays is
greater than energy of the light because energy is inversely
proportional to wavelength (E = hc/l) and wavelength of
X-rays are smaller than light waves.
9.
The speed of em-waves in vacuum determined by the
E
electric (E0) and magnetic fields (B0) is, c = 0 .
B0
58. (b) : Electromagnetic waves transport linear momentum
as well as energy. When electromagnetic waves strike a
surface, a pressure is exerted on the surface. If the intensity
of wave is I, the radiation pressure P (force per unit area)
exerted on the perfectly absorbing surface is P = I/c.
59. (b) : Infrared waves are sometimes called heat waves.
This is because water molecules present in most materials
readily absorb infrared waves. After absorption, their thermal
motion increases, that is, they heat up and heat their
surroundings.
60. (c) : The earth’s atmosphere is transparent to visible
light and radio waves, but absorbs X-rays. Therefore X-rays
telescope cannot be used on earth surface.
SUBJECTIVE TYPE QUESTIONS
1.
X-rays.
2.
Radio waves are the electromagnetic waves of frequency
ranging from 500 KHz to about 1000 MHz. These waves are
produced by oscillating electric circuits having inductor and
capacitor.
3.
U ltraviolet radiations produced during welding are
harmful to eyes. Special goggles or face masks are used to
protect eyes from UV radiations. UV radiations have a range
of frequency between 1014 Hz – 1016 Hz.
6.
If the Earth did not have atmosphere, then there would
be absence of green house effect of the atmosphere. Due
to this reason, the temperature of the earth would be lower
than what it is now.
When it is incident upon a surface it exerts pressure on it.
8.
Yes, electromagnetic waves carry energy and momentum.
10.
When an electromagnetic wave is propagating along the
x-axis then, electric field vector oscillates in y-axis and magnetic
field vector oscillates in z-axis.
11.
Electromagnetic waves like other waves carry energy and
momentum as they travel through empty space. If light didn’t
carry energy and momentum, it wouldn’t be able to heat stuff
up or generate photocurrent in photocells.
12.
The electric and magnetic field vectors E and B are
perpendicular to each other and also perpendicular to the
direction of propagation of the electromagnetic wave. If
a plane electromagnetic wave is propagating along the
z-direction, then the electric field is along x-axis, and magnetic
field is along y-axis.
13.
In figure the velocity of propagation of e.m. wave is along
X-axis v = vi and electric field E along Y-axis and magnetic
field B along Z-axis.
Y
E
X
Z
B
14.
The speed in vacuum (i.e., c = 3 × 108 m s–1) remains
same for both the given wavelengths. It is because both
microwaves and UV rays are electromagnetic waves.
CBSE Board Term-II Physics Class-12
14
15.
Here, E = 11 eV = 11 × 1.6 × 10–19 J = hu
\
υ=
11 × 1.6 × 10 −19 11 × 1.6 × 10 −19
=
= 2.6 × 1015 Hz
h
6.62 × 10 −34
This frequency belongs to ultraviolet region.
16.
Electric field intensity on a surface due to the incident
radiation is
U P

 U
E= =
∵ = P 
At A t
As, E ∝ P
(for the given area of the surface)
E ′ P ′ 50 1
= =
=
\
E P 100 2
E
E′ =
2
17. The energy density is
1
1
–8
–3
uav = ε0 E02 = × (8.85 × 10 −12 ) × (50)2 = 1.1 × 10 J m
2
2
The volume of the cylinder is V = 5 × 10–4 m3
The energy contained in cylinder is
U = (1.1 × 10–8 J m–3) × (5 × 10–4 m3) = 5.5 × 10–12 J.
18.
Infra red waves are produced by hot bodies and molecules.
They are produced due to the de-excitation of atoms.
Infrared waves incident on a substance increase the internal
energy and hence the temperature of the substance. That is
why they are called heat waves.
Infra red radiations play an important role in maintaining
the earth’s warmth or average temperature through the
greenhouse effect.
19.
A n oscillating or accelerated charge is supposed
to be source of an electromagnetic wave. An oscillating
charge produces an oscillating electric field in space which
further produces an oscillating magnetic field which in turn
is a source of electric field. These oscillating electric and
magnetic field hence, keep on regenerating each other and
an electromagnetic wave is produced.
The frequency of e.m. wave = Frequency of oscillating charge.
20.
(a) Gamma rays lie between 10–11 m to 10–14 m.
These rays are used in radiotherapy to treat certain cancers
and tumors.
(b) Infrared waves lie between 10–4 m to 10–6 m. These waves
are used in taking photographs during conditions of fog, smoke
etc., as these waves are scattered less than visible rays.
21. (i) Microwaves are used in radar system for aircraft
navigation. The frequency range is 3 × 108 Hz to 3 × 1011 Hz.
(ii) X-rays are used for studying crystals structure of solids.
Their frequency range is 3 × 1016 Hz to 3 × 1021 Hz.
22. Gamma rays : These rays are of nuclear origin and are
produced in the disintegration of radioactive atomic nuclei
and in the decay of certain subatomic particles. They are used
in the treatment of cancer and tumours.
Radio waves : These waves are produced by the accelerated
motion of charges in conducting wires or oscillating electric
circuits having inductor and capacitor. These are used in
satellite, radio and television communication
23.
Uses
Part of
electromagnetic
spectrum
(i)
In radar
system
Microwaves
(ii)
In eye surgery Ultraviolet
Frequency
range
3 × 108 Hz to
3 × 1011 Hz
8 × 1014 Hz to
8 × 1016 Hz
24. (i) Gamma rays has the highest frequency in
the electromagnetic waves. These rays are of the nuclear
origin and are produced in the disintegration of radioactive
atomic nuclei and in the decay of certain subatomic particles.
They are used in the treatment of cancer and tumours.
(ii) Ultraviolet rays lie near the high-frequency end of visible
part of e.m. spectrum. These rays are used to preserve food
stuff. The harmful effect from exposure to ultraviolet (UV)
radiation can be life threatening, and include premature aging
of the skin, suppression of the immune systems, damage to
the eyes and skin cancer.
25. (a) Microwaves are suitable for radar systems used in
aircraft navigation.
These waves are produced by special vacuum tubes, namely
klystrons, magnetrons and Gunn diodes.
(b) Infra-red waves are used to treat muscular pain. These
waves are produced by hot bodies and molecules.
(c) X-rays are used as a diagnostic tool in medicine. These
are produced when high energy electrons are stopped
suddenly on a metal of high atomic number.
26. In an electromagnetic wave, both E and B fields vary
sinusoidally in space and time. The average energy density
u of an e.m. wave can be obtained by replacing E and B by
their rms value
1
1 2
1
1 2
2
u = ε0 Erms
+
Brms or u = ε0 E02 +
B0
2
2µ 0
4
4µ 0
E0
B0 

∵ Erms = 2 , Brms = 2 


1
2
, therefore
Moreover, E0 = cB0 and c =
µ0 ε0
1
1
uE = ε0 E02 = ε0 ( cB0 )2
4
4
B02
1
1 2
uE = ε0 ⋅
=
B0 = uB
µ 0 ε0 4µ 0
4
15
Electromagnetic Waves
27. (i) Gamma rays
(ii) UV rays
(iii) Infra-red radiations
Infra-red waves are produced by hot bodies and molecules.
Infra-red waves are referred to as heat waves, because
water molecules present in most materials readily absorb
infra-red waves (many other molecules, for example, CO 2,
NH3 also absorb infra-red waves). After absorption, their
thermal motion increases, that is they heat up and heat their
surroundings.
31.
(a) (i) The e.m. wave propagates along z-axis.
28.
(a) Descending order of wavelengths for given
electromagnetic waves is:
Microwaves (10–3 – 10–1) m
Infra-red rays (7.5 × 10–7 – 10–3) m
Ultra-violet radiation (10–9 – 4 × 10–7) m
Gamma rays (< 10–12) m
(b) Electromagnetic waves or photons transport energy and
momentum. When an electromagnetic wave interacts with a
small particle, it can exchange energy and momentum with
the particle. The force exerted on the particle is equal to the
momentum transferred per unit time. Optical tweezers use this
force to provide a non-invasive technique for manipulating
microscopic-sized particles with light.
(b) Microwaves :
Frequency range → 3 × 108 Hz –3 × 1011 Hz. These are
suitable for the radar system, used in aircraft navigation.
Gamma rays :
Frequency range → > 3 × 1021 Hz.
These wave are used for the treatment of cancer cells.
32. (a) Consider a plane perpendicular to the direction of
propagation of the wave. An electric charge, on the plane will
be set in motion by the electric and magnetic fields of e.m.
wave, incident on this plane. This illustrates that e.m. waves
carry energy and momentum.
(b) Microwaves are produced by special vacuum tube like
the klystron, magnetron and Gunn diode.
The frequency of microwaves is selected to match the resonant
frequency of water molecules, so that energy is transformed
efficiently to the kinetic energy of the molecules.
(c) Uses of infra-red waves :
(i) They are used in night vision devices during warfare.
This is because they can pass through haze, fog and mist.
(ii) Infra-red waves are used in remote switches of household
electrical appliances.
29.
(a) The earth’s atmosphere is transparent to visible light
and radiowaves but it absorbs X-rays. X-ray astronomy is
possible only from satellites orbiting the earth. These satellites
orbit at a height of 36,000 km, where the atmosphere is very
thin and X-rays are not absorbed.
(b) Ozone layer absorbs ultraviolet radiation from the sun
and prevents it from reaching the earth and causing damage
to life.
30.
(a) The waves is propagating along negative y-direction
of its direction is – j .
(b) Comparing the given equation with the standard
equation.
 y

E = E0 cos 2π  + νt  


λ

We get,
\
2π
= 1.8
λ
Wavelength, λ =
2π 2 × 3.14
×
= 3.5 m.
1.8
1.8
(c) Also, 2pn = 5.4 × 106
∴ ν=
5.4 × 106 5.4 × 108
=
= 85.9 × 106 Hz.
2π
2 × 3.14
86 MHz
x
E
E0
z
B0
y
B
(ii) The speed of em-waves in vacuum determined by the
E
electric (E0) and magnetic fields (B0) is, c = 0
B0
33. In microwave oven, the frequency of the microwaves is
selected to match the resonant frequency of water molecules
so that energy from the waves get transferred efficiently to
the kinetic energy of the molecules. This kinetic energy raises
the temperature of any food containing water.
Microwaves are short wavelength radio waves, with frequency
of order of GHz. Due to short wavelength, they have high
penetrating power with respect to atmosphere and less
diffraction in the atmospheric layers. So these waves are
suitable for the radar systems used in aircraft navigation.
33.
(a) Intensity equals power per unit area carried by the wave.
P
⇒ Power per unit area = 2
πr
4W
=
3.14 × (0.2 × 10 −3 m)2
= 32 × 106 W/m2
CBSE Board Term-II Physics Class-12
16
(b) Intensity =
E02
⇒ Emax = E0 =
2µ 0 c
Total energy density of wave,
(2µ 0 c)( I )
=
= (8π × 10 −7 )(3 × 108 )(32 × 106 ) SI units
= 1.6 × 105 V/m
(c) Bmax = B0 =
(d) DP =
Now,
Now (uE)av =
E0
1.6 × 105 V/m
=
= 5.3 × 10–4 T
8
c
3 × 10 m/s
2U
where DP is the momentum delivered on the
c
sphere.
∆P
U 1
F=
= 2   .
 ∆t  c
∆t
1
1 B2
ε0E 2 +
2
2 µ0
ε0
= 2T
=
T
∫0 E 0 sin (kx − ωt )dt
Similarly, (uB)av =
\
2Iπr 2
⇒ m=
gc
W

2  3.2 × 107 2  (3.14 × 1 × 10 −10 m2 )

m 
108 m/s)
= 6.8 × 10–12 kg.
35. (a) Energy density of electric field E is
1
uE = ε0 E 2
2
Energy density due to magnetic field B is,
uB =
1 B2
2 µ0
2
T 2

 ∫ sin (kx − ωt )dt = T / 2 
0

2. I . πr 2
U
= I.A = I(pr2) ⇒ F =
c
∆t
(9.8 m/s2 )(3
2
ε0  2 T  1
2
E 0 ×  = ε0E 0
2T 
2
4
Now the force F must support the sphere mass mg.
=
1 T
1 T1
u εdt = ∫ ε0E 2
∫
T 0
T 02
uav =
1
1 B02
ε0E 02 +
4
4 µ0
(b) E = c B0
\
1 B02
4 µ0
and
c=
1
µ0 ε0
1 B02 1 E 02 / c 2 E 02
1
=
=
× µ0 ε0 = ε0 E 02
4
4 µ0 4 µ0
4 µ0
1
1 B02 1
1
ε0 E 02 +
= ε0E 02 + ε0E 02
4
4 µ0 4
4
1
1 B02
= ε0E 02 =
2
2 µ0
Time averaged intensity
1
Iav = uavc = c ε0 E 02
2
uav =

CHAPTER
9
Ray Optics and
Optical Instruments
Recap Notes
Optics : It is the branch of physics which
deals with the study of light and the
phenomena associated with it. It is divided
into two branches :
– Geometrical optics or ray optics
– Physical optics or wave optics
Refraction of light : When a ray of light
passes from one medium to another, in
which it has a different velocity, there occurs
a change in the direction of propagation
of light except when it strikes the surface
of separation of two media normally.
This bending of a ray of light is known as
refraction.
X The angles made by the incident ray and
the refracted ray with the normal to the
separating surface at the point of incidence
are known as the angle of incidence and of
refraction respectively.
X Laws of refraction : The two laws of
refraction are as follows :
– The incident ray, the normal and the
refracted ray all lie in the same plane.
– The ratio of the sine of angle of incidence
to the sine of angle of refraction for any
two media is constant for a light of
definite colour. This constant is denoted
by 1m2 or m21 called the refractive index
of the second medium with respect
to the first, the subscripts 1 and 2
indicating that the light passes from
medium 1 to medium 2.
sin i 1
= µ2
sin r
This is also known as Snell’s law.
where i = angle of incidence,
r = angle of refraction.
X
Absolute refractive index : Refractive
index of a medium with respect to vacuum
(or in practice, air) is known as absolute
refractive index of the medium.
µ=
c speed of light in vacuum
=
v speed of light in medium
General expression for Snell’s law
 c 
v  v
µ
1
µ2 = 2 =  2  = 1
µ1  c  v2
v 
 1
where c is the speed of light in air, v1
and v2 be the speeds of light in medium 1
and medium 2 respectively. According to
Snell’s law,
1
X
X
µ2 =
sin i µ2 sin i
;
=
sin r µ1 sin r
or m1 sin i = m2 sin r
When a light ray passes from a rarer
to denser medium (m2 > m1), it will bend
towards the normal as shown in the
figure.
When a light ray passes from a denser
medium to rarer medium (m1 > m2) it will
bend away from the normal as shown in
the figure.
CBSE Board Term-II Physics Class-12
18
X
If a light ray passes through a number
of parallel media and if the first and the
last media are same. The emergent ray is
parallel to the incident ray as shown in
figure.
1
µ2 =
sin r1
sin r2
sin i1 2
and 3µ1 =
µ =
sin i1
sin r1 , 3 sin r2
Hence,
1
X
µ2 × 2 µ3 × 3µ1 =
sin i1 sin r1 sin r2
×
×
=1
sin r1 sin r2 sin i1
Lateral shift : When the medium is
same on both sides of a glass slab, then
the deviation of the emergent ray is zero.
That is the emergent ray is parallel to
the incident ray but it does suffer lateral
displacement/shift with respect to the
incident ray and is given by
Lateral shift, d = t
– If a beaker is filled with water and a point
lying at its bottom is observed by someone
located in air, then the bottom point
appears raised. The apparent depth is
less than the real depth. It can be shown
that
real depth
apparent depth =
refractive index (µ)
If there is an ink spot at the bottom of
a glass slab, it appears to be raised by a
distance
t
1

d = t − = t 1 − 
µ
µ


where t is the thickness of the glass slab
and m is its refractive index.
– If a beaker is filled with immiscible
transparent liquids of refractive indices
m1, m2, m3 and individual depths d1, d2, d3
respectively, then the apparent depth of
d
d
d
the beaker is = 1 + 2 + 3
µ1 µ2 µ3
Total internal reflection
It is a phenomenon of reflection of light into
denser medium from the boundary of denser
medium and rarer medium. Two essential
conditions for the phenomenon of total
internal reflection are :
X Light should travel from a denser to a
rarer medium.
X Angle of incidence in denser medium
should be greater than the critical angle
for the pair of media in contact.
sin (i − r )
cos r
where t is the thickness of the slab.
X
X
Real depth and apparent depth : When
one looks into a pool of water, it does not
appear to be as deep as it really is. Also
when one looks into a slab of glass, the
material does not appear to be as thick
as it really is. This all happens due to
refraction of light.
Critical angle : It is that angle of
incidence for which the angle of refraction
1
becomes 90°. It is given by sin iC =
R
µD
If the rarer medium is air or vacuum,
1
then sin iC = . Critical angle for red light
µ
is more than that for blue light.
19
Ray Optics and Optical Instruments
X
X
X
Critical angle depends on
– nature of medium
– wavelength of light
A diver in water at a depth d sees the
world outside through a horizontal circle
d
of radius r = d tan iC =
2
µ −1
Applications
of
total
internal
reflection
– The brilliance of diamond is due to the
phenomenon of total internal reflection.
– Mirages in deserts are also due to total
internal reflection.
– The working of optical fibre is based
on the phenomenon of total internal
reflection.
Refraction from a spherical surface
The portion of a refracting medium, whose
curved surface forms the part of a sphere, is
known as spherical refracting surface. Sign
conventions for spherical refracting surface
are the same as those for spherical mirrors.
X Spherical refracting surfaces are of two
types :
– Convex refracting spherical surface
– Concave refracting spherical surface
X
When the object is situated in rarer
medium, the relation between m1
(refractive index of rarer medium), m2
(refractive index of the spherical refracting
surface) and R (radius of curvature) with
the object and image distances is given by
−
X
When the object is situated in denser
medium, the relation between m1, m2, R, u
and v can be obtained by interchanging m1
and m2. In that case, the relation becomes
−
X
µ1 µ2 µ2 − µ1
+
=
u
v
R
µ − µ1
µ2 µ1 µ1 − µ2
µ
µ
+
=
or − 1 + 2 = 2
u
v
R
v
u
R
These formulae are valid for both convex
and concave spherical surfaces.
Lens : A lens is a portion of a transparent
refracting medium bound by two spherical
surfaces or one spherical surface and the
other plane surface. Lenses are divided into
two classes :
X Convex lens or converging lens :
When a lens is thicker in the middle than
at the edges, it is known as convex lens or
converging lens. These are of three types :
– Double convex lens or biconvex lens
– Plano convex lens
– Concavo convex lens
X
Concave lens or diverging lens : When
the lens is thicker at the edges than in
the middle it is known as concave lens or
diverging lens. These are of three types :
– Double concave lens or biconcave lens
– Plano concave lens
– Convexo concave lens
X
Sign Conventions : The sign conventions
for thin lenses are the same as those of
spherical mirrors except that instead of
the pole of the mirror, we now use optical
centre of a lens.
Lens maker’s formula
1
1 
 1
= (µ − 1)  −

f
R
R
 1
2
X
where R1 and R2 are radii of curvature
of the two surfaces of the lens and m is
refractive index of material of lens w.r.t.
medium in which lens is placed. This
formula is valid for thin lenses. It is valid
for both convex and concave lenses. As
per sign convention, for a convex lens,
R1 is positive and R2 is negative and for
a concave lens, R1 is negative and R2 is
positive.
CBSE Board Term-II Physics Class-12
20
X
X
When the refractive index of the material
of the lens is greater than that of the
surroundings, then biconvex lens acts as
a converging lens and a biconcave lens
acts as a diverging lens as shown in the
figure :
When the refractive index of the material
of the lens is smaller than that of the
surrounding medium, then biconvex lens
acts as a diverging lens and a biconcave
lens acts as a converging lens as shown in
the figure.
– For a convex lens, P is positive.
– For a concave lens, P is negative.
When focal length (f) of lens is in cm, then
100
P=
dioptre
f (in cm)
X
X
Combination of thin lenses in contact:
When a number of thin lenses of focal
length f1, f2, ...etc., are placed in contact
coaxially, the equivalent focal length F of
the combination is given by
1 1
1
1
= +
+
+ ....
F f1 f2 f3
The total power of the combination is
given by
P = P1 + P2 + P3 + ...
When two thin lenses of focal lengths f1
and f2 are placed coaxially and separated
by a distance d, the focal length of a
combination is given by
1 1
1
d
= +
−
F f1 f2 f1 f2
In terms of power P = P1 + P2 – dP1P2.
X
X
X
Thin lens formula
1 1 1
− =
v u f
where
u = distance of the object from the optical
centre of the lens
v = distance of the image from the optical
centre of the lens
f = focal length of a lens
f is positive for converging or convex lens
and f is negative for diverging or concave
lens.
Linear magnification
size of image (I ) v
m=
=
size of object (O) u
where m is positive for erect image and m
is negative for inverted image.
Power of a lens
P=
1
focal length in metres
– The SI unit of power of lens is dioptre
(D).
– 1 D = 1 m–1
Refraction through a prism
X Prism : It is a homogeneous, transparent
medium enclosed by two plane surfaces
inclined at an angle. These surfaces are
called the refracting surfaces and angle
between them is known as the refracting
angle or the angle of prism. The angle
between the incident ray and the emergent
ray is known as the angle of deviation.
X For refraction through a prism it is found
that
d = i + e – A where A = r1 + r2
When A and i are small
d = (m – 1) A
In a position of minimum deviation
d = dm, i = e and r1 = r2 = r
 A + δm 
A
∴ i =
 and r =
 2 
2
21
Ray Optics and Optical Instruments
X
The refractive index of the material of the
prism is
 ( A + δm ) 
sin 

2

µ=
 A
sin  
2
This is known as prism formula where A
is the angle of prism and dm is the angle
of minimum deviation.
– When the final image is formed at
infinity (normal adjustment),
Optical Instruments
X Simple microscope : It is also known as
magnifying glass or simple magnifier. It
consists of a convergent lens with object
between its focus and optical centre and
eye close to it. The image formed by it is
erect, virtual, enlarged and on same side
of lens between object and infinity.
M=
vo  D 
uo  fe 
Length of tube, L = vo + fe
– When the final image is formed at least
distance of distinct vision,
M=
vo
uo
D

1 + f 
e 

here uo and vo represent the distance
w
of object and image from the objective
lens, fe is the focal length of an eye lens.
 f D 
Length of the tube, L = vo +  e

 fe + D 
– When the image is formed at infinity
(far point),
M=
D
f
– When the image is formed at the least
distance of distinct vision D (near point),
D
M =1+
f
X
Astronomical telescope (refracting
type)
It consists of two converging lenses. The
one facing the object is known as objective
or field lens and has large focal length
and aperture while the other facing the
eye is known as eye-piece or ocular and
has small focal length and aperture.
– Magnifying power
M=
=
X
angle subtended by image at the eye
angle subtended by thee object at the eye
tan β β
=
tan α α
where both the object and image are
situated at the least distance of distinct
vision.
Compound microscope : It consists
of two convergent lenses of short focal
lengths and apertures arranged co-axially.
Lens (of focal length fo) facing the object is
known as objective or field lens while the
lens (of focal length fe) facing the eye, is
known as eye-piece or ocular. The objective
has a smaller aperture and smaller focal
length than eye-piece. Magnifying power
of a compound microscope
M = mo × me
– When the final image is formed at
infinity (normal adjustment),
f
M= o
fe
Length of tube, L = fo + fe
– When the final image is formed at least
distance of distinct vision,
M=
fo 
f 
1 + e 
fe 
D
Length of tube, L = fo +
fe D
fe + D
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. From a point source, a light falls on a
spherical glass surface (m = 1.5 and radius of
curvature = 10 cm). The distance between point
source and glass surface is 50 cm. The position
of image is
(a) 25 cm
(b) 50 cm
(c) 100 cm
(d) 150 cm
2. A screen is placed 90 cm away from an
object. The image of the object on the screen is
formed by a convex lens at two different locations
separated by 20 cm. Find the focal length of lens.
(a) 42.8 cm
(b) 21.4 cm
(c) 10.7 cm
(d) 5.5 cm
3. A convergent beam of light passes through
a diverging lens of focal length 0.2 m and comes
to focus 0.3 m behind the lens. The position of
the point at which the beam would converge in
the absence of the lens is
(a) 0.12 m
(b) 0.6 m
(c) 0.3 m
(d) 0.15 m
4. The focal length of the lenses of an
astronomical telescope are 50 cm and 5 cm. The
length of the telescope when the image is formed
at the least distance of distinct vision is
(a) 45 cm
(b) 55 cm
325
275
cm
cm
(c)
(d)
6
6
5. Two lenses of power +10 D and –5 D are
placed in contact. Where should an object be held
from the lens, so as to obtain a virtual image of
magnification 2?
(a) 5 cm
(b) –5 m
(c) 10 cm
(d) –10 cm
6.
(a)
(b)
(c)
(d)
Mirage is a phenomenon due to
refraction of light
total internal reflection of light
diffraction of light
none of these.
7. A biconvex lens has a focal length 2/3 times
the radius of curvature of either surface. The
refractive index of the lens material is
(a) 1.75
(b) 1.33
(c) 1.5
(d) 1.0
8. An object approaches a convergent lens from
the left of the lens with a uniform speed 5 m s–1
and stops at the focus. The image
(a) moves away from the lens with an uniform
speed 5 m s–1.
(b) moves away from the lens with an uniform
acceleration.
(c) moves away from the lens with a nonuniform acceleration.
(d) moves towards the lens with a non-uniform
acceleration.
9. A convex lens of focal length 0.2 m and
made of glass (amg = 1.5) is immersed in water
(amw = 1.33). Find the change in the focal length
of the lens.
(a) 5.8 m
(b) 0.58 cm
(c) 0.58 m
(d) 5.8 cm
10. A small telescope has an objective lens of
focal length 144 cm and an eyepiece of focal
length 6.0 cm. What is the separation between
the objective and the eyepiece?
(a) 0.75 m
(b) 1.38 m
(c) 1.0 m
(d) 1.5 m
11. The far point of a near sighted person is
6.0 m from her eyes, and she wears contacts that
enable her to see distant objects clearly. A tree
is 18.0 m away and 2.0 m high. How high is the
image formed by the contacts?
(a) 1.0 m
(b) 1.5 m
(c) 0.75 m
(d) 0.50 m
12. A ray of light is incident on a thick slab of
glass of thickness t as shown in the figure. The
emergent ray is parallel to the incident ray but
23
Ray Optics and Optical Instruments
displaced sideways by a distance d. If the angles
are small then d is,
i

(a) t  1 − 

r
r

(c) it  1 − 

i

(b) rt  1 −

i

r
r

(d) t  1 − 

i
13. A converging lens is used to form an image
on a screen. When the upper half of the lens is
covered by an opaque screen,
(a) half the image will disappear
(b) complete image will disappear
(c) intensity of image will decreases
(d) intensity of image will increases.
14. A tank is filled with water to a height of
12.5 cm. The apparent depth of a needle lying
at the bottom of the tank is measured by a
microscope to be 9.4 cm. If water is replaced by
a liquid of refractive index 1.63 upto the same
height, by what distance would the microscope
have to be moved to focus on the needle again?
(a) 1.00 cm
(b) 2.37 cm
(c) 1.73 cm
(d) 3.93 cm
15. A mark placed on the surface of a sphere is
viewed through glass from a position directly
opposite. If the diameter of the sphere is
10 cm and refractive index of glass is 1.5. The
position of the image will be
(a) –20 cm
(b) 30 cm
(c) 40 cm
(d) – 10 cm
16. Two lenses of focal lengths 20 cm and –40 cm
are held in contact. The image of an object at
infinity will be formed by the combination at
(a) 10 cm
(b) 20 cm
(c) 40 cm
(d) infinity
17. The final image in an astronomical telescope
with respect to an object is
(a) virtual and erect
(b) real and erect
(c) real and inverted
(d) virtual and inverted
18. An object is placed at a distance of 1.5 m
from a screen and a convex lens is interposed
between them. The magnification produced is 4.
The focal length of the lens is
(a) 1 m
(b) 0.5 m (c) 0.24 m (d) 2 m
19. The image of the needle placed 45 cm from
a lens is formed on a screen placed 90 cm on the
other side of the lens. The displacement of the
image, if the needle is moved by 5.0 cm away
from the lens is
(a) 10 cm, towards the lens
(b) 15 cm, away from the lens
(c) 15 cm, towards the lens
(d) 10 cm, away from the lens
20. A real image of a distant object is formed
by a planoconvex lens on its principal axis.
Spherical aberration is
(a) absent.
(b) smaller, if the curved surfaces of the lens
face the object.
(c) smaller, if the plane surface of the lens faces
the object.
(d) same, whichever side of the lens faces the
object.
21. A convex lens is dipped in a liquid whose
refractive index is equal to the refractive index
of the lens. Then its focal length will
(a) become zero
(b) become infinite
(c) become small, but non-zero
(d) remain unchanged
22. A concave mirror of focal length f1 is placed
at a distance d from a convex lens of focal length
f 2 . A beam of light coming from infinity and
falling on this convex lens – concave mirror
combination returns to infinity. The distance d
must equal
(a) f1 + f2
(b) –f1 + f2
(c) 2f1 + f2
(d) –2f1 + f2
23. A ray incident at a point at an angle of
incidence of 60° enters a glass sphere of refractive
index 3 and is reflected and refracted at the
farther surface of the sphere. The angle between
the reflected and refracted rays at this surface is
(a) 50°
(b) 60°
(c) 90°
(d) 40°
24. For a glass prism (µ = 3 ) the angle of
minimum deviation is equal to the angle of the
prism. The angle of the prism is
(a) 45°
(b) 30°
(c) 90°
(d) 60°
CBSE Board Term-II Physics Class-12
24
25. Light travels in two media A and B with
speeds 1.8 × 10 8 m s –1 and 2.4 × 10 8 m s –1
respectively. Then the critical angle between
them is
−1  2 
−1  3 
(a) sin  
(b) tan  
 3
 4
(c) tan
−1  2 
 
3
(d) sin
−1  3 
 
4
3

26. Two identical glass  µ g =  equiconvex

2
lenses of focal length f are kept in contact. The
space between the two lenses is filled with water
4

 µ w =  . The focal length of the combination is
3
3f
f
4f
(a) f
(b)
(c)
(d)
4
2
3
27. A point source of light is placed at a depth
of h below the surface of water of refractive
index m. A floating opaque disc is placed on the
surface of water so that light from the source
is not visible from the surface. The minimum
diameter of the disc is
(a)
(c)
2h
2
(µ − 1)1 / 2
h
2(µ 2 − 1)1 / 2
(b) 2h(m2 – 1)1/2
(d) h(m2 – 1)1/2
28. Critical angle for light going from medium
(i) to (ii) is q. The speed of light in medium (i) is
v, then the speed of light in medium (ii) is
v
(a) v(1 – cosq)
(b)
sin θ
v
v
(c)
(d)
cosθ
(1 − sin θ)
31. Double convex lenses are to be manufactured
from a glass of refractive index 1.55, with both
faces of same radius of curvature. What is the
radius of curvature required if the focal length
is to be 20 cm?
(a) 11 cm (b) 22 cm (c) 7 cm
(d) 6 cm
32. A small angle prism (m = 1.62) gives a
deviation of 4.8°. The angle of prism is
(a) 5°
(b) 6.36°
(c) 3°
(d) 7.74°
33. The optical density of turpentine is higher
than that of water while its mass density is
lower. Figure shows a layer of turpentine floating
over water in a container. For which one of the
four rays incident
on turpentine in
figure, the path
shown is correct?
(a) 1
(b) 2
(c) 3
(d) 4
34. The power of a biconvex lens is 10 dioptre
and the radius of curvature of each surface is
10 cm. Then the refractive index of the material
of the lens is
4
5
3
9
(a)
(b)
(c)
(d)
3
3
2
8
35. A vessel of depth x is half filled with oil of
refractive index m1 and the other half is filled
with water of refractive index m2. The apparent
depth of the vessel when viewed from above is
x µ1µ 2
x(µ1 + µ 2 )
(a)
(b)
2(µ1 + µ 2 )
2µ1µ 2
(c)
x µ1µ 2
(µ1 + µ 2 )
(d)
2 x(µ1 + µ 2 )
µ1µ 2
29. A ray of light passes through an equilateral
prism (refractive index 1.5) such that angle of
incidence is equal to angle of emergence and the
latter is equal to 3/4th of the angle of prism. The
angle of deviation is
(a) 60°
(b) 30°
(c) 45°
(d) 120°
36. There are certain material developed in
laboratories which have a negative refractive
index. A ray incident from air (medium 1) into
such a medium (medium 2) shall follow a path
given by
30. A concave lens is placed in contact with
a convex lens of focal length 25 cm. The
combination produces a real image at a distance
of 80 cm, If an object is at a distance of 40 cm,
the focal length of concave lens is
(a) – 400 cm
(b) – 200 cm
(c) + 400 cm
(d) + 200 cm
(a)
(b)
(c)
(d)
25
Ray Optics and Optical Instruments
37. A point luminous object (O) is at a distance
h from front face of a glass slab of width d and
of refractive index m. On the back face of the slab
is a reflecting plane mirror. An observer sees
the image of object in mirror [figure]. Distance
of image from front face as seen by observer
will be
Observer
O
h
d
2d
(a) h +
µ
(b) 2h + 2d
(c) h + d
(d) h +
d
µ
38. You are given four sources of light each one
providing a light of a single colour – red, blue,
green and yellow. Suppose the angle of refraction
for a beam of yellow light corresponding to a
particular angle of incidence at the interface
of two media is 90°. Which of the following
statements is correct, if the source of yellow
light is replaced with that of other lights without
changing the angle of incidence?
(a) The beam of red light would undergo total
internal reflection.
(b) The beam of red light would bend towards
normal while it gets refracted through the
second medium.
(c) The beam of blue light would undergo total
internal reflection.
(d) The beam of green light would bend away
from the normal as it gets refracted through
the second medium.
39. An astronomical refractive telescope has an
objective of focal length 20 m and an eyepiece of
focal length 2 cm. Then
(a) the magnification is 1000.
(b) the length of the telescope tube is 20.02 m
(c) the image formed is inverted
(d) all of these.
40. A compound microscope consists of an
objective lens with focal length 1.0 cm and eye
piece of focal length 2.0 cm and a tube length
20 cm the magnification will be
(a) 100
(b) 200
(c) 250
(d) 300
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 41 to 45.
Refraction Through Lens
A convex or converging lens is thicker at the
centre than at the edges. It converges a parallel
beam of light on refraction through it. It has
a real focus. Convex lens is of three types :
(i) Double convex lens (ii) Plano-convex lens
(iii) Concavo-convex lens. Concave lens is thinner
at the centre than at the edges. It diverges a
parallel beam of light on refraction through it.
It has a virtual focus.
41. A point object O is placed at a distance of
0.3 m from a convex lens (focal length 0.2 m) cut
into two halves each of which is displaced by
0.0005 m as shown in figure.
What will be the location of the image?
(a) 30 cm right of lens (b) 60 cm right of lens
(c) 70 cm left of lens
(d) 40 cm left of lens
42. Two thin lenses are in contact and the focal
length of the combination is 80 cm. If the focal
length of one lens is 20 cm, the focal length of
the other would be.
(a) –26.7 cm
(b) 60 cm
(c) 80 cm
(d) 20 cm
43. A spherical air bubble is embedded in a piece
of glass. For a ray of light passing through the
bubble, it behaves like a
(a) converging lens
(b) diverging lens
(c) plano-converging lens
(d) plano-diverging lens
44.
(a)
(b)
(c)
(d)
Lens used in magnifying glass is
Concave lens
Convex lens
Both (a) and (b)
Neither concave lens nor convex lens
CBSE Board Term-II Physics Class-12
26
45. The magnification of an image by a convex
lens is positive only when the object is placed
(a) at its focus F
(b) between F and 2F
(c) at 2F
(d) between F and optical centre
Case II : Read the passage given below and
answer the following questions from 46 to 49.
Compound Microscope
A compound microscope is an optical instrument
used for observing highly magnified images of
tiny objects. Magnifying power of a compound
microscope is defined as the ratio of the angle
subtended at the eye by the final image to the
angle subtended at the eye by the object, when
both the final image and the object are situated
at the least distance of distinct vision from the
eye. It can be given that : m = me × mo, where me
is magnification produced by eye lens and mo is
magnification produced by objective lens.
Consider a compound microscope that consists
of an objective lens of focal length 2.0 cm and
an eyepiece of focal length 6.25 cm separated
by a distance of 15 cm.
46. The object distance for eye-piece, so that
final image is formed at the least distance of
distinct vision, will be
(a) 3.45 cm
(c) –1.29 cm
(b) –5 cm
(d) 2.59 cm
47. How far from the objective should an object
be placed in order to obtain the condition
described in Q.No. 46?
(a) 4.5 cm
(b) 2.5 cm
(c) 1.5 cm
(d) 3.0 cm
48. The intermediate image formed by the
objective of a compound microscope is
(a) real, inverted and magnified
(b) real, erect, and magnified
(c) virtual, erect and magnified
(d) virtual, inverted and magnified.
49. The magnifying power of a compound
microscope increases when
(a) the focal length of objective lens is increased
and that of eye lens is decreased.
(b) the focal length of eye lens is increased and
that of objective lens is decreased.
(c) focal lengths of both objective and eye-piece
are increased.
(d) focal lengths of both objective and eye-piece
are decreased.
Assertion & Reasoning Based MCQs
For question numbers 50-55, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
50. Assertion (A) : Higher is the refractive
index of a medium or denser the medium, lesser
is the velocity of light in that medium.
Reason (R) : Refractive index is inversely
proportional to velocity.
51. Assertion (A) : Convergent lens property of
converging remains same in all media.
Reason (R) : Property of lens whether the ray
is diverging or converging is independent of the
surrounding medium.
52. Assertion (A) : Endoscopy involves use of
optical fibres to study internal organs.
Reason (R) : Optical fibres are based on
phenomena of total internal reflection.
53. Assertion(A):Ifopticaldensityofasubstance
is more than that of water, then the mass density
of substance can be less than water.
Reason (R) : Optical density and mass density
are not related.
54. Assertion (A) : Microscope magnifies the
image.
Reason (R) : Angular magnification for image
is more than object in microscope.
55. Assertion (A) : A double convex lens
(m = 1.5) has focal length 10 cm. When the lens
is immersed in water (m = 4/3) its focal length
becomes 40 cm.
1 µ g − µm  1
1 
Reason (R) :
=
−
f
µ m  R1 R2 
27
Ray Optics and Optical Instruments
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. When a monochromatic light travels from
one medium to another its wavelength changes
but frequency remains the same. Explain.
2. When red light passing through a convex
lens is replaced by light of blue colour, how will
the focal length of the lens change?
3. Under what condition does a biconvex lens
of glass having a certain refractive index act as
a plane glass sheet when immersed in a liquid?
4. How does the angle of minimum deviation
of a glass prism vary, if the incident violet light
is replaced by red light?
5. A compound microscope is used because a
realistic simple microscope does not have
magnification.
6. Two thin lenses of power –4 D and 2 D are
placed in contact coaxially. Find the focal length
of the combination.
7. For the same value of angle of incidence, the
angles of refraction in three media A, B and C
are 15°, 25° and 35° respectively. In which media
would the velocity of light be minimum?
8. A biconvex lens made of a transparent
material of refractive index 1.25 is immersed
in water of refractive index 1.33. Will the lens
behave as a converging or a diverging lens? Give
reason.
9. An optical instrument uses an objective
lens of power 100 D and an eyepiece of power
40 D. The final image is formed at infinity when
the tube length of the instrument is kept at 20 cm.
Identify the optical instrument.
10. What is the angle of incidence for maximum
deviation through a prism?
Short Answer Type Questions (SA-I)
11. A ray of light falls on a transparent sphere
with centre C as shown in the figure. The ray
emerges from the sphere parallel to the line AB.
Find the angle of refraction at A if refractive
index of the material of the sphere is 3 .
Air
Air
B
A
60°
C
12. A concave lens of refractive index 1.5 is
immersed in a medium of refractive index 1.65.
What is the nature of the lens?
13. When a ray passes a glass slab of thickness t
at an angle i with an angle of refraction r, what
is the lateral shift of the emergent ray?
15. Draw the image, when an object is kept
beyond the focal point of a convex lens.
16. Draw the image, when an object is placed
between the focus and pole of a convex lens.
17. A thin prism of angle A = 6°° produces a
deviation d = 3°°. Find the refractive index of
the material of prism.
18. A parallel beam of light is incident on a thin
lens as shown. The radius of curvature of both
surfaces is R. Determine the focal length of this
system.
2
1
3
i
1
r
2
1
14. Calculate the focal length of a biconvex lens
if the radii of its surfaces are 60 cm and 15 cm,
and index of refraction of the lens glass = 1.5.
r
i
R
t
R
19. A point object O is kept in air as shown. The
radius of the curved part is 20 cm and medium
on the other side is of refractive index 1.33. Find
the image distance.
CBSE Board Term-II Physics Class-12
28
A fish in the tank is rising upward
along the same line with speed
8 cm/s.
[Take : mwater = 4/3] Find :
(i) speed of the image of fish as
seen by the bird directly
20. A bird in air is diving vertically over a tank
with speed 9 cm/s. Base of the tank is silvered.
(ii) speed of image of bird relative to the fish
looking upwards.
Short Answer Type Questions (SA-II)
21. A coin 2 cm in diameter is embedded in a
solid glass ball of radius 30 cm. The index of
refraction of the ball is 1.5, and the coin is 20 cm
from the surface. Find the position and height
of the image of the coin.
22. What is the difference in the construction
of an astronomical telescope and a compound
microscope? The focal lengths of the objective and
eyepiece of a compound microscope are 1.25 cm
and 5.0 cm, respectively. Find the position of the
object relative to the objective in order to obtain
an angular magnification of 30 when the final
image is formed at the near point.
23. Define power of a lens. Write its units.
1 1 1
Deduce the relation
=
+
for two thin
f
f1 f2
lenses kept in contact coaxially.
24. (a) State the necessary conditions for
producing total internal reflection of light.
(b) Draw ray diagrams to show how specially
designed prisms make use of total internal
reflection to obtain inverted image of the object
by deviating rays (i) through 90°° and (ii) through
180°.
25. Which two of the following lenses L1, L2,
and L3 will you select as objective and eyepiece
for constructing best possible (i) telescope
(ii) microscope? Give reason to support your
answer.
Lens
Power (P)
Aperture (A)
L1
6D
1 cm
L2
3D
8 cm
L3
10 D
1 cm
26. A small bulb (assumed to be a point source)
is placed at the bottom of a tank containing
water to a depth of 80 cm. Find out the area of
the surface of water through which light from
the bulb can emerge. Take the value of the
refractive index of water to be 4/3.
27. (a) Obtain relation between the critical
angle of incidence and the refractive index of
the medium.
(b) Explain briefly how the phenomenon of total
internal reflection is used in fibre optics.
28. Three rays of light red (R), green (G) and
blue (B) fall normally on one of the sides of
an isosceles right angled prism as shown. The
refractive index of prism for these rays is 1.39,
1.44 and 1.47 respectively. Find which of these
rays get internally reflected and which get only
refracted from AC. Trace the paths of rays.
Justify your answer with the help of necessary
calculations.
29. (a) A small telescope has an objective lens
of focal length 140 cm and an eyepiece of focal
length 5.0 cm. Find the magnifying power of the
telescope for viewing distant objects when
(i) the telescope is in normal adjustment,
(ii) the final image is formed at the least distance
of distinct vision.
(b) Also find the separation between the objective
lens and the eyepiece in normal adjustment.
29
Ray Optics and Optical Instruments
30. Find the equivalent focal length of the
combination of lens.
R
air
R R
w R
g g
air
32. A sphere of radius R is made of material
of refractive index m2. Where would an object
be placed so that a real image is formed at
equidistant from the sphere?
m1
glass
glass
O
water
31. An isosceles prism has one of the refracting
surface silvered. A ray of light is incident
normally on the refracting face AB. After two
reflections the ray emerges from the base of the
prism perpendicular to it. Find the angle of the
prism.
m2
R
m2
I
33. Draw a ray diagram to show the image
formation by a combination of two thin convex
lenses in contact. Obtain the expression for the
power of this combination.
Long Answer Type Questions (LA)
34. (a) A point object is placed in front of
a double convex lens (of refractive index
n = n2/n1 with respect to air) with its spherical
faces of radii of curvature R 1 and R 2 . Show
the path of rays due to refraction at first and
subsequently at the second surface to obtain the
formation of the real image of the object. Hence
obtain the lens-maker’s formula for a thin lens.
(b) A double convex lens having both faces of
the same radius of curvature has refractive
index 1.55. Find out the radius of curvature
of the lens required to get the focal length of
30 cm.
OBJECTIVE TYPE QUESTIONS
µ2 µ1 µ2 − µ1
− =
v
u
R
Here, m1 = 1, m2 = 1.5, u = – 50 cm, R = 10 cm
1.5
1
(1.5 − 1)
∴
−
=
v ( −50)
10
1.
(b) : Using,
1.5
= 0.05 − 0.02 = 0.03 ⇒ v = 50 cm
v
2. (b) : The image of the object can be located on the
screen for two positions of convex lens such that u and v are
exchanged.
The separation between two positions of the lens is d = 20 cm
or
35. (a) Plot a graph for angle of deviation as a
function of angle of incidence for a triangular
prism.
(b) Deduce the expression for the refractive
index of glass in terms of angle of prism and
angle of minimum deviation.
36. (a) Define magnifying power of a telescope.
Write its expression.
(b) A small telescope has an objective lens of focal
length 150 cm and an eye piece of focal length
5 cm. If this telescope is used to view a 100 m
high tower 3 km away, find the height of the
final image when it is formed 25 cm away from
the eye piece.
From the figure,
u1 + v1 = 90 cm
v1 – u1 = 20 cm
On solving,
v1 = 55 cm, u1 = 35 cm
From lens formula,
1 1 1
− =
v u f
1
1
1
1 1 1
= or
+ =
−
55 −35 f
55 35 f
⇒ f=
55 × 35
= 21.4 cm
90
CBSE Board Term-II Physics Class-12
30
3. (a) : Here, f = – 0.2 m, v = +0.3 m
The lens formula,
1 1 1 ∴ 1 = 1 − 1 = 1 + 1 = 0.5
− =
u v f 0.3 0.2 0.06
v u f
0.06
\ u=
= 0.12 m
0.5
4. (d) : Here, fo = 50 cm, fe = 5 cm, D = 25 cm
The length of the telescope when the image is formed at the
least distance of distinct vision is
fD
5 × 25
25 325
= 50 +
= 50 + =
cm
L = fo + e
fe + D
5 + 25
6
6
5. (d) : Here, P1 = 10 D and P2 = –5 D
Therefore, power of the combined lens is
P = P1 + P2 = + 10 + (–5) = + 5 D
f
Now, magnification, m =
u +f
1 1
Here, m = 2 and f = = = 0.2 m = 20 cm
P 5
20
∴ 2=
⇒ u + 20 = 10
u + 20
⇒ u = –10 cm
6. (b) : Mirage is phenomenon due to total internal
reflection of light.
Now,
10. (d) : The separation between the objective and the eye
piece = Length of the telescope tube i.e. f = fo + fe
Here, fo = 144 cm = 1.44 m
fe = 6.0 cm = 0.06 m
\ f = 1.44 + 0.06 = 1.5 m
11. (d) : The far point of 6.0 m tell us that the focal length
of the lens is f = –6.0 m, u = – 18 m and h = 2 m
1 1 1
1 1 1
1
1
Using, = −
⇒
= + =
−
f v u
v f u −6.0 18.0
⇒ v = – 4.5 m
 −v 
 −4.5 
\ The image size, h ’ = h   = 2 ×  −
= 0.50 m
 u 
 18.0 
12. (c) : From figure, in right angled DCDB, ∠CBD = (i – r)
CD d
=
BC BC d = BC sin(i – r) ...(i)
∴ sin(i − r ) =
1 1
1
9. (c) : Using = (a µg )  − 
fa
 R1 R2 
a
Here, fa = 0.2 m, mg = 1.50
1 1
1
1 1
∴
= (1.50 − 1)  −  = 0.50  − 
0.2
 R1 R2 
 R1 R2 
1 1
⇒
−
= 10
R1 R2
\
Consider fw be the focal length of the lens, when immersed
in water. If wmg be refractive index of glass w.r.t. water, then
a
µg 1.50
w
µg = a
=
= 1.128
µw 1.33
fw =
= 0.78 – 0.2 = 0.58 m
or
(a) :
( w µg − 1)  R11 − R12  = (1.128 − 1) × 10 = 1.28
1
= 0.78
1.28
Hence, change in focal length of the lens,
fw – fa
or
1 1
1
= (µ − 1)  − 
f
 R1 R2 
2
Here, f = R , R1 = +R , R2 = −R
3
1
 1 1  (µ − 1) × 2
∴
= (µ − 1)  +  =
R R 
(2 / 3)R
R
3
µ − 1 = = 0.75 ⇒ m = 1.75
4
8. (c) : When an object moving towards a convergent
lens from the left of the lens with a uniform speed of
5 m s–1, the image moves away from the lens with a non
uniform acceleration.
7.
1
=
fw
Also, in right angled DCNB
BN
t
=
BC
BC
t
or BC =
...(ii)
cos r
Substitute equation (ii) in equation (i), we get
t
d=
sin(i − r )
cos r
cosr =
For small angles sin(i – r) ≈ i – r; cosr ≈ 1
r

d = t(i – r), d = it 1 − 
i

13. (c) : Image formed is complete but has decreased
intensity.
14. (c) : Actual depth of the needle in water, h1 = 12.5 cm
Apparent depth of needle in water, h2 = 9.4 cm
h 12.5
∴ µ water = 1 =
= 1.33
h2 9.4
Hence, mwater = 1.33
When water is replaced by a liquid of refractive index
m′ = 1.63, the actual depth remains the same, but its apparent
depth changes.
Let H be the new apparent depth of the needle.
31
Ray Optics and Optical Instruments
h1
h 12.5
or H = 1 =
= 7.67 cm
H
µ' 1.63
Here, H is less than h2. Thus to focus the needle again, the
microscope should be moved up. Distance by which the
microscope should be moved up
= 9.4 – 7.67 = 1.73 cm
∴ µ’ =
15. (a) : As refraction occurs from denser to rarer medium
µ2 µ1 µ1 − µ2
∴
+ =
−u v
R
Here, m1 = 1, m2 = 1.5, R = 5 cm, u = – 10 cm
1 1 3 −1
1.5 1 1− 1.5 1
=
or
= − =
or v = –20 cm
+ =
10 v
v 10 20 20
−5 10
16. (c) : Equivalent focal length,
Here, f1 = 20 cm and f2 = –40 cm
1 1 1
= +
f f1 f2
1 1 1 2 −1
, f = 40 cm
= − =
f 20 40 40
1 1 1
From lens formuls, = −
f v u
Here, f = 40 cm, u = ∞
1 1 1
∴
= −
⇒ v = f = 40 cm
f v ∞
17. (d) : The final image in an astronomical telescope with
respect an object is virtual and inverted.
v
−v
18. (c) : Here, m = = −4 or u =
u
4
Also, |u| + |v| = 1.5
v
−1.2
+ v = 1.5 or v = 1.2 m and u =
= −0.3 m
4
4
uv
−0.3 × 1.2
∴ f=
=
= 0.24 m
u − v −0.3 − 1.2
19. (c) : Here, u = – 45 cm, v = 90 cm
1 1 1 1 1
1
\
⇒ f = 30 cm
= − = + =
f v u 90 45 30
When the needle is moved 5 cm away from the lens,
u = –(45 + 5) = – 50 cm
1 1 1 1
1
2
\
= + = +
=
v ′ f u ′ 30 −50 150
or v′ = 75 cm
\ Displacement of image = v – v′ = 90 – 75 = 15 cm,
towards the lens.
22. (c) :
\
d = 2f1 + f2
23. (c) : Refraction at P,
sin 60°
= 3
sin r1
1
sinr1 =
2
or r1 = 30°
Since, r2 = r1
\ r2 = 30°
sin r2
sin 30° 1
1
or
=
=
Refraction at Q,
or i2 = 60°
sin i2
sin i2
3
3
At point Q, r2′ = r2 = 30°
∴ α = 180° − (r2′ + i2 ) = 180° – (30° + 60°) = 90°
 A + δm 
sin 
 2 
24. (d) : Using, µ =
A
sin
2
According to question,
dm = A
A+A
sin 
 2 
(Given, µ = 3 )
3=
∴
A
sin
2
A
A
2 sin cos
sin A
2
2
3=
=
A
A
sin
sin
2
2
A
3
A
or cos =
= cos 30° ⇒
= 30° or A = 60°
2
2
2
20. (b) : When the curved surface of the lens faces the
object, the spherical aberration is smaller. The total deviation
is shared between the curved and the plane surfaces.
25. (d) : Here, vA = 1.8 × 108 m s–1,
vB = 2.4 × 108 m s–1
Light travels slower in denser medium. Hence medium A is a
denser medium and medium B is a rarer medium.
Here, light travels from medium A to medium B. Let C be the
critical angle between them.
1
∴ sinC = A µB =
B
µA
21. (b) : When refractive index of lens is equal to the
refractive index of liquid, the lens behave like a plane surface
with focal length infinity.
Refractive index of medium B w.r.t. to medium A is
Velocity of light in medium A vA
A
µB =
=
Velocity of light in medium B vB
CBSE Board Term-II Physics Class-12
32
∴ sinC =
vA 1.8 × 108 3
=
=
or C = sin −1  3 
4
vB 2.4 × 108 4
26. (d) : Let R be the radius of curvature of each surface.
For the combination of focal length
1 1 1
1 1 1
= + ⇒ = −
f f1 f2
f2 f f1
1 3 1 15 − 16
= − =
f2 80 25
400
1
1 1
= (1.5 − 1)  +  \ R = f
R R 
f
1 4  1 1 
For the water-lens, =  − 1  − −  f′ 3  R R 
1  2
= − 
3 f 
1
2
=−
f′
3f
1 1 1 1
Using,
= + + ,
F f1 f2 f3
or
1 1 1 1 2 2 4
= + + = − =
F f f f ′ f 3f 3f
∴
F=
3f
4
27. (a) : The figure shows incidence from water at critical
angle qc for the limiting case.
Disc
r
Air
1
Now, sinθc = Water
µ
θc
h
1
θ
c
so that tan θc = 2
(µ − 1)1/2
r
h
where r is the radius of the disc.
Therefore, diameter of the disc is
2h
2r = 2h tan θc = 2
(µ − 1)1/2
µ
v
28. (b) : sinθ = 2 =
µ1 v ′
From figure, tanθc =
Light
source
where v and v′ are the speeds of light in medium (i) and
medium (ii) respectively.
v
\ v′ =
sinθ
29. (b) : Using, A + d = i + e
Here, A = 60°, m = 1.5,
3
3
i = e = A = × 60° = 45°
4
4
\ 60° + d = 45° + 45° ⇒ d = 90° – 60° = 30°
1 1 1
30. (a) : = −
f v u
Here, u = –40 cm, v = +80 cm
1 1
1
1+ 2 3
∴
= −
=
=
f 80 −40 80 80
1 −1
=
⇒ f2 = −400 cm
f2 400
31. (b) : Lens maker formula,
1 1
1
= (µ − 1)  − 
f
 R1 R2 
Here, f = 20 cm, m = 1.55, R1 = R and R2 = –R
1
1 
1
1 1
∴
= (1.55 − 1)  −
 = 0.55 ×  + 

20
R R
R ( −R )
2
1
= 0.55 ×
or
20
R
\ R = 1.1 × 20 = 22 cm
32. (d) : Here, m = 1.62, d = 4.8°
Using, d = (m – 1)A
4.8
δ
A=
=
(µ − 1) (1.62 − 1)
A = 7.74°
33. (b) : In the figure, the path shown for the ray 2 is correct.
The ray suffers two refractions : At A, ray goes from air to
turpentine, bending towards normal. At B, ray goes from
turpentine to water (i.e., From denser to rarer medium),
bending away from normal.
34. (a) : Power of lens, P (in dioptre)
100
=
focal length f (in cm)
100
= 10 cm
10
According to lens maker’s formula
1
1
1
= (µ − 1)  − 
f
 R1 R2 
∴ f=
For biconvex lens, R1 = + R, R2 = – R
∴
1
1 1
= (µ − 1)  + 
R R 
f
1
2
1
2
= (µ − 1)  
= (µ − 1)   ⇒
 10 


10
f
R
1
3
1
(µ − 1) =
⇒ µ = + 1=
2
2
2
33
Ray Optics and Optical Instruments
and magnification, m =
35. (a) :
fo
20
=
= 1000
fe 0.02
The image formed is inverted with respect to the object.
As refractive index, µ =
40. (c) : Magnification, of compound microscope
Real depth
Apparent depth
\
Apparent depth of the vessel when viewed from above is
x 1
1 
x
x
=  + 
d apparent =
+
2 µ1 2 µ2 2  µ1 µ2 
 L  D 
m = mo × me =   ×  
 fo   fe 
36. (a) : Using Snell’s law
sin i
sin i
µ=
⇒ sin r =
sin r
µ
Here, L = 20 cm, D = 25 cm (near point), fo = 1 cm and
fe = 2 cm
20 25
∴ m = × = 250
1 2
41. (b) : Each half of the lens will form an image in the same
plane. The optic axes of the lenses are displaced,
1
1
1
= ; v = 60 cm
−
v ( −30) 20
According to question, m is negative
\ sin r is negative.
Hence, r is negative, therefore the ray follows path as shown
in option (a).
42. (a) : Here f1 = 20 cm ; f2 = ?
F = 80 cm
1 1 1
1 1 1
As + =
⇒
= −
f
F f1
f1 f2 F
2
37. (a) : As shown in figure glass
slab will form the image of bottom
d 
i.e., mirror MM’ at a depth  
µ
1 1 1 −3
= − =
f2 80 20 80
−80
f2 =
= −26.7 cm
3
43. (b) : The bubble behaves like a diverging lens.
=
x  µ2 + µ1  x (µ1 + µ2 )
=
2  µ1µ2 
2 µ1µ2
from its front face. So the distance
of object O from virtual mirror mm’
d 
will be h +   .
µ
Observer
O
h
m
M
d
h
d
m
M
d
h+
I
Now as a plane mirror forms image behind the mirror at the
same distance as the object is in front of it, the distance of
d 
image I from mm’ will be h +   and as the distance of
µ
d 
virtual mirror from the front face of slab is  µ  , the distance
44. (b) : Convex lens is used in magnifying glass.
45. (d) : Magnification of an image by a convex lens is
positive when object is placed between F and optical centre.
46. (b) : Here f0 = 2.0 cm, fe = 6.25 cm, u0 = ?
When the final image is obtained at the least distance of
distinct vision, ve = – 25 cm
As
1 1 1
− =
v e ue fe
of image I from front face as seen by observer will be
 d d
2d
= h +  + = h +
µ
µ
µ


\
1
1 1
1
1
−1 − 4 −5
1
= − =
−
=
=
=−
ue v e fe −25 6.25
25
25
5
or
ue = –5 cm
38. (c) : Here, for yellow light, r = 90° when i = C. As i is
kept same, C must be smaller for total internal reflection.
1
, C will be smallest for blue colour.
From µ =
sinC
Therefore, blue light would undergo total internal reflection.
47. (b) : Distance between objective and eye-piece
= 15 cm
\ Distance of the image from objective is
v0 = 15 – 5 = 10 cm
\ 1 = 1 − 1 = 1 − 1 = 1− 5 = − 2
u0 v 0 f0 10 2 10
5
39. (d) : In normal adjustment,
Length of telescope tube, L = fo + fe
Here, fo = 20 m and fe = 2 cm = 0.02 m
\ L = 20 + 0.02 = 20.02 m
or
\
5
u0 = − = −2.5 cm
2
Distance of object from objective = 2.5 cm
CBSE Board Term-II Physics Class-12
34
48. (a) : The intermediate image formed by the objective of
a compound microscope is real, inverted and magnified.
49. (d) : Magnifying power of a compound microscope
increases when fixed length of both objective and eye-piece
are decreased.
50. (a) : Both assertion and reason are true and reason is
the correct explanation of assertion.
According to Snell’s law,
µ
c / v2
v
sin i
= 2 =
= 1
sin r
µ1
c / v1
v2
or, m1v1 = m2v2
This shows that higher is the refractive index of a medium
or denser the medium, lesser is the velocity of light in that
medium.
51. (d) : A convex lens made of glass behaves as a
convergent lens when placed in air or water. However when
the same lens is immersed in carbon disulphide (m = 1.63),
it behaves as a divergent lens. Therefore when a convergent
lens is placed inside a transparent medium of refractive index
greater than that of material of the lens, it behaves as a
divergent lens. Behaviour of a lens depends on the refractive
index of a surrounding medium.
52. (a) : An endoscope is made of optical fibres. Its core
is made of optically denser material. Its outer cladding is
made of optically rarer material. It is based on total internal
reflection.
53. (a) : Optical density and mass density are not related to
each other. Mass density is mass per unit volume. It is possible
that mass density of an optically denser medium be less than
that of an optically rarer medium (optical density is the ratio
of the speed of light in two media). e.g., turpentine and water.
Mass density of turpentine is less than that of water but its
optical density is higher.
54. (a) : Microscope is an optical instrument which forms a
magnified image of a small nearby object and thus, increases
the visual angle subtended by the image at the eye so that
the object is seen to be bigger and distinct. Therefore, angular
magnification for image is more than object.
55. (a) :
 1 1  µ
1 w
 1 1 
= ( µg − 1)  −  =  g − 1  −  ...(i)
f′
R
R
µ
 1 2   w   R1 R2 
 1 1 
1  µg
=
− 1  −  ...(ii)
f  µα   R1 R2 
From (i) and (ii)
 µg

− 1

10 × 8
f ′  µα 
=
⇒ f′ =
= 40 cm
2
f
 µg 
 µ − 1
w
SUBJECTIVE TYPE QUESTIONS
1. Frequency being a characteristic of source of light, does
not change with change of medium.
Refractive index m of medium is defined as,
c (speed of light in vacuum)
µ= =
v (speed of light in medium)
As, v = ul
1
\ µ∝
(∵ υ is same in different media)
λ
Hence, wavelength of light is different in different media.
2. Focal length of the lens decreases when red light is
replaced by blue light.
3. When the refractive index of the biconvex lens is equal to
the refractive index of the liquid in which lens is immersed then
the biconvex lens behaves as a plane glass sheet. In this case,
1
= 0 or f → ∞.
f
1
4.
lred > lviolet and λ ∝
µ
\ mred < mviolet
 δ + A
sin  m
 2 
µ=
A
sin
2
\
(δm)violet > (δm)red
5. A compound microscope is used because a realistic
simple microscope does not have large magnification.
6. Net power P = P1 + P2 = –4 + 2 = –2 D
1
1
Focal length f = =
m = –0.5 m = –50 cm
P −2
c sini
7. Refractive index, µ = =
v sinr
As sin15° < sin25° < sin35°
So, vA < vB < vC
Hence in medium A, velocity of light is minimum.
8. The lens will act as a diverging lens as the refractive
index of water is greater than that of lens.
9. (a) Po = 100 D, \ fo = 1 cm, Pe = 40 D,
\ fe = 2.5 cm.
Since fo < fe, the instrument is a compound microscope.
10. The deviation is maximum when angle is 90°.
sin(i )
11. From Snell’s law, we have :
=µ
sin(r )
At A, i = 60°; m = 3
sin(i )
Now, sin(r ) =
µ
sin(60°) 1
 1
⇒ sin(r ) =
=
⇒ r = sin−1  
 2
3
2
\
r = 30°
35
Ray Optics and Optical Instruments
12. Focal length of a concave lens is negative.
Using lens maker’s formula,
1  µl
1
 1
=
− 1  − 
f  µ m   R1 R2 
Here, ml = 1.5, mm = 1.65
Also,
µl
µ

< 1, so  l − 1 is negative and focal length
µm
 µm 
of the given lens becomes positive. Hence, it behaves as a
converging lens.
13. Using Snell’s law, for ray 1 and its refraction,
sin i = m sin r and sin e = m sin r
2
⇒ ∠i = ∠e
e
N B
The emergent ray is parallel
d
r
to the incident ray.
C
Further,
)
r –r
(i
AN
t
AB =
=
cos r cos r
A
i
And BC = AB sin (i – r)
1
t
⇒ lateral shift (d) =
sin (i – r)
cos r
14.
R1 = 60 cm
µ1
R2 = 15 cm
µ1
µ2
Using the Lens maker’s Formula,
1  µ2   1
1
=  − 1  − 
f  µ1   R1 R2 
1  1.5   1
1 
 1 1
=
− 1  −
 = (0.5)  + 



f
1
60 −15
 60 15 
1
1 + 4 
= (0.5) 
⇒ f = +24 cm.
f
 60 
15. When the object is outside the focal point of a converging
lens, the image is real and inverted.
16. The object is between the pole and focal point of a
converging lens. The image is virtual, enlarged and erect.
I
f
f
O
O
17. We know that d = A(m – 1)
δ
or µ = 1 +
A
3
Here A = 6°, d = 3° therefore µ = 1 + = 1.5
6
18. for Ist surface, u1 = ∞
µ2 µ1 µ2 − µ1
µ2 µ1 µ2 − µ1
− =
⇒
− =
+R
v
u
R
v1 ∞
for 2nd surface, u2 = v1 , v2 = f
µ 3 µ2 µ 3 − µ2
−
=
f
v1
R
Adding eqn. (i) and (ii), we get
µ3 µ3 − µ1
µR
=
⇒ f= 3
f
R
µ3 − µ1
... (i)
... (ii)
19. u = –40 cm, R = – 20 cm, m = 1, m2 = 1.33
Applying equation for refraction through spherical surface,
we get
µ2 µ1 µ2 − µ1
1.33
1
1.33 − 1
−
=
− =
or
v
u
R
v
−40
−20
After solving, v = –32 cm
20. (i) Velocity of fish in air
3
8
=
= (8) = 6 cm s–1 (upwards)
4 /3 4
Velocity of fish w.r.t bird = 6 + 9 = 15 cm s–1 (upwards)
4
(ii) Velocity of bird in water = 9 × = 12 cm s–1 (downwards)
3
Velocity of bird w.r.t fish = 12 + 8 = 20 cm s–1 (downwards)
21. u = –20 cm,
v = ? ,
m1 = 1.5, m2 = 1,
20 cm
30 cm
R = –30 cm
C Coin
µ2 µ1 µ2 − µ1
⇒
− =
v
u
R
1 1.5 1 − 1.5
⇒
−
=
⇒ v = –17.1 cm.
v −20
−30
The positive sign indicates that the image is formed inside the
glass or the image is virtual.
v/µ 2
−17.11
/
−17.1 × 1.5
=
=
= +1.28.
m=
u/µ1 −20/1.5
−20
The positive value of m indicates it is an erect image.
(dia)image = m × (dia)coin = 1.28 × 2 = 2.56 cm.
22. Construction for astronomical telescope.
CBSE Board Term-II Physics Class-12
36
Construction for compound microscope :
u
v
A
B″
h
B
Objective
A″
30 =
v 0  25 
1 + 
u0 
5
From equation (iii) and equation (iv), we can write
1 1 1
= +
f f1 f2
24. (a) Essential conditions for total internal reflection:
(i) Light should travel from a denser medium to a rarer
medium.
(ii) Angle of incidence in denser medium should be greater
than the critical angle for the pair of media in contact.
(b) (i) To deviate a ray of light through 90° :
A′
D
Angular magnification, m =
⇒
Eyepiece
B′
h′ E
H
O
1 1 1
− = …(iv)
v u f
fe
v0  D 
1 + 
u 0  fe 
A
90°
P
⇒ v0 = 5u0...(i)
From lens formula,
u +v
1 1
−1
1
= −
⇒
= 0 0 ...(ii)
f 0 v 0 ( −u 0 )
1.25
v 0u 0
Substituting (i) in (ii), u0 = 1.5 cm
23. Power of lens : It is the reciprocal of focal length of a
lens.
1
P = (f is in metre)
f
Unit of power of lens : Dioptre
A
B
B
R
45°
45°
45°
Q
C
A totally reflecting prism is used to deviate the path of the
ray of light through 90°, when it is inconvenient to view the
direct light. In Michelson’s method to find velocity of light, the
direct light from the octagonal mirror is avoided from direct
viewing by making use of totally reflecting prism.
(ii) To deviate a ray of light through 180° : When the ray of
light comes to meet the hypotenuse face BC at right angles
to it, it is refracted out of prism as such along the path RS.
The path of the ray of light has been turned through 180°
due to two total internal reflections.
A
O
I
I1
v
u
90°
Q
P
v1
An object is placed at point O. The lens A produces an image
at I1 which serves as a virtual object for lens B which produces
final image at I.
Given, the lenses are thin. The optical centres (P) of the lenses
A and B coincide with each other.
For lens A, we have
1 1 1
− = …(i)
v 1 u f1
1 1 1
For lens B, we have − = …(ii)
v v 1 f2
Adding equations (i) and (ii),
1 1 1 1
− = + …(iii)
v u f1 f2
If two lenses are considered as equivalent to a single lens of
focal length f, then
B
45°
45°
R
45°
45°
P
C
S
25. An astronomical telescope should have an objective of
larger aperture and longer focal length while an eyepiece of
small aperture and small focal length. Therefore, we will use
L2 as an objective and L3 as an eyepiece.
For constructing microscope, L3 should be used as objective
and L1 as eyepiece because both the lenses of microscope
should have short focal lengths and the focal length of
objective should be smaller than the eyepiece.
26. The light rays starting from
bulb can pass through the surface if
angle of incidence at surface is less
than or equal to critical angle (C)
for water air interface. If h is the
depth of bulb from the surface, the
light will emerge only through a
r
h
C
S
37
Ray Optics and Optical Instruments
circle of radius r given by r =
Area of water surface =
πh 2
2
µ −1
h
µ2 − 1
=
22
(0.80)2
×
= 2.6 m2
7 (1.33)2 − 1
27. (a) Relation between refractive index and critical angle
When i = C, r = 90°
From Snell’s law, mb sinC = ma sin90° = ma × 1
µb
1 a
1
=
; µb =
µ a sinC
sinC
(b) Optical fibre is made up of very fine quality glass or
quartz of refractive index about 1.7.
A light beam incident on one end of an optical fibre at
appropriate angle refracts into the fibre and undergoes
repeated total internal reflection.
This is because the angle of incidence is greater than critical
angle. The beam of light is received at other end of fibre with
nearly no loss in intensity. To send a complete image, the
image of different portion is send through separate fibres
and thus a complete image can be transmitted through an
optical fibre.
29. (a) (i) Given f0 = 140 cm, fe = 5 cm
When final image is at infinity, magnifying power,
−f
140
m= 0 =−
5.0
fe
m = –28
Negative sign shows that the image is inverted.
(ii) When final image is at the least distance of distinct
vision,
−f  f 
magnifying power, m = 0 1 + e 
fe  D 
=
(b) Separation between objective and eye piece when final
image is formed at infinity,
L = f0 + fe
L = 140 cm + 5.0 cm
L = 145 cm
30.
r1
r2= c
glass core = 1.7
28. Critical angle for
(i)
Red light is sin c1 =
or
c1 = 46°
A
B
G
R
B G
1 1 1
+ +
f1 f2 f3
Second lens is water lens with surrounding air.
1  µw   1
1 
2
−
=
−1 
 = − (µw − 1)
f2  µ a   −R +R 
R
Third lens is glass lens with surrounding air.
1  µg   1
1  2
=
−1  −
 = (µ g − 1)
f3  µ a   R −R  R
1
= 0.6944
(ii) Green light is sin c2 =
1.44
or c2 = 44°
1
(iii) Blue light is sin c3 =
= 0.6802
1.47
or c3 = 43°
As angle of incidence i = 45° of red light ray on face AC is
less that its critical angle of 46°, so red light ray will emerge
out of face AC.
B
feq
=
1  µg   1
1  2
=
− 1  −
 = (µ g − 1)

f1  µ a  R −R  R
1
= 0.7194
1.39
C
1
First lens is glass lens with surrounding air.
cladding = 1.5
i1
−140  5.0 
1 +  = –33.6
5.0  25 
R
⇒
1 2
2
2
= (µ g − 1) − (µw − 1) + (µ g − 1)
feq R
R
R
R
feq =
4µ g − 2µw − 2
Equivalent lens will have a focal length feq with air surrounding
it.
31. The incident ray passes
without deviation from face AB. It
suffers reflections at P and Q. From
figure, incident ray and normal at
Q are parallel; therefore
a = 2A ...(i)
Also, 2a + A = 180° (Sum of
angles of isosceles prism) ...(ii)
On solving eqns. (i) and (ii), we get
A = 36°, a = 72°
A
(90° – A)
90°
A
P
A
Q
B
C
CBSE Board Term-II Physics Class-12
38
32. Let the object be placed at a distance x from the pole P1
of the sphere. If a real image is to be formed at equidistant
from the sphere, then the ray must pass symmetrically through
the sphere, as shown in the figure. Applying the equation at
the first surface, we get
1
1
P1
O
P2
2
x
I
x
2R
 µ1 
µ2 µ1 µ2 − µ1
−
=
or x = 
R
−R
+∞ − x
 µ2 − µ1 
Note that the real image is formed only when the refractive
index of the sphere is more than that of the surrounding,
i.e., m2 > m1.
33.
Consider two lenses placed close to each other. The focal
lengths of lens A and B is f1 and f2 respectively.
For lens A,
1 1 1
− = ... (i)
v ′ u f1
For lens B,
1 1 1
− = ...(ii)
v v ′ f2
Adding (i) and (ii),
1 1 1 1 1 1
− + − = +
v ′ u v v ′ f1 f2
Since
1 1 1
1 1 1
− = then, = +
v u f
fT f1 f2
1

∵ P = 
f
PT = P1 + P2
34. (a) Refraction at convex spherical surface :
When object is in rarer medium and image formed is real.
n1
O
N
A
i
u PN
By Snell’s law 1n2 =
or
n2 α + γ
=
or n2 γ − n2β = n1α + n1γ
n1 γ − β
or (n2 – n1)g = n1a + n2b...(i)
As a, b and g are small and P and N lie close to each other,
AN AN
So, α ≈ tan α =
≈
NO PO
β ≈ tanβ =
AN AN
≈
NI
PI
γ ≈ tan γ =
AN AN
≈
NC PC
On using them in equation (i), we get
AN
AN
AN
(n2 − n1 )
= n1
+ n2
PC
PO
PI
n −n
n
n
or 2 1 = 1 + 2 ...(ii)
PC
PO PI
where, PC = + R, radius of curvature
PO = – u, object distance
PI = + v, image distance
n −n
n
n
n −n n
n
So 2 1 = 1 + 2 or 2 1 = 2 − 1
R
v
R
v
u
−u
This gives formula for refraction at spherical surface when
object is in rarer medium.
1
1
 1
= (µ − 1)  − 
(b)
f
 R1 R2 
Here f = 30 cm, m = 1.55, R1 = – R2 = R
1
1
2
 1 1
\
= (1.55 − 1)  +  or
= 0.55 ×


30
R R
30
R
or
1 1 1 1
− = +
v u f1 f2
sin i i α + γ
≈ =
sin r r γ − β
\
R = 0.55 × 2 × 30 = 33 cm.
35. (a) If graph is plotted between angle of incidence i and
angle of deviation d, it is found that the angle of deviation d
first decreases with increase in angle of incidence i and then
becomes minimum ‘dm’ when i = e and then increases with
increase in angle of incidence i. Figure shows the path of a
ray of light suffering refraction through a prism of refracting
angle ‘A’.
n2
r
R C I
v
n1 < n 2
m
i
In DOAC, i = a + g
and in DAIC, g = r + b or r = g – b
i=e
(b) At minimum deviation, the inside beam travels parallel
to base of the prism.
39
Ray Optics and Optical Instruments
36. (a) Magnifying power of refracting telescope (M) is
defined as the ratio of the angle subtended by the image (b)
at the eye to the angle subtended by the distant object at the
unaided eye (a).
A
m
i
r
e=i
r
M=
i=e
r = r′
dm = (i + e) – (r + r′)
dm = 2i – 2r...(i)
Also r + r′ = A = 2r...(ii)
So, angle of incidence using equation (i)
A + δm
A
i=
, angle of refraction r =
2
2
Now refractive index of the material of prism
 A + δm 
sin 
 2 
sin i
a
=
µg =
A
sin r
sin
2
where A is the “refracting angle” of the prism and
A = 60° for an equiangular prism.
β
α
(b) Here, fo = 150 cm, fe = 5 cm
Angle subtended by 100 m tall tower at 3 km is
α=
100
1
= rad
3 × 1000 30
If h is the height of image formed by the objective, then
h
h
α= =
fo 150
\
h
1
150
=
or h =
cm = 5 cm
150 30
30
Magnification produced by eyepiece
 D   25 
me =  1 +  =  1 +  = 6
5
 fe  
\

Height of final image = h × me = 5 × 6 = 30 cm
CHAPTER
10
Wave Optics
Recap Notes
Wavefront : The locus of all particles of the
medium vibrating in the same phase at a
given instant is known as wavefront.
Depending on the shape of sources of light,
wavefront can be of three types
X Spherical wavefront : When the source
of light is a point source, the wavefront is
spherical.
X Cylindrical wavefront : When
the
source of light is linear, the wavefront is
cylindrical.
X Plane wavefront : When the point source
or linear source of light is at very large
distance, a small portion of spherical or
cylindrical wavefront appears to be plane.
Such a wavefront is known as plane
wavefront.
Huygens principle : According to Huygens
principle,
– Every point on given wavefront
(primary wavefront) acts as a fresh
source of new disturbance, called
secondary wavelets.
– The secondary wavelets spread out
in all the directions with the speed of
light in the medium.
– A surface touching these secondary
wavelets tangentially in the forward
direction at any instant gives the new
(secondary) wavefront at that instant.
Laws of reflection by Huygens principle
X Let us consider a plane wavefront AB
incident on the plane reflecting surface xy.
The tangent B′A′ represents reflected
wavefront after time t.
A
r
i
x
A′
i
r
B′
B
y
For every point on wavefront AB, a
corresponding point lies on the reflected
wavefront A′B′.
So, comparing two triangle DBAB′ and
DB′A′B.
We find that
AB′ = A′B = ct, BB′ = common
∠A = ∠A′ = 90°
Thus two triangles are congruent, hence
∠i = ∠r
This proves first law of reflection.
X Also incident rays, reflected rays and
normal to them all lie in the same plane.
This gives second law of reflection.
Laws of refraction by Huygens principle
X Let us consider a plane wavefront AB
incident on the plane refracting surface
xy. The tangent B′A′ represent refracted
wavefront after time ‘t’. For every point on
primary wavefront AB, a corresponding
point lies on the refracted wavefront A′B′.
vt
41
Wave Optics
From DABB′ and DA′B′B, Snell’s law can be
proved.
sin i ct / BB′ c a
=
= = µg
sin r vt / BB′ v
So, first law of refraction can be proved.
X Also, the incident ray, refracted rays
and normal to the rays, all lie in the
same plane. This gives the second law of
refraction.
– Effect on frequency, wavelength and
speed during refraction. When a wave
passes from one medium to another
then change in speed v takes place,
wavelength l also changes, whereas its
frequency u remains the same.
X Coherent and incoherent sources : The
sources of light, which emit continuous
light waves of the same wavelength, same
frequency and in same phase or having
a constant phase difference are known
as coherent sources. Two sources of light
which do not emit light waves with a
constant phase difference are called
incoherent sources.
Interference of light : It is the phenomenon
of redistribution of energy on account of
superposition of light waves from two coherent
sources. Interference pattern produce points
of maximum and minimum intensity. Points
where resultant intensity is maximum,
interference is said to be constructive and
at the points of destructive interference,
resultant intensity is minimum.
X Intensity distribution : If a, b are the
amplitudes of interfering waves due to two
coherent sources and f is constant phase
difference between the two waves at any
point P, then the resultant amplitude at
P will be
R = a2 + b2 + 2ab cos φ
Resultant intensity,
I = I1 +I2 + 2 I1I2 cosφ
– When cos φ = 1; I max =
(
– When cos φ = − 1, I min =
–
I max
=
I min
(
(
)
2
I2 )
I1 + I2
I1 −
2
I1 + I2
(
)
2
I1 − I2
)
2
Conditions for sustained interference
of light : The two sources should
continuously emit waves of the same
wavelength or frequency.
– The amplitudes of waves from two
sources should preferably be equal.
– The waves emitted by the two sources
should either be in phase or should
have a constant phase difference.
– The two sources must lie very close to
each other.
– The two sources should be very narrow.
Young’s double slit experiment : Young’s
double slit experiment was the first to
demonstrate the phenomenon of interference
of light. Using two slits illuminated by
monochromatic light source, he obtained
bright and dark bands of equal width placed
alternately. These were called interference
fringes.
X For
constructive interference (i.e.,
formation of bright fringes)
For nth bright fringe,
Path difference = xn d = nλ
D
where n = 0 for central bright fringe
n = 1 for first bright fringe,
n = 2 for second bright fringe and so on
d = distance between two slits
D = distance of slits from the screen
xn = distance of nth bright fringe from the
centre.
D
∴ xn = nλ
d
X For destructive interference (i.e. formation
of dark fringes).
For nth dark fringe,
Path difference = xn d = (2n − 1) λ
D
2
where
n = 1 for first dark fringe,
n = 2 for second dark fringe and so on.
xn = distance of nth dark fringe from the
centre
λD
∴ xn = (2n − 1)
2d
X Fringe width : The distance between
any two consecutive bright or dark fringes
is known as fringe width.
λD
Fringe width, β =
d
X
CBSE Board Term-II Physics Class-12
42
X
Angular fringe width : θ =
β λ
=
D d
If W1, W2 are widths of two slits, I1, I2 are
intensities of light coming from two slits;
a, b are the amplitudes of light from these
slits, then
W1 I1 a2
= =
W2 I2 b2
2
I max (a + b)
=
I min (a − b)2
– When entire apparatus of Young’s
double slit experiment is immersed in
a medium of refractive index m, then
fringe width becomes
λ ′D λD β
=
=
β′ =
d
µd µ
– When a thin transparent plate of
thickness t and refractive index m
is placed in the path of one of the
interfering waves, fringe width remains
unaffected but the entire pattern shifts
by
D
β
∆x = (µ − 1) t = (µ − 1) t
d
λ
This shifting is towards the side in which
transparent plate is introduced.
X Colour of thin films : A soap film or a
thin film of oil spread over water surface,
when seen in white light appears coloured.
This effect can be explained in terms of
phenomenon of interference.
Diffraction of light : It is the phenomenon
of bending of light around corners of an
obstacle or aperture in the path of light.
X Diffraction due to a single slit : The
diffraction pattern produced by a single slit
of width a consists of a central maximum
bright band with alternating bright and
dark bands of decreasing intensity on
both sides of the central maximum.
– Condition for nth secondary maximum is
Path difference = a sin θn = (2n + 1) λ
2
where n = 1, 2, 3,.......
– Condition for nth secondary minimum
is
Path difference = asinqn = nl
where n = 1, 2, 3,.......
– Width of secondary maxima or minima
λD
β=
a
where a = width of slit
D = distance of screen from the slit
2λD
– Width of central maximum =
a
– Angular fringe width of central
2λ
maximum =
a
– Angular fringe width of secondary
λ
maxima or minima =
a
X Fresnel distance : It is the minimum
distance a beam of light has to travel
before its deviation from straight line
path becomes significant.
Fresnel distance, Z F =
a2
λ
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. In a double slit experiment using light of
wavelength 600 nm, the angular width of a
fringe on a distant screen is 0.1°. The spacing
between the two slits is
(a) 3.44 × 10–4 m
(b) 1.54 × 10–4 m
(c) 1.54 × 10–3 m
(d) 1.44 × 10–3 m
2. A diffraction pattern is obtained by using
beam of red light what will happen, if red light
is replaced by the blue light?
(a) Bands disappear.
(b) Bands become broader and farther apart.
(c) No change will take place.
(d) Diffraction bands become narrow and
crowded together.
3. A screen is placed 50 cm from a single slit
which is illuminated with light of wavelength
6000 Å. If the distance between the first and
third minima in the diffraction pattern is
3.0 mm. The width of the slit is
(a) 1 × 10–4 m
(b) 2 × 10–4 m
(c) 0.5 × 10–4 m
(d) 4 × 10–4 m
4. Which of the following is correct for light
diverging from a point source?
(a) The intensity decreases in proportion for the
distance squared.
(b) The wavefront is parabolic.
(c) The intensity at the wavelength does not
depend on the distance.
(d) None of these.
5. A Young’s double slit experiment uses a
monochromatic source of light. The shape of
interference fringes formed on the screen is
(a) parabola
(b) straight line
(c) circle
(d) hyperbola
6. A pa ra lle l be a m o f so d iu m l igh t o f
wavelength 5890 Å is incident on a thin glass
plate of refractive index 1.5 such that the angle
of refraction in the plate is 60°. The smallest
thickness of the plate which will make it dark
by reflection
(a) 3926 Å
(c) 1396 Å
(b) 4353 Å
(d) 1921 Å
7. Consider the following statements in case of
Young’s double slit experiment.
(1) A slit S is necessary if we use an ordinary
extended source of light.
(2) A slit S is not needed if we use an ordinary
but well collimated beam of light.
(3) A slit S is not needed if we use a spatially
coherent source of light.
Which of the above statements are correct?
(a) (1), (2) and (3)
(b) (1) and (2) only
(c) (2) and (3) only
(d) (1) and (3) only
8. In a Young’s double slit experiment an
electron beam is used to obtain interference
pattern. If the spread of electron is decreases
then
(a) distance between two consecutive fringes
remains the same
(b) distance between two consecutive fringes
decreases
(c) distance between two consecutive fringes
increases
(d) none of these
9. The colours seen in the reflected white light
from a thin oil film are due to
(a) Diffraction
(b) Interference
(c) Polarisation
(d) Dispersion
10. In young’s double slit experiment using
monochromatic light of wavelengths l, the
intensity of light at a point on the screen with
path difference is l is M unit. The intensity of
light at a point where path difference is l/3 is
M
M
(a)
(b)
2
4
M
M
(c)
(d)
16
8
11. In a two slit experiment with monochromatic
light, fringes are obtained on a screen placed at
CBSE Board Term-II Physics Class-12
44
some distance from the plane of slits. If the
screen is moved by 5 × 10–2 m towards the slits,
the change in fringe width is 3 × 10–5 m. If the
distance between slits is 10–3 m, the wavelength
of light will be
(a) 3000 Å
(b) 4000 Å
(c) 6000 Å
(d) 7000 Å
12. In a double slit experiment the distance
between slits is increased ten times whereas
their distance from screen is halved then the
fringe width is
1
1
(a) becomes
(b) becomes
20
90
1
(c) it remains same
(d) becomes
10
13. The fringe width in a young’s double slit
interference pattern is 2.4 × 10 –4 m, when
red light of wavelength 6400 Å is used. How
much will it change, if blue light of wavelength
4000 Å is used ?
(a) 0.9 × 10–4 m
(b) 1.5 × 10–4 m
–4
(c) 4.5 × 10 m
(d) 0.45 × 10–4 m
14. Two slits in young’s double slit experiment
have widths in the ratio 81 : 1. The ratio of the
amplitudes of light waves is
(a) 3 : 1
(b) 3 : 2
(c) 9 : 1
(d) 6 : 1
15. In Young’s double slit experiment two
disturbances arriving at a point P have phase
π
difference of
. The intensity of this point
3
expressed as a fraction of maximum intensity
I0 is
1
3
I
I0
(a)
(b)
2 0
2
3
4
I
I0
(c)
(d)
4 0
3
16. Interference fringes were produced in
Young’s double slit experiment using light of
wavelength 5000 Å. When a film of material
2.5 × 10–3 cm thick was placed over one of the
slits, the fringe pattern shifted by a distance
equal to 20 fringe widths. The refractive index
of the material of the film is
(a) 1.25
(b) 1.33
(c) 1.4
(d) 1.5
17. Two beams of light having intensities I and
4I interfere to produce a fringe pattern on a
screen. The phase difference between the beams
is p/2 at point A and p at point B. Then the
difference between the resultant intensities at
A and B is
(a) 2I
(b) 4I
(c) 5I
(d) 7I
18. In the case of light waves from two coherent
sources S 1 and S 2, there will be constructive
interference at an arbitrary point P, the path
difference S1P – S2P is
1

(a)  n +  λ
(b) nl

2
λ
1

(c)  n −  λ
(d)

2
2
19. A plane wave passes through a convex lens.
The geometrical shape of the wavefront that
emerges is
(a) plane
(b) diversing spherical
(c) converging spherical
(d) none of these
20. In a Young’s double slit experiment, the
source is white light. One of the holes is covered
by a red filter and another by a blue filter. In this
case
(a) there shall be alternate interference patterns
of red and blue
(b) there shall be an interference pattern for red
distinct from that for blue
(c) there shall be no interference fringes
(d) there shall be an interference pattern for red
mixing with one for blue
21. In a double slit experiment, the distance
between the slits is d. The screen is at a distance
D from the slits. If a bright fringe is formed
opposite to one of the slits, its order is
(a)
(c)
22.
104
(a)
(b)
(c)
(d)
d
λ
(b)
λ2
dD
d2
D2
(d)
2D λ
2λd
Consider sunlight incident on a slit of width
Å. The image seen through the slit shall
be a fine sharp slit white in colour at the
centre
a bright slit white at the centre diffusing to
zero intensities at the edges
a bright slit white at the centre diffusing to
regions of different colours
only be a diffused slit white in colour
45
Wave Optics
23. Two so urce s o f lig ht o f w ave l e n gt h
2500 Å and 3500 Å are used in Young’s double
slit experiment simultaneously. Which orders
of fringes of two wavelength patterns coincide?
(a) 3rd order of 1st source and 5th of the 2nd
(b) 7th order of 1st and 5th order of 2nd
(c) 5th order of 1st and 3rd order of 2nd
(d) 5th order of 1st and 7th order of 2nd
24. Yellow light of wavelength 6000 Å produces
fringes of width 0.8 mm in Young’s double slit
experiment. If the source is replaced by another
monochromatic source of wavelength 7500 Å and
the separation between the slits is doubled then
the fringe width becomes
(a) 0.1 mm
(b) 0.5 mm
(c) 4.3 mm
(d) 1 mm
25. In Young’s double slit experiment the
distance d between the slits S1 and S2 is 1 mm.
What should the width of each slit be so as to
obtain 10 maxima of the double slit pattern
within the central maximum of the single slit
pattern?
(a) 0.9 mm
(b) 0.8 mm
(c) 0.2 mm
(d) 0.6 mm
26. The idea of secondary wavelets for the
propagation of a wave was first given by
(a) Newton
(b) Huygens
(c) Maxwell
(d) Fresnel
27. When interference of light takes place
(a) energy is created in the region of maximum
intensity
(b) energy is destroyed in the region of maximum
intensity
(c) conservation of energy holds good and
energy is redistributed
(d) conservation of energy does not hold good
28. Which of the following is the path difference
for distructive interference ?
λ
(a) n(l + 1)
(b) (2n + 1)
2
λ
(c) nl
(d) (n +1)
2
29. The phenomena which is not explained by
Huygens construction of wavefront
(a) reflection
(b) diffraction
(c) refraction
(d) origin of spectra
30. In a Young’s double slit experiment, (slit
distance d) monochromatic light of wavelength
l is used and the fringe pattern observed at a
distance D from the slits. The angular position
of the bright fringes are
 Nλ 
(a) sin −1 
 d 
 Nλ 
(c) sin −1 
 D 

 N
−1 
(b) sin 

  N
(d) sin −1 

1 
λ
2  


d
1 
+ λ
2


D
+
31. For what distance is ray optics a good
approximation when the aperture is 4 mm wide
and the wavelength is 500 nm?
(a) 22 m
(b) 32 m
(c) 42 m
(d) 52 m
32. A narrow slit of width 2 mm is illuminated
by monochromatic light of wavelength 500 nm.
The distance between the first minima on either
side on a screen at a distance of 1 m is
(a) 5 mm
(b) 0.5 mm
(c) 1 mm
(d) 10 mm
33. The two coherent sources with intensity
ratio b produce interference. The fringe visibility
will be
2 β
(a)
(b) 2b
1+β
2
β
(c)
(d)
(1 + β)
1+β
34. A slit of width d is illuminated by white light.
The first minimum for red light (l = 6500 Å) will
fall at q = 30° when d will be
(a) 3200 Å
(b) 6.5 × 10–4 mm
(c) 1.3 micron
(d) 2.6 × 10–4 cm
35. In a Young’s double slit experiment, let S1
and S2 be the two slits, and C be the centre of the
screen. If ∠S1CS2 = q and l is the wavelength,
the fringe width will be
λ
(a)
(b) lq
θ
(c)
2λ
θ
(d)
λ
2θ
36. Young’s experiment is performed with
light of wavelength 6000 Å wherein 16
fringes occupy a certain region on the screen.
If 24 fringes occupy the same region with another
light, of wavelength l, then l is
(a) 6000 Å
(b) 4500 Å
(c) 5000 Å
(d) 4000 Å
CBSE Board Term-II Physics Class-12
46
37. Wavefront is the locus of all points, where the
particles of the medium vibrate with the same
(a) phase
(b) amplitude
(c) frequency
(d) period
38. In a Young’s double slit experiment, the
angular width of a fringe formed on a distant
screen is 1°. The slit separation is 0.01 mm. The
wavelength of the light is
(a) 0.174 nm
(c) 0.174 mm
(b) 0.174 Å
(d) 0.174 × 10–4 m
39. Light from two coherent sources of the same
amplitude A and wavelength l illuminates the
screen. The intensity of the central maximum is
I0. If the sources were incoherent, the intensity
at the same point will be
(a) 4I0
(b) 2I0
I0
(c) I0
(d)
2
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 40 to 42.
Hnygen Principle
Huygen principle is the basis of wave theory of
light. Each point on a wavefront acts as a fresh
source of new disturbance, called secondary
waves or wavelets. The secondary wavelets
spread out in all directions with the speed light
in the given medium.
An initially parallel cylindrical beam travels
in a medium of refractive index m(I) = m0 + m2I,
where m0 and m2 are positive constants and I is
the intensity of the light beam. The intensity of
the beam is decreasing with increasing radius.
E
E
A C
C
a′
a
b
e
F B
A
D
b′
c
c′
d
d′
e′
F
a
a′
b
b′
c
c′
d
d′
e
e′
B
(b)
D
As the beam enters the medium, it will
travel as a cylindrical beam
diverge
converge
diverge near the axis and converge near the
periphery.
Case II : Read the passage given below and
answer the following questions from 43 to 45.
Diffraction
The phenomenon of bending of light around the
sharp corners and the spreading of light within
the geometrical shadow of the opaque obstacles is
called diffraction of light. The light thus deviates
from its linear path. The deviation becomes
much more pronounced, when the dimensions of
the aperture or the obstacle are comparable to
the wavelength of light.
42.
(a)
(b)
(c)
(d)
Incident
wave
Diffracted
wave
a
l
Screen
40. The initial shape of the wavefront of the
beam is
(a) planar
(b) convex
(c) concave
(d) convex near the axis and concave near the
periphery
43. In diffraction from a single slit the angular
width of the central maxima does not depends
on
(a) l of light used
(b) width of slit
(c) distance of slits from the screen
(d) ratio of l and slit width
41. According to Huygens Principle, the surface
of constant phase is
(a) called an optical ray (b) called a wave
(c) called a wavefront
(d) always linear in shape
44. For a diffraction from a single slit, the
intensity of the central point is
(a) infinite
(b) finite and same magnitude as the surrounding
maxima
(a)
47
Wave Optics
(c) finite but much larger than the surrounding
maxima
(d) finite and substantially smaller than the
surrounding maxima
45. In a single diffraction pattern observed on a
screen placed at D metre distance from the slit
of width d metre, the ratio of the width of the
central maxima to the width of other secondary
maxima is
(a) 2 : 1
(b) 1 : 2
(c) 1 : 1
(d) 3 : 1
Case III : Read the passage given below and
answer the following questions from 46 to 49.
Young’s Double Slit Experiment
A narrow tube is bent in the form of a circle of
radius R, as shown in figure. Two small holes S
and D are made in the tube at the positions at
right angle to each other. A source placed at S
generates a wave of intensity I0 which is equally
divided into two parts: one part travels along the
longer path, while the other travels along the
shorter path. Both the waves meet at point D
where a detector is placed.
R
S
D
46. If a maxima is formed at a detector, then the
magnitude of wavelength l of the wave produced
is given by
πR
(a) pR
(b)
2
πR
(c)
(d) all of these
4
47. The maximum intensity produced at D is
given by
(a) 4I0
(b) 2I0
(c) I0
(d) 3I0
48. In a Young’s double slit experiment, the
intensity at a point where the path difference
is l/6 (l – wavelength of the light) is I. If I0
denotes the maximum intensity, then I/I 0 is
equal to
1
2
1
(c)
2
(a)
3
2
3
(d)
4
(b)
49. Two identical light waves, propagating in the
same direction, have a phase difference d. After
they superpose the intensity of the resulting
wave will be proportional to
(a) cosd
(b) cos(d/2)
2
(c) cos (d/2)
(d) cos2d
Assertion & Reasoning Based MCQs
For question numbers 50-58, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
50. Assertion (A) : The maximum intensity in
interference pattern is four times the intensity
due to each slit.
Reason (R): Intensity is directly proportional to
square of amplitude.
53. Assertion (A) : We cannot get diffraction
pattern from a wide slit illuminated by
monochromatic light.
Reason (R) : In diffraction pattern, all the bright
bands are not of the same intensity.
51. Assertion (A) : Diffraction is common in sound
but not common in light waves.
Reason (R): Wavelength of light is more than the
wavelength of sound.
54. Assertion (A) : When a light wave travels
from a rarer to a denser medium, it loses speed.
The reduction in speed imply a reduction in
energy carried by the light wave.
Reason (R) : The energy of a wave is proportional
to velocity of wave.
52. Assertion (A) : Interference obeys the law of
conservation of energy.
Reason (R) : The energy is redistributed in case of
interference.
55. Assertion (A) : The film which appears bright
in reflected system will appear dark in the
transmitted light and vice-versa.
CBSE Board Term-II Physics Class-12
48
Reason (R) : The conditions for film to appear
bright or dark in reflected light are just reverse
to those in the transmitted light.
56. Assertion (A) : In Young’s double slit
experiment, the fringes become indistinct if one
of the slits is covered with cellophane paper.
Reason (R) : The cellophane paper decrease the
wavelength of light.
57. Assertion (A) : One of the condition for
interference is that the two source should be
very narrow.
Reason (R) : One broad source is equal to large
number of narrow sources.
58. Assertion (A) : When tiny circular obstacle is
placed in the path of light from some distance,
a bright spot is seen at the centre of the shadow
of the obstacle.
Reason (R) : Destructive interference occurs at the
centre of the shadow.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Define the term ‘coherent sources’ which
are required to produce interference pattern in
Young’s double slit experiment.
2. State Huygens principle of diffraction of
light.
3. How does the angular separation between
fringes in single-slit diffraction experiment
change when the distance of separation between
the slit and screen is doubled.
4. In a Young’s double slit experiment, the
fringe width is found to be 0.12 mm. If the whole
apparatus is immersed in water of refractive
index (4/3), without disturbing the geometrical
arrangement, what is the new fringe width?
5. In double-slit experiment using light of
wavelength 600 nm, the angular width of a
fringe formed on a distant screen is 0.1°.
What is the spacing between the two slits?
6. What is the shape of the wavefront on earth
for sunlight?
7. Is Huygens principle valid for longitudinal
sound waves?
8. Why is the diffraction of sound waves more
evident in daily experience than that of light
wave?
9. If one of the slits in Young’s double slit
experiment is fully closed, the new pattern has
______ central maximum in angular size.
10. Two slits are made one millimetre apart
and the screen is placed one metre away. What
is the fringe separation when blue-green light
of wavelength 500 nm is used?
Short Answer Type Questions (SA-I)
11. In a two-slit experiment with monochromatic
light, fringes are obtained on a screen placed
at some distance from the slits. If the screen
is moved by 5× × 10 –2 m towards the slits,
the change in fringe width is 3 ×× 10–5. If the
distance between the slits is 10–3 m, calculate
the wavelength of the light used.
12. A soap film of thickness 0.3 mm appears dark
when seen by the reflected light of wavelength
580 nm. What is the index of refraction of the
soap solution,if it is known to be between 1.3
and 1.5?
13. Light of wavelength 6 ×× 10–5 cm falls on a
screen at a distance of 100 cm from a narrow slit.
Find the width of the slit if the first minima lies
1 mm on either side of the central maximum.
14. The intensity of the light coming from one of
the slits in a YDSE is double the intensity from the
other slit. Find the ratio of maximum intensity
to minimum intensity in the interference fringe
pattern observed.
15. In a YDSE, the slits are 2 mm apart and are
illuminated with a mixture of two wavelengths
λ = 750 nm and λ′ = 900 nm. At what distance
from the common central bright fringe on a
screen 2 m from the slits will a bright fringe
from one interference pattern coincide with a
bright fringe from the other?
49
Wave Optics
16. Two wavelengths of sodium light 590 nm and
596 nm are used, in turn to study the diffraction
taking place at a single slit of aperture 2 ××10–4 m.
The distance between the slit and the screen
is 1.5 m. Calculate the separation between the
positions of the first maxima of the diffraction
pattern obtained in the two cases.
18. A slit of width a is illuminated by light of
wavelength 6000 Å. For what value of a will the
17. Laser light of wavelength 640 nm incident on
a pair of slits produces an interference pattern
in which the bright fringes are separated by
7.2 mm. Calculate the wavelength of another
source of light which produces interference fringes
separated by 8.1 mm using same arrangement.
Also find the minimum value of the order ( n)
of bright fringe of shorter wavelength which
coincides with that of the longer wavelength.
19. Yellow light (l = 6000 Å) illuminates a single
slit of width 1 ×× 10–4 m. Calculate the distance
between two dark lines on either side of the
central maximum, when the diffraction pattern
is viewed on a screen kept 1.5 m away from the
slit.
(i) First maximum fall at an angle of diffraction
of 30°
(ii) First minimum fall at an angle of diffraction
30°?
20. In a single slit diffraction experiment first
minimum for l1 = 660 nm coincides with first
maxima for wavelength l2. Calculate l2.
Short Answer Type Questions (SA-II)
21. A beam of light consisting of two wavelengths,
6500 Å and 5200 Å is used to obtain slit experiment
(1 Å = 10–10 m). The distance between the slits is
2.0 mm and the distance between the plane of
the slits and the screen is 120 cm.
S1
Source
C
S2
T1
P
O
OP = x
CO = D
S1C = CS2 = D
T2
Screen
(i) F
ind the distance of the third bright fringe
on the screen from the central maximum for
the wavelength 6500 Å.
24. (a) In a single slit diffraction pattern, how
does the angular width of the central maximum
vary, when
(ii) W
hat is the least distance from the central
maximum where the bright fringes due to
both the wavelengths coincide?
(i) aperture of slit is increased?
22. White coherent light (400 nm-700 nm) is
sent through the slits of a YDSE, the separation
between the slits is 0.5 mm and the screen is
50 cm away from the slits. There is a hole in
the screen at a point 1 mm away (along the
width of the fringes) from the central line. Which
wavelength will be absent in the light coming
from the hole?
S
0.5
mm
1mm
50 cm
O
23. Consider a two slit interference arrangements
such that the distance of the screen from the slits
is half the distance between the slits. Obtain
the value of D in terms of λ such that the first
minima on the screen falls at a distance D from
the centre O.
(ii) distance between the slit and the screen is
decreased?
(b) How is the diffraction pattern different from
the interference pattern obtained in Young’s
double slit experiment?
25. (a) Two monochromatic waves emanating
from two coherent sources have the displacements
represented by
y1 = a cos wt and y2 = a cos (wt + f)
where f is the phase difference between the
two displacements. Show that the resultant
intensity at a point due to their superposition is
given by I = 4 I0 cos2 f/2, where I0 = a2.
(b) Hence obtain the conditions for constructive
and destructive interference.
26. Why cannot two independent monochromatic
sources produce sustained interference pattern?
Deduce, with the help of Young’s arrangement
to produce interference pattern, an expression
for the fringe width.
CBSE Board Term-II Physics Class-12
50
27. The figure shows a modified Young’s double
slit experimental set-up. Here SS2 – SS1 = l/4.
(a) W
rite the condition for constructive
interference.
30. A plane wavefront propagating in a medium
of refractive index ‘ m1’ is incident on a plane
surface making the angle of incidence i as
shown in the figure. It enters into a medium of
refraction of refractive index ‘m2’ (m2 > m1). Use
Huygens’ construction of secondary wavelets to
trace the propagation of the refracted wavefront.
Hence verify Snell’s law of refraction.
(b) Obtain an expression for the fringe width.
28. The YDSE set-up is shown. On the lower slit
a slab of thickness 0.1 mm and refractive index
3/2 is placed. Find
31. Explain the following, giving reasons:
S1
d
S
O
S2
D
S
(i) Position of central maxima
(ii) Order of maxima at O and how many fringes
will cross O if slab is removed? l = 5000 Å,
d = 50 ×× 10–4 cm, f = 30°, D = 2 m
29. In a YDSE, D = 1 m, d = 1 mm, and l = 1/2
mm.
(i) Find the distance between the first and
central maxima on the screen.
(ii) Find the number of maximum and minimum
obtained on the screen.
(i) When light travels from a rarer to a denser
medium, the speed decreases. Does this decrease
in speed imply a reduction in the energy carried
by the wave ?
(ii) In the wave picture of light, intensity of light
is determined by the square of the amplitude of
the wave. What determines the intensity in the
photon picture of light?
32. (a) If one of two identical slits producing
interference in Young’s experiment is covered
with glass, so that the light intensity passing
through it is reduced to 50%, find the ratio of the
maximum and minimum intensity of the fringe
in the interference pattern.
(b) What kind of fringes do you expect to observe
if white light is used instead of monochromatic
light?
Long Answer Type Questions (LA)
33. (a) Use Huygen’s geometrical construction
to show how a plane wave-front at t = 0
propagates and produces a wave-front at a later
time.
decreased?
(b) Verify, using Huygen’s principle, Snell’s law
of refraction of a plane wave propagating from
a denser to a rarer medium.
(b) The intensity at the central maxima in
Young’s double slit experimental set-up is I0.
Show that the intensity at a point where the
path difference is l/3 is I0/4.
(c) When monochromatic light is incident on a
surface separating two media, the reflected and
refracted light both have the same frequency.
Explain why.
34. (a) What is the effect on the interference
fringes in a Young’s double slit experiment when
(i) the separation between the two slits is
(ii) the width of the source slit is increased?
Justify your answer in each case.
35. (a) U s i n g H u y g e n s ’ c o n s t r u c t i o n o f
secondary wavelets explain how a diffraction
pattern is obtained on a screen due to a narrow
slit on which a monochromatic beam of light is
incident normally.
(b) Show that the angular width of the first
51
Wave Optics
diffraction fringe is half that of the central
fringe.
1 λ

(c) Explain why the maxima at q =  n + 

2 a
become weaker and weaker with increasing n.
36. (a) Describe briefly how a diffraction
pattern is obtained on a screen due to a single
narrow slit illuminated by a monochromatic
source of light. Hence obtain the conditions for
OBJECTIVE TYPE QUESTIONS
1.
(a) : Angular width of fringe,
π
θ = 0.1° = 0.1 ×
rad
180
3.14
=
rad
1800
∴
d=
λ 600 × 10−9
=
= 3.44 × 10−4 m
θ 3.14 / 1800
2. (d) : When red light is replaced by blue light the
diffraction bands become narrow and crowded.
3.
is
(b) : The position of nth minima in the diffraction pattern
xn =
nDλ
d
∴
x 3 − x 1 = (3 − 1)
or
d=
4.
(a)
5.
(d)
Dλ 2Dλ
=
d
d
2 × 0.50 × 6000 × 10−10
2Dλ
=
= 2 × 10–4 m
x 3 − x1
3 × 10−3
5890 × 10−10
λ
=
2 µ cos r 2 × 1.5 × cos 60°
⇒ 3926 × 10–10 m = 3926 Å
7.
(a)
8.
(c) : Fringe width, β =
λD
d
h
mv
Here h is planck’s constant. This wavelength is inversily
proportional to the velocity. Hence, the fringe width increases
with decrease in electron speed.
Also,λ =
9. (b) : The colours of a thin oil film are due to interference.
10. (b) : Resultant intensity
IR = I1 + I2 + 2 I1I2 cos θ
If path difference = l, phase difference 2p
\ IR = I + I + 2 I × I cos 2π (Q I1 = I2 = I)
= 4I = M..(i)
λ
2π
If path difference , phase difference φ = rad
3
3
2π
IR ′ = I + I + 2 I × I cos
3
M
 1
= 2I + 2I  −  = I =
 2
4
[(Using (i)]
λD
d
where l is the wavelength of light, D is the distance between
screen and the slits and d is the distance between two slits
λ
∴ ∆β = ∆D
(As l and d are constants)
d
11. (c) : Fringe width, β =
6. (a) : The condition for minimum thickness corresponding
to a dark band is, 2µt cosr = l
∴ t=
the angular width of secondary maxima and
secondary minima.
(b) Two wavelengths of sodium light of
590 nm and 596 nm are used in turn to study
the diffraction taking place at a single slit of
aperture 2 ×× 10 –6 m. The distance between
the slit and the screen is 1.5 m. Calculate the
separation between the positions of first maxima
of the diffraction pattern obtained in the two
cases.
∆βd
∆D
Substituting the given values, we get
(3 × 10−5 m) (10−3 m)
λ=
= 6 × 10–7 m = 6000 Å
(5 × 10−2 m)
12. (a) : Fringe width,
Dλ
β=
... (i)
d According to the question,
D
D ′ = and d ′ = 10 d
2
D ′ λ (D / 2) λ 1 Dλ
∴ β′ =
=
=
d′
10 d
20 d
β
(Using (i))
⇒ β′ = 20
13. (a) : Here, b1 = 2.4 × 10–4 m,
l1 = 6400 Å, l2 = 4000 Å
or λ =
CBSE Board Term-II Physics Class-12
52
Q
β2 λ2 4000 5
=
=
=
β1 λ1 6400 8
= I + 4I + 2 I × 4I cos π p = 5I – 4I = I
5
or β2 = × β1
8
5
= × 2.4 × 10−4 = 1.5 × 10–4 m
8
Decrease in fringe width
Db = b1 – b2 = (2.4 – 1.5) × 10–4 = 0.9 × 10–4 m
14. (c) : Width ratio,
β1 I1 81
= =
β2 I2 1
\
A1
I
81
= 1 =
=9:1
1
A2
I2
Amplitude ratio,
15. (d) : The resultant intensity
φ
I = I 0 cos2
2
Given : S = 20b...(iii)
From equations (i), (ii) and (iii) we get,
(m – 1)t = 20l
or
20 λ 20 × 5000 × 10−8 cm
=
t
2.5 × 10−3 cm
m – 1 = 0.4 or
m = 1.4
17. (b) : Here, I1 = I, I2 = 4I, f1 =
IA = I1 + I2 + 2 I1I2 cos φ1
= I + 4I + 2 I × 4I cos
π
= 5I
2
IB = I1 + I2 + 2 I1I2 cos φ2
IA – IB = 5I – I = 4I
18. (b) : Constructive interference occurs when the path
difference (S1P – S2P) is an integral multiple of l.
or S1P – S2P = nl
where n = 0, 1, 2, 3, .....
19. (c) : Converging spherical.
20. (c) : The light from two slits of Young’s double slit
experiment is of different colours/wavelengths/frequencies,
Hence, there shall be no interference fringes.
21. (d) : When a bright fringe is formed opposite to one of
d
the slits, x =
2
xd d d d 2
path difference =
= × =
D 2 D 2D
If it is nth order bright fringe,
Here, I0 is the maximum intensity
π
and φ =
3
2
 π 
 3
2 π
∴ I = I 0 cos2 
cos
=
I
=
I
0

 3 × 2  0
6
 2 
3
I = I0
4
λD
16. (c) : Fringe width, β =
...(i)
d
where D is the distance between the screen and slit and d is
the distance between two slits.
When a film of thickness t and refractive index m is placed
over one of the slit, the fringe pattern is shifted by distance
S and is given by
(µ − 1)tD
S=
...(ii)
d
(µ − 1) =
\
π
,f =p
2 2
d2
d2
or n =
2D
2Dλ
22. (a) : Visible range, l = 3900 Å to 7000 Å
Given width of the slit, a = 104 Å = 10000 Å
Thus a > l, Hence, no diffraction occurs.
The image seen through the slit shall be a fine sharp slit white
in colour at the centre.
path differnce nλ =
23. (b) : Let nth fringe of 2500 Å coincide with (n – 2)th
fringe of 3500 Å.
\ 3500(n – 2) = 2500 × n
1000n = 7000, or n = 7
th
\ 7 order fringe of 1st source will coincide with 5th order
fringe of 2nd source.
24. (b) : Fringe width in first case
Dλ
β1 = 1
d Fringe width in second case
Dλ
β2 = 2
2d Divide equation (ii) by (i),
β2 Dλ2 / 2d 1 λ2
1 λ
∴
=
= .
or β2 = . 2 . β1
β1 Dλ1 / d
2 λ1
2 λ1
...(i)
...(ii)
1 7500 Å
×
× 0.8 mm = 0.5 mm = 0. 5 mm
2 6000 Å
25. (c) : The linear separation between n bright fringes in
nλD
an interference pattern on the screen is x n =
d
As xn < < D, the angular separation between n bright fringes
x
nλ
θn = n =
D
d
=
53
Wave Optics
For 10 bright fringes,
10 λ
θ10 =
d
The angular width of the central maximum in the diffraction
pattern due to slit of width a is
2λ
2θ1 =
a
d 1
10 λ 2 λ
or a ≤ = mm = 0. 2 mm
Now
<
5 5
d
a
26. (b)
28. (b) : For destructive interference, the path difference
λ
should be an odd multiple of 2 .
29. (d) : The Huygen’s construction of wavefront does not
explain the phenomena of origin of spectra.
30. (a) :
The condition for bright fringes is,
path difference, d = dsinqbright = Nl
where N = 0, ± 1, ± 2,
The angular position of the bright fringes is
 Nλ 
θbright = sin−1  
 d 
31. (b) : Fresnel distance,
\
a 2 (4 × 10−3 )2 4 × 4 × 10−6
=
=
λ 500 × 10−9
5 × 10−7
zF = 32 m
–3
32. (b) : Here, a = 2 mm = 2 × 10 m
l = 500 nm = 500 × 10–9 m = 5 × 10–7 m
D=1m
The distance between the first minima on either side on a
screen is
=
2λD 2 × 5 × 10−7 × 1
=
a
2 × 10−3
= 5 × 10–4 m = 0.5 × 10–3 m = 0.5 mm
I1 a 2
a
= 2 =β ∴ = β
I2 b
b
Fringe visibility is given by
33. (a) :
Imax − Imin (a + b )2 − (a − b )2
=
Imax + Imin (a + b )2 + (a − b )2
2(a / b ) 2 β
=
2(a + b )  a 2  β + 1
 2 + 1
b
34. (c) : For first minimum, a sinq = 1l
=
a=
4ab
2
2
=
6.5 × 10−7
λ
=
= 13 × 10–7 = 1.3 micron
sin θ
sin 30°
35. (a) : Fringe width, β =
27. (c)
zF =
V=
λD
d
and θ = ∴
d
D
β=
λ
θ
36. (d) : n1l1 = n2l2
n
16 × 6000 Å
λ2 = 1 λ1 =
= 4000 Å
n2
24
37. (a) : Wavefront is the locus of all points, where the
particles of the medium vibrate with the same phase.
λ
38. (c) : Angular fringe width θ =
d
\ l = qd
Here, d = 0.01 mm = 10–5 m,
π
q = 1° =
radians
180
π
λ=
× 10−5 = 1.74 × 10−7 m
180
= 0.174 × 10−6 m = 0.174 µm
39. (d) : If sources are coherent
IR = I1 + I2 + 2 I1I2 cos φ
I0 = I + I + 2I cos 0° = 4I
If sources are incoherent,
4I I
IR = I1 + I2 = 2I = = 0
2 2
40. (a) : As the beam is initially parallel, the shape of
wavefront is planar.
41 (c) : According to Huygens Principle, the surface of
constant phase is called a wavefront.
42. (c)
43. (c) : Angular width of central maxima, 2q = 2l/e.
Thus, q does not depend on screen i.e., distance between the
slit and the screen.
44. (c) : The intensity distribution of single slit diffraction
Intensity
pattern is shown in the figure.
From the graph it is clear that the
intensity of the central point is
finite but much larger than the
0 θ
surrounding maxima.
CBSE Board Term-II Physics Class-12
54
45. (a) : Width of central maxima = 2lD/e
width of other secondary maxima = lD/e
\ Width of central maxima : width of other secondary
maxima = 2 : 1
46. (d) : Path difference produced is
3
π
Dx = pR – R = pR
2
2
For maxima: Dx = nl
\ nl = pR
πR
⇒ l=
, n = 1, 2, 3, ....
n
πR πR
Thus, the possible values of l are pR,
,
,...
2 3
47. (b) : Maximum intensity, Imax =
I
Here, I1 = I2 = 0 (given)
2
2
 I0
I0 
∴ Imax = 
+
= 2I 0
2 
 2
(
I1 + I2
)
2
2π
× Path difference
λ
φ
I = Imax cos2
2
2
I 3
3 3
=
 = I 0 ⇒
I0 4
2
4
48. (d) : Phase difference φ =
2π λ π
× = = 60° As
λ 6 3

60°
∴ I = I 0 cos2
= I0 × 

2
φ=
49. (c) : Here A2 = a12 + a22 + 2a1a2 cos δ
Q a1 = a2 = a
δ 

∴ A2 = 2a 2 (1 + cos δ) = 2a 2 1 + 2 cos2 − 1

2 
δ
or A2 ∝ cos2
2
δ
δ
Now I ∝ A2 ∴ I ∝ A2 ∝ cos2
⇒ I ∝ cos2
2
2
50. (b) : Suppose the amplitude of waves from each slit is a2.
Therefore, intensity due to each slit = a2. When interference
is destructive, the resultant amplitude = a – a = 0.
\ Minimum intensity = 0
When interference is constructive, the resultant amplitude =
a + a = 2a.
\ Maximum intensity = (2a)2 = 4a2
= 4 times the intensity due to each slit.
51. (c) : For diffraction of a wave, size of an obstacle or
aperture should be comparable to the size of wavelength of
the wave. As wavelength of light is of the order of 10–6 m and
obstacle / aperture of this size are rare, therefore, diffraction
is not common in light waves. On the contrary, wavelength
of sound is of the order of 1 m and obstacle / aperture of this
size are readily available, therefore, diffraction is common in
sound.
52. (a) : In case of interference, intensity of maxima
is Imax =
(
(
Ia + Ib
)
2
and intensity of minima is
)
2
Imin = I a − I b . Thus, whatever energy disappears at
the minimum is actually appearing at the maximum. So, the
law of conservation of energy holds good in the phenomenon
of interference because in interference energy is neither
created nor destroyed but is redistributed.
53. (b) : When slit is wide (i.e. a >> l), bending of light
becomes so small that it cannot be detected upto a certain
distance of screen from the slit. Hence, practically, no
diffraction occurs.
54. (d) : When a light wave travel from a rarer to a denser
medium it loses speed, but energy carried by the wave does
not depend on its speed. Instead, it depends on the amplitude
of wave. The frequency also remain constant.
55. (a) : For reflected system of the film, the maxima or
constructive interference is 2mtcosr = while the maxima for
transmitted system of film is given by equation
2mt cosr = nl
where t is thickness of the film and r is angle of refraction.
From these two equations we can see that condition for
maxima in reflected system and transmitted system are just
opposite.
56. (c) : When one of slits is covered with cellophane paper,
the intensity of light emerging from the slit is decreased
(because this medium is translucent). Now the two interfering
beam have different intensities or amplitudes. Hence, intensity
at minima will not be zero and fringes will become indistinct.
57. (a) : As a broad source is equivalent to a large number
of narrow sources lying side by side. Each set of these sources
will produce an interference pattern of its own which will
overlap on another to such an extent that all traces of a fringe
system is lost and results in general illumination. Because of
this reason, for interference a narrow slit should be used.
58. (c) : The waves diffracted from the edges of circular
obstacle, placed in the path of light, interfere constructively
at the centre of the shadow resulting in the formation of a
bright spot.
SUBJECTIVE TYPE QUESTIONS
1. Two sources are said to be coherent, if they emit light
waves of same frequency or wavelength and of a constant
phase difference.
2. According to Huygens’ principle, each point on a
wavefront is a source of secondary waves, which add up to
give a wavefront at any later time.
3. In a single slit diffraction separation between fringes
nλ
q∝
a
So, there is no effects on angular separation 2q by changing
of the distance of separation ‘D’ between slit and the screen.
55
Wave Optics
λD
= 0.12mm
d
λ
⇒ by changing µ; λ becomes
µ
3
⇒ β ′ = × 0.12 = 0.9 mm.
4
λ
5. Angular width, θ =
d
4.
As β =
0 .1
6 × 10−7
0 .1 =
π rad =
d
180
\ d=
6 × 10−7 × 180
= 3.44 × 10–4 m
0 .1 × π
6. There would be spherical wave front on earth for sunlight
which is treated as point source, but radius is very large as
compared to radius of earth, so it is almost a plane wavefront.
7. Yes, Huygen’s principle is valid for longitudinal as well
as transverse waves and for all wave phenomena.
8. The diffraction effect is more pronounce if the size of the
aperture or the obstacle is of the order of wavelength of wave.
As wavelength of light ( 10–6 m) is much more smaller than
size of object around us so diffraction of light is not easily seen
but sound wave has large wavelength (15 mm < l < 15 m),
they get easily diffracted by objects around us.
9. In young’s double slit experiment, if one slit is fully
closed, the new pattern has larger central maximum in
angular size.
10. Here, d = 1 mm = 1 × 10–3 m
D = 1 m, l = 500 nm = 5 × 10–7 m
Fringe spacing,
β=
λD
5 × 10−7 × 1
=
= 5 × 10–4 m = 0.5 mm
d
1× 10−3
11. Initial fringe width bi = l × (Di/d)
Final fringe width bf = l × (Df /d)
λ
3 × 10−5 × 10−3
⇒ ∆β = (Df − Di )
⇒ λ=
= 600 nm
d
5 × 10−2
12. The path difference is 2mt.
Now for destructive interference it can be
λ
3λ
5λ
2µt = or
or
and so on ......
2
2
2
580 × 10−9 3 × 580 × 10−9
λ 3λ 5 λ
,
.... =
µ= ,
,
....
4t 4t 4t
4 × 0.3 × 10−6 4 × 0.3 × 10−6
= 0.4833, 3 × 0.4833 ....
So only m = 3 × 0.4833 = 1.45 is the answer {1.3 < m < 1.5}
13. Here n = 1, l = 6 × 10–5 cm
Distance of screen from slit = 100 cm
Distance of first minimum from central maxima = 0.1 cm
sinθ =
θ1 =
Distance of 1st minima from the central maxima
Distance of the screen from the slit
0 .1
1
=
100 1000
We know that asinq = nl
I
14. max =
Imin
Now, I1 = 2I2
⇒
(
(
I1 + I2
I1 − I2
)
)
⇒
a=
2
λ
= 0.06 cm
θ1
2
 2I2 + I2 
Imax
= 

Imin
 2I2 − I2 
2
2
 2 + 1
= 
  34
 2 − 1
15. The n th bright fringe of the λ pattern and the
n′ th bright fringe of the λ′ pattern are situated at
Dλ
Dλ ′
yn = n .
and yn′ = n ′
.
d
d
As this coincide, yn = yn′
n ′Dλ ′
λ ′ 900 6
nDλ
n
⇒
=
⇒
=
=
=
d
λ 750 5
d
n′
hence the first position where overlapping occur is
nDλ
6(2m)(750 × 10−9 m)
y5′ = y6 =
=
= 4.5 mm.
d
(2 × 10−3 m)
16. Given that: Wavelength of the light beam,
l1 = 590 nm = 5.9 ×10–7 m
Wavelength of another light beam,
l2 = 596 nm = 5.96 × 10–7 m
Distance of the slits from the screen = D = 1.5 m
Slits width = a = 2 × 10–4 m
For the first secondary maxima,
sin q =
x1 =
3λ1 x 1
=
2a
D
3λ D
3λ1D
and x 2 = 2
2a
2a
\ Separation between the positions of first secondary
maxima of two sodium lines,
3D
( λ − λ1)
2a 2
3 × 1 .5
=
(5.96 × 10–7 – 5.9 × 10–7)
2 × 2 × 10−4
x 2 − x1 =
= 6.75 × 10–5 m
CBSE Board Term-II Physics Class-12
56
Dλ
; β ∝λ
17. Fringe width β =
d
β1 λ1
β
8 .1
\
=
or λ2 = 2 λ1 =
× 640 nm
β2 λ 2
β1
7 .2
l2 = 720 nm
x = n1b 1 = n2b 2
n1Dλ1 n2 λ2D
or
or n1λ1 = n2 λ2
=
d
d
Q Bright fringes coincides at least distance x, if
n1 = n2 + 1
⇒ n1 × 640 = (n1 – 1) × 720
n1 − 1 640
or n1 = 9
=
n1
720
Q
18. l = 6000 Å = 6000 × 10–10 m = 6 × 10–7 m, q1 = 30°,
m=1
1

 m +  λ
2
(i) For first maximum, sinθm =
a
3λ
3 × 6 × 10−7
3λ
or a =
=
sinθ1 =
2 × sin 30°
2 sin θ1
2a
= 1.8 × 10–6 m = 1.8 mm
(ii) For first minimum,
λ
λ
mλ
6 × 10−7
sin θm =
∴ sin θ1 = ⇒ a =
=
sin θ1 sin 30°
a
a
= 1.2 × 10−6 m = 1.2 µm
19. (i) Here a = 1 × 10–4 m, D = 1.5 m
l = 6000 Å = 6000 × 10–10 m
The distance between the two dark bands on each side of
central band is equal to width of the central bright band,
2Dλ
2 × 1.5 × 6000 × 10−10
i.e.,
= 18 mm
=
a
1× 10−4
20. For minima in diffraction pattern, dsinq = nl
λ1
For first minima, dsinq1 = (1)l1 ⇒ sinθ1 =
d
3λ2
3
For first maxima, d sin θ2 = λ2 ⇒ sin θ2 =
2
2d
The two will coincide if, q1 = q2 or sinq1 = sinq2
λ
3λ
2
2
∴ 1 = 2 ⇒ λ2 = λ1 = × 660 nm = 440 nm
d
2d
3
3
21. (i)
3rd bright fringe will be at
3 × λD 3 × (6500 × 10−10 ) × 1.2
=
d
2 × 10−3
= 0.117cm = 1.17mm
=
(ii) Say mth bright fringe of 6500 Å coincides with nth bright
fringe of 5200 Å
n 5
=
m 4
Hence, for least distance, 5th bright fringe of 5200 Å coincides
with 4th bright fringe of 6500 Å.
⇒ nb1 = mb2
⇒ y ′ = 4β1 =
= 1.56 mm
⇒
4 × (6500 × 10−10 ) × 1.2
2 × 10−3
= 0.156 cm
22. At the hole, wavelength which satisfy the minima
condition will be absent.
1  Dλ
2y .d

⇒ y = n + 
or λ =

D (2n + 1)
2 d
=
2(1 × 10−3 m)(0.5 × 10−3 m)
(0.5m)(2n + 1)
2000 nm
2 × 10−6 m
=
(2n + 1)
2n + 1
2000
Now, 400 nm ≤ λ ≤ 700 nm or 400 ≤
≤ 700
2n + 1
n = 0, 1, 2, 3
λ = 667 nm, 400 nm.
667 nm and 400 nm have a complete destructive interference
at the hole and consequently will be absent in the light
coming from that hole.
23. From diagram
T1P = T1O – OP = (D – x)
T2P = T2O + OP = (D + x)
=
Now S1P =
=
S2P =
(S1T1)2 + (T1P )2
D 2 + (D − x )2
(S2T2 )2 + (T2P )2 =
D 2 + (D + x )2
λ
; for first minimum to occur
2
λ
D 2 + (D + x )2 − D 2 + (D − x )2 =
2
The first minimum falls at a distance D from the center, i.e.,
x = D.
λ
[D 2 + 4D 2 ]1/ 2 − D =
2
λ
D( 5 − 1) =
2
λ
λ
D(2.236 – 1) =
; D=
2
.
472
2
24. (a) The angular width of central maximum is given by
2λ
2θ0 =
, ...(i)
a
where the letters have their usual meanings.
(i) Effect of slit width : From the equations (i), it follows that
1
β0 ∝ . Therefore, as the slit width is increased, the width
a
of the central maximum will decrease.
Path difference, S2P – S1P =
57
Wave Optics
(ii) Effect of distance between slit and screen (D) : From the
equation (i), it follows that 2q0 is independent of D. So the
angular width will remain same whatever the value of D.
(b) Difference between interference and diffraction
experiment to observe diffraction pattern
Interference
Diffraction
1. Interference is caused by 1. Diffraction is caused by
superposition two waves superposition of a number of
starting from two coherent waves starting from the slit.
sources.
2. All bright and dark fringes 2. Width of central bright
are of equal width.
fringe is double of all other
maxima.
3. All bright fringes are of 3. Intensity of bright
same intensity.
fringes decreases sharply
as we move away from
central bright fringe.
4. Dark Fringes are perfectly 4. Dark fringes are not
dark.
perfectly dark.
25. (a) y1 = a cos wt, y2 = a cos (wt + f)
where f is phase difference between them. Resultant
displacement at point P will be,
y = y1 + y2 = a cos wt + a cos(wt + f)
= a [cos wt + cos (wt + f)]
( ωt + ωt + φ )
( ωt − ωt − φ ) 

= a 2 cos
cos


2
2
φ

φ
...(i)
y = 2a cos  ωt +  cos   
2
2
φ
Let y = 2a cos   = A , the equation (i) becomes
2
φ

y = A cos  ωt + 

2
where A is amplitude of resultant wave,
φ
Now, A = 2a cos  
2
φ
On squaring, A 2 = 4a 2 cos2  
2
Hence, resultant intensity,
φ
I = 4I 0 cos2  
2
(b) Condition for constructive interference,
cos Df = + 1
∆x
2π
= 0,2π, 4 π...
λ
or Dx = nl; n = 0, 1, 2, 3, ...
Condition for destructive interference, cos Df = – 1
∆x
2π
= π, 3π,5π...
λ
or Dx = (2n – 1) l/2
where n = 1, 2, 3...
26. (i) Two independent monochromatic sources cannot
produce sustained interference pattern because the phase
difference between the light waves from two independent
sources keeps on changing continuously.
(ii)
Consider a point P on the screen at distance x from the centre
O. The nature of the interference at the point P depends on
path difference,
p = S2P – S1P
From right-angled DS2BP and DS1AP,
(S2P)2 – (S1P)2 = [S2B2 + PB2] – [S1A2 + PA2]
2
2

d  
d 


= D 2 +  x +   − D 2 +  x −  


2  
2 

or (S2P – S1P)(S2P + S1P) = 2xd
2xd
or S2P − S1P =
S2P + S1P
In practice, the point P lies very close to O,
therefore S1P  S2P  D. Hence
2 xd
p = S2P − S1P =
2D
xd
or p =
D
Positions of bright fringes : For constructive interference,
xd
p=
= nλ
D
nDλ
or x =
where n = 0, 1, 2, 3, ....
d
Positions of dark fringes : For destructive interference,
CBSE Board Term-II Physics Class-12
58
p=
xd
λ
= (2n − 1)
D
2
Dλ
where n = 1, 2, 3
2d
Width of a dark fringe = Separation between two
consecutive bright fringes
nDλ (n − 1)Dλ Dλ
= x n − x n −1 =
−
=
d
d
d
or
x = (2n − 1)
Width of bright fringe = Separation between two
consecutive dark fringes
Dλ
Dλ Dλ
= x n′ − x n′ −1 = (2n − 1)
− [2(n − 1) − 1]
=
2d
2d
d
Clearly, both the bright and dark fringes are of equal width.
Hence the expression for the fringe width in Young’s double
slit experiment can be written as
Dλ
β=
d
λ
4
Now path difference between the two waves from slit S1 and
S2 on reaching point P on screen is
Dx = (SS2 + S2P) – (SS1 + S1P)
or Dx = (SS2 – SS1) + (S2P – S1P)
λ yd
or ∆x = +
, where d is the slits separation.
4 D
27. (a) Given :SS2 − SS1 =
Screen
28. (i) Dx2 = (m – 1)t = (3/2 – 1)(0.01) = 5.0 × 10–3 cm
Dx1 = dsinf = (50 × 10–4) sin30° = 2.5 × 10–3 cm
Since Dx2 > Dx1, central maxima will be obtained below O.
dsinq + Dx1 = Dx2
50 × 10–4sinφ + 2.5 × 10–3 = 5.0 × 10–3
1
sinφ =
or φ = 30°
2
(ii) At O
Dx = Dx1 = 2.5 × 10–3 = nl
n=
2.5 × 10−3
= 50
500 × 10−3
Number of fringes that will cross O if slab is removed
Path difference due to slab
5 × 10−3
=
= 100
λ
500 × 10−7
29. (i) D > > d
Hence, path difference at any angular position q on the screen
Dx = dsinq
The path difference for first maxima
λ 1
∆x = d sin θ = λ ⇒ sin θ = =
⇒ q = 30°
d 2
Hence, distance between central maxima and first maxima
1
y = D sinθ = m
2
(ii) Maximum path difference, Dxmax = d = 1 mm
d 
⇒ Highest order maxima, nmax =   = 2 and highest
λ
 d 1
order minimum nmin =  +  = 2
 λ 2
Total number of maxima = 2nmax + 1 = 5
Total number of minima = 2nmin = 4
30. Snell’s law of refraction : Let P1P2 represents the surface
separating medium 1 and medium 2 as shown in figure.
=
For constructive interference at point P, path difference,
λ yd
+
= nλ
Dx = nl or
4 D
yd 
1
or
=  n −  λ ...(i)
D 
4
where n = 0, 1, 2, 3,...,
1 λD
(b) From equation (i), y n =  n − 

4 d
1  λD

and yn – 1 =  n − 1− 

4 d
The fringe width is given by separation of two consecutive
bright fringes.
λD
1  λD 
1  λD

b = yn – yn – 1 =  n − 
−  n − 1− 
=




d
4 d
4 d
Let v1 and v2 represents the speed of light in medium 1 and
medium 2 respectively. We assume a plane wavefront AB
propagating in the direction A′A incident on the interface at
an angle i. Let t be the time taken by the wavefront to travel
the distance BC.
\ BC = v1t
[... distance = speed × time]
In order to determine the shape of the refracted wavefront,
we draw a sphere of radius v2t from the point A in the second
59
Wave Optics
medium (the speed of the wave in second medium is v2).
Let CE represents a tangent plane drawn from the point C.
Then
AE = v2t
\ CE would represent the refracted wavefront.
In DABC and DAEC, we have
BC v 1t
AE v 2t
sini =
=
=
and sinr =
AC AC
AC AC
where i and r are the angles of incident and refraction
respectively.
sin i v 1t AC
=
⋅
\
sin r AC v 2t
sin i v 1
=
sin r v 2
If c represents the speed of light in vacuum, then
c
c
µ1 = and µ2 =
v1
v2
c
c
⇒ v1 =
and v 2 =
µ1
µ2
Where m1 and m2 are the refractive indices of medium 1 and
medium 2.
\
sin i c / µ1 sin i µ2
⇒ m1 sin i = m2 sin r
=
⇒
=
sin r c / µ2 sin r µ1
This is the Snell’s law of refraction.
31. (i) Energy carried by a wave depends on the frequency
of the wave, not on the speed of wave propagation, hence
energy remains same.
(ii) For a given frequency, intensity of light in the photon
picture is determined by
Energy of photons
n × hυ
=
area × time
A ×t
where n is the number of photons incident normally on
crossing area A in time t.
I=
I
(a + a )2
32. (a) We know, max = 1 2
Imin (a1 − a2 )2
a2 =
C
E
E
a
a1
2
Hence,
2
Imax (a1 + a1 / 2 )2 (1+ 1 / 2 )2  2 + 1
=
=
=
 ≈ 34
Imin (a1 − a1 / 2 )2 (1− 1 / 2 )2  2 − 1
(b) The central fringes are white. On the either side of the
central white fringe the coloured bands (few coloured maxima
a
b
b
b
b
c
c
c
c
d
d
e
e
d
d
e
F
B
e
D
F
B
(a)
D
(b)
Huygens geometrical construction for the propagation of (a)
spherical, (b) plane wavefront.
According to Huygens principle, each point on AB becomes a
source of secondary disturbance, which takes with the same
speed c. To find the new wavefront after time t, we draw
spheres of radii ct, from each point on AB.
The forward envelope or the tangential surface CD of the
secondary wavelets gives the new wavefront after time t.
The lines aa′, bb′, cc′, etc., are perpendicular to both AB
and CD. Along these lines, the energy flows from AB to CD.
So these lines represent the rays. Rays are always normal to
wavefronts.
(b) A source of light sends the disturbance in all the
directions and continuous locus of all the particles vibrating
in same phase at any instant is called as wavefront.
Given figure shows the refraction of a plane wavefront at a
rarer medium i.e., v2 > v1
Incident
wavefront
B
v1t
i
i
r
A
r
v1 < v2
I ∝ a 2)
a
a
Rarer – v2
a22 = 0.5 a12(
C
A
A
Denser – v1
According to question, I2 = 50% of I1
I2 = 0.5I1;
and minima) will appear. This is because fringes of different
colours overlap.
33. (a) Consider a spherical or plane wavefront moving
towards right. Let AB be its position at any instant of time.
The region on its left has received the wave while region on
the right is undisturbed.
C
v2t
D
Refracted
wavefront
The incident and refracted wavefronts are shown in figure.
Let the angles of incidence and refraction be i and r
respectively.
From right DABC, we have,
BC
sin ∠BAC = sin i =
AC
From right DADC, we have,
sin ∠DCA = sin r =
AD
AC
CBSE Board Term-II Physics Class-12
60
sin i v 1 1
sin i BC v 1t
=
= µ2
=
=
or
sin r v 2
sin r AD v 2t
(a constant)
This verifies Snell’s law of refraction. The constant 1m2 is called
the refractive index of the second medium with respect to
first medium.
(c) Reflection and refraction arise through interaction
of incident light with atomic constituents of matter which
vibrate with the same frequency as that of the incident light.
Hence frequency remains unchanged.
λD
34. (a) (i) Fringe width (β) =
d
If d decreases then b increases.
s λ
(ii) For interference fringe, the condition is <
D d
where s = size of source, D = distance of source from slits.
If the source slit width increases, fringe pattern gets less sharp
or faint and fringe width decrease.
When the source slit is made wide which does not fullfil the
above condition and interference pattern not visible.
λD
(b) Fringe width (β) =
d
β λD
y= =
3 3d
yd
λD d λ
⇒ ∆p =
⋅ =
Path difference (Dp) =
D
3d D 3
2π
2π λ 2π
∆φ =
⋅ ∆p =
⋅ =
3
λ
λ 3
Intensity at point P = I0 cos2 Df
\
2
Before the wavefront strikes the barrier the wavefront
generates another forward moving wavefront. Once the
barrier blocks most of the wavefront the forward moving
wavefront bends around the slit because the secondary
waves they would need to interfere with to create a straight
wavefront have been blocked by the barrier.
According to Huygen’s principle, each point on the wavefront
moving through the slit acts like a point source. We can
think about some of the effect of this if we analyse what
happens when two point sources are close together and emit
wavefronts with the same wavelength and frequency. These
two point sources represent the point sources on the two
edges of the slit and we can call the source A and source B
as shown in the figure.
Each point source emits wavefronts from the edge of the
slit. In the diagram we show a series of wavefronts emitted
from each point source. The continuous lines show peaks in
the waves emitted by the point sources and the dotted lines
represent troughs. We label the places where constructive
interference (peak meets a peak or trough meets a trough)
takes place with a solid diamond and places where destructive
interference (trough meets a peak) takes place with a hollow
diamond. When the wavefronts hit a barrier there will be
places on the barrier where constructive interference takes
place and places where destructive interference happens.
2
I
2π 
 1

= I 0 cos  = I 0   = 0
2

3
4
Here D = 120 cm = 1.20 m
and d = 2 mm = 2 × 10–3 m
\ Least distance,
nDλ1 4 × 1.2 × 650 × 10−9
y min =
m
=
d
2 × 10−3
= 1.56 × 10–3 m = 1.56 mm
35. (a) Waves diffract when they encounter obstacles. A
wavefront impinging on a barrier with a slit in it, only the
points on the wavefront that move into the slit can continue
emitting forward moving waves but because a lot of the
wavefront has been blocked by the barrier, the points on the
edges of the hole emit waves that bend round the edges.
A
B
The measurable effect of the constructive or destructive
interference at a barrier depends on what type of waves we
are dealing with.
(b) Condition for nth secondary dark fringe :
1
A
I
1
a
P
O
screen
II
B
C
=
a sin 1
Light rays which on passing through the slit of width ‘a’
get diffracted by an angle q1, such that the path difference
between extreme rays on emerging from slit is a sin q1 = l
61
Wave Optics
Then the waves from first half and second half of slit have a
path difference of l/2, so they interfere destructively at point
P on screen, forming first secondary dark fringe.
Thus condition for first secondary dark fringe or first secondary
minimum is
λ
sin θ1 =
a
Similarly, condition for n th secondary dark fringe or n th
secondary minimum is
nλ
sin θn =
a
where n = 1, 2, 3, 4,...........
λ
Angular width of first diffraction fringe, θ1 =
a
Angular width of central maxima, q
λ
θ1 + θ1 = 2θ1 = 2
a
1
\ θ1 = θ
2
(c) On increasing the value of n, the part of slit contributing
to the maxima decreases. Hence, the maxima become weaker.
36. (a) Diffraction of light due to a narrow single slit
Consider a set of parallel rays from a lens L1 falling on a
slit, form a plane wavefront. According to Huygens principle,
each point on the unblocked point of plane wavefront AB
sends out secondary wavelets in all directions. The secondary
waves from points equidistant from the centre C of the slit
lying in the portion CA and CB of the wavefront travel the
same distance in reaching at O and hence the path difference
between them is zero. These secondary waves reinforce each
other, resulting maximum intensity at point O.
Position of secondary minima : The secondary waves travelling
in the direction making an angle q with CO, will reach a point
P on the screen. The intensity at P will depend on the path
difference between the secondary waves emitted from the
corresponding points of the wavefront. The wavelets from
points A and B will have a path difference equal to BN. If this
path difference is l, then P will be a point of minimum intensity.
This is because the whole wavefront can be considered to be
divided into two equal halves CA and CB. If the path difference
between secondary waves from A and B is l, then the path
difference between secondary waves from A and C will be l/2
and also the path difference between secondary waves from
B and C will again be l/2. Also for every point in the upper
half AC, there is a corresponding point in the lower half CB for
which the path difference between secondary waves reaching
P is l/2. Thus, at P destructive interference will take place.
From the right-angled DANB given in figure
BN = AB sin q
BN = a sin q
Suppose BN = l and q = q1
\ l = a sin q1
λ
sinθ1 =
a
Such a point on the screen will be the position of the first
secondary minimum.
If BN = 2l and q = q2, then,
2l = a sin q2
2λ
a
Such a point on the screen will be the position of the second
secondary minimum.
In general, for nth minimum at point P.
nλ
sinθn =
a
nλ
θ
For small n , θn =
...(i)
a
Position of secondary maxima :
If any other P′ is such that path difference at point is given by
3λ
a sinθ =
2
Then P1 will be position of first secondary maximum. Here,
we can consider the wavefront to be divided into three equal
parts, so that the path difference between secondary waves
from corresponding points in the 1st two parts will be l/2.
This will give rise to destructive interference. However, the
secondary waves from the third part remain unused and
therefore, they will reinforce each other and produce first
secondary maximum.
Similarly if the path difference at that points given by
5λ
a sinθ5 =
2
We get second secondary maximum of lower intensity.
In general, for nth secondary maximum, we have
λ
a sin θn = (2n + 1)
2
λ
θ
,
θ
=
For small n n (2n + 1)
2a
The diffraction pattern on the screen is shown below along
with intensity distribution of fringes
sinθ2 =
CBSE Board Term-II Physics Class-12
62
I
Central Maxima
3 2 a
a
a
O
a
2 3
a a
Width of the secondary maximum,
nDλ (n − 1)Dλ
β = y n − y n −1 =
−
a
a
Dλ
...(i)
a
\ b is independent of n, all the secondary maxima are of
the same width b.
3λ
If BN =
and θ = θ1′ , from above equation, we have
2
3λ
sin θ1′ =
2a
β=
Such a point on the screen will be the position of the first
secondary maximum.
Corresponding to path difference,
5λ
BN =
and θ = θ2′ , the second secondary maximum is
2
produced. In general, for the nth maximum at point P,
(2n + 1) λ
sin θn′ =
...(ii)
2a
If y′n is the distance of nth maximum from the centre of the
screen, then the angular position of the nth maximum is
given by
y′
tan θn′ = n ...(iii)
D
In case q′n is small,
sin θn′ ≈ tan θn′
(2n + 1)Dλ
\ y n′ =
2a
Width of the secondary minimum,
Dλ
β′ =
...(iv)
a
Since b′ is independent of n, all the secondary minima are
of the same width b′.
(b) Here, l1 = 590 nm = 590 × 10–9 m,
l2 = 596 nm = 596 × 10–9 m, d = 2 × 10–6 m, D = 1.5 m
Distance of first secondary maximum from the centre of the
screen is
3 λD
x=
2 d
For the two wavelengths,
3 Dλ 2
3 Dλ1
x1 =
and x 2 =
2 d
2 d
Spacing between the first two maximum of sodium lines,
3D
x 2 − x1 =
( λ − λ1)
2d 2
3 × 1 .5
=
(596 × 10−9 − 590 × 10−9 )
2 × 2 × 10−6
=
3 × 1.5 × 6 × 10−3
= 6.75 × 10−3 m = 6.75 mm
4

11
Dual Nature of Radiation
and Matter
Electron emission : It is the phenomenon
of emission of electrons from the surface of
a metal. The minimum energy needed by an
electron to come out from a metal surface is
known as “work function” of the metal. It is
denoted by f0 or W0 and measured in electron
volt (eV).
hc
Work function W = hυ0 =
λ0
The electron emission can be obtained from
the following physical processes :
X Thermionic emission : It is the
phenomenon of emission of electrons from
the metal surface when heated suitably.
X Photoelectric emission : It is the
phenomenon of emission of electrons from
the surface of metal when light radiations
of suitable frequency fall on it.
X Field emission or cold cathode emission :
It is the phenomenon of emission of
electrons from the surface of a metal under
the application of a strong electric field.
Photoelectric effect : It is the phenomenon
of emission of electrons from the surface of
metals, when light radiations of suitable
frequency fall on them.
X Photoelectric current :
Photoelectric
current
depends on the intensity
of incident light and
the potential difference
applied between the two Intensity of light
electrodes.
X Stopping potential : The minimum
negative potential given to anode plate
w.r.t. to cathode plate at which the
photoelectric current becomes zero is
known as stopping potential or cut off
potential. It is denoted by V0. If e is the
Photoelectric
current
CHAPTER
charge on the photoelectron, then
1 2
K max = eV0 = mvmax
2
where m is the mass of photoelectron and
vmax is the maximum velocity of emitted
photoelectrons.
– Variation of stopping potential V0 with
frequency u of incident radiation :
– Variation of photocurrent with collector
plate potential for different intensity of
incident radiation :
Photocurrent
Saturation
current
Stopping
potential
–V0
Retarding
potential
I3 > I 2 > I 1
I3
I2
I1
O
Collector plate
potential
– Variation of photocurrent with collector
plate potential for different frequencies
of incident radiation :
CBSE Board Term-II Physics Class-12
64
Einstein’s photoelectric equation :
If a light of frequency u is incident on a
photosensitive material having work function
(f0), then maximum kinetic energy of the
emitted electron is given as
Kmax = hu – f0
For u > u0
or eV0 = hu – f0 = hu – hu0
1 1 
or eV0 = K max = hc  −  .
 λ λ0 
where u0 = threshold frequency
l0 = threshold wavelength
l = incident wavelength
Einstein’s photoelectric equation is in
accordance with the law of conservation of
energy.
Dual nature of radiation : Wave theory
of electromagnetic radiation explains the
phenomenon of interference, diffraction and
polarisation.
On
the
other
hand,
photoelectric effect is supported by particle
nature of light. Hence, we assume dual
nature of light.
Photons : These are the packets of energy
(or energy particles) which are emitted by
a source of radiation. The photons emitted
from a source, travel through space with
the same speed c (equal to the speed of
light).
hc
X Energy of a photon E = hυ =
λ
where, u = frequency, l = wavelength
h = Planck’s constant, c = speed of the
light
X Momentum of photon is
E hυ
p= =
c
c
X The rest mass of photon is zero.
E hυ
X The moving mass m of photon is m =
= .
c2 c2
X All photons of light of a particular
frequency u or wavelength l have the
hc
same energy E  = hυ =  and momentum

λ
 hυ h 
p =
=  , whatever be the intensity of
 c
λ
X
radiation.
Photon energy is independent of intensity
of radiation.
X
X
X
Photons are not deflected by electric and
magnetic fields.
In a photon-particle collision (such as
photon-electron collision), the total energy
and total momentum are conserved.
Number of photons emitted per second of
frequency u from a lamp of power P is
P Pλ
n=
=
hυ hc
de-Broglie waves (Matter waves) :
Radiation has dual nature, wave and particle.
The nature of experiment determines whether
a wave or a particle description is best suited
for understanding the experimental result.
Reasoning that radiation and matter should
be symmetrical in nature, Louis Victor
de Broglie attributed a wave like character
to matter (material particles). The waves
associated with the moving material particles
are known as matter waves or de Broglie
waves.
X de-Broglie wavelength : The de Broglie
wavelength associated with a moving
particle is related to its momentum as
h
h
de-Broglie wavelength, λ = =
p mv
where m is the mass of the particle, v
is the velocity of the particle, p is the
momentum of the particle.
– de-Broglie wavelength is independent
of the charge and nature of the material
particle.
– In terms of kinetic energy K, de Broglie
h
wavelength is given by λ =
.
2mK
– If a particle of charge q is accelerated
through a potential difference V, its
de-Broglie wavelength is given by
λ=
h
2mqV
.
150 
For an electron, λ = 
 V 
1/2
Å.
– For a gas molecule of mass m at
temperature T kelvin, its de-Broglie
h
wavelength is given by λ =
,
3mkT
where k is the Boltzmann constant.
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. A particle is dropped from a height H. The
de Broglie wavelength of the particle as a
function of height is proportional to
(a) H
(b) H 1/2
(c) H0 (d) H–1/2
2. The threshold frequency of a certain
metal is 3.3 × 10 14 Hz. If light of frequency
8.2 × 10 24 Hz is incident on the metal, then
the cut off voltage for photoelectric emission is
(Given h = 6.63 × 10–34 J s)
(a) 2 V (b) 4 V
(c) 6 V (d) 8 V
3. A proton and an a-particle are accelerated
through the same potential difference. The ratio
of de Broglie wavelength lp to that of la is
(a)
(c)
2 :1
6 :1
(b)
4 :1
(d)
8 :1
4. The de Broglie wavelength of a particle of
kinetic energy K is l. What will be the wavelength
K
of the particle, if its kinetic energy is
?
4
λ
(a) l
(b) 2l
(c)
(d) 4l
2
5. When the velocity of an electron increases,
its de Broglie wavelength
(a) increases
(b) decreases
(c) remains same
(d) may increase or decrease
6. Which of the following figure represents
the variation of particle momentum (p) and
associated de Broglie wavelength (l)?
(a)
(c)
(b)
(d)
7. The de Broglie wavelength associated with
a ball of mass 150 g travelling at 30 m s–1 is
(a) 1.47 × 10–34 m
(c) 1.47 × 10–19 m
(b) 1.47 × 10–16 m
(d) 1.47 × 10–31 m
8. T h e w a v e l e n g t h o f m a t t e r w a v e i s
independent of
(a) mass
(b) velocity
(c) momentum
(d) charge
9. Photons absorbed in matter are converted to
heat. A source emitting n photons per second of
frequency u is used to convert 1 kg of ice at 0°C
to water at 0°C. Then, the time T taken for the
conversion
(a) decreases with increasing n, with u fixed.
(b) decreases with n fixed, u increasing.
(c) remains constant with n and u changing
such that nu = constant.
(d) All of these.
10. If alpha particle, proton and electron move
with the same momentum, then their respective
de Broglie wavelengths la, lp, le are related as
(a) la = lp = le
(b) la < lp < le
(c) la > lp > le
(d) lp > le > la
11. If the momentum of an electron is changed
by p, then the de Broglie wavelength associated
with it changes by 0.5%. The initial momentum
of electron will be
(a) 200p
(b) 400p
(c)
p
200
(d) 100p
12. Who established that electric charge is
quantised ?
(a) J.J. Thomson
(b) William Crookes
(c) R.A Millikan
(d) Wilhelm Rontgen
13. The phenomenon of photoelectric emission
was discovered in 1887 by
(a) Albert Einstein
(b) Heinrich Hertz
(c) Wilhelm Hallwachs (d) Philipp Lenard
14. The de Broglie wavelength of an electron in
a metal at 27°C is
(Given me = 9.1 × 10–31 kg, kB = 1.38 × 10–23 J K–1)
G:\MTG\Book2021\CBSETerm-2_ExamReady\Class12\Physics\Ch_11_DualNatureofRadiationandMatter\Ch_11_DualNatureofRadiationandMatter_Exe\Ch_11_DualNatureofRadiationandMatter_Exe.indd
CBSE Board Term-II Physics Class-12
66
(a) 6.2 × 10–9 m
(c) 6.2 × 10–8 m
(b) 6.2 × 10–10 m
(d) 6.2 × 10–7 m
15. A student performs an experiment on
photoelectric effect using two materials A and
B. A plot of stopping potential (V0)vs frequency (u)
(V0) frequency is as shown in the figure.
he value of h obtained from the experiment for
T
both A and B respectively is
(Given electric charge of an electron
= 1.6 × 10–19 C)
(a) 3.2 × 10–34 J s, 4 × 10–34 J s
(b) 6.4 × 10–34 J s, 8 × 10–34 J s
(c) 1.2 × 10 –34 J s, 3.2 × 10 –34 J s
(d) 4.2 × 10 –34 J s, 5 × 10 –34 J s
16. The maximum frequency and minimum
wavelength of X-rays produced by 30 kV electrons
respectively is
(a) 7.24 × 1018 Hz, 0.041 nm
(b) 3.21 × 1018 Hz, 0.211 nm
(c) 5.32 × 1018 Hz, 0.001 nm
(d) 2.13 × 1018 Hz, 0.011 nm
17. The matter-wave picture of electromagnetic
wave/radiation elegantly incorporated the
(a) Heinsenberg’s uncertainty principle
(b) correspondence principle
(c) cosmic theory
(d) Hertz’s observations
18. The minimum energy required for the
electron emission from the metal surface can
be supplied to the free electrons by which of the
following physical processes?
(a) Thermionic emission
(b) Field emission
(c) Photoelectric emission
(d) All of these
19. Assume that a molecule is moving with the
root mean square speed at temperature 300 K.
The de Broglie wavelength of nitrogen molecule is
(Atomic mass of nitrogen = 14.0076 u,
h = 6.63 × 10–34 J s, kB = 1.38 × 10–23 J K–1,
1 u = 1.66 × 10–27 kg)
–11
(a) 2.75 × 10 m
(b) 2.75 × 10–12 m
–11
(c) 3.24 × 10
m
(d) 3.24 × 10 –12 m
20. The photoelectric cut off voltage in a certain
experiment is 1.5 V. The maximum kinetic
energy of photoelectrons emitted is
(a) 2.4 eV
(b) 1.5 eV
(c) 3.1 eV
(d) 4.5 eV
21. Monochromatic light of frequency 6 × 1014 Hz
is produced by a laser. The power emitted is
2 × 10–3 W. The number of photons emitted per
second is (Given h = 6.63 × 10–34 J s)
(a) 2 × 1015
(b) 3 × 1015
(c) 4 × 1015
(d) 5 × 1015
22. The wavelength of a photon needed to
remove a proton from a nucleus which is bound
to the nucleus with 1 MeV energy is nearly
(a) 1.2 nm
(b) 1.2 × 10–3 nm
(c) 1.2 × 10–6 nm
(d) 1.2 × 10–5 nm
23. In an accelerator experiment on high
energy collisions of electrons with positrons,
a certain event is interpreted as annihilation
of an electron-positron pair of total energy
10.2 × 109 eV into two g-rays of equal energy. The
wavelength associated with each g-ray is
(a) 3.21 × 10–18 m
(b) 1.23 × 10–16 m
–16
(c) 2.44 × 10 m
(d) 4.21 × 10–14 m
24. A blue lamp mainly emits light of wavelength
4500 Å. The lamp is rated at 150 W and 8% of the
energy is emitted as visible light. The number
of photons emitted by the lamp per second is
(a) 3 × 1019
(b) 3 × 1024
(c) 3 × 1020
(d) 3 × 1018
25. A metallic surface is irradiated by a
monochromatic light of frequency u 1 and
stopping potential is found to be V1. If the light of
frequency u2 irradiates the surface, the stopping
potential will be
h
h
(a) V1 + (υ1 + υ2 )
(b) V1 + (υ2 − υ1 )
e
e
h
e
(c) V1 + (υ2 − υ1 )
(d) V1 − (υ1 + υ2 )
e
h
26. A particle of mass 4m at rest decays into
two particles of masses m and 3m having nonzero velocities. The ratio of the de Broglie
wavelengths of the particles 1 and 2 is
67
Dual Nature of Radiation and Matter
1
1
(b)
(c) 2
(d) 1
2
4
27. The photoelectric threshold frequency of a
metal is u. When light of frequency 4u is incident
on the metal, the maximum kinetic energy of the
emitted photoelectron is
5hυ
(a) 4hu (b) 3hu
(c) 5hu (d)
2
28. The de Broglie wavelength is given by
2π
(a) p = λ
(b) p = 2λ
2π
2π
(c) p = λ
(d) p =
λ
29. The de Broglie wavelength of an electron
with kinetic energy 120 eV is
(Given h = 6.63 × 10–34 J s, me = 9 × 10–31 kg,
1 eV = 1.6 × 10–19 J)
(a) 2.13 Å
(b) 1.13 Å
(c) 4.15 Å
(d) 3.14 Å
(a)
30. If h is Planck’s constant, the momentum of
a photon of wavelength 0.01 Å is
(a) 10–2h (b) h
(c) 102h (d) 1012h
31. According to Einstein’s photoelectric
equation, the graph between the kinetic energy
of photoelectrons ejected and the frequency of
incident radiation is
(a)
(b)
(c)
(d)
(c)
2m( E − φ0 )
eB
2e( E − φ0 )
(b)
(d)
34. The photoelectric threshold wavelength for
silver is l0. The energy of the electron ejected from
the surface of silver by an incident wavelength
l(l < l0) will be
hc
(a) hc(l0 – l)
(b)
λ0 − λ
h  λ0 − λ 
 λ − λ
(d) hc  0
c  λλ 0 
 λλ 0 
35. Electrons with de Broglie wavelength l
fall on the target in an X ray tube. The cut off
wavelength (l0)of the emitted X rays is
(c)
(a) λ 0 =
2mcλ 2
h
(b) λ 0 =
2h
mc
2m2 c2 λ 2
(d) l0 = l
h2
36. A and B are two metals with threshold
frequencies 1.8 × 10 14 Hz and 2.2 × 10 14 Hz.
Two identical photons of energy 0.825 eV each
are incident on them. Then photoelectrons are
emitted in
(Take h = 6.6 × 10–34 J s)
(a) B alone
(b) A alone
(c) neither A nor B
(d) both A and B
(c) λ 0 =
37. If the kinetic energy of the particle is
increased by 16 times, the percentage change
in the de Broglie wavelength of the particle is
(a) 25% (b) 75%
(c) 60% (d) 50%
32. A photon of energy E ejects a photoelectron
from a metal surface whose work function is f0.
If this electron enters into a uniform magnetic
field B in a direction perpendicular to the field
and describes a circular path of radius r, then
the radius r is (in the usual notation)
(a)
(a) remains constant (b) increases with time
(c) decreases with time
(d) increases and decreases periodically
2m( E − φ0 )eB
2m( E − φ0 )
mB
eB
33. An electron is moving with an initial
velocity
v = v0i and is in a magnetic field B = B0 j . Then
its de Broglie wavelength
38. The rest mass of photon is
hυ
hυ
(a)
(b) 2
c
c
hυ
(c)
(d) zero
λ
39. In photoelectric effect, the photoelectric
current is independent of
(a) intensity of incident light
(b) potential difference applied between the two
electrodes
(c) the nature of emitter material
(d) frequency of incident light
40. The work function for Al, K and Pt is 4.28 eV,
2.30 eV and 5.65 eV respectively. Their respective
threshold frequencies would be
(a) Pt > Al > K
(b) Al > Pt > K
(c) K > Al > Pt
(d) Al > K > Pt
CBSE Board Term-II Physics Class-12
68
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 41 to 44.
Photoelectric Effect
Photoelectric effect is the phenomenon of
emission of electrons from a metal surface, when
radiations of suitable frequency fall on them. The
emitted electrons are called photoelectrons and
the current so produced is called photoelectric
current.
41. With the increase of intensity of incident
radiations on photoelectrons emitted by a photo
tube, the number of photoelectrons emitted per
unit time is
(a) increases
(b) decreases
(c) remains same
(d) none of these
42. It is observed that photoelectron emission
stops at a certain time t after the light source
is switched on. The stopping potential (V) can
be represented as
(a) 2(KEmax/e)
(b) (KEmax/e)
(c) (KEmax/3e)
(d) (KEmax/2e)
43. A point source of light of power 3.2 × 10–3 W
emits monoenergetic photons of energy 5.0 eV
and work function 3.0 eV. The efficiency
of photoelectron emission is 1 for every
106 incident photons. Assume that photoelectrons
are instantaneously swept away after emission.
The maximum kinetic energy of photon is
(a) 4 eV
(b) 5 eV
(c) 2 eV
(d) Zero
44. If the frequency of incident light falling on
a photosensitive metal is doubled, the kinetic
energy of the emitted photoelectron is
(a) unchanged
(b) halved
(c) doubled
(d) more than twice its initial value
Case II : Read the passage given below and
answer the following questions from 45 to 47.
de-Broglie Wavelength
According to de-Broglie, a moving material
particle sometimes acts as a wave and sometimes
as a particle or a wave associated with moving
material particle which controls the particle in
every respect. The wave associated with moving
particle is called matter wave or de-Broglie wave
where wavelength called de-Broglie wavelength,
h
is given by λ =
.
mv
l
45. If a proton and an electron have the same
de Broglie wavelength, then
(a) kinetic energy of electron < kinetic energy
of proton
(b) kinetic energy of electron = kinetic energy
of proton
(c) momentum of electron = momentum of
proton
(d) momentum of electron < momentum of
proton
46. Which of these particles having the same
kinetic energy has the largest de Broglie
wavelength?
(a) Electron
(b) Alpha particle
(c) Proton
(d) Neutron
47. Two particles A1 and A2 of masses m1, m2
(m1 > m2) have the same de Broglie wavelength.
Then
(a) their momenta are the same.
(b) their energies are the same.
(c) momentum of A1 is less than the momentum
of A2.
(d) energy of A1 is more than the energy of A2.
Case III : Read the passage given below and
answer the following questions from 48 to 52.
Wave Theory of Light
According to wave theory, the light of any frequency
can emit electrons from metallic surface provided
the intensity of light be sufficient to provided
necessary energy for emission of electrons, but
according to experimental observations, the light
of frequency less than threshold frequency can
not emit electrons; whatever be the intensity
of incident light. Einstein also proposed that
electromagnetic radiation is quantised.
If photoelectrons are ejected from a surface when
light of wavelength l1 = 550 nm is incident on
it. The stopping potential for such electrons is
Vs = 0.19 V. Suppose the radiation of wavelength
l2 = 190 nm is incident on the surface.
69
Dual Nature of Radiation and Matter
48. Photoelectric effect supports quantum
nature of light because
(A) there is a minimum frequency of light below
which no photoelectrons are emitted.
(B) the maximum K.E. of photoelectric depends
only on the frequency of light and not on its
intensity.
(C) even when the metal surface is faintly
illuminated, the photo electrons leave the
surface immediately.
(D) electric charge of the photoelectrons is
quantized.
(a) A, B, C
(b) B, C
(c) C, D
(d) A, D, C
49. In photoelectric effect, electrons are ejected
from metals, if the incident light has a certain
minimum
(a) wavelength
(b) frequency
(c) amplitude
(d) angle of incidence
50. Calculate the stopping potential Vs of
2
surface.
(a) 4.47
(b) 3.16
(c) 2.76
(d) 5.28
51. Calculate the work function of the surface.
(a) 3.75
(b) 2.07
(c) 4.20
(d) 3.60
52. Calculate the threshold frequency for the
surface.
(a) 500 × 1012 Hz
(b) 480 × 1013 Hz
11
(c) 520 × 10 Hz
(d) 460 × 1013 Hz
Case IV : Read the passage given below and
answer the following questions from 53 to 55.
Hertz Observations
To study photoelectric effect, an emitting electrode
C of a photosensitive material is kept at negative
potential and collecting electrode A is kept at
positive potential in an evacuated tube. When light
of sufficiently high frequency falls on emitting
electrode, photoelectrons are emitted which travel
directly to collecting electrode and hence an electric
current called photoelectric current starts flowing
in the circuit, which is directly proportional to the
number of photoelectrons emitted by emitting
electrode C.
While
demonstrating
the
existence
of
electromagnetic waves, Hertz found that high
voltage sparks passed across the metal electrodes
of the detector loop more easily when the cathode
was illuminated by ultraviolet light from an arc
lamp. The ultraviolet light falling on the metal
surface caused the emission of negatively charged
particles, which are now known to be electrons,
into the surrounding space and hence enhanced
the high voltage sparks.
53.
(a)
(b)
(c)
(d)
Cathode rays were discovered by
Maxwell Clerk James
Heinrich Hertz
William Crookes
J. J. Thomson
54. Cathode rays consists of
(a) photons
(b) electrons
(c) pistons
(d) a-particles.
55.
the
(a)
(c)
Who discovered the charge on an electron for
frist time?
Millikan
(b) Thomson
Kelvin
(d) Coulomb
Assertion & Reasoning Based MCQs
For question numbers 56-60, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
56. Assertion (A) : Photoelectric effect
demonstrates the wave nature of light.
Reason (R) : The number of photoelectrons is
proportional to the frequency of light.
57. Assertion (A) : The de-Broglie wavelength
of particle having kinetic energy K is l. If
its kinetic energy becomes 4 K then its new
wavelength would be l/2.
CBSE Board Term-II Physics Class-12
70
Reason (R) : The de-Broglie wavelength l
is inversely proportional to square root of the
kinetic energy.
58. Assertion (A) : There is a physical
significance of matter waves.
Reason (R) : Both interference and diffraction
occurs in it.
59. Assertion (A) : A photon has no rest mass,
yet it carries definite momentum.
Reason (R) : Momentum of photon is due to its
energy and hence its equivalent mass.
60. Assertion (A) : Photosensitivity of a metal
is high if its work function is small.
Reason (R) : Work function = hu0, where u0 is
the threshold frequency.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. When a light of wavelength 400 nm falls on
a metal of work function 2.5 eV, what will be
the maximum magnitude of linear momentum
of emitted photoelectron?
2. Do all the electrons that absorb a photon
come out as photoelectron?
3. If light of wavelength 412.5 nm is incident
on each of the metals given in table, which one
will show photoelectric emission and why?
which absorb photons of larger wavelength and
emit light of shorter wavelength.
7. Name the phenomenon which shows the
quantum nature of electromagnetic radiation.
8. Write the relationship of de-Broglie
wavelength l associated with a particle of
mass m in terms of its kinetic energy E.
5. T h e t h r e s h o l d w a v e l e n g t h f o r t w o
photosensitive surfaces A and B are l 1 and
l2 respectively. What is the ratio of the work
functions of the two surfaces?
9. (i) In the explanation of photoelectric effect,
we assume one photon of frequency u collides
with an electron and transfers its energy. This
leads to the equation for the maximum energy
Emax of the emitted electron as
Emax = hu – φ0
where φ0 is the work function of the metal. If an
electron absorbs 2 photons (each of frequency u)
what will be the maximum energy for the emitted
electron?
(ii) Why is this fact (two photon absorption) not
taken into consideration in our discussion of the
stopping potential?
6. There are materials which absorb photons of
shorter wavelength and emit photons of longer
wavelength. Can there be stable substances
10. A proton and an electron have same kinetic
energy. Which one has greater de-Broglie
wavelength and why?
Metal
Na
K
Ca
Mo
Work Function (eV)
1.92
2.15
3.20
4.17
4. Define the term “Intensity” in photon picture
of electromagnetic radiation.
Short Answer Type Questions (SA-I)
11. A metallic surface is irradiated with
monochromatic light of variable wavelength.
Above a wavelength of 5000 Å, no photoelectrons
are emitted from the surface. With an unknown
wavelength, stopping potential is 3 V. Find the
unknown wavelength.
12. In an experiment on photoelectric effect,
light of wavelength 800 nm (less than threshold
wavelength) 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 every
106 photons is able to eject a photoelectron, find
the photo current in the circuit.
71
Dual Nature of Radiation and Matter
13. One milliwatt of light of wavelength
l = 4560 Å is incident on a cesium metal surface.
Calculate the electron current liberated. Assume
a quantum efficiency of h = 0.5%. [Work function
for cesium = 1.89 eV]. Take hc = 12400 eV Å.
14. When light of wavelength l is incident on a
metal surface, stopping potential is found to be x.
When light of wavelength nl is incident on the
same metal surface, stopping potential is found
x
to be
Find the threshold wavelength of
n +1
the metal.
15. The magnetic field at a point associated with
a light wave is B = 2 × 10–6 tesla 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?
16. A light beam of wavelength 400 nm is
incident on a metal plate of work function of
2.2 eV. An electron absorbs a photon and makes
some collisions before coming out of the metal.
Assuming that 10% of the instantaneous energy
is lost to the metal in each collision.
(i) Find the kinetic energy of electron which
makes two collisions as it comes out of the metal.
(ii) Under the same assumptions, find the
minimum number of collisions the electron can
suffer before it becomes unable to come out of
metal.
(Use hc = 12400 eV Å)
17. Consider figure for photoemission.
How would you reconcile with
momentum-conservation? Note light
(photons) have momentum in a
different direction than the emitted
electrons.
electron
light
metal
18. (a) Define the terms, (i) threshold frequency
and (ii) stopping potential in photoelectric effect.
(b) Plot a graph of photocurrent versus anode
potential for a radiation of frequency u and
intensities I1 and I2 (I1 < I2).
19. Using the graph shown in the figure for
stopping potential versus the incident frequency
of photons, calculate Planck’s constant.
20. X-rays fall on a photosensitive surface to
cause photoelectric emission. Assuming that the
work function of the surface can be neglected,
find the relation between the de-Broglie
wavelength (l) of the electrons emitted and the
energy (Ev) of the incident photons. Draw the
nature of the graph for l as a function of Ev.
Short Answer Type Questions (SA-II)
21. In an experiment on photoelectric emission,
following observations were made
1. Wavelength of the incident light = 2 × 10–7 m
2. Stopping potential = 3 V
Find
(i) kinetic energy of photoelectrons with
maximum speed
(ii) work function and
(iii) threshold frequency (h = 6.62 × 10–34 J s).
22. A beam of light consists of four wavelengths
4000 Å, 4800 Å, 6000 Å and 7000 Å, each of
intensity 1.5 × 10 –3 Wm –2 . The beam falls
normally on an area 10–4 m2, of a clean metallic
surface of work function 1.9 eV. Assuming no
loss of light energy (i.e., each capable photon
emits one electron), calculate the number of
photoelectrons liberated per second.
23. Two monochromatic beams A and B of equal
intensity I, hit a screen. The number of photons
hitting the screen by beam A is twice that by
beam B. Then what inference can you make
about their frequencies?
24. An electron and a photon each have a
wavelength 1.00 nm. Find
(i) their momenta,
(ii) the energy of the photon and
(iii) the kinetic energy of electron.
25. (a) Ultraviolet light of wavelength 2271 Å
from a 100 W mercury source is incident on a
photocell made of molybdenum metal. If the
stopping potential is 1.3 V, estimate the work
function of the metal.
CBSE Board Term-II Physics Class-12
72
(b) How would the photocell respond to high
intensity (105 W/m2) red light of wavelength
6328 Å produced by a He – Ne laser?
26. Two particles A and B of de–Broglie
wavelengths l 1 and l 2 combine to form a
particle C. The process conserves momentum.
Find the de–Broglie wavelength of the particle
C. (The motion is one dimensional).
27. (a) An electron and a proton are accelerated
through the same potential. Which one of the
two has
(i) greater value of de-Broglie wavelength
associated with it, and
(ii) lesser momentum?
Justify your answer in each case.
(b) How is the momentum of a particle related
with its de-Broglie wavelength? Show the variation
on a graph.
28. State three important properties of
photons which describe the particle picture of
electromagnetic radiation.
29. A neutron beam of energy E scatters from
atoms on a surface with a spacing d = 0.1 nm.
The first maximum of intensity in the reflected
beam occurs at q = 30°. What is the kinetic
energy E of the beam in eV?
30. Assuming an electron is confined to a 1 nm
wide region, find the uncertainty in momentum
using Heisenberg Uncertainty principle. You can
assume the uncertainty in position Dx as 1 nm.
Assuming p Dp, find the energy of the electron
in electron volts.
31. C o n s i d e r a m e t a l e x p o s e d t o l i g h t
of wavelength 600 nm. The maximum
energy of the electron doubles when light of
wavelength 400 nm is used. Find the work
function in eV.
32. (a) S t at e t w o i m p o r t an t fea tu res of
Einstein’s photoelectric equation.
(b) Radiation of frequency 1015 Hz is incident
on two photosensitive surfaces P and Q.
There is no photoemission from surface P.
Photoemission occurs from surface Q but
photoelectrons have zero kinetic energy.
Explain these observations and find the value
of work function for surface Q.
Long Answer Type Questions (LA)
33. Consider a 20 W bulb emitting light of
wavelength of 5000 Å and shining on a metal
surface kept at a distance 2 m. Assume that the
metal surface has work function of 2 eV and that
each atom on the metal surface can be treated
as a circular disk of radius 1.5 Å.
(i) Estimate no. of photons emitted by the bulb
per second. [Assume no other losses]
(ii) Will there be photoelectric emission?
(iii) How much time would be required by the
atomic disk to receive energy equal to work
function (2 eV)?
(iv) How many photons would atomic disk receive
within time duration calculated in (iii) above?
(v) Can you explain how photoelectric effect was
observed instantaneously?
34. Figure shows a plot of stopping potential
(V0) with frequency (u) of incident radiation for
two photosensitive material M1 and M2. Explain
(i) why the slope of both the lines is same?
(ii) for which material emitted electrons have
greater kinetic energy for the same frequency of
incident radiation?
35. Explain how Einstein’s photoelectric
equation is used to describe photoelectric effect
satisfactorily.
36. Consider a thin target (10 –2 m square,
10–3 m thickness) of sodium, which produces
a photocurrent of 100 mA when a light of
intensity 100 W m–2 (l = 660 nm) falls on it.
Find the probability that a photoelectron is
produced when a photon strikes a sodium atom.
[Take density of Na = 0.97 kg m–3].
37. A particle A with a mass mA is moving with
a velocity v and hits a particle B (mass mB) at
rest (one dimensional motion). Find the change
in the de Broglie wavelength of the particle A.
Treat the collision as elastic.
73
Dual Nature of Radiation and Matter
OBJECTIVE TYPE QUESTIONS
8.
1. (d) : Velocity acquired by a particle while falling from a
height H is
9.
v = 2gH ...(i)
h
h
(Using (i))
As λ =
=
mv m 2gH
1
or λ ∝
H
2. (a) : Given threshold frequency, u0 = 3.3 × 1014 Hz
Frequency of incident light, u = 8.2 × 1014 Hz
h ( υ − υ0 )
As eV0 = h(u – u0) or V0 =
e
3.
∴
6.63 × 10 −34 (8.2 × 1014 − 3.3 × 1014 )
=2V
=
1.6 × 10 −19
h
(d) : As λ =
2mqV
λ∝
=
1
mq
4m p × 2e
mp × e
∴
λp
λα
= 8
=
m αq α
mpq p
(∵ m α = 4mp , q α = 2qp )
h
4. (b) : de Broglie wavelength, λ =
...(i)
2mK
K
When the kinetic energy is
then
4
h
2h
λ′ =
=
= 2λ
(Using (i))
2m (K / 4)
2mK
5.
(b) : The de Broglie wavelength is given by
h
h
λ= =
p mv
1
λ∝
v
So if the velocity of the electron increases, the de Broglie
wavelength decreases.
h
6. (d) : de Broglie wavelength, l =
p
1
or l ∝
p
Hence, curve (d) is the correct option.
7. (a) : Mass of the ball, m = 150 g = 0.15 kg,
Speed of the ball, v = 30 m s–1
Momentum, p = mv = 0.15 × 30 = 4.5 kg m s–1
h 6.63 × 10 −34
de Broglie wavelength, λ = =
= 1.47 × 10–34 m
p
4.5
(d)
(d) : Energy spent to convert ice into water
= mL = (1000 g) × 80 cal g –1 = 80000 cal
Energy of photons used = nT × E = nT × hu
\
nThu = mL
T=
or
mL
nhυ
T ∝ 1/n when u is constant, T ∝ 1/u when n fixed,
T ∝ 1/nu. Thus T is constant if nu is constant. Thus options
(a), (b) and (c) are correct.
h
10. (a) : de-Broglie wavelength, λ =
p
where symbols have their usual meaning.
\
∵
pα = pp = pe
\
la = lp = le
11. (a) : de Broglie wavelength associated with an electron is
h
h
λ=
or p =
p
λ
∴
∆p
∆λ
=−
⇒
p
λ
0.5
p
=
pinitial 100
pinitial = 200p
12. (c) : Millikan’s experiment established that electric
charge is quantised.
13. (b) : The phenomenon of photoelectric emission was
discovered by Heinrich Hertz in 1887.
14. (a) : Here, T = 27 + 273 = 300 K
For an electron in a metal, momentum p = 3mk BT
de Broglie wavelength of an electron is
h
h
λ= =
p
3mk BT
6.63 × 10 −34
=
3 × (9.1 × 10 −31 ) × (1.38 × 10 −23 ) × 300
= 6.2 × 10 –9 m
15. (b) :
CBSE Board Term-II Physics Class-12
74
Since, eV0 = hn
v
h
∴ slope of = slope of 0
e
ν
=
22. (b) : Here, E = 1 MeV = 106 eV, h = 6.63 × 10–34 J s,
c = 3 × 108 m s–1
For metal A,
2−0
h
Slope = =
e (10 − 5) × 1014
or
h=
or
2×e
2 × 1.6 × 10 −19
=
= 6.4 × 10 −34 J s
14
14
5 × 10
5 × 10
For metal B,
2.5 − 0
h
Slope = =
e (15 − 10) × 1014
2.5 × e 2.5 × 1.6 × 10 −19
=
h=
5 × 1014
5 × 1014
= 8 × 10–34 J s
16. (a) : Here, V = 30 kV, h = 6.63 × 10–34 J s
As eV = humax
or
υmax =
As
eV =
eV 1.6 × 10 −19 × 3 × 104
= 7.24 × 1018 Hz
=
h
6.63 × 10 −34
hc
λ min
18. (d)
19. (a) : Mean kinetic energy of a molecule
3
1 2
= k BT
= mv rms
2
2
3k BT
m
Here, m = 2 × 14.0076 = 28.02 u
h
h
As λ =
=
mvrms
3mkB T
∴
v rms =
=
hc =
As E =
or
λ=
6.63 × 10 −34 × 3 × 108
≈ 1240 eV nm
1.6 × 10 −19
hc
λ
hc 1240 eV nm
=
= 1.24 × 10 −3 nm
E
106 eV
23. (c) : Here, total energy of 2 g-rays = 10.2 × 109 eV
1
9
−19
\ Energy of each g-rays = = (10.2 × 10 × 1.6 × 10 ) J
2
= 8.16 × 10 –10 J
hc
As E = hυ =
λ
or
λ=
hc 6.63 × 10 −34 × 3 × 108
=
= 2.44 × 10–16 m
E
8.16 × 10 −10
24. (a) : Number of photons emitted per second =
hc 6.63 × 10 −34 × 3 × 108
or λ min =
=
= 0.041 × 10 −9 m
eV 1.6 × 10 −19 × 3 × 104
= 0.041 nm
17. (a)
2 × 10 −3
= 5 × 1015 photons per second
3.98 × 10 −19
6.63 × 10 −34
3 × (28.02 × 1.66 × 10 −27 ) × 1.38 × 10 −23 × 300
= 2.75 × 10 –11 m
20. (b) : Here, V0 = 1.5 V,
Maximum kinetic energy = eV0 = 1.5 eV
21. (d) : Here, u = 6 × 1014 Hz, h = 6.63 × 10–34 J s
As E = hu = 6.63 × 10–34 × 6 × 1014 = 3.98 × 10–19 J
Number of photons emitted per second,
P
n=
E
=
8% of P
E
8 × 150 × 4500 × 10 −10
8Pλ
=
100 hc 100 × 6.63 × 10 −34 × 3 × 108
= 3 × 1019 photons per second
1
25. (b) : Maximum kinetic energy K max = mv 2 = eVo
2
where Vo is the stopping potential.
According to Einstein’s photoelectric equation
hu1 = f0 + eV1
... (i)
hu2 = f0+ eV2
... (ii)
\ h(u1 – u2) = e(V1 – V2)
h
h
( υ1 − υ2 ) = V1 − V2 or V2 = V1 + ( υ2 − υ1 )
e
e
26. (d) : According to law of conservation of linear
momentum, two particles will have equal and opposite
momentum.
The de Broglie wavelength is given by
λ1
h
λ=
∴
=1
p
λ2
27. (b) : The maximum kinetic energy of the emitted electron
is given by
Kmax = hu – f0 = h(4u) – h(u) = 3hu
28. (a) : de Broglie wavelength is given by
75
Dual Nature of Radiation and Matter
λ=
or p =
h
λ
Since, =
As this force is perpendicular to v and B , so the magnitude
of v will not change. i.e momentum (= mv) will remains
constant in magnitude.
 h 
Therefore, de Broglie wavelength, λ  =
remains
 mv 
constant.
h , where p is the momentum
p
h
2π
∴ p=
2π
λ
34. (d) : According to Einstein’s photoelectric equation
p = 2mK
29. (b) : As
K=
= 2 × 9 × 10 −31 × 120 × 1.6 × 10 −19
= 5.88 × 10–24 kg m s–1
de Broglie wavelength,
h 6.63 × 10 −34
λ= =
= 1.13 × 10–10 m = 1.13 Å
−24
p 5.88 × 10
E hυ
30. (d) : Momentum of photon, p = =
c
c
where E is the energy of a photon and c is the velocity of light.
hc
c

∴ p=
∵ υ = λ 
cλ
h
h
= =
= 1012 h
λ 0.01 × 10 −10
31. (d) : The maximum kinetic energy of photoelectron
ejected is given by
35. (a) : Let K be the kinetic energy of the incident electron.
Its linear momentum, p = 2mK
The de Broglie wavelength is related to the linear momentum
as
λ=
2
mv
= evB
r
mv
p
2mK
r=
=
=
eB eB
eB
p2
( As K =
)
2m
From Einstein’s photoelectric equation
K = E – f0
∴
r=
2m (E − φ0 )
eB
^
^
33. (a) : Here, v = v 0 i , B = B0 j
Force on moving electron due to magnetic field is
^
^
^
F = − e (v × B ) = − e (v 0 i × B0 j ) = − ev 0B0 k
h
h
h2
=
or K =
p
2mλ2
2mK
The cut off wavelength of the emitted X rays is related to
the kinetic energy of the incident electron as
hc
h2
=K =
λ0
2mλ2
36. (b) : φ0 A =
K.E. = hu – f0
= hu – hu0
where work function depends on the type of material.
If the frequency of incident radiation is greater than
u0 only then the ejection of photoelectrons start. After that as
frequency increases kinetic energy also increases.
32. (d) : As the electron describes a circular path of radius
r in the magnetic field, therefore
hc hc
 λ − λ
1 1
−
= hc  −  = hc  0
λ λ0
 λλ 0 
 λ λ0 
or λ 0 =
2mcλ2
h
hυ0
eV
e
=
(6.6 × 10 −34 ) × (1.8 × 1014 )
eV = 0.74 eV
1.6 × 10 −19
φ0B =
(6.6 × 10 −34 ) × (2.2 × 1014 )
eV = 0.91 eV
1.6 × 10 −19
Since the incident energy 0.825 eV is greater than 0.74 eV
and less than 0.91 eV, so photoelectrons are emitted from
metal A only.
37. (b) : As λ =
∴
h
h
1
=
or λ ∝
mv
K
2mK
λ′
K
1 1
=
=
=
λ
K′
16 4
 λ − λ′ 
× 100
% change in de Broglie wavelength = 
 λ 
 1
 λ′ 
= 1 −  100 = 1 −  × 100 = 75%
 4

λ
38. (d) : The rest mass of photon is zero.
39. (d) : Photoelectric current depends on
(i) the intensity of incident light.
(ii) the potential difference applied between the two
electrodes.
(iii) the nature of the emitter material.
CBSE Board Term-II Physics Class-12
76
40. (a) : Work function, f0 = hu0
where u0 is the threshold frequency
So, f0 ∝ u0
Hence Pt > Al > K
41. (a) : With the increase of intensity of the incident
radiation the number of photoelectrons emitted per unit
time increases.
49. (b) : For photoelectric emission, the incident light
must have a certain minimum frequency, called threshold
frequency.
50. (a) : From Einstein’s relation
eVs = hυ – W
As work function is a constant for a surface.
e(Vs – Vs ) = h(υ2 – υ1)
2
42. (b) : As eV = KEmax
KE
\ V =  max 
 e 
43. (c) : From Einstein’s photoelectric equation,
KEmax = hυ – φ = (5 – 3) = 2 eV
44. (d) : According to Einstein’s photoelectric equation, the
kinetic energy of the emitted photoelectron is
K = hu – f0
... (i)
where u is the frequency of incident radiation and f0 is a
work function of the metal.
If the frequency of incident radiation is doubled, then
K′ = 2hu – f0 = 2(hu – f0) + f0 = 2K + f0 (Using (i))
K′ > 2K
h
45. (c) : de Broglie wavelength, λ =
p
where p is the momentum of the particle
h
For electron, λ e =
pe
h
For proton, λ p =
pp
As le = lp ⇒ pe = pp(Given)
or Momentum of electron = Momentum of proton
h
1
46. (a) : As λ =
so λ ∝
2mK
m
Out of the given particles m is least for electron, therefore
electron has the largest value of de Broglie wavelength.
h
h
1
47. (a) : As λ = or p = or p ∝
λ
λ
p
p1 λ2 λ
\
=
= = 1 or p1 = p2
p 2 λ1 λ
h

1 p2
1 h2
=
Also E =
∵ p = 
2
λ
2 m 2m λ
or
1
E∝
m
E m
∴ 1 = 2 < 1 or E1 < E 2
E 2 m1
48. (a) : The existence of the frequency and the
instantaneous emission of photo electrons support the
quantum nature of light.
1
h
v s2 = Vs1 + ( υ2 − υ1 )
e
1 
 1
−
= 0.19 + 1240 
= 4.47 V
 190 550 
51. (b) : W =
hc
1240
− eVs1 =
− 0.19 = 2.07 eV
λ1
550
52. (a) : hυc = W
υc =
W (2.07)(1.602 × 10 −19 )
=
≈ 500 × 1012 Hz
6.626 × 10 −34
h
53. (c) 54. (b)
55. (a)
56. (d) : Photoelectric effect can be explained on the basis
of quantum theory or particle nature of light where wave
nature of light fails to explain the photoelectric effect. The
number of photoelectrons is proportional to the intensity of
incident light.
I = nhυ where n is the number of photons emitted/absorbed
per unit area per second. n and hυ are independent factors.
57. (a) 58. (a)
59. (a) : Equivalent mass of photon (m) is given from
equation
hυ
E = mc2 = hυ ∴ m = 2
c
where E is energy, m is mass, c is speed of light, h is Planck’s
constant, υ is frequency.
hυ
hυ
∴ Momentum of photon = 2 × c =
c
c
60. (b) : Less work function means less energy is required
for ejecting out the electrons.
SUBJECTIVE TYPE QUESTIONS
1.
K max =
hc
−φ
λ

p 2  1.24 × 104
=
− 2.5 eV = 0.6 eV

2m  4000
–25
p = 2 × 9.1 × 10 −31 × 0.6 × 1.6 × 10 −19 = 4.2 × 10 kg m/s
77
Dual Nature of Radiation and Matter
2. Not all the electrons that absorb a photon come out as
photoelectrons because most of electrons get scattered into
the metal. Only those electrons come out as photoelectrons
whose energy becomes greater than work function of metal.
3. Wavelength of incident light, l = 412.5 nm
hc 1242 eV nm
Energy of incident light, E = =
= 3 eV
412.5 nm
λ
Metals Na and K will show photoelectric emission because
their work functions are less than the energy of incident light.
4. The amount of light energy or photon energy, incident per
unit area per unit time is called intensity of electromagnetic
radiation.
hc
5. Work function = hυ =
λ
φA hc λB λ2
\ The ratio,
=
×
=
φB λ A hc λ1
6. In case of stable material, this is not possible because,
to absorb a photon of larger wavelength and emit a photon
of shorter wavelength, energy has to be supplied by the
material.
7. Photoelectric effect shows the quantum nature of
electromagnetic radiation.
h
8. λ =
2mE
9. Given, Emax = hu – f0
As energy of 2 photons = 2hu
E′max = 2hu – f0
As there is one to one interaction, probability of absorbing
2 photons by same electron is very low. Thus, emission of
two photon absorption is negligible.
h
10. We know the relation, λ =
p
p2
kinetic energy, K =
2m
h
Then, λ =
2mK
Kp = Ke ⇒ λ ∝ 1
m
 mp >> me \ lp << le
Hence for same kinetic energy wavelength associated with
electron will be greater.
11. According to question, l th = 5000 Å and V s = 3V.
Using equation of photoelectric equation,
Kmax = E – W (Kmax = eVs)
12400 12400 12400
−
=
− 2.48 eV
λ
5000
λ
\
3 eV =
or
l = 2262 Å
P Pλ
=
E λ hc
where P is power of light and El is energy of photon.
Pλ 1
Number of photoelectron emitted per second =
⋅
hc 106
Pλ
\ Photocurrent =
⋅e
hc × 106
12. Number of photons falling in one second =
=
5 × 800 × 10 −9 × (1.6 × 10 −19 )
6.63 × 10 −34 × 3 × 108 × 106
= 3.2 µA
hc 12400
13. Since
=
= 2.7 eV > 1.89 eV so photocurrent
4560
λ
will flow.
Number of photons incident per second
P (Power)
P Pλ
=
=
=
E λ (Energy of photon) hc hc
λ
0.5  Pλ 
Number of photoelectron emitted per second =
 
100  hc 
0.5  Pλ 
−6
Photocurrent =
  e = 1.84 × 10 A.
100  hc 
14. Let l0 is the threshold wavelength, the work function is
hc
φ=
λ0
hc hc
Now, by photoelectric equation, eVs = −
λ λ0
hc hc
ex = − ...(i)
λ λ0
ex
hc hc
...(ii)
=
−
n + 1 nλ λ 0
From (i) and (ii),
or
hc hc
hc
hc
−
= (n + 1) − (n + 1)
λ0
λ λ0
nλ
nhc hc
=
⇒ λ 0 = n2λ
λ 0 nλ
15. B = 2 × 10–6 tesla sin[(3.0 × 1015 s–1)t] sin[(6.0 × 1015 s–1)t]
= 1 × 10–6 tesla [cos{(3.0 × 1015 s–1)t} – cos{(9.0 × 1015 s–1)t}]
9 × 1015 −1
s
2π
Maximum kinetic energy of the photoelectrons
 9 × 1015

h − 2 eV = 3.93 eV
=
 2πe

Maximum frequency of the wave =
hc 1.24 × 104
=
eV = 3.1 eV
λ
4000
(i) K.E. of emitted e– = (3.1 × (0.9)2 – 2.2) eV = 0.31 eV
(ii) For e– not be able to come out, its energy should be less
than 2.2 eV
i.e., (3.1) × (0.9)n < 2.2, n > 4
16. E l = Energy of the photon =
CBSE Board Term-II Physics Class-12
78
17. The momentum of incident photon is transferred to the
metal during photoelectric emission at macroscopic level.
At microscopic level, atoms of metal absorb photons and
momentum is transferred mainly to the nuclei and the
electron. Excited electrons are emitted in the process and
thus momentum is conserved.
18. (a) (i) Threshold Frequency : The minimum frequency
of incident light which is just capable of ejecting electrons
from a metal is called the threshold frequency. It is denoted
by u0.
(ii) Stopping Potential : The minimum retarding potential
applied to anode of a photoelectric tube which is just capable
of stopping photoelectric current is called the stopping
potential. It is denoted by V0 (or VS).
(b)
Photoelectric
Higher intensity
current
I=
Applied voltage
ε
R +r
19 Using Einstein’s photoelectric equation,
eV = hu – f
on differentiation we get eDV = hDu
or h =
6.625 × 10
=
(1.5 × 10 −3 )(10 −4 )  1
1
1 
+
 +
 = 1.12 × 1012
−19
3.1 2.58 2.05 
1.6 × 10
h 6.6 × 10 −34
=
= 6.6 × 10 −25 kg m s −1
−9
λ
1 × 10
Momentum of electron
p=
p=
6.6 × 10 −34
1 × 10 −9
= 6.6 × 10 −25 kg m s −1
(ii) Energy of photon
E=
hc 6.6 × 10 −34 × 3 × 108
=
= 1.98 × 10 −16 J
λ
1 × 10 −9
Ee =
–7
−34
I1A1 I2 A2 I3 A3
1
 1 1
+
+
= IA  + + 
E1
E2
E3
 E1 E 2 E 3 
(iii) Kinetic energy of electron
m = 2000 Å and hc = 12400 eVÅ.
hc 12400
=
eV = 6.20 eV
Energy of incident photon =
λ
2000
W = E – Kmax = 3.2 eV
(iii) huth = W = 3.2 × 1.6 × 10–19 J
υth =
=
EV
21. (i) Since Vs = 3 V and Kmax = eVs, so Kmax = 3 eV
3.2 × 1.6 × 10 −19
\ Number of photoelectrons emitted per second = Number
of photons incident per second
24. (i) Momentum of photon
20. According to Einstein’s photoelectric effect
1
E = W + mv 2
2
Since work function of the surface is negligible, the above
equation becomes
1
E = mv 2
2
mv = 2 mE
(ii) l = 2 × 10
emit photoelectrons.
i.e, frequency of beam B is twice of that of beam A.
e∆V 1.6 × 10 −19 × (1.23 − 0)
=
= 6.56 × 10 −34 J s
14
∆υ
(8 − 5) × 10
If l is de-Broglie wavelength of the
emitted electrons, then
h
h
λ=
=
mv
2mE
light of wavelengths 4000 Å, 4800 Å,and 6000 Å can only
23. Let n1, n2 be the number of photons hitting the screen
per second by beam A and B respectively
Intensity of beam of photon, I = nhu
\ n1u 1 = n2u 2
n1 υ2
=
n2 υ1
n1
υ
= 2 \ 2 = 2 , u2 = 2u1
As
n2
υ1
I2
I1
–V0
12400
12400
= 3.1 eV, E 2 =
= 2.58 eV ,
4000
4800
12400
12400
= 2.06 eV and E 4 =
= 1.77 eV
E3 =
6000
7000
Energy of photon (E4) is less than work function. Therefore,
22. E1 =
14
= 7.76 × 10
Hz
p 2 (6.6 × 10 −25 )2
=
= 2.39 × 10–19 J
2m 2 × 9.1 × 10 −31
25.
(a) From Einstein’s equation
hu = f0 + K = f0 + eVs
hc
–eVs
λ
(Equation is independent of the power of the source)
or
f0 = hu – eVs =
φ0 =
6.6 × 10 −34 × 3 × 108
2271 × 10 −10
− 1.3 eV
79
Dual Nature of Radiation and Matter
 6.6 × 10 −34 × 3 × 108

=
− 1.3 eV
−10
−19
 2271 × 10 × 1.6 × 10

= 5.5 eV – 1.3 eV = 4.2 eV
=
4.2 × 1.6 × 10 −19
6.6 × 10 −34
φ0
h
= 1.0 × 1015 Hz
and the frequency of red light from the source is 105 W/m2.
c
3 × 108
= 4.7 × 1014 Hz
υ= =
λ 6328 × 10 −10
Since frequency of red light is less than threshold frequency
so photocell will not respond to red light, however high
(105 W/m2) be the intensity of light.
26. For one dimensional motion,
→
→
→
pC = p A + pB
If pA, pB > 0 or
direction).
pC = pA + pB
Kmax
(b) Threshold frequency υ0 =
28. Photons : According to Planck’s quantum theory of
radiation, an electromagnetic wave travels in the form of
discrete packets of energy called quanta.
The main features of photons are as follows:
pA, pB < 0, i.e., pA and pB are in same
h
h
h
 λ + λB 
= h A
=
+
λC λ A λ B
 λ A λB 
λ A λB
lC = λ + λ
A
B
If pA > 0, pB < 0 or pA < 0, pB > 0
(pA and pB are in opposite direction)
pC = |pA – pB|
h | λ A − λB |
h
h
h
=
−
=
λ A λB
λC λ A λ B
λ A λB
lC =
| λ A − λB |
27. For same accelerating potential, a proton and a deutron
have same kinetic energy.
(a) de Broglie wavelength is given by
h
h
h
λ= =
=
p
2 mK
2 m (qV )
1
m
Mass of a deutron is more than that of a proton. So, proton
will have greater value of de-Broglie wavelength.
(b) de-Broglie wavelength of a particle
h
λ=
p
or λp = h = constant
So, λ ∝
It shows a rectangular hyperbola.
0
0
Frequency of incident
radiation
(i) In the interaction of photons with free electrons, the
entire energy of photon is absorbed.
(ii) Energy of photon is directly proportional to frequency.
Intensity of incident radiation depends on the number of
photons falling per unit area per unit time for a given
frequency.
(iii) In photon electron collision, the total energy and
momentum remain constant.
Einstein’s photoelectric equation is
Kmax = hu – f0
29. Given q = 30°, d = 0.1 nm, n = 1
According to Bragg’s law,
2d sin q = nl
2 × 0.1 × sin 30° = 1 × l
l = 0.1 nm = 1 × 10–10 m.
h
As we know l =
2mE
2
(6.63 × 10 −34 )2
h
E=
=
J
2(1.67 × 10 −27 )(1 × 10 −10 )2
2mλ2
=
(6.63 × 10 −34 )2
2(1.67 × 10
−27
)(1 × 10
−10 2
)
×
1
1.6 × 10 −19
eV
= 8.2 × 10–2 eV
30. Given, Dx = 1 nm = 1 × 10–9m, Dp = ?
DxDp = 6.62 × 10 −34 J s
h
∴ ∆p =
=
=
∵ = h 

∆x 2π∆x
 22 
−9
2π 
2 ×   × 10 m
 7
= 1.05 × 10–25 kg m s–1
Energy of electron,
E=
=
=
p2
(Taking Dp ≈ p)
2m
(1.05 × 10 −25 )2
2 × 9.1 × 10 −31
0.06 × 10 −19
1.6 × 10 −19
= 0.06 × 10 −19 J
= 0.0375 eV = 3.75 × 10–2 eV
CBSE Board Term-II Physics Class-12
80
31. Given, l = 600 nm, l′ = 400 nm
E′max = 2Emax, f0 = ?
hc
− φ0 …(i)
Emax = hu – f0 =
λ
hc
2Emax =
− φ0 …(ii)
λ′
Dividing (ii) by (i)
2E max hc / λ ′ − φ0
=
E max
hc / λ − φ0
⇒
\
2hc
hc
− 2φ0 = − φ0 ⇒ hc  2 − 1  = φ 0
 λ λ′ 
λ′
λ
1 
 2
−
 = 1.03 eV
600 400 
f0 = 1240 

(
(iii)
r
Disk of
radius r
P
energy emitted by the bulb in time Dt = PDt
energy falling on the disk i.e.,
2
 P∆t  2  Pr 
(
)
=
E= 
π
r
 2  Dt
 4 πd 2 
4d
According to given condition, E = f0
 Pr2 
∴  2  ∆t = φ0
 4d 
Dt =
hc = 1240 eV nm)
32. (a) Two features of Einstein’s photoelectric equation:
(a) Below threshold frequency uo corresponding to Wo, no
emission of photoelectrons takes place.
(b) As the number of photons in light depend on its intensity,
and one photon liberates one photo electron. So number
of emitted photoelectrons depend only on the intensity of
incident light for a given frequency.
(b) Below threshold frequency no emission takes place. As
there is no photoemission from surface P i.e., the frequency
of incident radiation is less than the threshold frequency for
surface P.
From surface Q photoemission is possible i.e., the frequency
of incident radiation is equal or greater than threshold
frequency. As the kinetic energy of photo electrons is zero
i.e., the energy of incident radiation is just sufficient to pull
out the electron from the surface Q.
Work function for surface Q, WQ= hu.
As K.E. = 0 ; u = u0 = 1015 Hz
WQ = 6.6 × 10–34 × 1015 = 6.6 × 10–19 J = 4.125 eV
d
∆t =
4φ0d 2
Pr 2
(4)(2 × 1.6 × 10 −19 )(2)2
= 11.4 s
(20)(1.5 × 10 −10 )2
(iv) Number of photons received by atomic disk in time Dt is
 πr 2 
nr 2 ∆t
×
∆
t
N = n
=

4d 2
 4π d 2 
=
5 × 1019 × (1.5 × 10 −10 )2 × 11.4
(v) when
Dt =
Dt =
= 0.8 ≈ 1
4(2)2
r = 1 cm = 1 × 10–2 m
4φ0d 2
Pr 2
4 × 2 × 1.6 × 10 −19 × (2)2
−2
= 2.56 × 10–15 s
20 × (10 )
Dt is exceedingly small, photo emission is instantaneous.
∆V
[∵e∆V = h∆υ]
34. (i) Slope of line =
∆υ
h
Slope of line =
e
33. Given P = 20 W, l = 5000 Å = 5 × 10–7 m.
d = 2 m, f0 = 2 eV, r = 1.5 Å = 1.5 × 10–10 m.
(i) Number of photons emitted by bulb per second is
n=
P
Pλ
20 × 5 × 10 −7
=
=
(hc / λ ) hc 6.62 × 10 −34 × 3 × 108
≈ 5 × 1019 s–1
(ii) Energy of incident photon, E =
=
(6.62 × 10 −34 )(3 × 108 )
hc
λ
= 2.48 eV.
5 × 10 −7 × 1.6 × 10 −19
As E > f0 (2.48 eV > 2 eV)
hence photoelectric emission will take place.
It is a constant quantity and does not depend on nature of
metal surface.
(ii) Maximum kinetic energy of emitted photoelectron,
KE = eV0 = hu – hu0,
...(i)
For a given frequency V1 > V2 (from the graph)
So from equation (i),
(KE)1 > (KE)2
Since the metal M1 has smaller threshold frequency i.e.,
smaller work function. It emits electrons having a larger
kinetic energy.
81
Dual Nature of Radiation and Matter
35. Einstein’s photoelectric equation is given below.
1 2
hυ = mv max
+ W0
2
where u = frequency of incident radiation
1 2
mv max = maximum kinetic energy of an emitted electron
2
W0 = work function of the target metal
Three salient features observed are
(i) Below threshold frequency u0 corresponding to W0, no
emission of photoelectrons takes place.
(ii) As energy of a photon depends on the frequency of light,
so the maximum kinetic energy with which photoelectron
is emitted depends only on the energy of photon or on the
frequency of incident radiation.
(iii) For a given frequency of incident radiation, intensity
of light depends on the number of photons per unit area
per unit time and one photon liberates one photoelectron,
so number of photoelectrons emitted depend only on its
intensity.
36. Given d = 10–3 m, I = 100 × 10–6 A = 10–4 A
A = (10–2 m)2 = 10–4 m2
Intensity, I = 100 W m–2, l = 660 nm = 660 × 10–9 m
Volume of 6.02 × 1026 sodium atoms
23 kg
=
= 23.7 m3
0.97 kg/m3
Volume of target = (10–2) (10–2) (10–3) = 10–7m3
Number of sodium atoms in the target
37. From the law of conservation of momentum,
mAv = mAvA + mBvB
or mA(v – vA) = mBvB…(i)
(as particle B is at rest, its initial velocity is zero and vA
and vB are the velocities of particles A and B after collision)
Since the collision is elastic, kinetic energy is conserved
during collision
1
1
1
\
m Av 2 = m Av 2A + mB v B2
2
2
2
mA(v2 – vA2 ) = mBvB2…(ii)
Dividing eqn. (ii) by eqn. (i), we obtain
m A (v 2 − v 2A ) mB v B2
=
m A (v − v A ) mB v B
or v + vA = vB…(iii)
From eqns. (i) and (iii),
mA(v – vA) = mB(v + vA)
or (mA – mB)v = (mA + mB)vA
or
6.02 × 1026
× 10 −7 = 2.54 × 1018
23.7
Let n be the number of photons falling per second on the
target.
hc
Energy of each photon =
λ
Total energy falling per second on target
nhc
=
= IA
λ
=
IAλ 100 × 10 −4 × 660 × 10 −9
= 3.3 × 1016
=
−
34
8
hc (6.62 × 10 ) × (3 × 10 )
Number of electrons emitted per second by all the atoms in
the target if one electron is emitted by each atom for one
incident photon.
= (2.54 × 1018) (3.3 × 1016) = 8.4 × 1034
Expected photocurrent
= (8.4 × 1034) (1.6 × 10–19) = 1.34 × 1016A
∴ n=
Observed photocurrent = 100 mA
Probability of photo emission by single photon incident on
a single atom
100 µ A
P=
= 7.5 × 10 −21
16
1.34 × 10 A
Thus the probability of emission by single photon on a single
atom is very much less than 1, the probability of absorption
of two photons by single atoms is negligible.
v  m A + mB 
…(iv)
=
v A  m A − mB 
Initial wavelength of the particle A, i.e., (lA)i
h
=
m Av
Final wavelength of the particle B, i.e., (lA)f
h
=
m Av A
Change in wavelength,
h
h
Dl = (lA)f – (lA)i =
−
m Av A m Av
or
Dl =
h
m Av
v

 v − 1 …(v)
 A 
From eqns. (iv) and (v),
Dl =

h   m A + mB  
−1
m Av  m A − mB  
CHAPTER
12
Atoms
Recap Notes
Rutherford’s a-scattering experiment
X Rutherford and his two associates, Geiger
and Marsden, studies the scattering of the
a-particles from a thin gold foil in order to
investigate the structure of the atom.
at various angles from 0 to p. This
shows that atom has a small positively
charged core called ‘nucleus’ at centre
of atom, which deflects the positively
charged a-particles at different angles
depending on their distance from
centre of nucleus.
– Very few a-particles (1 in 8000) suffers
deflection of 180°. This shows that size
of nucleus is very small, nearly 1/8000
times the size of atom.
N()
Thomson’s model of
atom : It was proposed
by J. J. Thomson in
1898. According to this
model, the positive charge of the atom is
uniformly distributed throughout the volume
of the atom and the negatively charged
electrons are embedded in it like seeds in a
watermelon.
0°
= 180° his graph shows deflection of number of
T
particles with angle of deflection q.
X
Rutherford’s observations and results :
– Most of the a-particles pass through
the gold foil without any deflection.
This shows that most of the space in an
atom is empty.
– Few a-particles got scattered, deflecting
Rutherford’s a-scattering formulae
X Number of a particles scattered per unit
area, N(q) at scattering angle q varies
inversely as sin4(q/2),
1
i.e., N(θ) ∝ 4
sin (θ / 2)
X Impact parameter : It is defined as
the perpendicular distance of the initial
velocity vector of the alpha particle from
the centre of the nucleus, when the
particle is far away from the nucleus of
the atom.
– The scattering angle q of the a particle
and impact parameter b are related as
Ze 2 cot(θ / 2)
b=
4 πε0 K
where K is the kinetic energy of the
a-particle and Z is the atomic number of
the nucleus.
83
Atoms
X
– Smaller the impact parameter, larger
the angle of scattering q.
Distance of closest approach : At the
distance of closest approach whole kinetic
energy of the alpha particles is converted
into potential energy.
– Distance of closest approach
2Ze 2
r0 =
4 πε0 K
=
eV
n2
The potential energy of the electron in
the nth orbit
2
 1  4 π2 me 4 Z 2
1 Ze 2
Un = −
= −
4 πε0 rn
 4 πε0 
n2 h2
=
Rutherford’s nuclear model of the atom :
According to this model the entire positive
charge and most of the mass of the atom is
concentrated in a small volume known as the
nucleus with electrons revolving around it
just as planets revolve around the sun.
X
Bohr’s model : Bohr combined classical and early
quantum concepts and gave his theory of hydrogen
and hydrogen-like atoms which have only one orbital
electron. His postulates are
X An electron can revolve around the nucleus
only in certain allowed circular orbits of definite
energy and in these orbits it does not
radiate. These orbits are known as
stationary orbits.
X Angular momentum of the electron in a
stationary orbit is an integral multiple of
h/2p.
nh
nh
i.e., L =
or mvr =
2π
2π
This is known as Bohr’s quantisation condition.
X The emission of radiation takes place
when an electron makes a transition from
a higher to a lower orbit. The frequency of
the radiation is given by
E − E1
υ= 2
h
where E2 and E1 are the energies of the
electron in the higher and lower orbits
respectively.
X
Bohr’s formulae
X Radius of nth orbit
4 πε0n2 h2
0.53n2
; rn =
rn =
Å
Z
4 π2 mZe 2
X Velocity of the electron in the nth orbit
1 2 πZe 2 2.2 × 106 Z
m/s
vn =
=
4 πε0 nh
n
The kinetic energy of the electron in the
nth orbit
2
 1  2 π2 me 4 Z 2
1 Ze 2
Kn =
=
4 πε0 2rn
 4 πε0 
n2 h 2
13.6Z 2
−27.2Z 2
n2
eV
Total energy of electron in the nth orbit
2
 1  2 π2 me 4 Z 2
En = U n + K n = − 
 4 πε0 
n2 h2
2
13.6Z
=−
eV
n2
Kn = –En, Un = 2En = –2Kn
Frequency of the electron in the nth orbit
2
 1  4 π2 Z 2 e 4 m 6.62 × 1015 Z 2
=
υn = 
 4 πε0 
n3h3
n3
X
Wavelength of radiation in the transition
1
1
1
from n2 → n1 is given by = RZ 2  2 − 2 
λ
 n1 n2 
where R is called Rydberg’s constant.
 1  2 π2 me 4
R=
= 1.097 × 107 m −1

3
 4 πε0  ch
Spectral series of hydrogen atom :
When the electron in a H-atom jumps from
higher energy level to lower energy level,
the difference of energies of the two energy
levels is emitted as radiation of particular
wavelength, known as spectral line. Spectral
lines of different wavelengths are obtained
for transition of electron between two
different energy levels, which are found to
fall in a number of spectral series given by
X
Lyman series
– Emission spectral lines corresponding
to the transition of electron from higher
energy levels (n2 = 2, 3, ...,∞) to first energy
level (n1 = 1) constitute Lyman series.
1
1
1
= R  2 − 2  where n2 = 2, 3, 4, ......,∞
λ
1 n2 
CBSE Board Term-II Physics Class-12
84
– Series limit line (shortest wavelength)
of Lyman series is given by
1 
1
1
1
= R  2 − 2  = R or λ =
λ
R
1
∞ 
– The first line (longest wavelength) of
the Lyman series is given by
1
1  3R
4
1
=R 2 − 2=
or λ =
λ
3R
1
2  4
X
– Lyman series lie in the ultraviolet
region of electromagnetic spectrum.
– Lyman series is obtained in emission
as well as in absorption spectrum.
Balmer series
– Emission spectral lines corresponding
to the transition of electron from higher
energy levels (n2 = 3, 4, ....∞) to second energy
level (n1 = 2) constitute Balmer series.
1
1
1
= R  2 − 2  where n2 = 3, 4, 5.........,∞
λ
n2 
2
– Series limit line (shortest wavelength)
of Balmer series is given by
1
1  R
4
1
= R  2 − 2  = or λ =
λ
R
2
∞  4
– The first line (longest wavelength) of
the Balmer series is given by
1
1  5R
36
1
=R 2 − 2=
or λ =
λ
5R
3  36
2
X
– Balmer series lie in the visible region of
electromagnetic spectrum.
– This series is obtained only in emission
spectrum.
Paschen series
– Emission spectral lines corresponding
to the transition of electron from
higher energy levels (n2 = 4, 5, .....,∞)
to third energy level (n1 = 3) constitute
Paschen series.
1
1
1
= R  2 − 2  where n2 = 4, 5, 6.........,∞
λ
3
n

2
– Series limit line (shortest wavelength)
of the Paschen series is given by
1
1  R
9
1
=R 2 − 2=
or λ =
λ
R
3
∞  9
The first line (longest wavelength) of the
Paschen series is given by
1
1  7R
144
1
=R 2 − 2=
or λ =
λ
144
7R
3
4 
– Paschen series lie in the infrared region
of the electromagnetic spectrum.
– This series is obtained only in the
emission spectrum.
Brackett series
– Emission spectral lines corresponding
to the transition of electron from higher
energy levels (n2 = 5, 6, 7, ..... ,∞) to
fourth energy level (n1 = 4) constitute
Brackett series.
1
1
1
= R  2 − 2  where n2 = 5, 6, 7..........,∞
λ
n2 
4
Series limit line (shortest wavelength) of
Brackett series is given by
1
1  R
16
1
=R 2 − 2=
or λ =
λ
R
4
∞  16
– The first line (longest wavelength) of
Brackett series is given by
1
1  9R
400
1
=R 2 − 2=
or λ =
9R
λ
4
5  400
– Brackett series lie in the infrared region
of the electromagnetic spectrum.
– This series is obtained only in the
emission spectrum.
X Pfund series
– Emission spectral lines corresponding
to the transition of electron from higher
energy levels (n2 = 6, 7, 8,.......,∞) to
fifth energy level (n1 = 5) constitute
Pfund series.
1
1
1
= R  2 − 2  where n2 = 6, 7,...........,∞
λ
n2 
5
X
– Series limit line (shortest wavelength)
of Pfund series is given by
1
1  R
25
1
=R 2 − 2=
or λ =
λ
R
5
∞  25
– The first line (longest wavelength) of
the Pfund series is given by
1
1  11R
900
1
=R 2 − 2=
or λ =
λ
11R
5
6  900
– Pfund series also lie in the infrared
region of electromagnetic spectrum.
– This series is obtained only in the
emission spectrum.
X Number of spectral lines due to transition
of electron from nth orbit to lower orbit is
n(n − 1)
N=
2
Ionization energy and ionization potential
X Ionisation : The process of knocking
an electron out of the atom is called
13.6
ionisation. ionisation energy = 2 eV
n
X Ionisation energy : The energy required, to
knock an electron completely out of the
atom.
13.6Z 2
X Ionisation potential =
V
n2
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. The first model of atom in 1898 was proposed
by
(a) Ernest Rutherford (b) Albert Einstein
(c) J. J. Thomson
(d) Niels Bohr
2. The transition from the state n = 3 to n = 1
in a hydrogen like atom results in ultraviolet
radiation. Infrared radiation will be obtained in
the transition from
(a) 2 → 1
(b) 3 → 2
(c) 4 → 2
(d) 4 → 3
3. In a hydrogen atom the total energy of
electron is
e2
4πε 0 r
−e2
(c)
8πε 0 r
(a)
(b)
−e2
4πε 0 r
(d)
e2
8πε 0 r
4. The relation between the orbit radius and
the electron velocity for a dynamically stable
orbit in a hydrogen atom is (where, all notations
have their usual meanings)
(a) v =
4 π ε0
me2 r
(b) r =
e2
4πε 0 v
(c) v =
e2
4πε 0 mr
(d) r =
v e2
4πε 0 m
5. In the Geiger-Marsden scattering experiment
the number of scattered particles detected are
maximum and minimum at the scattering angles
respectively at
(a) 0°° and 180°°
(b) 180°° and 0°°
(c) 90°° and 180°°
(d) 45°° and 90°°
postulate applies to a satellite just as it does
to an electron in the hydrogen atom, then the
quantum number of the orbit of satellite is
(a) 5.3 × 1040
(b) 5.3 × 1045
48
(c) 7.8 × 10
(d) 7.8 × 1050
8. Which of the following is not correct about
Bohrs model of the hydrogen atom?
(a) An electron in an atom could revolve in
certain stable orbits without the emission
of radiant energy.
(b) Electron revolves around the nucleus only in
those orbits for which angular momentum
nh
Ln =
.
2π
(c) When electron make a transition from one of
its stable orbit to lower orbit then a photon
emitted with energy hu = Ef – Ei.
(d) Bohr model is applicable to all atoms.
9. The shortest wavelength present in the
Paschen series of spectral lines is
(a) 720 nm
(b) 790 nm
(c) 800 nm
(d) 820 nm
10. In a Geiger-Marsden experiment. Find the
distance of closest approach to the nucleus of a
7.7 MeV a-particle before it comes momentarily
to rest and reverses its direction.
(Z for gold nucleus = 79)
(a) 10 fm
(b) 20 fm
(c) 30 fm
(d) 40 fm
6. Rutherford’s experiments suggested that the
size of the nucleus is about
(a) 10–14 m to 10–12 m (b) 10–15 m to 10–13 m
(c) 10–15 m to 10–14 m (d) 10–15 m to 10–12 m
11. If an electron in hydrogen atom is revolving
in a circular track of radius 5.3 × 10–11 m with
a velocity of 2.2 × 106 m s–1 around the proton
then the frequency of electron moving around
the proton is
(a) 6.6 × 1012 Hz
(b) 3.3 × 1015 Hz
12
(c) 3.3 × 10 Hz
(d) 6.6 × 1015 Hz
7. A 10 kg satellite circles earth once every
2 h in an orbit having a radius of 8000 km.
Assuming that Bohr ’s angular momentum
12. The Rydberg formula, for the spectrum of
the hydrogen atom where all terms have their
usual meaning is
CBSE Board Term-II Physics Class-12
86
4
(a) hυif = me
2 2
8ε 0 h
4
(b) hυif = me
8 ε 0 2 h2
 1 1
n −n 
 f
i
 1
1 
 2 − 2 
ni 
 nf
(c) hυif =
8 ε 0 2 h2
me4
 1 1
n −n 
 f
i
(d) hυif =
8 ε 0 2 h2
me4
 1 1
 2 − 2
 nf ni 
13. In which of the following systems will the
radius of the first orbit (n = 1) be minimum ?
(a) Doubly ionized lithium.
(b) Singly ionized helium.
(c) Deuterium atom.
(d) Hydrogen atom.
14. An electron in a hydrogen atom makes a
transition from n = n1 to n = n2. The time period
of the electron in the initial state is eight times
that in the final state. The possible values of n1
and n2 are
(a) n1 = 4, n2 = 2
(b) n1 = 8, n2 = 2
(c) n1 = 8, n2 = 1
(d) n1 = 6, n2 = 2
15. Energy is absorbed in the hydrogen atom
giving absorption spectra when transition takes
place from
(a) n = 1 → n′ where n′ > 1
(b) n = 2 → 1
(c) n′ → n
(d) n → n′ = ∞
16. The value of ionisation energy of the
hydrogen atom is
(a) 3.4 eV
(b) 10.4 eV
(c) 12.09 eV
(d) 13.6 eV
17. The moment of momentum for an electron
in second orbit of hydrogen atom as per Bohr’s
model is
2h
π
h
(a)
(b) 2h
(c)
(d)
π
h
π
18. In the Geiger-Marsden scattering experiment,
in case of head-on collision the impact parameter
should be
(a) maximum
(b) minimum
(c) infinite
(d) zero
19. In an atom the ratio of radius of orbit of
electron to the radius of nucleus is
(a) 103
(c) 105
(b) 104
(d) 106
20. The radius of nth orbit rn in terms of Bohr
radius (a0) for a hydrogen atom is given by the
relation
(a) na0
(b) n a0
2
(c) n a0
(d) n3a0
21. In the Bohr model of the hydrogen atom, the
lowest orbit corresponds to
(a) infinite energy
(b) maximum energy
(c) minimum energy
(d) zero energy.
22. In an experiment on a-particle scattering,
a-particles are directed towards a gold foil and
detectors are placed in position P,Q and R. What
is the distribution of a-particles as recorded at
P, Q and R ?
(a)
(b)
(c)
(d)
P
all
none
a few
most
Q
none
none
some
some
R
none
all
most
a few
23. From quantisation of angular momentum,
one gets for hydrogen atom, the radius of the
2
 n2   h   4 π 2 ε 0 
nth orbit as rn = 
   2 
 me   2π   e 
For a hydrogen like atom of atomic number Z,
(a) the radius of the first orbit will be the same
(b) rn will be greater for larger Z values
(c) rn will be smaller for larger Z values
(d) none of these.
24. Bohr’s basic idea of discrete energy levels
in atoms and the process of emission of photons
from the higher levels to lower levels was
experimentally confirmed by experiments
performed by
(a) Michelson–Morley (b) Millikan
(c) Joule
(d) Franck and Hertz
25. The binding energy of an electron in the
ground state of He is equal to 24.6 eV. The
energy required to remove both the electrons is
(a) 49.2 eV
(b) 54.4 eV
(c) 79 eV
(d) 108.8 eV
87
Atoms
26. Out of the following which one is not a
possible energy for a photon to be emitted
by hydrogen atom according to Bohr’s atomic
model?
(a) 0.65 eV
(b) 1.9 eV
(c) 11.1 eV
(d) 13.6 eV
27. The ratio of the speed of the electron in the
ground state of hydrogen atom to the speed of
light in vacuum is
1
2
1
1
(b)
(c)
(d)
137
237
2
237
28. The diagram shows the energy levels for an
electron in a certain atom.
Which transition shown represents the emission
of a photon with the most energy ?
(a)
(a) I
(b) II
(c) III
(d) IV
29. If u1 is the frequency of the series limit of
Lyman series, u2 is the frequency of the first line
of Lyman series and u3 is the frequency of the
series limit of the Balmer series, then
(a) u1 – u2 = u3
(b) u1 = u2 – u3
(c)
1
1
1
=
+
υ2 υ1 υ3
(d)
1
1
1
=
+
υ1 υ2 υ3
30. Suppose an electron is attracted towards the
origin by a force k/r, where k is a constant and r
is the distance of the electron from the origin. By
applying Bohr model to this system, the radius
of nth orbit of the electron is found to be rn and
the kinetic energy of the electron is found to be
Tn. Then which of the following is true?
1
(a) Tn ∝ 2
n
(b) Tn is independent of n; rn ∝ n
1
(c) Tn ∝ ; rn ∝ n
n
1
(d) Tn ∝ and rn ∝ n2
n
31. The first line of the Lyman series in a
hydrogen spectrum has a wavelength of 1210 Å.
The corresponding line of a hydrogen-like atom
of Z = 11 is equal to
(a) 4000 Å (b) 100 Å (c) 40 Å
(d) 10 Å
32. The de-Broglie wavelength of an electron in
the first Bohr orbit is
(a) equal to one-fourth the circumference of the
first orbit
(b) equal to half the circumference of first orbit
(c) equal to twice the circumference of first
orbit
(d) equal to the circumference of the first orbit.
33. The excitation energy of Lyman last line is
(a) the same as ionisation energy
(b) the same as the last absorption line in
Lyman series
(c) both (a) and (b)
(c) different from (a) and (b)
34. If the wavelength of the first line of the Balmer
series of hydrogen is 6561 Å, the wavelength of the
second line of the series should be
(a) 13122 Å
(b) 3280 Å
(c) 4860 Å
(d) 2187 Å
35. The electric current I created by the electron
in the ground state of H atom using Bohr model
in terms of Bohr radius (a 0 ) and velocity of
electron in first orbit v0 is
v0
2πa
ev0
2πa
(a)
(b)
(c)
(d)
ev
2πa
v0
2πa0
0
36. If the radius of inner most electronic orbit
of a hydrogen atom is 5.3 × 10–11 m, then the
radii of n = 2 orbit is
(a) 1.12 Å
(b) 2.12 Å
(c) 3.22 Å
(d) 4.54 Å
37. The wavelength limit present in the Pfund
series is (R = 1.097 × 107 m–1)
(a) 1572 nm
(b) 1898 nm
(c) 2278 nm
(d) 2535 nm
38. An electron is revolving in the nth orbit of
radius 4.2 Å, then the value of n is (r1 = 0.529 Å)
(a) 4
(b) 5
(c) 6
(d) 3
39. A hydrogen atom initially in the ground
level absorbs a photon and is excited to n = 4
level then the wavelength of photon is
(a) 790 Å
(b) 870 Å
(c) 970 Å
(d) 1070 Å
40. If muonic hydrogen atom is an atom in which
a negatively charged muon (m) of mass about
207 me revolves around a proton, then first Bohr
radius of this atom is (re = 0.53 × 10–10 m)
(a) 2.56 × 10–10 m
(b) 2.56 × 10–11 m
(c) 2.56 × 10–12 m
(d) 2.56 × 10–13 m
CBSE Board Term-II Physics Class-12
88
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 41 to 43.
Electron Transitions for the Hydrogen Atom
Read the Bohr’s model explains the spectral lines
of hydrogen atomic emission spectrum. While the
electron of the atom remains in the ground state,
its energy is unchanged. When the atom absorbs
one or more quanta of energy, the electrons
moves from the ground state orbit to an excited
state orbit that is farther away.
the orbit and n = 1, 2, 3, .... When transition takes
place from Kth orbit to Jth orbit, energy photon is
emitted. If the wavelength of the emitted photon
1
1 
 1
= R  2 − 2  , where R is
is l, we find that
λ
K 
J
Rydberg’s constant.
On a different planet, the hydrogen atom’s
structure was somewhat different from ours. The
angular momentum of electron was P = 2n(h/2p),
i.e., an even multiple of (h/2p).
e–
m
Ze
The given figure shows an energy level diagram
of the hydrogen atom. Several transitions are
marked as I, II, III and so on. The diagram is
only indicative and not to scale.
41. In which transition is a Balmer series photon
absorbed?
(a) II
(b) III
(c) IV
(d) VI
42. The wavelength of the radiation involved in
transition II is
(a) 291 nm
(b) 364 nm
(c) 487 nm
(d) 652 nm
43. Which transition will occur when a hydrogen
atom is irradiated with radiation of wavelength
1030 nm?
(a) I
(b) II
(c) IV
(d) V
Case II : Read the passage given below and
answer the following questions from 44 to 48.
Second Postulate of Bohr’s Theory
Hydrogen is the simplest atom of nature. There
is one proton in its nucleus and an electron moves
around the nucleus in a circular orbit. According
to Niels Bohr, this electron moves in a stationary
orbit. When this electron is in the stationary
orbit, it emits no electromagnetic radiation. The
angular momentum of the electron is quantized,
i.e., mvr = (nh/2p), where m = mass of the electron,
v = velocity of the electron in the orbit, r = radius of
F
vn
rn
44. The minimum permissible radius of the
orbit will be
ε 0 h2
4 ε 0 h2
ε 0 h2
2ε 0 h 2
(a)
(b)
(c)
(d)
mπe2
mπe2
2mπe2
mπe2
45. In our world, the velocity of electron is v0
when the hydrogen atom is in the ground state.
The velocity of electron in this state on the other
planet should be
(a) v0
(b) v0/2
(c) v0/4
(d) v0/8
46. In our world, the ionization potential energy
of a hydrogen atom is 13.6 eV. On the other
planet, this ionization potential energy will be
(a) 13.6 eV
(b) 3.4 eV
(c) 1.5 eV
(d) 0.85 eV
47. Check the correctness of the following
statements about the Bohr model of hydrogen
atom.
(i) The acceleration of the electron in n = 2 orbit
is more than that in n = 1 orbit.
(ii) The angular momentum of the electron in
n = 2 orbit is more than that in n = 1 orbit.
(iii) The kinetic energy of the electron in n = 2
orbit is less than that in n = 1 orbit.
89
Atoms
(a)
(b)
(c)
(d)
Only (iii) and (i) are correct.
Only (i) and (ii) are correct.
Only (ii) and (iii) are correct.
All the statements are correct.
48. In Bohr’s model of hydrogen atom, let PE
represent potential energy and TE the total
energy. In going to a higher orbit
(a) PE increases, TE decreases
(b) PE decreases, TE increases
(c) PE increases, TE increases
(d) PE decreases, TE decreases
Case III : Read the passage given below and
answer the following questions from 49 to 52.
Hydrogen Emission Spectrum
Hydrogen spectrum consists of discrete bright
lines in a dark background and it is specifically
known as hydrogen emission spectrum. There
is one more type of hydrogen spectrum that
exists where we get dark lines on the bright
background, it is known as absorption spectrum.
Balmer found an empirical formula by the
observation of a small part of this spectrum and
it is represented by
1
1
 1
= R  2 − 2  , where n = 3, 4, 5, ..... .

λ
n 
2
For Lyman series, the emission is from first state
to nth state, for Paschen series, it is from third
state to nth state, for Brackett series, it is from
fourth state to nth state and for Pfund series, it is
from fifth state to nth state.
49. Number of spectral lines in hydrogen atom
is
(a) 8
(b) 6
(c) 15
(d) ∞
50. W h i c h s e r i e s o f h yd r o ge n s pectru m
corresponds to ultraviolet region?
(a) Balmer series
(b) Brackett series
(c) Paschen series
(d) Lyman series
51. Which of the following lines of the H-atom
spectrum belongs to the Balmer series?
(a) 1025 Å
(b) 1218 Å
(c) 4861 Å
(d) 18751 Å
52.
(a)
(b)
(c)
(d)
Rydberg constant is
a universal constant
same for same elements
different for different elements
none of these.
Assertion & Reasoning Based MCQs
For question numbers 53-60, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
53. Assertion (A) : The force of repulsion
between atomic nucleus and a-particle varies
with distance according to inverse square law.
Reason (R) : Rutherford did a-particle scattering
experiment.
54. Assertion (A) : According to classical theory,
the proposed path of an electron in Rutherford
atom model will be circular.
Reason (R) : According to electromagnetic theory
an accelerated particle continuously emits
radiation.
55. Assertion (A) : Between any two given energy
levels, the number of absorption transitions
is always less than the number of emission
transitions.
Reason (R) : Absorption transitions start from
the lowest energy level only and may end at any
higher energy level. But emission transitions
may start from any higher energy level and end
at any energy level below it.
56. Assertion (A) : Electrons in the atom are held
due to coulomb forces.
Reason (R) : The atom is stable only because
the centripetal force due to Coulomb’s law is
balanced by the centrifugal force.
57. Assertion (A) : Total energy of revolving
electron in any stationary orbit is negative.
Reason (R) : Energy is a scalar quantity. It can
have positive or negative value.
58. Assertion (A) : Balmer series lies in the
visible region of electromagnetic spectrum.
CBSE Board Term-II Physics Class-12
90
1
1 
 1
Reason (R) :
= R  2 − 2  , where K = 3, 4,

λ
2
K 
5, ...
59. Assertion (A) : For the scattering of a-particles
at large angles, only the nucleus of the atom is
responsible.
Reason (R) : Nucleus is very heavy in comparison
to electrons.
60. Assertion (A) : Fraunhofer lines are observed
in the spectrum of the sun.
Reason (R) : The different elements have different
spectra.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. Why is the classical (Rutherford) model for
an atom of electron orbiting around the nucleus
not able to explain the atomic structure?
2. When is Hα line of the Balmer series in the
emission spectrum of hydrogen atom obtained?
3. What is the maximum number of spectral
lines emitted by a hydrogen atom when it is in
the third excited state?
4. Find the radius and energy of a He+ ion in
the states n = 2.
5. State Bohr ’s quantization condition of
angular momentum.
6.
Find the wavelength of the IInd line of
Balmer series.
7.
What is the wavelength of the electronic de
Broglie wave in the 3rd orbit of hydrogen?
8.
Find the radius of the ground state orbit of
hydrogen atom.
9.
Find the speed in the ground state
10. Calculate the shortest wavelength of the
Brackett series and state to which part of the
electromagnetic spectrum does it belong.
Short Answer Type Questions (SA-I)
11. Find the number of unique radiations that
can be emitted for a sample of hydrogen atoms
excited to the nth level.
12. An electron in the hydrogen atom makes
a transition from n = 2 energy state to the
ground state (corresponding to n = 1). Find the
wavelength and frequency of the emitted photon.
13. Use the Bohr ’s model to estimate the
wavelength of the Kα line in the X-ray spectrum
of platinum (Z = 78).
14. The radius of the innermost electron orbit
of a hydrogen atom is 5.3 × 10–11 m. What are
the radii of the n = 2 and n = 3 orbits?
15. If Bohr’s quantisation postulate (angular
momentum = nh/2π) is a basic law of nature,
it should be equally valid for the case of planetary
motion also. Why then do we never speak of
quantisation of orbits of planets around the sun?
16. Would the Bohr formula for the H-atom
remain unchanged if proton had a charge
(+4/3)e and electron a charge (–3/4)e, where
e = 1.6 × 10–19C. Give reasons for your answer.
17. Consider a gas consisting Li ++ (which is
hydrogen like ion).
(i) Find the wavelength of radiation required
to excite the electron in Li++ from n = 1 and
n = 3.
(ii) How many spectral lines are observed in
the emission spectrum of the above excited
system?
18. Using Bohr model, calculate the electric
current created by the electron when the H-atom
is in the ground state.
19. The kinetic energy of the electron orbiting in
the first excited state of hydrogen atom is 3.4 eV.
Determine the de Broglie wavelength associated
with it.
20. Calculate the shortest wavelength in the
Balmer series of hydrogen atom. In which
region (infrared), visible, ultraviolet of hydrogen
spectrum does this wavelength lie?
91
Atoms
Short Answer Type Questions (SA-II)
21. Assume that their is no repulsive force
between the electrons in an atom but the
force between positive and negative charges is
given by Coulomb’s law as usual. Under such
circumstances, calculate the ground state energy
of a He-atom.
22. Show that the first few frequencies of light
that is emitted when electrons fall to the n th
level from levels higher than n, are approximate
harmonics (i.e. in the ratio 1 : 2 : 3...) when
n > >1.
23. What is the minimum energy that must be
given to a H atom in ground state so that it can
emit an Hγ line in Balmer series. If the angular
momentum of the system is conserved, what
would be the angular momentum of such H γ
photon?
24. The electron in a given Bohr orbit has a total
energy of –1.5 eV. Calculate its
(i) kinetic energy. (ii) potential energy.
(iii) wavelength of radiation emitted, when this
electron makes a transition to the ground
state.
[Given : Energy in the ground state = –13.6 eV
and Rydberg’s constant = 1.09 × 107 m–1]
25. A 12.5 eV electron beam is used to bombard
gaseous hydrogen at room temperature. Upto
which energy level the hydrogen atoms would
be excited?
Calculate the wavelengths of the first member of
Lyman and first member of Balmer series.
26. The value of ground state energy of hydrogen
atom is –13.6 eV.
(i) Find the energy required to move an electron
from the ground state to the first excited
state of the atom.
(ii) Determine (a) the kinetic energy and
(b) orbital radius in the first excited state of
the atom.
(Given the value of Bohr radius = 0.53 Å).
27. (i)
How does de-Broglie hypothesis explain
Bohr ’s quantization condition for
stationary orbits ?
(ii) Find the relation between the three
wavelengths l1, l2 and l3 from the energy
level diagram shown in the figure.
28. Show that the radius of the orbit in hydrogen
atom varies as n 2 , where n is the principal
quantum number of the atom.
29. Direction : Read the following passage and
answer the questions given below.
In 1911, Rutherford, along with his assistants,
H. Geiger and E. Marsden, performed the alpha
particle scattering experiment. H. Geiger and
E. Marsden took radioactive source (214
83Bi) for
a-particles. A collimated beam of a-particles
of energy 5.5 MeV was allowed to fall on
2.1 × 10–7 m thick gold foil. Observations of this
experiment are as follows
(I) Most of the a-particles passed through the
foil without deflection.
(II) Only about 0.14% of the incident a-particles
scattered by more than 1°.
(III) Only about one a-particle in every 8000
a-particles deflected by more than 90°.
Most -particles pass
straight through
-particles
About 1 in 8000 -particles
is repelled back
(i) Gold foil used in Geiger-Marsden experiment
is about 10–8 m thick. What does it ensures?
(ii) On which factor, the trajectory traced by an
a-particle depends?
(iii) In Rutherford scattering experiment the
fact that only a small fraction of the number
of incident particles rebound back. What it
indicates?
30. Positronium is just like a H-atom with
the proton replaced by the positively charged
antiparticle of the electron (called the positron
which is as massive as the electron). What would
be the ground state energy of positronium?
CBSE Board Term-II Physics Class-12
92
31. The photons from Balmer series in hydrogen
spectrum having wavelength between 450 nm to
700 nm are incident on a metal surface of work
function 2 eV. Find the maximum kinetic energy
of one photoelectron.
32. A s m a l l p a r t i c l e o f m a s s m m o v e s
in such a way that the potential energy
U = ar2 where a is a constant and r is the distance
of the particle from the origin. Assuming Bohr’s
model of quantisation of angular momentum for
circular orbits, find the radius of nth allowed
orbit.
33. A particle known as µ-meson, has a charge
equal to that of an electron and mass 208 times
the mass of the electron. It moves in a circular
orbit around a nucleus of charge +3e. Take the
mass of the nucleus to be infinite. Assuming that
the Bohr’s model is applicable to this system,
(i) derive an expression for the radius of the
nth Bohr orbit,
(ii) find the value of n for which the radius of
the orbit is approximately the same as that
of the first Bohr orbit for a hydrogen atom.
34. A gas of hydrogen like atoms can absorb
radiations of 68 eV. Consequently, the atoms emit
radiations of only three different wavelength. All
the wavelengths are equal or smaller than that
of the absorbed photon.
(i) Determine the initial state of the gas atoms.
(ii) Identify the gas atoms.
(iii) Find the minimum wavelength of the
emitted radiations.
Long Answer Type Questions (LA)
35. In the Auger process an atom makes a
transition to a lower state without emitting a
photon. The excess energy is transferred to an
outer electron which may be ejected by the atom.
(This is called an Auger electron). Assuming the
nucleus to be massive, calculate the kinetic
energy of an n = 4 Auger electron emitted by
Chromium by absorbing the energy from a
n = 2 to n = 1 transition.
37. Using Bohr ’s postulates, derive the
expression for the frequency of radiation emitted
when electron in hydrogen atom undergoes
transition from higher energy state (quantum
number ni) to the lower state, (nf). When electron
in hydrogen atom jumps from energy state
ni = 4 to nf = 3, 2, 1. Identify the spectral series
to which the emission lines belong.
36. (a) Write two important limitations of
Rutherford model which could not explain the
observed features of atomic spectra.
(b) How were these explained in Bohr’s model
of hydrogen atom?
38. The first four spectral lines in the Lyman
series of a H-atom are λ = 1218 Å, 1028 Å, 974.3 Å
and 951.4 Å. If instead of Hydrogen, we consider
Deuterium, calculate the shift in the wavelength
of these lines.
OBJECTIVE TYPE QUESTIONS
1. (c) : Sir J. J. Thomson proposed the first model of
atom called plum pudding model of atom. According to
this model, the positive charge of the atom is uniformly
distributed throughout the volume of the atom and the
negatively charged electrons are embedded in it like seeds
in a watermelon.
2.
(d)
3.
(c) : The kinetic energy of the electron in hydrogen atom are
e2
1
K = mv 2 =
2
8 πε 0 r

e2 
2
∵ v =

4 πε 0 mr 

Electrostatic potential energy,
− e2
and U =
4 π ε0 r
⇒
The total energy E of the electron in a hydrogen atom is
E=K+U
 −e 2 
e2
e2
=
−
E =
+
8 π ε 0 r  4 π ε 0 r 
8 π ε0 r
Here negative sign shows that electron is bound to the
nucleus.
4. (c) : In hydrogen atom electrostatic force of attraction
(Fe) is acting between the revolving electrons and the nucleus
93
Atoms
provides the requisite centripetal force (Fc) to keep them in
their orbits. Thus,
Fe = Fc
mv 2
1 e2
∴
=
r
4 π ε0 r 2
e2
e2
⇒ v =
4 π ε0 m r
4 π ε0 m r
5. (a) : The number of scattered particles detected will be
maximum at the angle of scattering q = 0° and minimum at
q = 180°.
6. (c)
7. (b) : Here, m = 10 kg, r n = 8 × 106 m
T = 2 × 60 × 60 = 7200 s
nh
2π rn
Velocity of nth orbit, v n =
and from m v n rn =
2π
T
2π
2π rn
n=
×m×
× rn
h
T
m
(2π × 8 × 106 )2 × 10
= (2 π rn )2 ×
=
= 5.3 × 1045.
T × h 7200 × 6.64 × 10 −34
or v 2 =
8. (d) : The first three options (a), (b) and (c) are Bohr’s
postulates of atomic model whereas option (d) is not correct
as Bohr’s model is applicable to hydrogen atom only.
1
1 1
9. (d) : The wavelength for Paschen series, = R  2 − 2 
λ
3 n 
For shortest wavelength n = ∞
1
1 R
1
= R − 2 =
\
λ
9 ∞  9
9
9
λ= =
R 1.097 × 107
= 8.20 × 10–7 m = 820 nm.
10. (c) : Let d be the distance of closest approach then by
the conservation of energy,
Initial kinetic energy of incoming a-particle, K
= Final electric potential energy U of the system
(2e )( Ze )
1
×
As K =
4π ε0
d
1 2 Z e2
...(i)
4 π ε 0 K 1
= 9 × 109 N m2 C−2 , Z = 79, e = 1.6 × 10 −19 C.
Here,
4 π ε0
∴
d =
K = 7.7 MeV = 7.7 × 106 × 1.6 × 10 –19 J = 1.2 × 10–12 J
Substituting these values in (i)
2 × 9 × 109 × (1.6 × 10 −19 )2 × 79
d=
1.2 × 10 −12
d = 3 × 10–14 m
= 30 fm
(∵ 1 fm = 10 −15 m)
11. (d) : Frequency of the electron moving around the
proton is
velocity of electron (v )
υ=
circumference (2 π r )
Here, v = 2.2 × 106 m s–1 and r = 5.30 × 10–11 m
∴
υ=
2.2 × 106
2 × 3.14 × 5.3 × 10 −11
u = 6.6 × 1015 Hz.
12. (b)
1
,
Z
For doubly ionized lithium, Z (= 3) will be maximum, hence
for doubly ionized lithium, r will be minimum.
14. (a) : In the nth orbit, let rn = radius and vn = speed of
electron.
2πr n rn
Time period, Tn =
∝ .
vn
vn
13. (a) : Radius of first orbit, r ∝
Now, rn ∝ n 2 and v n ∝
1
n
rn
∝ n 3 or Tn ∝ n 3 .
vn
3
n 
Here, 8 =  1 
 n2 
∴
or
n1
= 2 or n1 = 2n2
n2
15. (a) : Absorption is from the ground state n = 1 to n′
where n′ > 1.
16. (d) : The minimum energy required to free the electron
from the ground state of the hydrogen atom is called the
ionisation energy of hydrogen atom and its value is 13.6 eV.
17. (a) : According to Bohr’s second postulate
nh
Angular momentum, L =
2π
Angular momentum is also called a moment of momentum.
For second orbit, n = 2
2h h
L=
=
2π π
18. (b) : At minimum impact parameter a particles rebound
back (q ≈ p) and suffers large scattering.
19. (c) : The radius of orbit of electrons = 10–10 m
radius of nucleus = 10–15 m
\
Ratio =
10 −10
= 105
10 −15
Hence the radius of electron orbit is 105 times larger than
the radius of nucleus.
CBSE Board Term-II Physics Class-12
94
20. (c) : The radius of nth orbit
2 4 π ε 0
rn = n 2
me 2
4 π ε0
= a0 (Bohr radius)
m e2
2
where
Hence, rn = n 2 a0 .
21. (c) : In hydrogen atom, the lowest orbit corresponds to
minimum energy.
22. (c)
23. (c) : For an atom with a single electron, Bohr atom
model in applicable.
As the value of attraction force between a proton and electron
e2
is proportional to e2, for an ion with a single electron,
4πε 0
Ze 2
is replaced by
4πε 0
2
n
i.e. rn ∝ .
Z
24. (d) : Franck and Hertz showed that exciting mercury
vapour by electrons of energy 4.9 eV and more, mercury lines
of energy 4.9 eV were obtained.
First they were excited to level B and then thec atoms emitted
spectral lines of 2530 Å i.e. 4.9 eV, Bohr’s concepts were
verified.
25. (c) : The energy needed to remove one electron from the
ground state of He = 24.6 eV.
As the He + is now hydrogen-like, ionisation energy
22
= | −13.6 | 2 eV ⇒ E = 54.4 eV
1
\ To remove both the electrons, energy needed
= (54.4 + 24.6) eV = 79 eV
26. (c) : The energy of nth orbit of hydrogen atom is given
as
13.6
E n = − 2 eV
n
\ E1 = –13.6 eV
E2 = −
13.6
= −3.4 eV
22
\
E3 = −
13.6
= −1.5 eV
32
E4 = −
13.6
= −0.85 eV
42
E3 – E2 = –1.5 – (–3.4) = 1.9 eV
E4 – E3 = –0.85 – (–1.5) = 0.65 eV
27. (c) : Speed of the electron in the ground state of
hydrogen atom is
2πe 2
c
=
= cα
4 πε 0 h 137
where, c = speed of light in vacuum,
v=
e2
is the fine structure constant. It is a pure number
2ε 0 hc
1
.
whose value is
137
v
1
∴ =
c 137
α=
28. (c) : Ist transition is showing absorption of a photon.
From rest of three transitions, III is having maximum energy
from level n = 2 to n = 1
 1 1
∆E ∝  2 − 2 
 n1 n2 
29. (a) : For Lyman series
1 1 
υ = Rc  2 − 2 
1 n 
where n = 2, 3, 4,.......
For the series limit of Lyman series, n = ∞
1
1
∴ υ1 = Rc  2 − 2  = Rc
∞
1


For the first line of Lyman series, n = 2
1 1 3
∴ υ2 = Rc  2 − 2  = Rc
1 2  4 For Balmer series
1 1
υ = Rc  2 − 2 
2 n 
where n = 3, 4, 5....
For the series limit of Balmer series, n = ∞
1  Rc
1
∴ υ3 = Rc  2 − 2  =
 2 ∞  4 From equations (i), (ii) and (iii), we get
u1 = u2 + u3 or u1 – u2 = u3
...(i)
...(ii)
...(iii)
30. (b) : Applying Bohr model to the given system,
mv 2 k
=
rn
rn ...(i)
95
Atoms
and mvrn =
nh
nh
or v =
2π
2πmrn
Put in (i),
m
n 2h 2
k
× 2 22=
rn 4π m rn rn
n 2h 2
...(ii)
rn2 = 2
4 π mk
\ rn2 ∝ n 2 or rn ∝ n
1
n 2h 2
n 2h 2
1
K.E. of the electron, Tn = mv 2 = m 2 2 2 = 2 2
2
2 4 π m rn 8π mrn
Using (ii), we get
n 2h 2 4 π2mk k
Tn =
=
8π2mn 2h 2
2
\ Tn is independent of n.
31. (d) : By Bohr’s formula
1
1 1
= Z 2R  2 − 2 
λ
 n1 n2 
For first line of Lyman series n1 = 1, n2 = 2
1
3
∴
= Z 2R
λ
4
In the case of hydrogen atom, Z = 1
1
3
=R
λ
4
For hydrogen-like atom, Z = 11
1
3
λ ′ 3R
4
1
= 121R
=
×
=
⇒
λ′
4
λ
4 121R × 3 121
λ 1210
λ′ =
=
= 10 Å
121 121
nh
32. (d) : Angular momentum =
2π
nh
⇒ Moment of momentum =
2π
nh
⇒ p × rn =
2π
h
nh
2πr
rn =
⇒ λ= n
λ
2π
n
For 1st orbit, n = 1, λ = 2pr1
⇒ λ = circumference of 1st orbit.
33. (c)
34. (c) : For Balmer series, n1 = 2, n2 = 3 for 1st line and n2
= 4 for second line.
 1 1
−
λ1  22 42  3 / 16 3 36 27
=
= × =
=
λ2  1 1  5 / 36 16 5 20
−
 2

2 32 
λ2 =
20
20
λ1 = × 6561 = 4860 A°
27
27
35. (a) : In the ground state of hydrogen atom, suppose,
a0 = Bohr radius
v0 = velocity of electron in first orbit
\ Time taken by electron to complete one revolution,
2πa0
T=
v0
charge (e ) ev 0
=
\ Current created, I =
time (T ) 2πa0
36. (b) : As, r n = n2a0
Here, a0 = 5.3 × 10–11 m
n = 2
\
r2 = (2)2a0
= 4a0= 4 × 5.3 × 10–11 m
= 21.2 × 10–11 m = 2.12 Å.
37. (c) : The wavelength for Pfund series is given by
1
1
1
= R  2 − 2
n
λ
5


for series limit, n = ∞
1
1 R
1
∴
=R  − 2 =
λ
 25 ∞  25
∴ λ=
25
25
=
= 2278 nm.
R 1.097 × 107
38. (d) : Since r n ∝ n2
rn
n2
= 2
r1 (1)
\
r n = n2 r1
or n 2 =
rn
⇒ n=
r1
rn
=
r1
4.2
= 7.939
0.529
= 2.81 ≈ 3
39. (c) : Here, n1 = 1, and n2 = 4
Energy of photon absorbed, E = E2 – E1
13.6
Since, E n = − 2 eV
n
13.6  13.6 
Then, E 2 − E1 = − 2 −  − 2 
 (1) 
(4)
=−
13.6
+ 13.6 = 12.75 eV
16
= 12.75 × 1.6 × 10–19 J = 20.4 × 10–18 J
hc
E 2 − E1 =
λ
∴ λ=
6.6 × 10 −34 × 3 × 108
hc
=
E 2 − E1
20.4 × 10 −18
= 9.70 × 10–8 m = 970 × 10–10 = 970 Å.
CBSE Board Term-II Physics Class-12
96
1
40. (d) : According to Bohr’s atomic model, r ∝
m
rµ me
⇒
=
re
mµ
Here, re = 0.53 × 10–10 m
mµ = 207 me
me
∴ rµ =
× 0.53 × 10 −10
207 me
...(i)
49. (d) : Number of spectral lines in hydrogen atom is ∞.
50. (d) : Lyman series lies in the ultraviolet region.
(using (i))
= 2.56 × 10–13 m.
41. (d) : For Balmer series, n1 = 2; n2 = 3, 4,…
(lower)
(higher)
Therefore, in transition (VI), photon of Balmer series is
absorbed.
42. (c) : In transition II,
E2 = –3.4 eV, E4 = –0.85 eV,
hc
hc
DE = 2.55 eV ⇒ DE =
⇒l=
= 487 nm
λ
∆E
43. (d) : Wavelength of radiation = 1030 Å
12400
DE =
= 12.0 eV
1030 Å
So, difference of energy should be 12.0 eV (approx.)
Hence for n1 = 1 to n2 = 3
E n3 − E n1 = −1.51 eV − ( −13.6 eV) ≈ 12 eV
Therefore, transition V will occur.
44. (b) : On other planet : mvr = 2n
h
nh
⇒v =
2π
πmr
mv 2
1 e2
mn 2h 2
1 e2
=
⇒ 2 23=
2
r
4 πε0 r
nmr
4 πε0 r 2
Putting n = 1, we get r =
4h 2 ε 0
mπe 2
e2
45. (b) : On our planet : v 0 =
2ε 0 nh
v
e2
On other planet : v =
= 0
2ε 0 (2n )h 2
46. (b) : On our planet : E n = −
48. (c) : Potential energy = –C/r2 and
total energy = –Rhc/n2. With higher orbit, both r and
n increase. So, both become less negative; hence both
increase.
13.6
n2
13.6
On other planet : E n′ = −
(2n )2
E
⇒ E n′ = n = −3.4 eV
4
47. (c) : Centripetal acceleration = mv2/r
Further, as n increases, r also increases. Therefore, centripetal
acceleration for n = 2 is less than that for n = 1. So,
statement (i) is wrong. Statement (ii) and (iii) are correct.
51. (c) : The shortest Balmer line has energy
= |(3.4 – 1.51)| eV = 1.89 eV
and the highest energy = |(0 – 3.4)| = 3.4 eV
The corresponding wavelengths are
12400 eV Å
12400 eV Å
= 6561 Å and
= 3647 Å
1.89 eV
3.4 eV
Only 4861 Å is between the first and last line of the Balmer
series.
52. (a)
53. (b) : In Rutherford’s α-particle scattering experiment,
some of α-particles were found to be scattered at very large
angles inspite of having very high kinetic energy. This shows
that there are α-particles which will be passing very close
to nucleus. Rutherford confirmed the repulsive force on
α-particles due to nucleus varies with distance according
to inverse square law and that the positive charges are
concentrated at the centre and not distributed throughout
the atom. This is the nuclear model of Rutherford.
54. (b) : According to classical electromagnetic theory, an
accelerated charge continuously emits radiation. As electrons
revolving in circular paths are constantly experiencing
centripetal acceleration, hence they will be losing their energy
continuously and the orbital radius will go on decreasing and
form spiral and finally the electron will fall into the nucleus.
55. (a) : Absorption transition
C
B
Absorb
radiation
A
Two possibilities in absorption transition.
Emission transition
C
B
A
Three possibilities in emission transition. Therefore number of
absorption transition < number of emission transition.
For any two states A and B such that EA < EB we have
absorption spectrum for A → B transition and emission B → A.
But most of the time atoms are in ground state, absorption is
only from the ground state.
97
Atoms
56. (c) : According to postulates of Bohr’s atomic model,
the electron revolve around the nucleus in fixed orbit of
definite radii. As long as the electron is in a certain orbit it
does not radiate any energy. Not only the centripetal force
has to be the centrifugal force, even the stable orbits are
fixed by Bohr’s theory.
These lines correspond to Lyman series.
4.
r = 0.529
n2
Z
22
= 1.058 Å
2
2
2
E (n = 2) = −13.6 × 2 = −13.6 eV
2
r (n = 2) = 0.529 ×
57. (b) : The reason is correct, but does not explain the
assertion properly. Negative energy of revolving electron
indicates that it is bound to the nucleus. The electron is not
free to leave the nucleus.
58. (b) : When we put R = 107 m–1 and K = 3, 4, 5 in
the given formula, values of l calculated lie between 4000 Å
and 8000 Å, which is the visible region. The reason is true,
but does not explain the assertion properly.
59. (a) : We know that an electron is very light particle as
compared to an α-particle. Hence electron cannot scatter the
α-particle at large angles, according to law of conservation of
momentum. On the other hand, mass of nucleus is comparable
with the mass of α-particle, hence only the nucleus of atom
is responsible for scattering of α-particles.
60. (b) : When white light from the photosphere (central
portion of the sun) passes through vapours of various
elements present in the outer chromosphere, then these
elements absorb those wavelengths which they themselves
emit to bring incandescent. Hence dark lines (absence of light)
appear in the continuous solar spectrum, due to absorption
of these lines. Absorption is possible in the sun, not only
from the ground state but also in higher states because of
the high temperature of the sun.
SUBJECTIVE TYPE QUESTIONS
1. According to electromagnetic theory, electron revolving
around the nucleus are continuously accelerated. Since an
accelerated charge emits energy, the radius of the circular path
of a revolving electron should go on decreasing and ultimately
it should fall into the nucleus. So, it could not explain the
structure of the atom. As matter is stable, we cannot expect
the atoms to collapse.
2. Hα line of the Balmer series in the emission spectrum
of hydrogen atoms obtained when the transition occurs from
n = 3 to n = 2 state.
3. Number of spectral lines obtained due to transition
of electron from n = 4 (3rd excited state) to n = 1 (ground
state) is
(4)(4 − 1)
N=
=6
2
5. Bohr’s quantization condition : The electron can revolve
round the nucleus only in those circular orbits in which
angular momentum of an electron is an integral multiple
h
of 2π
nh
i.e., mvr = , n = 1, 2, 3,....
2π
1
1
1 
 1 1
= RZ 2  2 −
= RZ 2  − 
6.
2
4 9
λB 2
 2 ( 2 + 1) 
⇒ λB 2 =
7.
36
5RZ 2
2πr = nλ
n2
Here, n = 3 ⇒ 2πr = 3λ and r = 0.53 Å
Z
9
or r3 = 0.53   Å
1
⇒λ=
2πr3 2π
=
(0.53 × 9) Å ≈ 9.99 Å.
3
3
8.
rn =
9.
v=
n 2h 2
=
n2
(0.053 nm)
Z
4π2mekZe 2
For our case, n = 1 and Z = 2, and the result is
r1 = 0.027 nm.
Ze 2
c Z m
=
2 ε 0nh 137  n  s
2c
m/s.
137
10. The shortest wavelength of Brackett series is given as
Here n = 1, Z = 2, v1 =
1
1  1.097 × 107
1
= 1.097 × 107  2 − 2  =
16
λ
4
∞ 
–6
⇒ λ = 1.4585 × 10 m
This wavelength lies in the infrared region of electromagnetic
spectrum.
11. The first excited level is 2nd line.
From the 2nd level electron can go to level 1 ⇒ one radiation
CBSE Board Term-II Physics Class-12
98
3rd level electron can go to levels 1, 2 ⇒ three radiations
4th level electron can go to levels 1, 2, 3 ⇒ six radiations
nth level electron can go to levels 1, 2, 3, ...(n – 1)
∴ Total number of radiations
(n − 1) ⋅ n
= 1 + 2 + ...... + (n – 1) =
.
2
hc
12. λ =
(E 2 − E1)
1242 eV - nm
λ=
= 122 nm.
(−3.4 eV) − (−13.6 eV)
This wavelength lies in the ultraviolet region.
Since, c = υ λ, the frequency of the photon is
υ=
13.
8
c 3.00 × 10 m / s
=
= 2.46 × 1015 Hz
λ 1.22 × 10 −7 m
1 1
1
= R (Z )2  2 − 2 
λ
n n 
1
2
For characteristic X-rays, we replace Z by Z – a or (Z – 1)
here. For K α , the electron jumps to the K-shell, hence,
n1 = 1 and n2 = 2.
1
1 1
⇒ = R (Z − 1)2  2 − 2 
λ
1 2 
Force ∝ (–e) (e) = –e2
 4 
If charge on proton is  + e  and charge on electron is
 3 
 3 
2
4  3 
 − 4 e  , then their product  e   − e  = –e . Thus
3  4 
Bohr Formula remains the same.
1 1 
eV = 13.6 × 8 eV
 1 32 
17. (i) As, ∆E = 13.6 × 32 ×  −
⇒
λ=
hc
12400
=
Å = 113.7 Å
∆E 13.6 × 8
(ii) Number of spectral lines =
n(n − 1) 3(3 − 1)
=
=3
2
2
18. Let a0 = Bohr radius.
v0 = velocity of electron in first orbit.
∴ Time taken by electron to complete one revolution
2πa 0
T=
.
v0
e
∴ Current created by electron, I =
T
e
ev
=
= 0 .
 2πa 0  2πa 0
 v 
0
1
1 1
= (1.097 × 107 m–1) (78 – 1)2  2 − 2 
λ
1 2 
1
= 4.9 × 1010 m–1 or λ = 2 × 10–11 m = 0.2 Å.
λ
19. Kinetic energy in the first excited state of hydrogen atom
EK = 3.4 eV = 3.4 × 1.6 × 10–19 J
h
de-Broglie wavelength, λ =
2m E K
14. Radius of innermost electron
=
r=
n 2h 2ε 0
πme 2
For n = 1, r1 =
h 2ε 0
2
= 5.3 × 10–11 m
πme
For n = 2, r2 = (2)2 r1 = 2.12 × 10–10 m
For n = 3, r3 = (3)2 r1 = 4.77 × 10–10 m.
h
associated with
2π
planetary motion are incomparably large relative to h.
For example angular momentum of earth in its orbital motion
h
is of the order of 1070 .
2π
For such large value of n, the difference in successive energies
and angular momenta of the quantised levels of the Bohr
model are so small that one can predict the energy level
continuous.
16. In Bohr’s formula,
15. Angular momentum mvr = n
mv 2
1
=
(e)(−e)
r
4π ε 0
6.63 × 10 −34
2 × 9.1 × 10 −31 × 3.4 × 1.6 × 10 −19
= 0.67 nm
20. Wavelength (l) of Balmer series is given by
1 1
1
= R H  2 − 2  where ni = 3, 4,5,...
λ
 2 ni 
For shortest wavelength, when transition of electrons take
place from ni = ∞ to nf = 2 orbit, wavelength of emitted
photon is shortest.
7
 1 1  1.097 × 10
= RH  2 −  =
4
λ min
2 ∞ 
–7
\ lmin = 3.646 × 10 m = 3646 Å
This wavelength lies in visible region of electromagnetic
spectrum.
21. There are two protons and two neutrons in helium atom.
Two electrons are revolving around the nucleus in first orbit.
It is assumed that there is no interaction between the two
electrons to He-atom.
so, replacing Z = 1 by Z = 2
1
 −13.6 Z 2 
En = 
 eV,
2
 n

99
Atoms
For ground state (n = 1) of helium atom, energy
 −13.6 
E1 =  2 (2)2  eV = −54.4 eV
 1

Helium has two electrons in ground state, so total energy
E = 2E1 = –108.8 eV.
1
1
2 1
22. As λ = RZ  2 − 2  ,
 nf n i 
1
1 
υ = cRZ 2  2 −
2
 n (n + p ) 
[as c = ul, nf = n, ni = (n + p), where p = 1, 2, 3...]
 (n + p )2 − n 2 
or υ = cRZ 2  2
2 
 n (n + p ) 
 (n 2 + p 2 + 2pn) − n 2 
= cRZ 2 

n 2(n + p )2


2
 2pn   2cRZ 
u ≈ cRZ2  4  ≈  3  p (∴ n >> 1, p << n)
n
 n 
or
obviously u ∝ p, i.e., the values of υ are approximately in
the ratio 1 : 2 : 3.
23. In Balmer series, Hγ line corresponds to transition from
state ni = 5 to state nf = 2.
Energy required, E = E5 – E1
 −13.6   −13.6 
=  2  −  2  = 13.06 eV.
 5   1 
If angular momentum of system is conserved,
change in angular momentum of electron = change in angular
momentum of photon
 h   h  3h
= 5  −2  =
 2π   2π  2π
=
3 × 6.6 × 10 −34
= 3.17 × 10 −34 J s
2 × 3.14
24. (i) The kinetic energy (Ek) of the electron in an orbit is
equal to negative of its total energy (E).
Ek = –E = – (–1.5) = 1.5 eV
(ii) The potential energy (Ep) of the electron in an orbit is
equal to twice of its total energy (E).
Ep = 2E = –1.5 × 2 = –3.0 eV
(iii) Here, ground state energy of the H-atom = –13.6 eV
When the electron goes from the excited state to the ground
state, energy emitted is given by
E = –1.5 – (–13.6) = 12.1 eV = 12.1 × 1.6 × 10–19 J
hc
Now,
E=
λ
λ=
hc 6.62 × 10 −34 × 3 × 108
=
E
12.1 × 1.6 × 10 −19
λ = 1.025 × 10–7 = 1025 Å
25. Here, ∆E = 12.5 eV
Energy of an electron in nth orbit of hydrogen atom is,
13.6
E n = − 2 eV
n
In ground state, n = 1
E1 = –13.6 eV
Energy of an electron in the excited state after absorbing a
photon of 12.5 eV energy will be
En = –13.6 + 12.5 = –1.1 eV
−13.6 −13.6
∴ n2 =
=
= 12.36 ⇒ n = 3.5
En
−1.1
Here, state of electron cannot be in fraction.
So, n = 3 (2nd excited state).
The wavelength λ of the first member of Lyman series is
given by
1
1 1 3
=R  2 − 2 = R
λ
1 2  4
4
4
⇒ λ=
=
3R 3 × 1.097 × 107
⇒ λ = 1.215 × 10–7 m
⇒ λ = 121 × 10–9 m ⇒ λ = 121 nm
The wavelength λ′ of the first member of the Balmer series
is given by
1
1 1 5
=R  2 − 2 = R
λ′
 2 3  36
⇒ λ′ =
36
36
=
5R 5 × (1.097 × 107 )
= 6.56 × 10–7 m = 656 × 10–9 m = 656 nm
−13.6
26. (i) ∵ E n = 2 eV
n
Energy of the photon emitted during a transition of the
electron from the first excited state to its ground state is,
∆E = E2 – E1
−13.6  −13.6  −13.6 13.6
= 2 − 2 =
+
= −3.40 + 13.6
4
1
 1 
2
= 10.2 eV
This transition lies in the region of Lyman series.
(ii) (a) The energy levels of H-atom are given by
Rhc
13.6
E n = − 2 = − 2 eV
n
n
For first excited state n = 2
13.6
E 2 = − 2 eV = −3.4 eV
(2)
Kinetic energy of electron in (n = 2) state is
K2 = – E2 = + 3.4 eV
CBSE Board Term-II Physics Class-12
100
(b) Radius in the first excited state
r1 = (2)2 (0.53) Å
r1 = 2.12 Å
From Bohr’s quantization condition
27. (i) de-Broglie hypothesis may be used to derive Bohr’s
formula by considering the electron to be a wave spread over
the entire orbit, rather than as a particle which at any instant
is located at a point in its orbit. The stable orbits in an atom
are those which are standing waves. Formation of standing
waves require that the circumference of the orbit is equal in
length to an integral multiple of the wavelength. Thus, if r is
the radius of the orbit
nh
h

2πr = nλ =
s
∵ λ= 


p
p
which gives the angular momentum quantization.
h
L = pr = n
2π
1 e2
m ⋅ n 2h 2
1 e2
 nh 
=
=
m⋅
or
 2πmr  4πε 0 r
4π2m 2r 2 4πε 0 r
mvr =
nh
nh
...(ii)
or v =
2π
2πmr
Using equation (ii) in (i), we get
2
or
n 2h 2ε 0
...(iii)
πme 2
where n = 1, 2, 3, ... is principal quantum number.
Equation (iii), gives the radius of nth orbit of H-atom. So
the radii of the orbits increase proportionally with n2 i.e.,
[r ∝ n2]. Radius of first orbit of H-atom is called Bohr radius
a0 and is given by
a0 =
r=
h 2ε 0
πme 2
for n = 1 or a0 = 0.529 Å
So, radius of nth orbit of H-atom then becomes
r = n2 × 0.529 Å
(ii)
C
EC
1
EB
3
B
2
A
EA
Clearly, from energy level diagram,
EC – EA = (EC – EB) + (EB – EA)
(On the basis of energy of emitted photon).
hc hc hc
= +
\
λ 3 λ1 λ2
⇒
1
1
1
λλ
= +
⇒ λ3 = 1 2
λ1 + λ2
λ 3 λ1 λ2
which is the required relation between the three given
wavelengths.
28. Radius of nth orbit of hydrogen atom : In H-atom, an
electron having charge –e revolves around the nucleus of
charge +e in a circular orbit of radius r, such that necessary
centripetal force is provided by the electrostatic force of
attraction between the electron and nucleus.
2
1 e2
i.e., mv = 1 e.e or mv 2 =
...(i)
2
4πε r
r
4πε 0 r
0
+e r
F
–e
mv2
r
29. (i) As the gold foil is very thin, it can be assumed that
a-particles will suffer not more than one scattering during
their passage through it. Therefore, computation of the
trajectory of an a-particle scattered by a single nucleus is
enough.
(ii) Trajectory of a-particles depends on impact parameter
which is the perpendicular distance of the initial velocity
vector of the a particles from the centre of the nucleus. For
small impact parameter, a particle close to the nucleus suffers
larger scattering.
(iii) In case of head-on-collision, the impact parameter is
minimum and the a -particle rebounds back. So, the fact
that only a small fraction of the number of incident particles
rebound back indicates that the number of a -particles
undergoing head-on collision is small. This in turn implies
that the mass of the atom is concentrated in a small volume.
30. According to Bohr’s formula,
En =
−me 4
8ε20n 2h 2
where m is called reduced mass.
In case of hydrogen, m = me = mass of electron.
For positronium,
m × me me
m= e
=
me + me
2
Since for H-atom, E1 =
m ee 4
8ε20n 2h 2
So, for positronium E 1′ =
= −13.6 eV
−13.6
= −6.8 eV.
2
101
Atoms
31. Wavelengths corresponding to minimum wavelength
(λmin) or maximum energy will emit photoelectrons having
maximum kinetic energy.
Wavelengths belonging to Balmer series and lying in the given
range (450 nm to 750 nm) corresponds to transition from
(n = 4 to n = 2). Here,
E4 =
13.6
(4)2
= −0.85 eV and E 2 = −
13.6
(2)2
= −3.4 eV
∴ ∆E = E4 – E2 = 2.55 eV
Kmax = Energy of photon – Work function
= 2.55 – 2.0 = 0.55 eV
32. The force at a distance r is
F =−
Suppose r be the radius of nth orbit. The necessary centripetal
force is provided by the above force. Thus,
mv 2
= 2ar ...(i)
r
Further, the quantisation of angular momentum gives,
nh
...(ii)
mvr =
2π
1/ 4
 n 2h 2 
Solving, equations (i) and (ii) for r, we get r = 
 8amπ2 
33. (i) We have
mv 2
Ze 2
Ze 2
2
=
=
or
v
r
4πε 0m
r
4πε 0r 2
nh
The quantisation rule is vr =
2πm
(vr )2
rh =
0
2
2 2
h 2ε 0
πme e 2
For rµ = rh ,
or n 25
n 2h 2ε 0
624πme e 2
=
h 2ε 0
πme e 2
or n2 = 624
n(n − 1)
=3 ∴ n = 3
2
i.e. after excitation atom jumps to second excited state.
Hence, nf = 3. So ni can be 1 or 2.
If ni = 1 then energy emitted is either equal to, greater than or
less than the energy absorbed. Hence, the emitted wavelength
34. (i)
35. As the nucleus is massive, recoil momentum of the atom
can be ignored. We can assume that the entire energy of
transition is transferred to the Auger electron.
As there is a single valence electron in chromium (Z = 24),
the energy states may be thought of as given by Bohr model.
The energy of the nth state is
RhcZ 2
where R is Rydberg constant.
n2
In the transition from n = 2 to n = 1, energy released,
1  3
∆E = −RhcZ 2  − 1 = RhcZ 2
4  4
En = −
The energy required to eject a n = 4 electron
2
(2nhπm ) 4πεZe m = Znπhmeε
2
( )
2
 1  RhcZ
= RhcZ 2   =
16
4
0 2
v 2r
For the given system, Z = 3 and m = 208 me
n 2h 2ε 0
Thus, rµ =
624πme e2
(ii) The radius of the first Bohr orbit for a hydrogen atom is
=
(ii) E3 – E2 = 68 eV
1 1
∴ (13.6)(Z 2 ) − = 68
4 9
∴ Z=6
The gas atoms correspond to carbon.
12400
12400
(iii) λ min =
=
E 3 − E1
1
(13.6)(6)2 1 −
9
12400
=
= 28.49 Å
435.2
( )
dU
= −2ar
dr
The radius is r =
is either equal to, less than or greater than the absorbed
wavelength.
Hence, ni ≠ 1
If ni = 2, then Ee ≥ Ea. Hence λe ≤ λ0
∴
KE of Auger electron =
3RhcZ 2 RhcZ 2
−
4
16
 3 1  11
KE = RhcZ 2  −  = RhcZ 2
 4 16  16
11
= (13.6 eV) × 24 × 24 = 5385.6 eV [ Rhc = 13.6 eV]
16
36. (a) (i) Limitation of Rutherford’s model :
e–
Rutherford’s atomic model is inconsistent
with classical physics. According to
+
electromagnetic theory, an electron is a
charged particle moving in the circular orbit
around the nucleus and is accelerated, so
it should emit radiation continuously and thereby loose
energy. Due to this, radius of the electron would decrease
continuously and also the atom should then produce
continuous spectrum, and ultimately electron will fall into
the nucleus and atom will collapse in 10–8 s. But the atom is
fairly stable and it emits line spectrum.
(ii) Rutherford’s model is not able to explain the spectrum
of even most simplest H-spectrum.
(b) Bohr’s postulates to resolve observed features of atomic
spectrum :
CBSE Board Term-II Physics Class-12
102
(i) Quantum condition: Of all the possible circular orbits
allowed by the classical theory, the electrons are permitted to
circulate only in those orbits in which the angular momentum
h
of an electron is an integral multiple of
, h being Planck’s
2π
constant. Therefore, for any permitted orbit,
nh
L = mvr = , n = 1, 2, 3 .....,
2π
where n is called the principal quantum number, and this
equation is called Bohr’s quantisation condition.
(ii) Stationary orbits: While revolving in the permissible
orbits, an electron does not radiate energy. These nonradiating orbits are called stationary orbits.
(iii) Frequency condition: An atom can emit or absorb
radiation in the form of discrete energy photons only when
an electron jumps from a higher to a lower orbit or from a
lower to a higher orbit, respectively.
hυ = Ei – Ef
where υ is frequency of radiation emitted, Ei and Ef are
the energies associated with stationary orbits of principal
quantum number ni and nf respectively (where ni > nf).
37. ∵
mv 2
1 e2
=
…(i)
rn
4πε 0 r 2
n
nh
…(ii)
2π
From eqn. (i) and (ii)
and mvrn =
∴
rn =
ε 0h 2n 2
πme 2
Total energy
1
1 e2
1 e2
1 e2
E n = mv n2 −
=
−
2
4πε 0 rn 8πε 0 rn 4πε 0 rn
me 4
−Rhc where, Rydberg constant R =
8ε20h 3c
n2
Energy emitted ∆E = Ei – Ef
1 1
∆E = Rhc  2 − 2 
 nf n i 
But ∆E = hυ
1
1
υ = Rc  2 − 2 
 nf n i 
or
38. Let µH and µD are the reduced masses of electron for
hydrogen and deuterium respectively.
1 1
1
=R  2 − 2 
λ
n n 
We know that
f
λ∝
\
RH =
RD =
i
1
λ
R
or D = H …(i)
R
λH RD
m ee 4
8 ε 0ch 3
m ee 4
8 ε 0 ch
=
3
µH e 4
8ε 0ch 3
=
µD e 4
8 ε 0ch
3
∴
R H µH
=
…(ii)
R D µD
From equation (i) and (ii)
λ D µH …(iii)
=
λ H µD
Reduced mass for hydrogen,
me
 m 
µH =
me 1− e 
1+ me / M
 M 
Reduced mass for deuterium,
2M ⋅ me
 m 
me 1 − e 
µD =
 2M 
 me 
2M 1 +
 2M 
where M is mass of proton.
 m 
me 1 − e 
−1
 M   me   me 
µH
−
= 1 −
=
1


µD
 m   M   2M 
me 1 − e 
 2M 
 m  m 
= 1 − e  1 + e 
 M   2M 
1 e2
1 me 4
En = −
=− 2 2 2
8πε 0 rn
8ε 0 h n
En =
When electron in hydrogen atom jumps from energy state
ni = 4 to nf = 3, 2, 1, the Paschen, Balmer and Lyman spectral
series are found.
me 4  1 1 
υ= 2 3  2 − 2
8ε 0h  nf ni 
⇒
µH  me 
= 1−
µD  2M 
1 
µ

or H = 1 −
= 0.99973
µD  2 × 1840 
From (iii) and (iv)
λD
= 0.99973 , λD = 0.99973 λH.
λH
(∴
…(iv)
M = 1840 me)
Using λH = 1218 Å, 1028 Å, 974.3 Å and 951.4 Å, we get
λD = 1217.7 Å, 1027.7 Å, 974.04 Å, 951.1 Å
Shift in wavelength (λH – λD) ≈ 0.3 Å.

CHAPTER
13
Nuclei
Recap Notes
Composition of the Nucleus
X The nucleus of an atom contains protons
and neutrons which are collectively known
as nucleons. The number of protons in a
nucleus is called its atomic number and is
denoted by Z. The total number of protons
and neutrons in a nucleus is called its
mass number and is denoted by A.
– Number of protons in an atom = Z
– Number of electrons in an atom = Z
– Number of nucleons in an atom = A
– Number of neutrons in an atom = N
= A – Z.
X Nuclide : It is a specific nucleus of
an atom which is characterised by its
atomic number Z and mass number A.
It is represented by ZXA where X is the
chemical symbol of the element.
X Nuclear Radius : Nuclear radius R = R0
A1/3 where R0 is a constant and A is the
mass number.
Nuclear radius is measured in fermi.
1 fm = 10–15 m
X Nuclear Density :
Mass of nucleus
Nuclear density, ρ =
Volume of nucleus
Nuclear density is independent of A and
is in order of the 1017 kg m–3.
Isotopes : Isotopes of an element are the
atoms of the element which have the same
atomic number but different mass numbers,
e.g., 1H1, 1H2, 1H3, are the three isotopes of
hydrogen.
Isobars : Isobars are the atoms of different
elements which have the same mass number
but different atomic numbers, e.g., 11Na 22
and 10Ne22.
Isotones : Isotones are the nuclides which
contain the same number of neutrons, e.g.,
37
39
17Cl and 19K .
Nuclear forces : Nuclear forces are the
strong forces of attraction which hold
together the nucleons (neutrons and protons)
in the tiny nucleus of an atom, inspite
of strong electrostatic forces of repulsion
between protons.
X Nuclear forces are strongest forces in
nature.
X Nuclear forces are short range forces.
X Nuclear forces do not obey inverse square law.
X Nuclear forces are charge independent.
Mass defect : The difference in mass of a
nucleus and its constituents is known as the
mass defect and is given by
Dm = [Zmp + (A – Z)mn – mN]
where mp is the mass of the proton and mn is
the mass of the neutron and mN is the mass
of the nucleus.
Binding Energy : The binding energy of
nucleus is given by
Eb = Dmc2 = [Zmp + (A – Z)mn – mN]c2
= [Zmp + (A – Z)mn – mN] × 931.49 MeV/u.
Nuclear reaction : A nuclear reaction is
represented by
A+a→B+b+Q
where A is the target nucleus, a is the
impinging particle, B and b are the products,
Q is the energy released in the process.
The nuclear reaction is represented by
notation A(a, b)B.
Q value of nuclear reaction,
Q = (mA + ma – mB – mb)c2
CBSE Board Term-II Physics Class-12
104
X
X
X
If Q is positive, the reaction is exothermic
and if Q is negative the reaction is
endothermic.
Conservation laws obeyed by every
nuclear reaction are
– Conservation of charge number
– Conservation of mass number
– Conservation of linear momentum
– Conservation of energy
Nuclear fission : It is the phenomenon of
splitting a heavy nucleus into two or more
smaller nuclei.
The nuclear fission of 92U235 is represented
as
235
+ 0n1 → 56Ba141 + 36Kr92 + 3 0n1 + Q
92U
The value of the Q is 200 MeV per fission
reaction.
X Nuclear chain reaction : Under suitable
conditions, the three secondary neutrons
may cause further fission of U235 nuclei
and start what is known as nuclear chain
reaction. The nuclear chain reaction is
controlled by
Neutron reproduction factor, (K)
Rate of production of neutrons
=
Rate of loss of neutrons
X
X
X
X
Uncontrolled nuclear chain reaction is the
basis of an atom bomb. Controlled nuclear
chain reaction is the basis of a nuclear
reactor.
Nuclear reactor : Nuclear reactor uses
nuclear energy for peaceful purposes. It
is based on the phenomenon of controlled
nuclear chain reaction. Moderators like
heavy water, graphite, paraffin and
deuterium slow down neutrons. Rods of
cadmium and boron serve as control rods.
Ordinary water and heavy water serve as
coolants.
Nuclear fusion : It is the phenomenon of
fusing two or more lighter nuclei to form a
single heavy nucleus.
– The nuclear fusion reaction of two
deutrons is represented as
– 1H2 + 1H2 → 2He4 + 24 MeV
– Temperature ≈ 107 K are required for
fusion to take place.
– Nuclear fusion is a basis of hydrogen
bomb.
Stellar energy : It is the energy obtained
from the sun and stars. The source of
stellar energy is nuclear fusion.
Practice Time
OBJECTIVE TYPE QUESTIONS
Multiple Choice Questions (MCQs)
1. If 200 MeV energy is released in the fission
of a single nucleus of 235
92U, the fissions which are
required to produce a power of 1 kW is
(a) 3.125 × 1013
(b) 1.52 × 106
12
(c) 3.125 × 10
(d) 3.125 × 1014
The equation 4 11 H +
4
2+
2 He
2.
→
represents
(a) both fusion and fission
(b) neither fusion nor fission
(c) only fusion
(d) only fission
3.
(a)
(b)
(c)
(d)
−
+ 2e + 26 MeV
In nuclear reaction, there is conservation of
mass only
energy only
momentum only
mass, energy and momentum
4. The fission properties of 239
94 Pu are very
similar to those of 235
U.
The
average energy
92
released per fission is 180 MeV. If all the atoms
in 1 kg of pure 239
94Pu undergo fission, then the
total energy released in MeV is
(a) 4.53 × 1026 MeV
(b) 2.21 × 1014 MeV
13
(c) 1 × 10 MeV
(d) 6.33 × 1024 MeV
5. 1 MeV positron encounters a 1 MeV
electron travelling in opposite direction.
What is the wavelength of photons produced?
(Given rest mass energy of electron or positron
= 0.512 MeV and h = 6.62 × 10–34 Js)
(a) 8.2 × 10–11 m
(b) 8.2 × 10–13 m
(c) 8.2 × 10–12 m
(d) 8.2 × 10–9 m
8. The mass number of iron nucleus is 55.854
and A = 56, the nuclear density is
(a) 2.29 × 1016 kg m–3 (b) 2.29 × 1017 kg m–3
(c) 2.29 × 1018 kg m–3 (d) 2.29 × 1015 kg m–3
9. Order of magnitude of density of uranium
nucleus is
(a) 1020 kg m–3
(b) 1017 kg m–3
(c) 1014 kg m–3
(d) 1011 kgm–3
10. How much mass has to be converted into
energy to produce electric power of 500 MW for
one hour?
(a) 2 × 10–5 kg
(b) 1 × 10–5 kg
(c) 3 × 10–5 kg
(d) 4 × 10–5 kg
11. Two stable isotopes 36 Li and 37 Li have
respective abundances of 7.5% and 92.5%. These
isotopes have masses 6.01512 u and 7.01600 u
respectively. The atomic weight of lithium is
(a) 6.941 u
(b) 3.321 u
(c) 2.561 u
(d) 0.621 u
12. The ratio of the nuclear radii of the gold
197
isotope 79 Au and silver isotope 107
47Ag is
(a) 1.23
(b) 0.216
(c) 2.13
(d) 3.46
13. Let mp be the mass of a proton, mn the mass
20
of a neutron, M1 the mass of a 10
Ne nucleus and
40
M2 the mass of a 20
Ca nucleus. Then
(a) M2 = M1
(b) M2 > 2M1
(c) M2 = 2M1
(d) M1 < 10 (mn + mp)
Light energy emitted by star is due to
breaking of nuclei
joining of nuclei
burning of nuclei
reflection of solar light
6. In nuclear reactors, the control rods are
made of
(a) cadmium
(b) graphite
(c) krypton
(d) plutonium
14.
(a)
(b)
(c)
(d)
7. The equivalent energy of 1 g of substance is
(a) 9 × 1013 J
(b) 6 × 1012 J
13
(c) 3 × 10 J
(d) 6 × 1013 J
15. The mass of proton is 1.0073 u and that of
neutron is 1.0087 u (u = atomic mass unit). The
4
binding energy of 24 He, if mass of 2He is 4.0015 u
CBSE Board Term-II Physics Class-12
106
(a) 0.0305 erg
(c) 28.4 MeV
(b) 0.0305 J
(d) 0.061 u
16. The set which represents the isotope, isobar
and isotone respectively is
198
3
2
(a) (12H, 13H), (197
79Au, 80Hg) and (2 He, 1 H)
(b)
198
(23He, 11H), (197
79Au, 80Hg)
and
(11H, 31H)
198
(c) (23He, 13H), (12H, 13H) and (197
79Au, 80Hg)
198
(d) (12H, 13H), (23He, 13H) and (197
79Au, 80Hg)
17. The radius of a spherical nucleus as measured
by electron scattering is 3.6 fm. What is the mass
number of the nucleus most likely to be?
(a) 27
(b) 40
(c) 56
(d) 120
18. If in a nuclear fusion reaction, mass defect is
0.3%, then energy released in fusion of 1 kg mass
is
(a) 27 × 1010 J
(b) 27 × 1011 J
10
(c) 27 × 10 J
(d) 27 × 1013 J
19. If the nucleus 13Al27 has a nuclear radius of
about 3.6 fm, then 52Te125 would have its radius
approximately as
(a) 9.6 fm
(b) 12 fm
(c) 4.8 fm
(d) 6 fm
20. Boron has two stable isotopes, 105B and 115B.
Their respective masses are 10.01294 u and
11.00931 u, and the atomic mass of boron is
10.811 u. Find the abundances of 115B.
(a) 90.1%
(b) 80.1%
(c) 85.5%
(d) 95%
Case Based MCQs
Case I : Read the passage given below and answer
the following questions from 21 to 25.
Discovery of Nucleus
The nucleus was first discovered in 1911 by Lord
Rutherford and his associates by experiments
on scattering of a-particles by atoms. He found
that the scattering results could be explained,
if atoms consist of a small, central, massive and
positive core surrounded by orbiting electrons.
The experimental results indicated that the size
of the nucleus is of the order of 10–14 m and is
thus 10000 times smaller than the size of atom.
21. Ratio of mass of nucleus with mass of atom is
approximately
(a) 1
(b) 10
(c) 103
(d) 1010
22. Masses of nuclei of hydrogen, deuterium and
tritium are in ratio
(a) 1 : 2 : 3
(b) 1 : 1 : 1
(c) 1 : 1 : 2
(d) 1 : 2 : 4
23. Nuclides with same neutron number but
different atomic number are
(a) isobars
(b) isotopes
(c) isotones
(d) none of these
24. If R is the radius and A is the mass number,
then log R versus log A graph will be
(a) a straight line
(b) a parabola
(c) an ellipse
(d) none of these.
25. The ratio of the nuclear radii of the mercury
107
isotope 198
80Hg and silver isotope 47Ag is
(a) 1.23
(b) 0.216
(c) 2.13
(d) 3.46
Case II : Read the passage given below and
answer the following questions from 26 to 30.
Nuclear Fission
In the year 1939, German scientist Otto Hahn
and Strassmann discovered that when an
uranium isotope was bombarded with a neutron,
it breaks into two intermediate mass fragments.
It was observed that, the sum of the masses of
new fragments formed were less than the mass
of the original nuclei. This difference in the mass
appeared as the energy released in the process.
Thus, the phenomenon of splitting of a heavy
nucleus (usually A > 230) into two or more lighter
nuclei by the bombardment of proton, neutron,
a-particle, etc with liberation of energy is called
nuclear fission.
235
236
+ 0n1 →
→ 56Ba144 + 36Kr89
92U
92U
26.
of
(a)
(b)
(c)
(d)
Unstable nucleus
+ 30n1 + Q
Nuclear fission can be explained on the basis
Millikan’s oil drop method
Liquid drop model
Shell model
Bohr’s model.
107
Nuclei
27. For sustaining the nuclear fission chain
reaction in a sample (of small size) of 235
92U, it is
desirable to slow down fast neutrons by
(a) friction
(b) elastic damping/scattering
(c) absorption
(d) cooling
(a) more mass per nucleon than either of the two
fragments
(b) more mass per nucleon as the two fragment
(c) exactly the same mass per nucleon as the
two fragments
(d) less mass per nucleon than either of two
fragments.
28. Which of
reaction(s)?
33. When 92U235 undergoes fission, about 0.1%
of the original mass is converted into energy. The
energy released when 1 kg of 92U235 undergoes
fission is
(I)
1
0n
the
following
is/are
fission
236
133
99
1
+ 235
92U → 92U → 51Sb + 41Nb + 4 0n
1.40
94
1
(II) 10n + 235
92U → 54 Xe + 38Sr + 2 0n
(III) 21H + 21H → 32He + 10n
(a) Both II and III
(c) Only II
(b) Both I and III
(d) Both I and II
29. On an average, the number of neutrons and
the energy of a neutron released per fission of a
uranium atom are respectively
(a) 2.5 and 2 keV
(b) 3 and 1 keV
(c) 2.5 and 2 MeV
(d) 2 and 2 keV
30. In any fission process, ratio of mass of
daughter nucleus to mass of parent nucleus is
(a) less than 1
(b) greater than 1
(c) equal to 1
(d) depends on the mass of parent nucleus.
Case III : Read the passage given below and
answer the following questions from 31 to 35.
Nuclear Energy
A heavy nucleus breaks into comparatively
lighter nuclei which are more stable compared to
the original heavy nucleus. When a heavy nucleus
like uranium is bombarded by slow moving
neutrons, it splits into two parts releasing large
amount of energy. The typical fission reaction of
235
.
92U
235
92U
+ 0n1 → 56Ba141 + 36Kr92 + 30n1 + 200 MeV
The fission of 92U235 approximately released
200 MeV of energy.
31. If 200 MeV energy is released in the fission
of a single nucleus of 235
92U, the fissions which are
required to produce a power of 10 kW is
(a) 3.125 × 1013
(b) 1.52 × 106
12
(c) 3.125 × 10
(d) 3.125 × 1014
32. The release in energy in nuclear fission is
consistent with the fact that uranium has
(a) 9 × 1011 J
(c) 9 × 1015 J
(b) 9 × 1013 J
(d) 9 × 1018 J
34. A nuclear fission is said to be critical when
multiplication factor or K
(a) K = 1
(b) K > 1
(c) K < 1
(d) K = 0
35. Einstein’s mass-energy conversion relation
E = mc2 is illustrated by
(a) nuclear fission
(b) atomic transition
(c) rocket propulsion
(d) steam engine
Case IV : Read the passage given below and
answer the following questions from 36 to 40.
Nuclear Force
Neutrons and protons are identical particle in
the sense that their masses are nearly the same
and the force, called nuclear force, does into
distinguish them. Nuclear force is the strongest
force. Stability of nucleus is determined by the
neutron proton ratio or mass defect or packing
fraction. Shape of nucleus is calculated by
quadrupole moment and spin of nucleus depends
on even or odd mass number. Volume of nucleus
depends on the mass number. Whole mass of the
atom (nearly 99%) is centred at the nucleus.
36. The correct statements about the nuclear
force is/are
(a) charge independent
(b) short range force
(c) non-conservative force
(d) all of these.
37. The range of nuclear force is the order of
(a) 2 × 10–10 m
(b) 1.5 × 10–20 m
–4
(c) 1.2 × 10 m
(d) 1.4 × 10–15 m
38. A force between two protons is same as the
force between proton and neutron. The nature of
the force is
CBSE Board Term-II Physics Class-12
108
(a) electrical force
(c) gravitational force
(b) weak nuclear force
(d) strong nuclear force.
39. Two protons are kept at a separation of 40 Å.
Fn is the nuclear force and Fe is the electrostatic
force between them. Then
(a) Fn << Fe
(b) Fn = Fe
(c) Fn >> Fe
40.
(a)
(b)
(c)
(d)
(d) Fn ≈ Fe
All the nucleons in an atom are held by
nuclear forces
Van der Waal’s forces
tensor forces
coulomb forces
Assertion & Reasoning Based MCQs
For question numbers 41-49, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
41. Assertion (A) : Two protons can attract
each other.
Reason (R) : The distance between the protons
within the nucleus is about 10–15 m.
46. Assertion (A) : There is a chain reaction
when uranium is bombarded with slow neutrons.
Reason (R) : When uranium is bombarded with
slow neutrons more neutrons are produced.
42. Assertion (A) : The nuclear force becomes
weak if the nucleus contains too many protons
compared to neutrons.
Reason (R) : The electrostatic forces weaken the
nuclear force.
47. Assertion (A) : Cadmium rods used in a
nuclear reactor, control the rate of fission.
Reason (R) : Cadmium rods speed up the slow
neutrons.
43. Assertion (A) : For the fission of heavy
nuclei, neutrons are more effective than protons.
Reason (R) : Neutrons are heavier than protons.
44. Assertion (A) : Energy is released in a
nuclear reaction.
Reason (R) : In any nuclear reaction the
reactants and resultant products obey the law of
conservation of charge and mass only.
45. Assertion (A) : Density of all the nuclei is same.
Reason (R) : Radius of nucleus is directly
proportional to the cube root of mass number.
48. Assertion (A) : A fission reaction can be
more easily controlled than a fusion reaction.
Reason (R) : The percentage of mass converted
to energy in a fission reaction is 0.1% whereas
in a fusion reaction it is 0.4%.
49. Assertion (A) : Thermonuclear fusion
reactions may become the source of unlimited
power for the mankind.
Reason (R) : A single fusion event involving
isotopes of hydrogen produces more energy than
energy from nuclear fission of a single uranium.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1.
What is the radius of the nucleus of
2.
Find
64
29 Cu
the
27
and 13
Al .
ratio
of
nuclear
3.
64
29 Cu?
radius
of
Define the nuclear force?
4. The nuclear radius of a nucleus with nucleon
number 16 is 3 × 10–15 m. Then, find the nuclear
radius of a nucleus with nucleon number 128?
109
Nuclei
5. Assume that a neutron breaks into a proton
and an electron. Find the energy required during
this process ?
(Mass of neutron = 1.6725 × 10–27 kg, Mass of proton
= 1.6725 × 10–27 kg, mass of electron = 9 × 10–31 kg)
6. A neutron is absorbed by a 63 Li nucleus with
the subsequent emission of an alpha particle.
Calculate the energy released, in MeV, in this
reaction.
and mass (triton) = 3.0100000 u.
Take 1 u = 931 MeV/c2]
7. From the relation R = R0 A1/3, where R0 is a
constant and A is the mass number of a nucleus,
show that the nuclear matter density is nearly
constant (i.e., independent of A).
8.
The three stable isotopes of neon :
21
10 Ne
and
22
10 Ne
20
10 Ne
,
have respective abundances of
90.51%, 0.27% and 9.22%. The atomic masses
of the three isotopes are 19.99 u, 20.99 u and
21.99 u, respectively. Obtain the average atomic
mass of neon.
[Given : mass 63 Li = 6.015126 u;
mass (neutron) = 1.0086654 u;
mass (alpha particle) = 4.0026044 u;
Short Answer Type Questions (SA-I)
9. Find the disintegration energy Q for the
fission event represented by equation
235
94
1
+ 0n1 → 92U236 → 140
92U
58Ce + 40Zr + 20n
If mass of 92U235 = 235.0439 u, 0n1 = 1.00867 u,
140
94
58Ce = 139.9054 u and 40Zr = 93.9063 u, find
energy released in the process.
10. 92U235 absorbs a slow neutron (thermal
neutron) and undergoes a fission represented by
235
+ 0n1 → 92U236 → 56Ba141 + 36Kr92 + 3 0n1 + E.
92U
Calculate :
(i) The energy released E per fission.
(ii) The energy released when 1 g of
undergoes complete fission.
Given : 92U235 = 235.1175 amu (atom),
235
92U
is mass converted into energy (or vice versa)?
Explain giving one example.
12. Calculate the energy released in fusion
reaction : 12 H +12 H → 32 He + n , where B.E. of
2
1H
= 2.23 MeV and of
3
2 He
= 7.73 MeV.
13. Which is more, the density of lead nuclei or
the density of oxygen nuclei?
14. Consider a particle or
nucleus, which
contains 2 protons and 2 neutrons. Find its
binding energy.
mp = 1.007276 u, mn = 1.008665 u
mHe = 4.001508 u
1 amu = 931 MeV/c
15. A nuclide 1 is said to be the mirror isobar of
nuclide 2 if Z1 = N2 and Z2 = N1. (a) Which nuclide
is a mirror isobar of 23
11Na ? (b) Which nuclide out
of the two mirror isobars have greater binding
energy and why?
11. If both the number of protons and neutrons
in a nuclear reaction is conserved, in what way
16. Why do stable nuclei never have more
protons than neutrons?
141
= 140.9577 amu (atom)
56Ba
92
1
36Kr = 91.9264 amu (atom), 0n =
2
1.00868 amu,
Short Answer Type Questions (SA-II)
17. Calculate for how many years will the fusion
of 2.0 kg deuterium keep 800 W electric lamp
glowing. Take the fusion reaction as
2
2
3
1
1H + 1H → 2He + 0n + 3.27 MeV
18. Consider the fusion reaction:
4
He + 4He → 8Be
For the reaction, find (i) mass defect (ii) Q-value
(iii) Is such a fusion energetically favourable?
Atomic mass of 8Be is 8.0053 u and that of 4He
is 4.0026 u.
19. A nuclear reactor using 235U generates 250 MW
of electrical power. The efficiency of the reactor
(i.e., efficiency of conversion of thermal energy
into electrical energy) is 25%. What is the amount
of 235U used in the reactor per year? The thermal
energy released per fission of 235U is 200 MeV.
CBSE Board Term-II Physics Class-12
110
20. The deuteron is bound by nuclear forces
just as H-atom is made up of p and e bound by
electrostatic forces. If we consider the force
between neutron and proton in deuteron as given
in the form of a Coulomb potential but with an
effective charge e′:
1 e′ 2
F=
4 πε0 r
22. When four hydrogen nuclei combine to form
a helium nucleus estimate the amount of
energy in MeV released in this process of fusion
(Neglect the masses of electrons and neutrons).
Given:
Estimate the value of (e′/e) given that the binding
energy of a deuteron is 2.2 MeV.
23. In a typical nuclear reaction, e.g.
21. Distinguish between the phenomena of
nuclear fission and fusion.
although number of nucleons is conserved, yet
energy is released. How? Explain.
(i) Mass of 11 H = 1.007825 u
(ii) mass of helium nucleus = 4.002603 u,
1u = 931 MeV/c2
2
1H
+ 12 H →32 He + 10 n +3.27 MeV,
Long Answer Type Questions (LA)
24. Deuteron is a bound state of a neutron and a
proton with a binding energy B = 2.2 MeV. A g-ray
of energy E is aimed at a deuteron nucleus to try
to break it into a (neutron + proton) such that
the n and p move in the direction of the incident
g-ray. If E = B, show that this cannot happen.
Hence, calculate how much E should be bigger
than B for such a process to happen.
OBJECTIVE TYPE QUESTIONS
6. (a) : In nuclear reactors, cadmium rods are used as
control rods.
7. (a) : Using, E = mc2
Here, m = 1 g = 1 × 10–3 kg,
c = 3 × 108 m s–1
\ E = 10–3 × 9 × 1016 = 9 × 1013 J
8. (b)
9. (b) : Order of magnitude of nuclear density = 1017 kg m–3
10. (a) : Here, P = 500 MW = 5 × 108 W, t = 1 h = 3600 s
Energy produced, E = P × t = 5 × 108 × 3600 = 18 × 1011 J
As E = mc2
1. (a) : Let the number of fissions per second be n.
Energy released = n × 200 MeV = n × 200 × 1.6 × 10–13 J
Energy required = power × time = 1 kW × 1 s = 1000 J
\ n × 200 × 1.6 × 10–13 = 1000
1000
10
or n =
=
× 1013 = 3.125 × 1013
−11 3.2
3.2 × 10
2. (c) : During nuclear fusion, two or more lighter nuclei
combine to form a heavier nucleus.
3. (d) : In any nuclear reaction mass, energy and momentum
all are conserved.
4. (a) : Number of atoms in 1 kg of pure 239Pu
6.023 × 1023
=
× 1000 = 2.52 × 1024
239
As average energy released per fission is 180 MeV
\ Total energy released = 2.52 × 1024 × 180 MeV
= 4.53 × 1026 MeV
5. (b) : –10e + +10e → 2g
The total energy of the positron = 1 + 0.512 MeV
The total energy of the electron = 1 + 0.512 MeV
Energy of each photon
2(1+ 0.512)
=
= 1.512 MeV = 1.512 × 1.6 × 10 −13 J
2
hc
hc 6.62 × 10 −34 × 3 × 108
E=
⇒λ=
=
λ
E
1.512 × 1.6 × 10 −13
–13
= 8.2 × 10 m
\
E
18 × 1011 18 × 1011
= 2 × 10 −5 kg
=
m= 2=
8 2
16
c
(3 × 10 )
9 × 10
11. (a) : Atomic weight = Weighted average of the isotopes
6.01512 × 7.5 + 7.01600 × 92.5
=
(7.5 + 92.5)
45.1134 + 648.98
=
= 6.941u
100
12. (a) : Here, A1 = 197 and A2 = 107
1/ 3
\
R1  A1 
= 
R2  A2 
1/ 3
 197 
=

 107 
= 1.225 1.23
13. (d) : Due to mass defect, the rest mass of a nucleus is
always less than the sum of the rest masses of its constituent
nucleons.
111
Nuclei
20
10Ne
nucleus consists of 10 protons and 10 neutrons.
\ M1 < 10 (mp + mn)
14. (b) : Light energy emitted by stars is due to fusion of light
nuclei.
15. (c) : Dm = 2mp + 2mn – m(42He)
= 2 × 1.0073 + 2 × 1.0087 – 4.0015 = 0.0305 u
Binding energy = 0.0305 × 931 MeV = 28.4 MeV
16. (d) : Nuclides with same atomic number Z but different
mass number A are known as isotopes.
Nuclides with same mass number A but different atomic
number Z are known as isobars.
Nuclides with same neutron number N = (A – Z ) but different
atomic number Z are known as isotones.
2
3
1H and 1H are isotopes
3
3
2He and 1H are isobars
197
and 80Hg198 are isotones.
79Au
17. (a) : Nuclear radius, R = R0(A)1/3
where A is the mass number of a nucleus.
Given, R = 3.6 fm
∴ 3.6 fm = (1.2 fm)( A1/ 3 )
or
[∵R0 = 1.2 fm]
3
A = (3) = 27
0 .3
kg = 3 × 10 −3 kg
100
\ E = (Dm) c2 = 3 × 10–3 × (3 × 108)2 = 27 × 1013 J
19. (d) : Here, A1 = 27, A2 = 125
R1 = 3.6 fm
18. (d) : Here, Dm = 0.3% of 1 kg =
1/ 3
1/ 3
R2  A2 
5
 125 
=  =
=
 27 
3
R1  A1 
\ R = 5 R = 5 × 3.6 = 6 fm
2
3 1 3
20. (b) : Let abundance of 105B is x % than abundance of 115B
will be (100 – x)%.
Atomic mass of boron
x [10.01294 u] + (100 − x )[11.00931u]
=
100
⇒ 100 × 10811 u = 1100.931 u – 0.99637x u
19.831
Solving we get, x =
= 19.9%
0.99637
As,
So, relative abundance of 105B isotope = 19.9%
Relative abundance of 115B isotope = 80.1%
21. (a) : As nearly 99.9% mass of atom is in nucleus
Mass of nucleus 99.9
∴
=
= 0.99 ≈ 1
Mass of atom
100
22. (a) : Since, the nuclei of deuterium and tritium are
isotopes of hydrogen, they must contain only one proton each.
But the masses of the nuclei of hydrogen, deuterium and
tritium are in the ratio of 1 : 2 : 3, because of presence of
neutral matter in deuterium and tritium nuclei.
23. (c)
24. (a) : R = R0A1/3
1
log R = log R0 +
log A
3
Which is of form, y = mx + c. So, the graph between log A and
log R is a straight line.
25. (a) : Here, A1 = 198 and A2 = 107
1/ 3
1/ 3
R1  A1 
 198 
=  =
 = 1.23
R2  A2 
 107 
26. (b)
27. (b) : Fast neutrons are slowed down by elastic scattering
with light nuclei as each collision takes away nearly 50% of
energy.
28. (d) : Reactions I and II represent fission of uranium
isotope 235
92U, when bombarded with neutrons that breaks
it into two intermediate mass nuclear fragments. However,
reaction III represents two deuterons fuses together to from
the light isotope of helium.
29. (c) : On an average 2.5 neutrons are released per fission
of the uranium atom.
The energy of the neutron released per fission of the uranium
atom is 2 MeV.
30. (a) : In fission process, when a parent nucleus breaks
into daughter products, then some mass is lost in the form of
energy. Thus,
mass of fission products < mass of parent nucleus.
Mass of fission products
⇒
<1
Mass of parent nucleus
\
31. (d) : Let the number of fissions per second be n.
Energy released = n × 200 MeV = n × 200 × 1.6 × 10–13 J
Energy required = power × time = 10 kW × 1 s = 10000 J
\ n × 200 × 1.6 × 10–13 = 10000
10000
or n =
= 3.125 × 1014
−11
3.2 × 10
32. (a)
33. (b) : As only 0.1% of the original mass is converted into
energy, hence out of 1 kg mass 1 g is converted into energy.
\ Energy released during fission, E = Dmc2
= 1 g × (3 × 108 m s–1)2 = 10–3 × 9 × 1016 J = 9 × 1013 J
34. (a)
35. (a)
36. (d) : All options are basic properties of nuclear forces. So,
all options are correct.
37. (d) : The nuclear force is of short range and the range of
nuclear force is the order of 1.4 × 10–15 m.
Now, volume ∝ R3 ∝ A
CBSE Board Term-II Physics Class-12
112
38. (d)
39. (a) : Nuclear force is much stronger than the electrostatic
force inside the nucleus i.e., at distances of the order of fermi.
At 40 Å, nuclear force is ineffective and only electrostatic
force of repulsion is present. This is very high at this distance
because nuclear force is not acting now and the gravitational
force is very feeble. Fnuclear << Felectrostatic in this case.
40. (a)
41. (a) : Due to electrostatic forces between two protons
(like charges) there is a force of repulsion. However, when
the distance between them is ~ 10–15 m they come under
the influence of the short range, strong nuclear forces. (The
range of the nuclear forces is ~ 10–15 m). These forces are
attractive forces and charge independent. The net force on the
protons is attractive as nuclear forces are much stronger than
electrostatic forces. The protons attract each other.
42. (c) : Nuclear forces are strongest when the number of
protons equals the number of neutrons. An excess of protons
compared to neutrons weakens the nuclear force. Also too
many neutrons compared to protons inside the nucleus
weaken the nuclear forces. The electrostatic force which is a
hundred times less than the nuclear force is not the cause.
43. (b) : A neutron is slightly heavier than a proton. As
neutrons are chargeless particles they penetrate matter more
than protons. Therefore, they are more effective than protons
in fission reactions.
44. (c) : In both fission and fusion large amount of energy is
released. Assertion is correct. Charge, mass, momentum and
energy, all are conserved.
45. (a) : Experimentally, it is found that the average radius of
a nucleus is given by
R = R0A1/3 where R0 = 1.1 × 10–15 m = 1.1 fm
and A = mass number
4
4
The volume of a nucleus is V = p R3 = p R03A.
3
3
Now as the masses of a proton and a neutron are roughly
equal, say m, the mass of a nucleus is also roughly proportional
to the mass number A, M = mA. Hence density within a
M
mA
m
nucleus, ρ = =
is independent of the
=
4 3
4 3
V
πR
πR A
3 0
3 0
mass number A.
46. (a) : When uranium is bombarded by slow neutrons the
reaction is represented as
235
1
92 U + 0n
92
1
→ 141
56 Ba + 36Kr + 30n + Energy.
As more neutrons are produced, the reason is correct. These
additional neutrons strike other uranium nuclei to produce
even more neutrons. Thus a chain reaction is established.
47. (c) : Cadmium rods are used in a nuclear reactor to
control the rate of fission. The cadmium rods do not slow
down or speed up the neutrons produced in a fission reaction
of 235U. Instead they absorb the neutrons thereby regulating
the power level of the reactor.
48. (b) : Percentage of mass converted to energy in a fission
reaction is 0.1% whereas in a fusion reaction it is 0.4%.
Consequently the amount of energy released is more in a
fusion than in a fission reaction.
It is not easy to control a fusion reaction.
49. (c) : When fusion is achieved by raising the temperature
of the system so that particles have enough kinetic energy
to overcome the coulomb repulsive behaviour, it is called
thermonuclear fusion. It is clean source of energy but energy
released in one fusion is much less than a single uranium
fission.
SUBJECTIVE TYPE QUESTIONS
1.
As, R = R0A1/3 = (1.2 × 10–15)(64)1/3 = 4.8 × 10–15 m
( R0 = 1.2 × 10–15 m)
2.
As, R = R0 A 3
1
1
1
RCu  Acu  3  64  3 4
⇒
=
 =  =
R Al  AAl 
3
 27 
3. Nuclear force is the strongest attractive force which binds
the protons and neutrons together inside a tiny nucleus.
4. Radius R of a nucleus changes with the nucleon number
A of the nucleus as
R
A 
Hence, 2 =  2 
R1  A1 
\
1/ 3
 128 
=
 16 
1/ 3
= (8)1/3 = 2
R2 = 2R1 = 2(3 × 10–15)m = 6 × 10–15 m
5. 0n1 → 1H1 + –1e0 + ν + Q
Dm = mn – ma – me
= (1.6725 × 10–27 – 1.6725 × 10–27 – 9 × 10–31) kg = – 9 × 10–31 kg
Energy = Dmc2 = 9 × 10–31 × (3 × 108)2 = 0.511 MeV
6. 3Li6 + 0n1 → 2He4 + 1H3 + Q
Total initial mass = 6.015126 + 1.0086654 = 7.0237914 amu
Total final mass = 4.0026044 + 3.01 = 7.0126044 amu
Mass defect, Dm = 7.0237914 – 7.0126044 = 0.0111870 amu
Energy released, Q = 0.0111870 × 931 = 10.415 MeV.
7.
Density of nuclear matter =
Mass of nucleus
Volume
A × 1 amu
, where R = R0 A1/ 3 ( R0 = 1.2 × 10–15 m)
4
πR 3
3
A × 1 amu 1 amu 3 amu
Density, ρ =
=
=
3
4
4
πR03A
πR03 4 πR0
3
3
3 × 1.66 × 10 −27
=
4 × 3.14 × (1.2 × 10 −15 )3
ρ=
= 2.27 × 1017 kg m–3
113
Nuclei
As R0 is constant, r is constant
So, nuclear density is constant irrespective of mass number or size.
8.
Average atomic mass of neon with the given abundances,
90.51 (19.99 u) + 0.27 (20.99 u) + 9.22 (21.99 u)
A=
100
2017.7
A=
u = 20.18 u
100
14.
B.E. = (2mp + 2mn – mHe)u × 931 MeV
= [2(1.007276 u + 1.008665 u) – 4.001508 u] × 931 MeV
= [4.031882 u – 4.001508 u] × 931 MeV
= 0.030374 × 931 MeV
B.E. = 28.3 MeV
9.
The mass lost in the process,
Dm = 235.0439 + 1.00867 – (139.9054 + 93.9063 + 2.01734)
= 0.22353 u
The corresponding energy released
= Dmc2 = 0.22353 × 931 MeV = 208 MeV
10.
(i) E = [MU + mn – MBa – MKr – 3mn] × 931 = 200.57 MeV
NA
×E
(ii) Energy released =
235
200.57 × 106 × 1.6 × 10−19 × 6.023 × 1023
=
= 22.84 MWh
235 × 3.6 × 106
11.
A certain number of neutrons and protons are brought
together to form a nucleus of a certain charge and mass, an
energy DEb will be released in this process.
The energy DEb is called the binding energy of the nucleus.
If we separate a nucleus into its nucleons we would have to
transfer a total energy equal to DEb, to the nucleons.
23
For mirror isobar of 11
Na, Z 2 = N1 = 12 and N2 = Z1 = 11.
92
1
Example : 92 U235 + 0 n1 → 141
56 Ba + 36Kr + 30 n + Q
The energy (Q) released was estimated to be 200 MeV per
fission and is equivalent to the difference in masses of the
nuclei before and after the fission.
12. Given fusion reaction,
Energy released = final B.E. – initial B.E.
= 7.73 – (2.23 + 2.23) = 3.27 MeV.
13.
We know that the density of solid lead is much greater
than the density of gaseous oxygen. But here we are checking
about the densities of their nuclei.
If the mass of a neutron is m, then mass of nuclei
M = Am.
4
Now, volume V = pR3
3
4
1/3
But R = R0A or V = p(R0A1/3)3
3
4
V = pR03A
3
M
The density of the nucleus r =
V
Am
3m
=
=
= constant.
3
4 3
πR0 A 4 πR0
3
Nuclear density is almost a constant, whether it is lead or
oxygen.
23
15. (a) For 11 Na , Z1 = 11, N1 = 12
23
23
Thus mirror isobar of 11
Na is 12
Mg .
23
(b) As 12
Mg contains even number of protons (12) against
23
11 Na
which contains odd number of protons (11), hence
23
12 Mg
has greater binding energy comparatively.
16.
The stability of a nucleus depends on its neutron to proton
ratio. More is the number of protons in the nucleus, greater is
the electrical forces between them. Therefore more neutrons
are needed to provide the strong attractive force necessary to
keep the nucleus stable.
17.
Given m = 2 kg, P = 800 W.
Here two deuterium nuclei produce 3.27 MeV energy
= 3.27 × 106 × 1.6 × 10–19 = 5.232 × 10–13 J
5.232 × 10 −13
= 2.616 × 10–13 J
2
No. of deuterium atom in 2 kg
\ Energy per nuclei =
6.023 × 1023 × 2000
= 6.023 × 1026 atom
2
\ Total energy = 6.023 × 1026 × 2.616 × 10–13
= 15.75 × 1013 J
Energy
Total energy
Power =
⇒t =
Time
Power
13
15.75 × 10
⇒ t=
= 1.96 × 1011s
800
=
=
1.96 × 1011
= 6.2 × 103 years
365 × 24 × 60 × 60
4
18.
He + 4He → 8Be + Q
(i) Dm = 2 × 4.0026 – 8.0053 = 8.0052 – 8.0053
= – 0.0001 amu
(ii) Q = (2mHe – mBe) c 2
MeV
= (2 × 4.0026 – 8.0053)c2 × 931 2
c
= –0.0931 MeV = –93.1 keV
(iii) Since Q is negative, the fusion is not energetically
favourable.
19.
Rate of electrical energy generation is 250 MW
= 250 × 106 W (or J s–1).
CBSE Board Term-II Physics Class-12
114
So, electrical energy generation is 250 MW = 250 × 106 W
(or J s–1).
Therefore, electrical energy generated in 1 year is
(250 × 106 J s–1) × (365 × 24 × 60 × 60 s) = 7.884 × 1015 J
Thermal energy from fission of one 235U nucleus is 200 MeV =
200 × 1.6 × 10–13 = 3.2 × 10–11 J
Since the efficiency is 25%, the electrical energy obtained from
the fission one 235U nucleus,
25
E1 = 3.2 × 10 −11 ×
= 8.0 × 10 −12 J
100
\ The number of fissions of 235U required in one year,
N=
15
7.884 × 10
−12
9.855 × 1026
23
∆m = 4m (11H) − m (24 He)
Energy released
Q = [4m (11H) − m (24 He)] × 931MeV
(21H) may be greater than the product nucleus 3He and
2
the outgoing neutron 01n. So from the law of conservation of
mass-energy some energy (3.27 MeV) is released due to mass
defect in the nuclear reaction. This energy is called Q-value
of the nuclear reaction.
24. Applying principle of conservation of energy,
,4
Binding energy, E ′ = 918 me = 2.2 MeV ...(ii)
8 ∈20 h 2
Dividing eq. (ii) by (i), we get
\
4
6
2.2 × 10
 e′
918   =
e
13.6
e′
= (176.21)1/ 4 = 3.64
e
Nuclear Fission
1. The process of splitting of 1.
a heavy nucleus into two
nuclei of nearly comparable
masses with liberation of
energy is called nuclear
fission.
Example:
n
22.
Energy released = Dm × 931 MeV
= 1.636 × 103
Replacing e by e′ and m by m′, reduced mass of neutron –
proton,
M × M M 1836m
m′ =
= =
= 918 m 2
M +M 2
(M = mass of neutron/proton)
235
1
92 U +0
3. The products of nuclear fission 3. The products of nuclear
reaction are radioactive.
fusion are not radioactive.
= [4 × 1.007825 – 4.002603] × 931 MeV = 26.72 MeV.
23.
I n a given nuclear reaction, the sum of the masses
of the target nucleus (21H) and the bombarding particle
6.02 × 10
Therefore, mass of 235U required per year,
m = 1.636 × 103 × 235 = 3.844 × 105 g = 384.4 kg
20.
We know that binding energy of hydrogen atom in
ground state,
me 4
= 13.6 eV …(i)
E =
8 ∈20 h 2
21.
2. The lighter nuclei have
to be brought very close
to each other against
electrostatic repulsion.
= 9.855 × 1026
8.0 × 10
Number of moles of 235U required per year,
N=
2. A suitable bullet or
projectile like neutron is
needed.
→141
56
Ba
1
+ 92
36 Kr + 30 n
+Q
Nuclear Fusion
When two or more
than two light nuclei
fuse together to form
heavy nucleus with the
liberation of energy, the
process is called nuclear
fusion.
Example:
2
2
4
1 H +1 H →2
He
+ 24 MeV
( )
2
pn2 p p
+
…(i)
2m 2m
From law of conservation of momentum,
pn + pp = E/c
when E = B,
from equation (i), pn = pp = 0
\ Process cannot take place.
For process to take place, let E be very slightly bigger than B so
that E = B + l, (l << B.)
E – B = Kn + Kp =
2
p2 pp
λ= n +
2m 2m
1 2
λ=
[p + ( p p − E /c )2 ]
2m p
pp =
 E2

E
E
m
±
−
−
λ


2c
4c 2  2c 2

For pp (momentum of proton) to be real, the determinant must
be positive.
 E2

E2
B2
m
−
−
λ
≥
0
or
λ
=


4c 2  2c 2
4mc 2 4mc 2
E2
Therefore, for the given process to occur the value of energy E
must be greater than binding energy B by a factor l which is
equal to

B2
4mc 2
.
CHAPTER
14
Semiconductor Electronics :
Materials, Devices and
Simple Circuits
Recap Notes
Classification of solids on the basis of
their conductivity : On the basis of the
relative values of electrical conductivity
(s) and resistivity (r = 1/s), the solids are
broadly classified as,
X Metals : Those solids which have high
conductivity and very low resistivity. The
value of conductivity for metals lies in
between 102 to 108 S m–1 and of resistivity
in between 10–2 to 10–8 W m.
X Insulators : Those solids which have
low conductivity and high resistivity.
The value of conductivity for insulators
lies between 10–11 to 10–19 S m–1 and of
resistivity between 1011 to 1019 W m.
X Semiconductors : Those solids which
have
conductivity
and
resistivity
intermediate to metals and insulators. The
value of conductivity for semiconductors
lies in between 105 to 10–6 S m–1 and of
resistivity between 10–5 to 106 W m.
Energy bands of solids or band theory of
solids
X Valence band : This band contains
valence electrons. This band may be
partially or completely filled with
electrons. This band is never empty.
Electrons in this band do not contribute
to electric current.
X Conduction band : In this band,
electrons are rarely present. This band is
either empty or partially filled. Electrons
in the conduction band are known as free
electrons. These electrons contribute to
the electric current.
X
X
X
X
X
Forbidden energy gap or forbidden
band : The energy gap between the
valence band and conduction band
is known as forbidden energy gap or
forbidden band. No electrons are present
in this gap. It is a measure of energy band
gap.
The minimum energy required for shifting
electrons from valence band to conduction
band is known as energy band gap.
If l is the wavelength of radiation used in
shifting the electron from valence band to
conduction band, then energy band gap is
hc
E g = hυ =
λ
where h is called Planck’s constant and c
is the speed of light.
The forbidden energy gap Eg in a
semiconductor depends upon temperature.
Fermi energy : It is the maximum possible
energy possessed by free electrons of
a material at absolute zero temperature
(i.e., 0 K)
Differences between metals, insulators
and semiconductors on the basis of
band theory
X Metals
– In metals either the conduction band is
partially filled or conduction band and
valence band partially overlap each
other.
– In metals, there is no forbidden
energy gap between the valence and
conduction bands.
CBSE Board Term-II Physics Class-12
116
– Hole : It is a seat of positive charge
which is produced when an electron
breaks away from a covalent bond in
a semiconductor. Hole has a positive
charge equal to that of electron.
Mobility of hole is smaller than that of
electron.
X
Insulators
– In insulators, valence band is
completely filled and conduction band
is completely empty.
– In insulators, there is a very wide
forbidden energy gap between the
valence and conduction bands. It is of
the order of 5 eV or more.
Empty
X
Semiconductors
– In semiconductors, valence band is
completely filled and the conduction
band is empty.
– In semiconductors, there is a small
forbidden energy gap between the
valence and the conduction bands. It is
of the order of 1 eV. For silicon, it is 1.1
eV and for germanium it is 0.72 eV.
– At absolute zero, semiconductors behave
as a perfect insulator.
Intrinsic semiconductor : A pure semiconductor which is free from every impurity
is known as intrinsic semiconductor.
Germanium (Ge) and silicon (Si) are
the important examples of intrinsic
semiconductors.
X In intrinsic semiconductor, ne = nh = ni
where ne , nh are number density of
electrons in conduction band and number
density of holes in valence band, ni is the
intrinsic carrier concentration.
X When an electric field is applied across
an intrinsic semiconductor, electrons
and holes move in opposite directions so
that total current (I) through the pure
semiconductor is given by
I = Ie + Ih
where Ie is the free electron current and
Ih is the hole current.
X Effect of temperature on conductivity
of intrinsic semiconductor
– An intrinsic semiconductor will behave
as a perfect insulator at absolute zero.
– With increasing temperature, the
density of hole-electron pairs increases
and hence the conductivity of an
intrinsic semiconductor increases with
increase in temperature. In other words,
the resistivity (inverse of conductivity)
decreases as the temperature increases.
– The semiconductors have negative
temperature coefficient of resistance.
Doping : It is a process of deliberate
addition of a desirable impurity to a pure
semiconductor in order to increase its
conductivity. The impurity atoms added are
known as dopants.
Extrinsic semiconductor : A doped
semiconductor is known as extrinsic
semiconductor. Extrinsic semiconductors are
of two types :
117
Semiconductor Electronics : Materials, Devices and Simple Circuits
X
X
n-type semiconductor
– When a pure semiconductor of Si or
Ge (tetravalent) is doped with a group
V pentavalent impurities like arsenic
(As), antimony (Sb), phosphorus (P)
etc, we obtain a n-type semiconductor.
The pentavalent impurity atoms are
known as donor atoms.
– It is called n-type semiconductor
because the conduction of electricity in
such semiconductor is due to motion of
electrons i.e., negative charges.
– It is called donor type semiconductor,
because
the
doped
impurity
atom donates one free electron to
semiconductor for conduction.
– In n-type semiconductor electrons
are majority carriers and holes are
minority carriers.
– The
representation
of
n-type
semiconductor is as shown in the figure.
– n-type semiconductor is neutral.
– In n-type semiconductor
ne ≈ Nd > > nh
where Nd is the density of donor atoms.
p-type semiconductor : When a pure
semiconductor of Si or Ge (tetravalent) is
doped with a group III trivalent impurities
like aluminium (Al), boron (B), indium
(In) etc, we obtain a p-type semiconductor.
The trivalent impurity atoms are known
as acceptor atoms.
– It is called p-type
because the
conduction of electricity in such
semiconductor is due to motion of holes
i.e., positive charges.
– It is called acceptor type semiconductor
because the doped impurity atom
creates
a hole in semiconductor
which accepts the electron, resulting
conduction in p-type semiconductor.
– In p-type semiconductor, holes are
majority carriers and electrons are
minority carriers.
– The
representation
of
p-type
semiconductor is as shown in the figure.
– p-type semiconductor is neutral.
– In p-type semiconductor
nh ≈ Na > > ne
where Na is the density of acceptor atoms.
Mass action law : Under thermal equilibrium,
the product of the free negative and positive
concentrations is a constant independent of
the amount of donor and acceptor impurity
doping. This relationship is known as the
mass action law and is given by
nenh = ni2
where ne , nh are the number density of
electrons and holes respectively and ni is the
intrinsic carriers concentration.
X Electrical conductivity in semiconductor :
The conductivity of the semiconductor is
given by s = e(neme + nhmh)
where me and mh are the electron and hole
mobilities, ne and nh are the electron and
hole densities, e is the electronic charge.
– The conductivity of an intrinsic
semiconductor is si = nie(me + mh)
– The conductivity of n-type
semiconductor is sn = eNdme
– The conductivity of p-type
semiconductor is sp = eNamh
p-n junction : When donor impurities
are introduced into one side and acceptors
into the other side of a single crystal of an
intrinsic semiconductor, a p-n junction is
formed. It is also known as junction diode.
The most important characteristic of a p-n
junction is its ability to conduct current in
one direction only. In the other (reverse)
direction it offers very high resistance. It is
symbolically represented by
X
Depletion region : In the vicinity
of junction, the region containing the
uncompensated acceptor and donor ions
is known as depletion region. There is a
depletion of mobile charges (holes and
free electrons) in this region. Since this
region has immobile (fixed) ions which
CBSE Board Term-II Physics Class-12
118
X
X
X
are electrically charged it is also known as
the space charge region. The electric field
between the acceptor and the donor ions is
known as a barrier. The physical distance
from one side of the barrier to the other
is known as the width of the barrier. The
difference of potential from one side of the
barrier to the other side is known as the
height of the barrier.
– For a silicon p-n junction, the barrier
potential is about 0.7 V, whereas
for a germanium p-n junction it is
approximately 0.3 V.
– The width of the depletion layer and
magnitude of potential barrier depend
upon the nature of the material of
semiconductor and the concentration
of impurity atoms. The thickness of the
depletion region is of the order of one
tenth of a micrometre.
Forward biasing of a p-n junction :
When the positive terminal of external
battery is connected to p-side and negative
to n-side of p-n junction, then the p-n
junction is said to be forward biased.
– In forward biasing, the width of the
depletion region decreases and barrier
height reduces.
– The resistance of the p-n junction
becomes low in forward biasing.
Reverse biasing of a p-n junction :
When the positive terminal of the
external battery is connected to n-side
and the negative terminal to p-side of a
p-n junction, then the p-n junction is said
to be reverse biased.
– In reverse biasing, the width of the
depletion region increases and barrier
height increases.
– The resistance of the p-n junction
becomes high in reverse biasing.
Breakdown voltage : A very small
current flows through p-n junction, when
it is reverse biased. The flow of the current
is due to the movement of minority charge
carriers. The reverse current is almost
independent of the applied voltage.
However, if the reverse bias voltage is
continuously increased, for a certain
reverse voltage, the current through the
X
p-n junction will increase abruptly. This
reverse bias voltage is thus known as
breakdown voltage. There can be two
different causes for the breakdown. One is
known as zener breakdown and the other
is known as avalanche breakdown.
I-V characteristics of a p-n junction :
The I-V characteristics of a p-n junction
do not obey Ohm’s law. The I-V
characteristics of a p-n junction are as
shown in the figure.
– Knee voltage : In forward biasing,
the voltage at which the current starts
to increase rapidly is known as cut-in
or knee voltage. For germanium it is
0.3 V while for silicon it is 0.7 V.
– Dynamic resistance : It is defined as
the ratio of a small change in voltage (DV)
applied across the p-n junction to a
small change in current DI through the
junction.
∆V
rd =
∆I
Ideal diode : A diode
permits only unidirectional
conduction. It conducts well
in the forward direction
and poorly in the reverse
direction. It would have been ideal if a diode
acts as a perfect conductor (with zero voltage
across it) when it is forward biased, and as
a perfect insulator (with no current flows
through it) when it is reverse biased. The I-V
characteristics of an ideal diode as shown in
figure.
X An ideal diode acts like an automatic
switch.
119
Semiconductor Electronics : Materials, Devices and Simple Circuits
X
In forward bias, it acts as a closed switch
whereas in reverse bias it acts as an open
switch as shown in the figure.
r=
rms value of the components of wave
average or dc value
2
I

r =  rms  − 1
 I 
dc
X
Rectifier : It is a device which converts
ac voltage to dc voltage. Diode is used as a
rectifier. Rectifier is based on the fact that,
a forward bias p-n junction conducts and a
reverse bias p-n junction does not conduct.
X Half wave rectifier : Diode conducts
corresponding to positive half cycle and
does not conduct during negative half
cycle. Hence, AC is converted by diode into
undirectional pulsating DC. This action is
known as half-wave rectification.
X
X
Special Purpose p-n Junction Diodes :
X Light emitting diode (LED) : It converts
electrical energy into light energy. It is a
heavily doped p-n junction which operates
under forward bias and emits spontaneous
radiation.
– The I-V characteristics of a LED is
similar to that of Si junction diode. But
the threshold voltages are much higher
and slightly different for each colour.
The reverse breakdown voltages of
LEDs are very low, typically around 5 V.
– The semiconductor used for fabrication
of visible LEDs must at least have
a band gap of 1.8 eV. The compound
semiconductor
gallium
arsenide
phosphide (GaAsP) is used for making
LEDs of different colours. GaAs is used
for making infrared LED.
– The symbol of a LED is shown in the
figure.
X
Photodiode : A photodiode is a special
type p-n junction diode fabricated with a
transparent window to allow light to fall
on the diode. It is operated under reverse
bias. When it is illuminated with light of
photon energy greater than the energy
gap of the semiconductor, electron-hole
pairs are generated in near depletion
region.
– The symbol of a photodiode is shown in
the figure below.
X
Solar cell : It converts solar energy into
electrical energy. A solar cell is basically
a p-n junction which generates emf when
solar radiation falls on the p-n junction. It
works on the same principle (photovoltaic
effect) as the photodiode, except that no
external bias is applied and the junction
area is kept large.
Full wave rectifier : The circuit diagram,
input and output waveforms for a full
wave rectifier are as shown in the figure.
Ripple factor : The ripple factor is a
measure of purity of the dc output of a
rectifier, and is defined as
1. The equivalent resistance of the circuit
shown in figure between the points A and B if
VA < VB is 20 (a) 10 W
(b) 20 W
A
B
(c) 5 W
(d) 40 W
20 2. Which of the following statements is correct?
(a) Hole is an antiparticle of electron.
(b) Hole is a vacancy created when an electron
leaves a covalent bond.
(c) Hole is the absence of free electrons.
(d) Hole is an artificially created particle.
3. Which of the following statements is
incorrect for the depletion region of a diode?
(a) There are mobile charges exist.
(b) Equal number of holes and electrons exist,
making the region neutral.
(c) Recombination of holes and electrons has
taken place.
(d) None of these.
4. In an unbiased p-n junction, holes diffuse
from the p-region to n-region because
(a) free electrons in the n-region attract them
(b) they move across the junction by the
potential difference
(c) hole concentration in p-region is more as
compared to n-region
(d) all of these.
5. A potential barrier of 0.3 V exists across
a p-n junction. If the depletion region is 1 mm
wide, what is the intensity of electric field in
this region?
(a) 2 × 105 V m–1
(b) 3 × 105 V m–1
5
–1
(c) 4 × 10 V m
(d) 5 × 105 V m–1
6. The dominant mechanism for motion of
charge carriers in forward and reverse biased
silicon p-n junction are
(a) drift in forward bias, diffusion in reverse
bias
(b) diffusion in forward bias, drift in reverse
bias
(c) diffusion in both forward and reverse bias
(d) drift in both forward and reverse bias.
7. When the voltage drop across a p-n
junction diode is increased from 0.65 V to
0.70 V, the change in the diode current is
5 mA. The dynamic resistance of the diode is
(a) 20 W
(b) 50 W (c) 10 W
(d) 80 W
8. In the circuit shown if drift current for the
diode is 20 mA, the potential
4V
difference across the diode is
(a) 2 V
15 (b) 4.5 V
(c) 4 V
(d) 2.5 V
9. If a small amount of antimony is added to
germanium crystal
(a) its resistance is increased
(b) it becomes a p-type semiconductor
(c) there will be more free electrons than holes
in the semiconductor
(d) none of these.
10. A sinusoidal voltage of rms value 220 V is
applied to a diode and a resistor R in the circuit
shown in figure, so that half wave rectification
occurs. If the diode is ideal, what is the rms
voltage across R1?
(a) 55 2 V
(c) 110 2 V
(b) 110 V
(d) 220 2 V
11. If the energy of a proton of sodium light
(l = 589 nm) equals the band gap of
121
Semiconductor Electronics : Materials, Devices and Simple Circuits
semiconductor, the minimum energy required to
create hole electron pair
(a) 1.1 eV
(b) 2.1 eV (c) 3.2 eV (d) 1.5 eV
12. Which of the following circuits provides full
wave rectification of an ac input?
(a)
(b)
(c)
(c) Eg(T) = 1.10 – 3.60 × 10–4T eV
(d) Eg(T) = 1.10 + 3.60 × 10–4T eV
16. The maximum wavelength of electromagnetic
radiation, which can create a hole-electron pair
in germanium. (Given that forbidden energy gap
in germanium is 0.72 eV)
(a) 1.7 × 10–6 m
(b) 1.5 × 10–5 m
(c) 1.3 × 10–4 m
(d) 1.9 × 10 –5 m
17. The circuit shown
D1
in the figure contains
D2
two diodes each with a
forward resistance of
30 W and with infinite
backward resistance. If
the battery is 3 V, the
current through the 50 W resistance (in ampere)
is
(a) zero
(b) 0.01
(c) 0.02
(d) 0.03
18. Which of the junction diodes shown below
are forward biased?
(a)
(b)
(c)
(d)
(d)
13. Carbon, silicon and germanium have four
valence electrons each. These are characterised
by valence and conduction bands separated by
energy band gap respectively equal to (E g) C,
(E g ) Si and (E g ) Ge . Which of the following
statements is true?
(a) (Eg)Si < (Eg)Ge < (Eg)C
(b) (Eg)C < (Eg)Ge < (Eg)Si
19. A potential barrier of 0.50 V exists in a p-n
junction. If the depletion region is 5.0 × 10–7 m
thick, what is the electric field in this region?
(a) 1 × 103 V m–1
(b) 1.0 × 106 V m–1
2
–1
(c) 1 × 10 V m
(d) 1 × 104 V m–1
(d)
20. The breakdown in a reverse biased p-n
junction diode is more likely to occur due to
(a) large velocity of the minority charge carriers
if the doping concentration is small
(b) large velocity of the minority charge carriers
if the doping concentration is large
(c) strong electric field in a depletion region if
the doping concentration is small
(d) none of these
15. Which of the following equations correctly
represents the temperature variation of energy
gap between the conduction and valence bands
for Si?
(a) Eg(T) = 0.70 – 2.23 × 10–4T eV
(b) Eg(T) = 0.70 + 2.23 × 10–4T eV
21. In a full wave junction diode rectifier the
input ac has rms value of 20 V. The transformer
used is a step up transformer having primary
and secondary turn ratio 1 : 2. The dc voltage
in the rectified output is
(a) 12 V
(b) 24 V (c) 36 V
(d) 42 V
(c) (Eg)C > (Eg)Si > (Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
14. A forward biased diode is
(a)
(b)
(c)
CBSE Board Term-II Physics Class-12
122
23. A semiconductor has equal electron and
hole concentration of 6 × 108 per m3. On doping
with certain impurity, electron concentration
increases to 9 × 10 12 per m 3 . The new hole
concentration is
(a) 2 × 104 per m3
(b) 2 × 102 per m2
4
3
(c) 4 × 10 per m
(d) 4 × 102 per m3
24. A p-n photodiode is made of a material with
a band gap of 2 eV. The minimum frequency
of the radiation that can be absorbed by the
material is nearly (Take hc = 1240 eV nm)
(a) 1 × 1014 Hz
(b) 20 × 1014 Hz
14
(c) 10 × 10 Hz
(d) 5 × 1014 Hz
25. A p-n photodiode is fabricated from a
semiconductor with a band gap of 2.5 eV. The
signal wavelength is
(a) 6000 Å
(b) 6000 nm
(c) 4000 nm
(d) 5000 Å
26. Potential barrier developed in a junction
diode opposes the flow of
(a) minority carrier in both regions only
(b) majority carriers only
(c) electrons in p region
(d) holes in p region
27. In pure semiconductor, the number of
conduction electrons is 6 × 1018 per cubic metre.
How many holes are there in a sample of size
1 cm × 1 cm × 1 mm ?
(a) 3 × 1010
(b) 6 × 1011
11
(c) 3 × 10
(d) 6 × 1010
28. Mobilities of electrons and holes in a sample
of intrinsic germanium at room temperature are
0.54 m2 V –1 s –1 and 0.18 m2 V –1 s–1 respectively.
If the electron and hole densities are equal to
3.6 × 10 19 m–3, the germanium conductivity is
(a) 4.14 S m–1
(b) 2.12 S m–1
–1
(c) 1.13 S m
(d) 5.6 S m–1
29. The probability of electrons to be found in the
conduction band of an intrinsic semiconductor of
finite temperature
(a) increases exponentially with increasing
band gap.
(b) decreases exponentially with increasing
band gap.
(c) decreases with increasing temperature.
(d) is independent of the temperature and band
gap.
30. The electrical conductivity of a semiconductor
increases when electromagnetic radiation of
wavelength shorter than 2480 nm is incident on
it. The band gap (in eV) for semiconductor is
(a) 0.9
(b) 0.7
(c) 0.5
(d) 1.1
31. The following table provides the set of values
of V and I obtained for a given diode. Let the
characteristics to be nearly linear, over this
range, the forward and reverse bias resistance
of the given diode respectively are
Forward biasing
Reverse biasing
(a) 10 W, 8 × 106 W
(c) 20 W, 8 × 106 W
V
2.0 V
2.4 V
0V
–2 V
I
60 mA
80 mA
0 mA
–0.25 mA
(b) 20 W, 4 × 105 W
(d) 10 W, 10 W
32. The mean free path of conduction electrons
in copper is about 4 × 10–8 m. The electric field
which can give on an average 2 eV energy to a
conduction electron in a block of copper is
(a) 4 × 106 V m–1
(b) 5 × 107 V m–1
(c) 10 × 107 V m–1
(d) 2.5 × 107 V m–1
33. The value of ripple factor for full wave
rectifier is
(a) 41%
(b) 141%
(c) 48.2%
(d) 121%
34. Current through the ideal diode as shown
in figure is
100 9
22. In a circuit as shown in the
figure, if the diode forward voltage
drop is 0.3 V, the voltage difference
between A and B is
(a) 1.3 V
(b) 2.3 V
(c) 0
(d) 0.5 V
2V
(a) zero
(b) 20 A
5V
(c)
1
A
20
(d)
1
A
50
35. In a half wave rectifier circuit operating
from 50 Hz mains frequency, the fundamental
frequency in the ripple would be
(a) 25 Hz
(b) 50 Hz (c) 70.7 Hz (d) 100 Hz
Semiconductor Electronics : Materials, Devices and Simple Circuits
123
Case I : Read the passage given below and answer
the following questions from 36 to 40.
Biasing of Diode
When the diode is forward biased, it is found
that beyond forward voltage V = Vk, called knee
voltage, the conductivity is very high. At this value
of battery biasing for p-n junction,the potential
barrier is overcome and the current increases
rapidly with increase in forward voltage.
When the diode is reverse biased, the reverse
bias voltage produces a very small current about
a few microamperes which almost remains
constant with bias. This small current is reverse
saturation current.
39. In the case of forward biasing of a p-n
junction diode, which one of the following figures
correctly depicts the direction of conventional
current (indicated by an arrow mark)?
36. In which of the following figures, the p-n
diode is forward biased.
(a)
(b)
(c)
(d)
37. Based on the V-I characteristics of the diode,
we can classify diode as
(a) bi-directional device
(b) ohmic device
(c) non-ohmic device
(d) passive element
38. The V-I characteristic of a diode is shown in
the figure. The ratio of forward to reverse bias
resistance is
I (mA)
20
15
10
–10
1 µA
(a) 100
(b) 106
0.7
0.8 V(volt)
(c) 10
(d) 10–6
(a)
–+ n
p –
+
–+
(c)
–+ n
p –
+
–+
(b)
–+ n
p –
+
–+
(d)
–+ n
p –
+
–+
40. If an ideal junction diode is connected as
shown, then the value of the current I is
I
(a) 0.013 A
(c) 0.01 A
(b) 0.02 A
(d) 0.1 A
Case II : Read the passage given below and
answer the following questions from 41 to 43.
Photodiode
A photodiode is an optoelectronic device in
which current carriers are generated by photons
through photo-excitation i.e., photoconduction
by light. It is a p-n junction fabricated from a
photosensitive semiconductor and provided with
a transparent window so as to allow light to fall
on its junction. A photodiode can turn its current
ON and OFF in nanoseconds. So, it can be used
as a fastest photo-detector.
Anode
hu
p n
Cathode
41.
(a)
(b)
(c)
Photodiode is a device
which is always operated in reverse bias.
which is always operated in forward bias.
in which photocurrent is independent of
intensity of incident radiation.
(d) which may be operated in both forward or
reverse bias.
42. To detect light of wavelength 500 nm,
the photodiode must be fabricated from a
semiconductor of minimum bandwidth of
(a) 1.24 eV
(b) 0.62 eV
(c) 2.48 eV
(d) 3.2 eV
CBSE Board Term-II Physics Class-12
124
43. Photodiode can be used as a photodetector
to detect
(a) optical signals
(b) electrical signals
(c) both (a) and (b)
(d) none of these.
Case III : Read the passage given below and
answer the following questions from 44 to 48.
p-n Junction Diode
A silicon p-n junction diode is connected to a
resistor R and a battery of voltage VB through a
milliammeter (mA) as shown in figure. The knee
voltage for this junction diode is VN = 0.7 V. The
p-n junction diode requires a minimum current
of 1 mA to attain a value higher than the knee
point on the I-V characteristics of this junction
diode. Assuming that the voltage V across the
junction is independent of the current above the
knee point.
A p-n junction is the basic building block of many
semiconductor devices like diodes. Important
process occurring during the formation of a p-n
junction are diffusion and drift. In an n-type
semiconductor concentration of electrons is more
as compared to holes. In a p-type semiconductor
concentration of holes is more as compared to
electrons.
48. Which of the below mentioned statement is
false regarding a p-n junction diode?
(a) Diodes are uncontrolled devices.
(b) Diodes are rectifying devices.
(c) Diodes are unidirectional devices.
(d) Diodes have three terminals.
Case IV : Read the passage given below and
answer the following questions from 49 to 52.
Potential Barrier
The potential barrier in the p-n junction diode
is the barrier in which the charge requires
additional force for crossing the region. In other
words, the barrier in which the charge carrier
stopped by the obstructive force is known as the
potential barrier.
When a p-type semiconductor is brought into a
close contact with n-type semiconductor, we get
a p-n junction with a barrier potential of 0.4 V
and width of depletion region is 4.0 × 10–7 m.
This p-n junction is forward biased with a
battery of voltage 3 V and negligible internal
resistance, in series with a resistor of resistance
R, ideal millimeter and key K as shown in figure.
When key is pressed, a current of 20 mA passes
through the diode.
+
+
mA
–
mA
–
R
+
VB
44.
the
(a)
(c)
If VB = 5 V, the maximum value of R so that
voltage V is above the knee point voltage is
40 kW
(b) 4.3 kW
5.0 kW
(d) 5.7 kW
45. If VB = 5 V, the value of R in order to establish
a current to 6 mA in the circuit is
(a) 833 W
(b) 717 W
(c) 950 W
(d) 733 W
–
3V
K
49. The intensity of the electric field in the
depletion region when p-n junction is unbiased
is
(a) 0.5 × 106 V m–1
(b) 1.0 × 106 V m–1
6
–1
(c) 2.0 × 10 V m
(d) 1.5 × 106 V m–1
50. The resistance of resistor R is
(a) 150 W
(b) 300 W
(c) 130 W
(d) 180 W
46. If V B = 6 V, the power dissipated in the
resistor R, when a current of 6 mA flows in the
circuit is
(a) 30.2 mW
(b) 30.8 mW
(c) 31.2 mW
(d) 31.8 mW
51. If the voltage of the potential barrier is
V0. A voltage V is applied to the input, at what
moment will the barrier disappear?
(a) V < V0
(b) V = V0
(c) V > V0
(d) V << V0
47. When the diode is reverse biased with a
voltage of 6 V and Vbi = 0.63 V. Calculate the
total potential.
(a) 9.27 V
(b) 6.63 V
(c) 5.27 V
(d) 0.63 V
52. If an electron with speed 4.0 × 105 m s–1
approaches the p-n junction from the n-side, the
speed with which it will enter the p-side is
(a) 1.39 × 105 m s–1
(b) 2.78 × 105 m s–1
6
–1
(c) 1.39 × 10 m s
(d) 2.78 × 106 m s–1
125
Semiconductor Electronics : Materials, Devices and Simple Circuits
For question numbers 53-60, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
53. Assertion (A) : The conductivity of a semiconductor increases with rise of temperature.
Reason (R) : On rising temperature covalent
bonds of semiconductor breaks.
54. Assertion (A) : p-n junction diode can be
used even at ultra high frequencies.
Reason (R) : Capacitative reactance of a p-n
junction diode increases as frequency increases.
55. Assertion (A) : The resistance of p-n
junction is low when forward biased and is high
when reverse biased.
Reason (R) : In reversed biased, the depletion
layer is reduced.
56. Assertion (A) : The direction of diffusion
current in a junction diode is from n-region to
p-region.
Reason (R) : The majority current carriers diffuse
from a region of lower concentration to a region of
higher concentration.
57. Assertion (A) : The resistivity of a semiconductor increases with temperature.
Reason (R) : The atoms of a semiconductor
vibrate with larger amplitude at higher
temperatures thereby increasing its resistivity.
58. Assertion (A) : The half-wave rectifier work
only for positive half cycle of ac.
Reason (R) : In half-wave rectifier only one
diode is used.
59. Assertion (A) : The ratio of free electrons to
holes in intrinsic semiconductor is greater than
one.
Reason (R) : The electrons are lighter particles
and holes are heavy particles.
60. Assertion (A) : In a semiconductor diode,
the reverse biased current is due to drift of free
electrons and holes.
Reason (R) : The drift of electrons and holes is
due to thermal excitations.
SUBJECTIVE TYPE QUESTIONS
Very Short Answer Type Questions (VSA)
1. What is the difference between an n-type
and a p-type extrinsic semiconductor?
2. What happens to the width of depletion
layer of a p-n junction when it is (i) forward
biased, (ii) reverse biased ?
3. Why cannot we use Si and Ge in fabrication
of visible LEDs?
4.
What is the function of a photodiode ?
6. In an n-type semiconductor, where does the
donor energy level lies.
7. Can the potential barrier across a p-n
junction be measured by simply connecting a
voltmeter across the junction?
8. Draw the output waveform across the
resistor.
5. Name the junction diode whose I-V
characteristics are drawn below :
A
+1 V
I
0
V
–1V
Input waveform at A
V0
CBSE Board Term-II Physics Class-12
126
Why are elemental dopants for Silicon or
10. How does an increase in doping concentration
Germanium usually chosen from group XIII or
affect the width of depletion layer of a p-n
group XV?
junction diode?
11. The number densities of electrons and hole
in pure Si at 27°C is 2 × 1016 m–3. When it is
doped with indium, the hole density increases to
4 × 1022 m–3, find the electron density in doped
silicon.
16. (a) Mention the important considerations
required while fabricating a p-n junction diode
to be used as a light emitting diode (LED).
(b) What should be the order of band gap of an
LED if it is required to emit light in the visible
range?
9.
12. A 3 V b a t t e r y m a y b e
connected across the points
A and B as shown. Assuming
ideal diode, find the current
supplied by battery if the
positive terminal of the battery
is connected to the point A.
J1
10 W
30 W
17. (a) In the following diagram, which bulb out
of B1 and B2 will glow and why?
J2
A
B
13. Why photodiodes are required to operate in
reverse bias ? Explain.
14. The circuit shown in the figure has two
oppositely connected ideal diodes connected in
parallel. Find the current flowing through each
diode in the circuit.
(b) If the forward voltage in a semiconductor
diode is changed from 0.5 V to 0.7 V, then the
forward current changes by 1.0 mA. Find the
forward resistance of diode junction.
18. Three photo diodes D1, D2 and D3 are made
of semiconductors having band gaps of 2.5 eV,
2 eV and 3 eV, respectively. Which ones will be
able to detect light of wavelength 6000 Å?
19. A p-n photodiode is fabricated from a
semiconductor with band gap of 2.8 eV. Can it
detect a wavelength of 6000 nm?
12 V
15. Draw V-I characteristics of a p-n junction
diode. Explain, why the current under reverse
bias is almost independent of the applied voltage
up to the critical voltage.
20. What is the current flowing in R 2 in the
circuit shown in figure? Given : R1 = 500 W and
R2 = 1 kW
R1
10 V
5V
R2
Short Answer Type Questions (SA-II)
21. (a) Distinguish between n-type and p-type
semiconductors on the basis of energy band
diagrams.
(b) Compare their conductivities at absolute
zero temperature and at room temperature.
22. Name the important process that occur
during the formation of a p-n junction. Explain
briefly, with the help of a suitable diagram, how
a p-n junction is formed. Define the term ‘barrier
potential’.
23. If each diode in figure has a forward bias
resistance of 25 Ω and infinite resistance in
reverse bias, what will be the values of the
current I1, I2, I3 and I4?
127
Semiconductor Electronics : Materials, Devices and Simple Circuits
A
C
E
I1
G
I4
125 Ω
I3
125 Ω
I2
125 Ω
based on a certain Ga-As-P semiconducting
material whose energy gap is 1.9 eV. Identify the
colour of the emitted light.
(iv) Assuming the ideal diode, draw the output
waveform for the circuit given in figure. Explain
the waveform.
B
D
F
25 Ω
H
5V
24. Suppose a ‘n’-type wafer is created by doping
Si crystal having 5 × 1028 atoms/m3 with 1 ppm
concentration of As. On the surface 200 ppm
Boron is added to create ‘p’ region in this wafer.
Considering ni = 1.5 × 1016 m–3,
(a) Calculate the densities of the charge carriers
in the n and p regions.
(b) Comment which charge carriers would
contribute largely for the reverse saturation
current when diode is reverse biased.
25. With the help of a simple diagram, explain
the working of a silicon solar cell giving all three
basic processes involved.
Forward current I (mA)
50
40
30
20
10
0
1
2
3
4
5
5V
27. In half-wave rectification, what is the output
frequency if the input frequency is 50 Hz. What
is the output frequency of a full-wave-rectifier
for the same input frequency?
28. In the following diagram ‘S’ is a semiconductor.
Would you increase or decrease the value of R
to keep the reading of the ammeter A constant
when S is heated? Give reason for your answer.
V
– +
A
S
R
29. (a) Why are Si and GaAs preferred materials
for fabrication in solar cells?
(b) Draw V-I characteristic of solar cell and
mention its significance.
30. Explain, with the help of a circuit diagram,
the working of a photodiode. Write briefly how
it is used to detect the optical signals.
31. The number of silicon atoms per m 3 is
5 × 10 28 . This is doped simultaneously with
5 × 10 22 atoms per m 3 of Arsenic and
5 × 10 20 per m 3 atoms of Indium. Calculate
the number of electrons and holes. Given that
ni = 1.5 × 1016 m–3. Is the material n-type or
p-type?
Yellow
Green
Blue
White
Infrared
Red
Amber
26. Direction : Read the following and answer
the questions given below.
Light emitting diode is a photoelectric device which
converts electrical energy into light energy. It is
a heavily doped p-n junction diode which under
forward biased emits spontaneous radiation. The
general shape of the I-V characteristics of an
LED is similar to that of a normal p-n junction
diode, as shown. The barrier potentials are much
higher and slightly different for each colour.
20 sin ωt
V
(i) Draw the I-V characteristic of an LED.
(ii) Draw the schematic symbol of light emitting
diode (LED).
(iii) An LED is constructed from a p-n junction
32. In the case of n-type Si-semiconductor, the
donor energy level is slightly below the bottom of
conduction band whereas in p-type semiconductor,
the acceptor energy level is slightly above the top
of valence band. Explain, giving examples, what
role do these energy levels play in conduction and
valence bands.
CBSE Board Term-II Physics Class-12
128
33. An a.c. signal is fed into two circuits ‘X’ and
‘Y’ and the corresponding output in the two cases
have the waveforms as shown in figure.
(a)Identify the circuits ‘X’ and ‘Y’. Draw their
labelled circuit diagrams.
(b) Briefly explain the working of circuit Y.
(c) How does the output waveform from circuit
Y get modified when a capacitor is connected
across the output terminals parallel to the load
resistor?
36. (a) In the following diagram, is the junction
diode forward biased or reverse biased?
.
(b)The V-I characteristic of a silicon diode is
as shown in the figure. Calculate the
resistance of the diode at (i) I = 15 mA and
(ii) V = –10 V
34. Draw the circuit arrangement for studying
the V-I characteristics of a p-n junction diode in
(i) forward and (ii) reverse bias. Briefly explain
how the typical V-I characteristics of a diode are
obtained and draw these characteristics.
35. Draw the circuit diagram of a p-n diode used
as a half-wave rectifier. Explain its working.
OBJECTIVE TYPE QUESTIONS
1. (b) : When VA < VB, the diode gets reverse biased and
offers infinite resistance. No current flows through the upper
branch
\ R = 20 W
2. (b) : A vacancy created when an electron leaves a
covalent bond. This vacancy is known as hole.
3.
(a)
4. (c) : In an unbiased p-n junction, the diffusion of charge
carriers across the junction takes place from higher
concentration to lower concentration. Thus, option (c) is
correct.
5.
(b) : Electric field
V
0.3
E= =
= 3 × 105 V m–1
d 1 × 10−6
(c) : Dynamic resistance, rd =
−3
=
9. (c) : Adding fifth group element to germanium makes it
an n-type semiconductor. Antimony is a fifth group element
and so germanium becomes n-type semiconductor.
10. (c)
hc 6.6 × 10−34 × 3 × 108
11. (b) : Using, E = Eg =
=
J
λ
589 × 10−9
=
6.6 × 10−34 × 3 × 108
589 × 10−9 × 1.6 × 10−19
eV = 2.1 eV
12. (d) 13. (c)
6. (b) : In p-n junction, the diffusion of majority carriers
takes place when junction is forward biased and drifting of
minority carriers takes place across the junction, when it is
reverse biased.
7.
0.7 V − 0.65 V
0.05 × 1000
Ω = 10 Ω
5
5 × 10 A
8. (c) : Since the diode is reversed biased, only drift current
exists in circuit which is 20 mA. Potential drop across 15 W
resistor
= 15 W × 20 mA = 300 mV = 0.0003 V
Potential difference across the diode = 4 – 0.0003
= 3.99 4 V
rd =
∆V
∆I
14. (a) : A diode is said to be forward biased if p-type
semiconductor of p-n junction is at high potential with respect
to n-type semiconductor of p-n junction. It is so for circuit (a).
15. (c) : The energy gap Eg depends on the temperature.
For silicon, Eg(T) = 1.10 – 3.60 × 10–4T eV
For germanium, Eg(T) = 0.70 – 2.23 × 10–4T eV
129
Semiconductor Electronics : Materials, Devices and Simple Circuits
16. (a) : Here, Eg = 0.72 eV = 0.72 × 1.6 × 10–19 J
If l is the maximum wavelength of electromagnetic radiation
which can create a hole-electron pair in germanium, then
hc
Eg =
λ
hc 6.62 × 10−34 × 3 × 108
or λ =
=
= 1.7 × 10−6 m
Eg
0.72 × 1.6 × 10−19
17. (c) : In the circuit, the
upper diode D 1 is reverse
biased and the lower diode D2
is forward biased. Thus there will
be no current across upper diode
junction. The effective circuit will
be as shown in figure.
Total resistance of circuit
R = 50 + 70 + 30 = 150 W
3V
V
= 0.02 A
Current in circuit , I = =
R 150 Ω
18. (a) : The p-n junction diode is forward biased when p is
at high potential with respect to n. Hence option (a) is correct.
V 0.50 V
= 1.0 × 106 V m −1
19. (b) : Electric field, E = =
d 5 × 10−7
20. (b) : In reverse biasing, the minority charge carriers will
be accelerated due to reverse biasing, which on striking with
atoms cause ionisation resulting in secondary electrons and
thus produce more number of charge carriers.
When doping concentration is large, there will be large
number of ions in the depletion region, which will give rise
to a strong electric field.
21. (c) : Here, input Vrms = 20 V
Peak value of input voltage
Vo = 2 Vrms = 2 × 20 = 28.28 V
Since the transformer is a step up transformer having
transformer ratio 1 : 2, the maximum value of output voltage
of the transformer applied to the diode will be
V0′ = 2 × Vo = 2 × 28.28 V
2V0′ 2 × 2 × 28.28
=
= 36 V
π
22 / 7
22. (b) : Let V be the potential difference between A and
B, then
V – 0.3 = (5 + 5) × 103 × (0.2 × 10–3) = 2
or V = 2 + 0.3 = 2.3 V
23. (c) : As, nenh = ni2
\
dc voltage =
Here, ni = 6 × 108 per m3 and ne = 9 × 1012 per m3
∴ nh =
8 2
ni (6 × 10 )
=
= 4 × 104 per m3
12
ne
9 × 10
24. (d) : Here, Eg = 2 eV
Wavelength of radiation corresponding to this energy is
hc 1240 eV nm
λ=
=
= 620 nm
Eg
2 eV
c 3 × 108 m s−1
14
Frequency υ = =
λ 620 × 10−9m = 5 × 10 Hz
25. (d) : The detection occurs only when the energy of
incident photon greater than or equal to the energy band gap
hc
= 2.5 eV
λ
1240 eV
hc
nm = 496 nm 5000 Å
∴ λ=
=
2.5 eV 2.5 eV
26. (b) : Potential barrier developed in a junction diode
opposes the majority carriers only.
27. (b) : Here, ne = 6 × 1018 m–3
Volume of the sample = 1 cm × 1 cm × 1 mm = 10–7 m3
Number of holes in the sample = Number of electrons in
the sample
= ne × V = 6 ×1018 × 10 –7 = 6 × 1011
28. (a) : As s = e(neme + nhmh) = eni(me + mh)
= 1.6 × 10–19 × 3.6 × 1019(0.54 + 0.18) = 4.147 S m–1
29. (b)
hc 1240 eV nm
=
= 0.5 eV
2480 nm
λ
31. (c) : For forward biasing,
DV = 2.4 – 2.0 = 0.4 V; DI = 80 – 60 = 20 mA
0.4
∆V
∴ rfb =
=
= 20 Ω
∆I 20 × 10−3
30. (c) : Eg =
For reverse biasing, DV = – 2 – 0 = –2 V
DI = – 0.25 – 0 = – 0.25 mA
−2
rrb =
= 8 × 106 Ω
−6
−0.25 × 10
32. (b) : The work done on an electron when it moves
through distance, d = eEd
Work done is equal to the energy transferred to the electron
\ eEd = 2 eV
2V
2V
⇒ E=
=
= 5 × 107 V m−1
d 4 × 10−8 m
33. (c) : Ripple factor for full wave rectifier = 0.482.
Expressed in %, it is 48.2%.
34. (a) : Here, p-n junction is reverse biased. Therefore, the
current flowing through p-n junction is zero.
35. (b) : Since the output voltage obtained in a half-wave
rectifier circuit has single variation in one cycle of ac voltage,
hence the fundamental frequency in the ripple of output
voltage would be 50 Hz.
CBSE Board Term-II Physics Class-12
130
36. (c) : The p-n diode is forward biased when p-side is at
a higher potential than n-side.
37. (c)
38. (d) : Forward bias resistance,
∆V
0.8 − 0.7
0.1
R1 =
=
=
= 10
−
3
∆I (20 − 10) × 10
10 × 10−3
10
= 107
Reverse bias resistance, R2 =
−6
1 × 10
Then, the ratio of forward to reverse bias resistance,
R1 10
=
= 10−6
R2 107
39. (d) : In p-region the direction of conventional current is
same as flow of holes.
In n-region the direction of conventional current is opposite
to the flow of electrons.
40 (c) : In the given circuit the junction diode is forward
biased and offers zero resistance.
3 V− 1 V
2V
=
= 0.01 A
\ The current, I =
200 Ω 200 Ω
41. (a) : Photodiode is a device which is always operated in
reverse bias.
42. (c) : Let Eg be the required bandwidth. Then
hc
Eg =
λ
Here, hc = 1240 eV nm, l = 500 nm
1240 eV nm
∴ Eg =
= 2.48 eV
500 nm
43. (a) : A photodiode is a device which is used to detect
optical signals.
44. (b) : Voltage drop across R,
VR = VB – VN = 5 – 0.7 = 4.3 V
Here, Imin = 1 × 10–3 A
V
4.3
\ Rmax = R =
= 4.3 × 103 Ω = 4.3 kW
Imin 1 × 10−3
45. (b) : I = 6 mA = 6 × 10–3 A;
VR = VB – VN = 5 – 0.7 = 4.3 V
V
4.3
R= R =
= 717 Ω
I 6 × 10−3
46. (d) : Here, VB = 6 V; VN = 0.7 V,
VR = 6 – 0.7 = 5.3 V
Power dissipated in R = I × VR
= (6 × 10–3) × 5.3 = 31.8 × 10–3 W = 31.8 mW
47. (b) : Vt = Vbi + VR = 0.63 + 6 = 6.63 V
48. (d) : Diode is two terminal device, anode and cathode
are the two terminals.
49. (b) : E =
VB
0.4
=
= 1.0 × 106 V m−1
d 4.0 × 10−7
50. (c) : Potential difference across R = 3 – 0.4 = 2.6 V
2.6
Potential difference
=
= 130 Ω
Resistance R =
Current
20 × 10−3
51. (b) : When the voltage will be the same as that of the
potential barrier, the potential barrier will disappear resulting
in flow of current.
1
1
52. (a) : mv12 = eVB + mv22
2
2
1
−31
⇒
× (9.1 × 10 ) × (4 × 105 )2
2
1
= 1.6 × 10−19 × (0.4) + × 9.1 × 10−31 × v22
2
On solving, we get
v2 = 1.39 × 105 m s–1
53. (c) : At 0 K, all semiconductors are insulators. The
valence band at 0 K is completely filled and there are no
free electrons in conduction band. At room temperature due
to thermal energy, the electron jump to the conduction band.
When the temperature increases, a large number of electrons
cross over the forbidden gap and jump from valence band to
conduction band. Thus with rise in temperature conductivity
increases. The covalent bonds of semiconductor breaks
only when it is heated up extremely either by increasing
temperature or by supplying strong current. After which it
behaves like conductor and no longer possesses the property
of low conduction, hence it is said to be damaged.
54. (c) : As capacitative reactance,
1
1
1
XC =
=
i.e., XC ∝
ωC 2πυC
υ
Thus, capacitative reactance decreases on increase in
frequency.
55. (c) : A small increase in forward voltage across p-n
junction shows large increase in forward current. Hence the
resistance (= voltage / current) of p-n junction is low when
forward biased. Also the width of depletion layer of p-n
junction decreases in forward bias.
A large increase in reverse voltage across p-n shows small
increase in reverse current. Hence the resistance of p-n
junction is high when reverse biased. Also the width of the
depletion layer of p-n junction increases in reverse biased.
56. (d) : The direction of diffusion current is that when
positively charged particles move from p-type to n-type of
diode.
57. (d) : With the increase of temperature, the average
energy exchanged in a collision increases and so more valence
electrons can cross the energy gap, thereby increasing the
electron-hole pairs. As in a semiconductor, conduction occurs
131
Semiconductor Electronics : Materials, Devices and Simple Circuits
mainly through electron-hole pairs, so conductivity increases
with increase of temperature. Which in turn implies that
the resistivity of a semiconductor decreases with rise in
temperature.
8.
58. (a) : In half wave rectifier, the one diode is biased only
when ac is in positive half of its cycle. For negative half of the
ac cycle the diode is reversed biased and there is no output
corresponding to that. Since for only one-half cycle we get
a voltage output, because of which it is called half wave
rectifier.
The diode acts as half wave rectifier, it offers low resistance
when forward biased and high resistance when reverse
biased.
59. (b) : In intrinsic semiconductor n e/n h = 1 and holes
are not particles but vacancies created due to breakage of
covalent bond.
60. (b) : A reverse bias on a p-n junction opposes the
movement of the majority charge carriers thus stopping the
diffusion current. It makes the free electrons and holes to
drift cross the junction. Therefore a small current in mA flows
even when the p-n junction is reverse biased. The drift current
is due to the thermal excitations of the electrons and holes.
SUBJECTIVE TYPE QUESTIONS
1.
n-type
Semiconductor
p-type
Semiconductor
(i)
It is formed by doping
pentavalent impurities.
It is formed by doping
trivalent impurities.
(ii)
The electrons are majority
carriers and holes are
minority carriers.
(ne >> nh)
The holes are majority
carriers and electrons
are minority carriers.
(nh >> ne)
2. (i) Forward biased : As forward voltage opposes the
potential barrier and effective barrier potential decreases. It
makes the width of the depletion layer smaller.
(ii) Reverse biased : As reverse voltage supports the potential
barrier and effective barrier potential increases. It makes the
width of the depletion layer larger.
+1V
0
9. Size of the dopant atom should be compatible in the pure
semiconductor and contribute a charge carrier by forming
covalent bond with Si or Ge atoms. Elemental dopants from
group XIII and group XV fulfil this condition.
10. When there is an increase in doping concentration, the
applied potential difference causes an electric field which
acts opposite to the potential barrier. This results in reducing
the potential barrier and hence the width of depletion layer
decreases.
11. For extrinsic or doped semiconductor
ne ⋅ nh = ni2 ⇒ ne =
Here ni = 2 × 1016 m–3 and nh = 4 × 1022 m–3
( 2 × 1016 m−3 ) 2
= 10 −10 m−3 .
4 × 10 22 m−3
12. If A is positive and B is
negative, J1 is forward biased and
J2 is reverse biased, so effective
current is
⇒ ia = = 0.3 A.
⇒
ne =
2Ω
4. Photodiode is used to detect the light signal and to
measure light intensity.
4Ω
6. (b) : In n-type semiconductor, the donor energy level lies
just below the conduction band.
7. No, the voltmeter should have a very high resistance as
compared to the resistance of p-n junction, which is nearly
infinite.
A
B
14. Diode D 1 is reverse biased,so it offers an infinite
resistance. So no current flows in the branch of diode D1.
D1
The junction diode is solar cell.
J1
13. In reverse bias condition of photodiode, the change
in saturation reverse current is directly proportional to the
change in the incident light flux or light intensity, which can
be measured accurately. It is not so when photodiode is
forward biased.
3. LED’s must have band gap in the order of 1.8 eV to
3 eV but Si and Ge have band gap less than 1.8 eV.
5.
ni2
nh
D2
3Ω
+
12 V
Diode D2 is forward biased, and offers negligible resistance
in the circuit. So current in the branch
V
12 V
I=
=
= 2A
R eq 2 Ω + 4 Ω
CBSE Board Term-II Physics Class-12
132
15.
20. The diode is reverse biased, but the voltage across it is
given as 5 V. R2 is in parallel with the diode, so current in R2
5V
=
⇒ Current in R2 = 5 mA.
1000 Ω
z
21. (a) The required energy band diagrams are given below:
The reverse current is due to minority charge carriers and
even a small voltage is sufficient to sweep the minority
carriers from one side of the junction to the other side of the
junction. Here the current is not limited by the magnitude of
the applied voltage but is limited due to the concentration of
the minority carrier on either side of the junction.
16. (a) (i) There is very little resistance to limit the current
in LED. Therefore, a resistor must be used in series with the
LED to avoid any damage to it.
(ii) The reverse breakdown voltages of LEDs are very low,
typically around 5 V. So care should be taken while fabricating
a p-n-junction diode so that the p side should only attached
to the positive of battery and vice versa as LED easily get
damaged by a small reverse voltage.
(b) The semiconductor used for fabrication of visible LEDs
must have at least a band gap of 1.8 eV because spectral
range of visible light is about 0.4 mm to 0.7 mm, i.e., about
3 eV to 1.8 eV.
17. (a) Bulb B1 will glow, as diode D1 is forward biased.
Bulb B2 will not glow as diode D2 is reverse biased.
∆V
(b) Forward resistance = ∆I
0.7 − 0.5
∴ ∆V =
−3 = 200 Ω.
∆I
T>0K
T>0K
(b)
at T = 0 K
At absolute zero temperature (0 K), conduction band of
semiconductor is completely empty, i.e., s = 0. Hence the
semiconductor behaves as an insulator. At room temperature,
some valence electrons acquire enough thermal energy
and jump to the conduction band where they are free to
conduct electricity. Thus the semiconductor acquires a small
conductivity at room temperature.
22. Two processes that take place in the formation of a p-n
junction are diffusion and drift.
1.0 × 10
hc
18. We know that, energy of incident photon, E =
λ
l = 6000 Å = 600 nm (given)
1242 eV nm
= 2.07 eV
600 nm
D2 will detect these radiations because energy of incident
radiation is greater than the band gap.
E =
19. Energy of the incident photon with a band gap of
6000 nm.
E=
6.63 × 10 −34 × 3 × 10 8
hc
hc
J = eV =
= 0.207 eV
λ
eλ
1.6 × 10 −19 × 6 × 10 −6
The photodiode need an energy of 2.8 eV to give response
to incident light.
As E < Eg, the given photodiode cannot detect the radiation
of wavelength 6000 nm.
When p-n junction is formed, then at the junction, free
electrons from n-type diffuse over to p-type, thereby filling
in the holes in p-type. Due to this a layer of positive charge is
built on n-side and a layer of negative charge is built on p-side
of the p-n junction. This layer sufficiently grows up within a
very short time of the junction being formed, preventing any
further movement of charge carriers (i.e., electrons and holes)
across the junction. Thus a potential difference V0 of the
order of 0.1 to 0.3 V is set up across the p-n junction called
potential barrier or junction barrier. The thin region around the
133
Semiconductor Electronics : Materials, Devices and Simple Circuits
junction containing immobile positive and negative charges
is known as depletion layer.
23. Let R be the effective resistance of the circuit, then
R = RAB || REF + 25
RAB = 125 + 25 = 150 W
REF = 125 + 25 = 150 W
150
\ R = 25 +
= 100 W
2
Since diode in the branch CD is reverse biased. I3 = 0.
5
Current, I1 =
= 0.05 A
100
According to Kirchhoff’s current rule,
I1 = I2 + I3 + I4 or I2 + I4 = I1 = 0.05
RAB = REF, so, I4 = I2
2I4 = 2I2 = 0.05
0.05
I4 = I2 =
= 0.025 A
2
24. (a) For n-type region,
1
ne = ND = 6 × 5 × 10 28 = 5 × 1022 m–3
10
1 

∵ 1 ppm = 6 
10
As nenh = ni2,
ni2 (1.5 × 1016 m−3 )2
= 0.45 × 1010 m–3
=
ne
5 × 10 22 m−3
For p-type region,
200
nh = NA = 6 × 5 × 10 28 = 1 × 1025 m–3
10
n i2 (1.5 × 1016 m−3 )2
Now, ne =
= 2.25 × 107 m–3
=
25
3
−
nh
1 × 10 m
(b) The minority carrier holes of n-region wafer
(nh = 0.45 × 1010 m–3) would contribute more to reverse
saturation current than minority carrier electrons of p–region
wafer (ne = 2.25 × 107 m–3) when p – n junction is reverse
biased.
nh =
25. Principle : A solar cell works on the principle of photovoltaic effect according to which when light photons of
energy greater than energy band gap of a semiconductor are
incident on p-n junction of that semiconductor, electron-hole
pairs are generated which give rise to an emf.
Generation of emf : Three basic processes are involved in the
generation of emf by a solar cell when solar radiations are
incident on it. These are:
(i) The generation of electron-hole pairs close to the
junction due to incidence of light with photo energy hu ≥ Eb.
(ii) The separation of electrons and holes due to the electric
field of the depletion region. So, electrons are swept to n-side
and holes to p-side.
(iii) The electrons reaching the
n-side are collected by the front
contact and holes reaching p-side
are collected by the back contact.
Thus, p-side becomes positive and
n-side become negative giving rise
to a photovoltage.
RL
When an external load RL is connected as shown in figure, a
photocurrent IL begins to flow through the load.
26. (i) The I-V characteristics of an
LED is similar to that of a Si junction
diode. But the threshold voltages are
much higher and slightly different for
each colour.
(ii)
hc
hc
\ λ=
λ
Eg
Here, Eg = 1.9 eV, hc = 1240 eV nm
1240 eV nm
\ λ=
= 652.6 nm
1.9 eV
Hence, the emitted light is of red colour.
(iv) For the voltage less than 5 V, the diode is reverse biased
and circuit will act as open circuit.
When input voltage is greater than 5 V, diode is in conducting
state.
(iii) As Eg =
27. In half wave rectification, only one ripple is obtained per
cycle in the output.
CBSE Board Term-II Physics Class-12
134
Output frequency of a half wave rectifier = input frequency
= 50 Hz
In full wave rectification, two ripples are obtained per cycle
in the output.
Output frequency = 2 × input frequency
= 2 × 50 = 100 Hz
28. We will increase the value of R. On heating a
semiconductor, its resistance decreases with rise in
temperature. As the semiconductor, S is in series, so net
resistance of the circuit also decreases. So by increasing the
value of R we can keep the resistance of circuit constant and
hence the current in the circuit or the reading of ammeter A
can be kept constant.
29. (a) The energy for the maximum intensity of the solar
radiation is nearly 1.5 eV. In order to have photo excitation
the energy of radiation (hu) must be greater than energy
band gap (Eg), i.e., hu > Eg. Therefore, the semiconductor
with energy band gap about 1.5 eV or lower and with higher
absorption coefficient, is likely to give better solar conversion
efficiency.
The energy band gap for Si is about 1.1 eV, while for GaAs, it
is about 1.53 eV. The gas GaAs is better inspite of its higher
bandgap than Si because it absorbs relatively more energy
from the incident solar radiations being of relatively higher
absorption coefficient.
(b)
(open circuit
voltage)
(i) V-I curve is drawn in the fourth quadrant, because a solar
cell does not draws current but supply current to the load.
(ii) In V-I curve, the point A indicates the maximum voltage
VOC being supplied by the given solar cell when no current is
being drawn from it. VOC is called the open circuit voltage.
(iii) In V-I curve, the point B indicates the maximum current
ISC which can be obtained by short circuiting the solar cell
without any load resistance. ISC is called the short circuit
current.
30. Working of photodiode :
h
A junction diode made from
light sensitive semiconductor is
c a l l e d a p h o t o d i o d e. A
A
photodiode is a p-n junction
p-side n-side
diode arranged in reverse
R
biasing.
The number of charge carriers increases when light of suitable
frequency is made to fall on the p-n junction, because new
electron-hole pairs are created by absorbing the photons of
suitable frequency. Intensity of light controls the number of
charge carriers. Due to this property photodiodes are used
to detect optical signals.
31. We know that for each atom doped with Arsenic, one
free electron is received. Similarly, for each atom doped with
indium, a vacancy is created.
So, the number of free electrons introduced by pentavalent
impurity added,
ne = NAs = 5 × 1022 m–3
The number of holes introduced by trivalent impurity added
nh = NIn = 5 × 1020 m–3
We know the relation, nenh = ni2...(i)
Now net electrons,
n′e = ne – nh = 5 × 1022 – 5 × 1020
= 4.95 × 1022 m–3...(ii)
Now using equation (i), net holes
ni 2 (1.5 × 1016 )2
= 4.5 × 10 9 m−3
=
22
n ′e
4.95 × 10
So, n′e >> n′h, the material is of n-type.
32. In n-type extrinsic semiconductors, the number of free
electrons in conduction band is much more than the number
of holes in valence band. The donor energy level lies just
below the conduction band. In p-type extrinsic semiconductor,
the number of holes in valence band is much more than the
number of free electrons in conduction band. The acceptor
energy level lies just above the valence band.
33. (a) X = Half wave rectifier
Y = Full wave rectifier
n′h =
(Half wave rectifier)
135
Semiconductor Electronics : Materials, Devices and Simple Circuits
through RL remains same, but its magnitude changes with
time, so it is called pulsating d.c.
(b) Two p–n junction diodes can be used to make full wave
rectifier which is used to convert alternating current into
direct current.
Centre- Tap
Transformer
Diode D1
A
Centre Tap
B
X
RL Output
Diode D2
Y
A full wave rectifier consists of two diodes connected in
parallel across the ends of secondary winding of a center
tapped step down transformer. The load resistance RL is
connected across secondary winding and the diodes between
A and B as shown in the circuit.
During positive half cycle of input a.c., end A of the secondary
winding becomes positive and end B negative. Thus diode D1
becomes forward biased, whereas diode D2 reverse biased. So
diode D1 allows the current to flow through it, while diode
D2 does not, and current in the circuit flows from D1 and
through load RL from X to Y.
During negative half cycle of input a.c., end A of the secondary
winding becomes negative and end B positive, thus diode D1
becomes reverse biased, whereas diode D2 forward biased.
So diode D1 does not allow the current to flow through it
but diode D2 does, and current in the circuit flows from D2
and through load RL from X to Y.
34. Forward biased characteristics : The circuit diagram for
studying forward biased characteristics is shown in the figure.
Starting from a low value, forward bias voltage is increased
step by step (measured by voltmeter) and forward current is
noted (by ammeter). A graph is plotted between voltage and
current. The curve so obtained is the forward characteristic of
the diode.
Current
(Full wave rectifier)
(c) A capacitor of large capacitance is connected in parallel
to the load resistor RL. When the pulsating voltage supplied
by the rectifier is rising, the capacitor C gets charged. If
there is no external load, the capacitor would have remained
charged to the peak voltage of the rectified output. However,
when there is no load and the rectified voltage starts falling,
the capacitor gets discharged through the load and the
voltage across capacitor begins to fall slowly.
Voltage (V)
Input-Output waveforms
At the start when applied voltage is low, the current through
the diode is almost zero. It is because of the potential barrier,
which opposes the applied voltage. Till the applied voltage
exceeds the potential barrier, the current increases very slowly
with increase in applied voltage (OA portion of the graph).
With further increase in applied voltage, the current increases
very rapidly (AB portion of the graph), in this situation, the
diode behaves like a conductor. The forward voltage beyond
which the current through the junction starts increasing
rapidly with voltage is called threshold or cut-in voltage. If
line AB is extended back, it cuts the voltage axis at potential
barrier voltage.
Since in both the half cycles of input a.c., electric current
through load R L flows in the same direction, so d.c. is
obtained across RL. Although direction of electric current
Reverse biased characteristics : The circuit diagram
for studying reverse biased characteristics is shown in the
figure.
Vin
O
t
Vout
t
O
CBSE Board Term-II Physics Class-12
136
Primary of transformer is connected to a.c. supply. During
positive half cycle of input a.c., end A of the secondary
winding becomes positive and end B negative. Thus, diode
D becomes forward biased and conducts the current through
it. So, current in the circuit flows from A to B through load
resistor RL.
Current Voltage (V)
In reverse biased, the applied voltage supports the flow of
minority charge carriers across the junction. So, a very small
current flows across the junction due to minority charge
carriers.
Motion of minority charge carriers is also supported by
internal potential barrier, so all the minority carriers cross
over the junction.
Therefore, the small reverse current remains almost constant
over a sufficiently long range of reverse bias, increasing very
little with increasing voltage (OC portion of the graph). This
reverse current is voltage independent upto certain voltage
known as breakdown voltage and this voltage independent
current is called reverse saturation current.
35. Half wave rectifier:
During negative half cycle of input a.c., end A of the secondary
winding becomes negative and end B positive. Thus, diode D
becomes reverse biased and does not conduct any current. So,
no current flows in the circuit. Since electric current through
load RL flows only during positive half cycle, in one direction
only i.e., from A to B, so d.c. is obtained across RL.
36. (a) Voltage at p side is less than voltage at n side of
the diode so it is in reverse bias.
(b) (i) From the given curve, we have
voltage, V = 0.8 volt for current, I = 20 mA
voltage, V = 0.7 volt for current, I = 10 mA
⇒ DI = (20 – 10)mA = 10 × 10–3 A
⇒ DV = (0.8 – 0.7) = 0.1 V
0.1
∆V
\ Resistance, R =
⇒ R=
⇒ R = 10 W
∆I
10 × 10 −3
(ii) For V = – 10 V, we have
It consists of a diode D connected in series with load resistor
RL across the secondary windings of a step-down transformer.
I = – 1 mA = –1 × 10–6 A ⇒ R =

10
1 × 10
−6
= 1.0 × 10 7 Ω
PRACTICE PAPER 1
Time allowed : 2 hours
Maximum marks : 35
General Instructions :
(i) All questions are compulsory. There are 16 questions in all.
(ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
(iii)Section A contains five very short answer questions and two assertion reasoning MCQs of 1 mark each. Section
B has one case based question of 4 marks, Section C contains four short answer questions of 2 marks each,
Section D contains two short answer questions of 3 marks each and Section E contains two long answer questions
of 5 marks each.
(iv)There is no overall choice. However internal choice is provided. You have to attempt only one of the choices in
such questions.
SECTION - A
All questions are compulsory. In case of internal choices, attempt any one of them.
1. Ultraviolet radiations of different frequencies u1 and u2 are incident on two photosensitive materials having work
functions W1 and W2 (W1 > W2) respectively. The kinetic energy of the emitted electrons is same in both the
cases. Which one of the two radiations will be of higher frequency?
2. State de Broglie hypothesis. Is it applicable for moving electron of hydrogen atom ?
OR
Calculate the de-Broglie wavelength of the electron orbiting in the n = 2 state of hydrogen atom.
3. Plot a graph showing variation of current versus voltage for the material Gas.
4. If A is the angle of prism, r angle of refraction, then what is the condition for minimum deviation?
OR
Find the magnifying power of a convex lens of focal length 10 cm when the image is formed at the near point.
5. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima
formed on a screen placed 1 m away, was found to be 0.2°. What will be the angular width of the first minima, if
the entire experimental apparatus is immersed in water? (mwater = 4/3)
For question numbers 6-7, two statements are given-one labelled Assertion (A) and the other labelled
Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
6. Assertion (A) : The depletion layer in the p-n junction is free from mobile charge carriers.
Reason (R) : There is no electric field across the junction barrier.
7. Assertion (A) : Young’s double slit experiment can be performed using a source of white light.
Reason (R) : The wavelength of red light is less than the wavelength of other colours in white light.
SECTION - B
Question 8 is a Case Study based question and it is compulsory. Attempt any 4 sub parts from this question. Each
part carries 1 mark.
Refraction through Spherical Surfaces
8.
Refraction of light is the change in the path of light as it passes obliquely from one transparent medium to another
medium. According to law of refraction sin i = 1µ2 , where 1m2 is called refractive index of second medium with
sin r
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
CBSE Board Term-II Physics Class-12
138
µ2 µ1 µ2 − µ1
− =
. Similarly from
v
u
R
µ
µ µ − µ1
refraction at a concave spherical surface when object lies in the rarer medium, we have 2 − 1 = 2
and when
v
u
R
µ µ
µ − µ2
object lies in the denser medium, we have 1 − 2 = 1
.
v
u
R
(i) Refractive index of a medium depends upon
(a) nature of the medium (b) wavelength of the light used
(c) temperature
(d) all of these
14
(ii) A ray of light of frequency 5 × 10 Hz is passed through a liquid. The wavelength of light measured inside the
liquid is found to be 450 × 10–9 m. The refractive index of the liquid is
(a) 1.33
(b) 2.52
(c) 2.22
(d) 0.75
(iii) A ray of light is incident at an angle of 60° on one face of a rectangular glass slab of refractive index 1.5. The angle
of refraction is
(a) sin–1(0.95)
(b) sin–1(0.58)
(c) sin–1(0.79)
(d) sin–1(0.86)
(iv) A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1.5. The distance of the
virtual image from the surface of sphere is
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 12 cm
(v) In refraction, light waves are bent on passing from one medium to the second medium because in the second medium
(a) the frequency is different
(b) the co-efficient of elasticity is different
respect to first medium. From refraction at a convex spherical surface, we have
(c) the speed is different
(d) the amplitude is smaller.
SECTION - C
All questions are compulsory. In case of internal choices, attempt anyone.
9. Calculate the energy in fusion reaction : 12 H +12 H → 32He + n , where B.E. of 12 H = 2.23 MeV and of 32 He = 7.73 MeV.
10. The work function for a certain metal is 4.2 eV. Will metal give photoelectric emission for incident radiation of
wavelength 330 nm ?
OR
If the momentum of an electron is changed by p, then the de Broglie wavelength associated with it changes by
0.5%. What will be the initial momentum of electron ?
11. A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the direction of
electric and magnetic field vectors?
12. The forbidden energy gap in semiconductors, insulators and metals are Es, Ei, and Em respectively. Arrange these
in descending order. The band gap in silicon is 1.12 eV. What is the maximum wavelength of light that can be
emitted by it?
OR
In the following diagram, is the junction diode forward biased or reverse biased?
SECTION - D
All questions are compulsory. In case of internal choices, attempt any one.
13. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy
band diagrams.
14. A convex lens made of a material of refractive index m1 is kept in a medium of refractive index m2. Parallel rays of
light are incident on the lens. Complete the path of rays of light emerging from the convex lens if
(a) m1 > m2
(b) m1 = m2
(c) m1 < m2
139
Practice Paper - 1
OR
The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material
of the lens be 1.5, then show that it will act as a convex lens irrespective of the side on which the object lies.
SECTION - E
All questions are compulsory. In case of internal choices, attempt any one.
15. A parallel beam of monochromatic light falls normally on a narrow slit of width ‘a’ to produce a diffraction
pattern on the screen placed parallel to the plane of the slit.
Use Huygens principle to explain that
(i) the central bright maxima is twice as wide as the other maxima.
(ii) the intensity falls as we move to successive maxima away from the centre of on either side.
OR
(a)Two monochromatic waves emanating from two coherent sources have the displacements represented by
y1 = a cos wt and y2 = a cos (wt + f)
where f is the phase difference between the two displacements. Show that the resultant intensity at a point
due to their superposition is given by I = 4 I0 cos2 f/2, where I0 = a2.
(b) Hence obtain the conditions for constructive and destructive interference.
16. (a)Using Bohr’s atomic model, derive the expression for the radius of nth orbit of the revolving electron in a
hydrogen atom.
(b) Write two important limitations of Rutherford nuclear model of the atom.
OR
(a)Using de Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of
quantization of energy levels in a hydrogen atom.
(b)The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the
electron in this state?
1.
or
or
or
so
1 2
mv
= hυ1 − W1 = hυ2 − W2
2 max
h(u1– u2 ) = W1 – W2
h(u1 – u2) = positive
u1 – u2 = positive
u1 > u2
13.6 eV
= 3.4 × 1.6 × 10–19 J
4
n
h
de Broglie wavelength λ =
2mEK
EK =
[∵ W1 > W2]
2. de-Broglie hypothesis : It states that a moving particle
sometimes acts as a wave and sometimes as a particle or
a wave is associated with moving particle which controls
the particle in every respect. The wave associated with
moving particle is called matter wave or de-Broglie wave
whose wavelength is given by
h
mv
where m and v are the mass and velocity of the particle
and h is Planck’s constant.
Yes, it is applicable for moving electron of hydrogen atom.
λ=
OR
Kinetic energy of the electron in the second state of
hydrogen atom
=
13.6 eV
2
=
6.63 × 10−34
2 × 9.1 × 10−31 × 3.4 × 1.6 × 10−19
= 0.67 nm
3. Variation of current versus voltage for Gas.
CBSE Board Term-II Physics Class-12
140
4.
As r1 + r2 = A
At minimum deviation, r1 = r2 = r
\ 2r = A
OR
(d) : When the final image is formed at the near point,
the magnifying power is m = 1 + D
f
where D is the least distance of distinct vision and f is the
focal length of the convex lens.
Here, D = 25 cm, f = 10 cm
25 cm
∴ m = 1+
= 1 + 2.5 = 3.5
10 cm
5. Angular width for first minima in Young’s double slit
experiment, θ =
λ
a
For given value of a, q ∝ l
θ
λ
λ
θ 0. 2°
=
= = µ ⇒ θw = =
= 0.15°
θw λw λ
4
µ
µ
3
6. (c) : Due to diffusion of holes from the p-region to
the n-region and of electrons from the n-region to the
p-region an electric field is set up across the junction
barrier. Once the depletion layer is formed it is in
equilibrium and becomes free of mobile charge carriers.
7. (c) : When source in Young’s double slit experiment
is of white light, the central fringe is white as all colours
meet there in phase.
8. (i) (d) : Refractive index of a medium depends upon
nature and temperature of the medium, wavelength of
light.
(ii) (a) : Here u = 5 × 1014 Hz; l = 450 × 10–9m
c = 3 × 108 m s–1
Refractive index of the liquid,
c
c
3 × 108
=
=
v υλ 5 × 1014 × 450 × 10−9
m = 1.33
(iii) (b) : Here i = 60° ; m = 1.5
sin i
By snell’s law, m =
sin r
sin i sin 60° 0.866
sin r =
=
=
µ
1. 5
1. 5
sin r = 0.5773 or r = sin–1(0.58)
µ=
(iv) (c): As object is at the centre of the sphere, the image
must be at the centre only.
\ Distance of virtual image from centre of sphere = 6 cm.
(v) (c) : Speed of light in second medium is different
than that in first medium
9. Fusion reaction,
Energy released = final B.E. – initial B.E.
= 7.73 – (2.23 + 2.23) = 3.27 MeV.
10. Let us calculate the energy associated with incident
photons
E=
−34
× 3 × 108
hc 6.63 × 10
=
= 6.027 × 10−19 J
−9
λ
330 × 10
or E =
6.027 × 10−19
J = 3.77 eV
1.6 × 10−19
So, energy of incident photons i.e., 3.77 eV is less than
work function, hence emission will not take place.
OR
de Broglie wavelength associated with an electron is
h
h
λ=
or p =
p
λ
∆p
p
0. 5
∆λ ⇒
∴
=−
=
p
λ
pinitial 100
pinitial = 200 p


11. The electric and magnetic field vectors E and B are
perpendicular to each other and also perpendicular to
the direction of propagation of the electromagnetic wave.
If a plane electromagnetic wave is propagating along the
z-direction, then the electric field is along x-axis, and
magnetic field is along y-axis.
12. Here, Ei > Es > Em
As, E = hυ = hc
λ
6.67 × 10−34 × 3 × 108
or λ = hc =
E
1.12 × 1.6 × 10−19
l = 1.11 × 10–6 m
OR
Voltage at p side is less than voltage at n side of the diode
so it is in reverse bias.
13. The band diagram for conductors, semiconductors
and insulators are given as follows:
141
Practice Paper - 1
15. (i)
Two distinguishing features :
(i) In conductors, the valence band and conduction
band tend to overlap (or nearly overlap) while in
insulators they are separated by a large energy gap and in
semiconductors they are separated by a small energy gap.
(ii) The conduction band of a conductor has a large
number of electrons available for electrical conduction.
However, the conduction band of insulators is almost
empty while that of the semiconductor has only a (very)
small number of such electrons available for electrical
conduction.
14. In case (a) m1 > m2, the lens behaves as convergent lens.
In case (b) m1 = m2, the lens behaves as a plane plate.
In case (c) m1 < m2, the lens behaves as a divergent lens.
The path of rays in all the three cases is shown in the figure.
2
1
2
2
F
2
1
2
1
F
(i) 1 > 2
(ii) 1 = 2
(iii) 1 < 2
OR
Here, m = 1.5
If object lies on plane side;
R1 = ∞, R2 = – 20 cm
1
1
1 
= (µ − 1)  − 
f
 R1 R2 
1 1
1
= (1.5 − 1)  +  =
 ∞ 20  40
f = + 40 cm. The lens behaves as convex lens.
If object lies on its curved side. Then
R1 = 20 cm, R2 = ∞
O
(a)
Width of central maximum =
O
1
1
1 
 1 1 1
= (µ − 1)  −  = (1.5 − 1)  −  =
 20 ∞  40
f′
R
R
 1
2
f ′ = 40 cm. The lens behaves as convex lens.
Consider a parallel beam of monochromatic light is
incident normally on a single slit AB of width a as shown
in the figure. According to Huygens principle every point
of slit acts as a source of secondary wavelets spreading in
all directions. The mid point of the slit is O. A straight
line through O perpendicular to the slit plane meets the
screen at C. At the central point C on the screen, the angle
q is zero. All path differences are zero and hence all the
parts of the slit contribute in phase. This gives maximum
intensity at C.
Consider a point P on the screen.
The observation point is now taken at P.
Secondary minima : Now we divide the slit into two equal
a
halves AO and OB, each of width . For every point, M1
2
in AO, there is a corresponding point M2 in OB, such that
a
M1M2 = . The path difference between waves arriving
2
at P and starting from M1 and M2 will be a sin θ = λ .
2
2
asinq = l
In general, for secondary minima
asinq = nl where n = ±1, ±2, ±3...... Secondary maxima
: Similarly it can be shown that for secondary maxima
λ
a sinθ = (2n + 1) where n = ±1, ±2........
2
The intensity pattern on the screen is shown in the given figure.
(b)
2D λ
a
(ii) The reason is that the intensity of the central
maximum is due to the constructive interference of
wavelets from all parts of the slit, the first secondary
maximum is due to the contribution of wavelets from
one third part of the slit (wavelets from remaining two
parts interfere destructively), the second secondary
maximum is due to the contribution of wavelets from
the one fifth part only (the remaining four parts interfere
destructively) and so on. Hence the intensity of secondary
CBSE Board Term-II Physics Class-12
142
maximum decreases with the increase in the order n of
the maximum.
OR
(a) y1 = a cos wt, y2 = a cos (wt + f)
From Bohr’s quantization condition
nh
nh
mvr =
or v =
...(ii)
2π
2 πmr
Using equation (ii) in (i), we get
2
1 e2
1 e2
nh 
m ⋅ n2h2
=
=
or
m ⋅ 

2
2
2
 2 πmr 
4 πε0 r
4 πε0 r
4π m r
or r =
where f is phase difference between them.
Resultant displacement at point P will be,
y = y1 + y2 = a cos wt + a cos(wt + f)
= a [cos wt + cos (wt + f)]
(ωt − ωt − φ) 
(ωt + ωt + φ)

= a 2 cos
cos


2
2
φ
φ

y = 2a cos  ωt +  cos  
...(i)
2


2
φ
Let y = 2a cos   = A, the equation (i) becomes
2
φ

y = A cos  ωt + 

2
where A is amplitude of resultant wave,
φ
Now, A = 2a cos  
2
φ
On squaring, A2 = 4a2 cos2  
2
φ
Hence, resultant intensity, I = 4 I0 cos2  
2
(b) Condition for constructive interference, cos Df = + 1
∆x
2π
= 0, 2 π, 4 π... or Dx = nl; n = 0, 1, 2, 3, ...
λ
Condition for destructive interference, cos Df = – 1
∆x
2π
= π, 3π, 5π... or x = (2n – 1) l/2
λ
where n = 1, 2, 3...
th
16. (a) Radius of n orbit of hydrogen atom : In
H-atom, an electron having charge –e revolves around the
nucleus of charge +e in a circular orbit of radius r, such
that necessary centripetal force is provided by the electrostatic
force of attraction between the electron and nucleus.
i.e.,
mv 2
1 e.e
=
r
4 πε0 r 2
or mv 2 =
2
1 e
4 πε0 r
+e r
...(i)
F
–e
mv2
r
n2h2ε0
...(iii)
πme 2
where n = 1, 2, 3, ... is principal quantum number.
Equation (iii), gives the radius of nth orbit of H-atom. So
the radii of the orbits increase proportionally with n2 i.e.,
[r ∝ n2]. Radius of first orbit of H-atom is called Bohr
radius a0 and is given by
h 2 ε0
a0 =
for n = 1 or a0 = 0.529 Å
πme 2
So, radius of nth orbit of H-atom then becomes
r = n2 × 0.529 Å
(b) The two important limitations of Rutherford nuclear
model of the atom are :
(i) This model cannot explain about the stability of
matter.
(ii) It cannot explain the characteristic line spectra of
atoms of different elements.
OR
(a) According to de Broglie, a stationary orbit is that
which contains an integral number of de Broglie waves
associated with the revolving electron.
For an electron revolving in nth circular orbit of radius rn,
Total distance covered = Circumference of the orbit = 2prn
\ For the permissible orbit, 2prn = nl
According to de-Broglie,
l=
h
mvn
where vn is speed of electron revolving in nth orbit.
nh
nh
h
= n 
\ 2prn =
or mvnrn =

mvn
2π
2π 
(b) For ground state, n = 1
−13.6 −13.6
E = 2 = 2 = −13.6 eV
n
1
\ K.E. = – E = – (–13.6) = 13.6 eV
Q P.E. = 2E
\ P.E. = 2 (–13.6) = – 27.2 eV

PRACTICE PAPER 2
Time allowed : 2 hours
Maximum marks : 35
General Instructions :
(i) All questions are compulsory. There are 16 questions in all.
(ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
(iii)Section A contains five very short answer questions and two assertion reasoning MCQs of 1 mark each. Section
B has one case based question of 4 marks, Section C contains four short answer questions of 2 marks each,
Section D contains two short answer questions of 3 marks each and Section E contains two long answer questions
of 5 marks each.
(iv)There is no overall choice. However internal choice is provided. You have to attempt only one of the choices in
such questions.
SECTION - A
All questions are compulsory. In case of internal choices, attempt any one of them.
1.
Do electromagnetic waves carry energy and momentum?
2.
The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential?
OR
What is the difference between kinetic energies of photoelectrons emitted from a surface by light of wavelength
2500 Å and 5000 Å ?
3.
Two coherent monochromatic light beams of intensities I and 4I are superimposed. What is the maximum
and minimum possible resulting intensities?
4.
The 6563 Å Ha line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Find the speed with
which the star is receding from the earth.
OR
How is the radius of a nucleus related to its mass number A?
5.
How is the speed of em-waves in vacuum determined by the electric and magnetic fields?
For question numbers 6, 7 two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is NOT the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
6.
Assertion (A) : Graph of density of nuclei with mass number is a straight line parallel to mass number axis.
Reason (R) : Radius of nucleus is directly proportional to the cube root of mass number.
7.
Assertion (A) : Energy is released in nuclear fusion and fission.
Reason (R) : In any nuclear reaction the reactants and resultant products obey the law of conservation of
mass and energy only.
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
CBSE Board Term-II Physics Class-12
144
SECTION - B
Questions 8 is a Case Study based question and is compulsory. Attempt any 4 sub parts from this question.
Each part carries 1 mark.
Fringe Width
8.
Distance between two successive bright or dark fringes is called fringe width.
(n + 1)λD nλD λD
b = Yn+1 – Yn =
−
=
d
d
d
Fringe width is independent of the order of the maxima. If whole apparatus is immersed in liquid of refractive
λD
index m then b =
(fringe width decreases). Angular fringe width (q) is the angular separation between
µd
two consecutive maxima or minima
β λ
θ=
=
D d
(i) The maximum number of possible interference maxima for slit separation equal to twice the wavelength in
Young’s double-slit experiment, is
(a) infinite
(b) five
(c) two
(d) zero
(ii) In Young’s double – slit experiment if yellow light is replaced by blue light, the interference fringes become
(a) wider
(b) brighter
(c) narrower
(d) darker
(iii) In Young’s double slit experiment, if the separation between the slits is halved and the distance between the
slits and the screen is doubled, then the fringe width compared to the unchanged one will be
(a) Unchanged
(b) Halved
(c) Doubled
(d) Quadrupled
(iv) When the complete Young’s double slit experiment is immersed in water, the fringes
(a) remain unaltered
(b) become wider
(c) become narrower
(d) disappear
(v) In a two slit experiment with white light, a white fringe is observed on a screen kept behind the slits. When
the screen is moved away by 0.05 m, this white fringe
(a) does not move at all
(b) gets displaced from its earlier position
(c) becomes coloured
(d) disappears.
SECTION - C
All questions are compulsory. In case of internal choices, attempt anyone.
9.
If light ray with small angle of incidence falls on glass surface having speed of light in glass as 4/5 of speed of
light in air, find the angle of deviation after refraction from first surface.
OR
Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviation dm for
a triangular prism. A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum
deviation. If the angle of prism is 60°, find the angle of minimum deviation.
10. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important
conclusions which you can draw regarding the nature of nuclear forces. A ray of light falling at an angle of
incidence of 50° is refracted through a prism and suffers minimum deviation. It the angle of prism is 60° then
find the angle of minimum deviation.
145
Practice Paper - 2
11. The V-I characteristic of a diode is shown in the figure. Find the ratio of forward to reverse bias resistance.
I (mA)
20
15
10
–10
1 A
.7
.8
V(volt)
OR
Find the resistance of a germanium junction diode whose V – I is shown in figure. (Vk = 0.3 V)
12. Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and
3 eV respectively. Which of them will not be able to detect light of wavelength 600 nm?
SECTION - D
All questions are compulsory. In case of internal choices, attempt any one.
13. An equiconvex lens with radii of curvature of magnitude R each, is put over a liquid layer poured on top of
a plane mirror. A small needle, with its tip on the principal axis of the lens, is moved along the axis until its
inverted real image coincides with the needle itself. The distance of the needle from the lens is measured to
be a. On removing the liquid layer and repeating the experiment the distance is found to be b.
Liquid
Given that two values of distances measured represent the focal length values in the two cases, obtain a formula
for the refractive index of the liquid.
OR
A compound microscope uses an objective lens of focal length 4 cm and eye lens of focal length 10 cm. An
object is placed at 6 cm from the objective lens
(a) Calculate the length of the compound microscope.
(b)Calculate magnifying power of the compound microscope, if the final image is formed at the near point.
14. In a full wave junction diode rectifier the input ac has rms value of 20 V. The transformer used is a step up
transformer having primary and secondary turn ratio 1 : 2. What would be the dc voltage in the rectified
output?
CBSE Board Term-II Physics Class-12
146
SECTION - E
All questions are compulsory. In case of internal choices, attempt any one.
15. (a) Draw a ray diagram to show the formation of the image of an object placed on the axis of a convex
refracting surface of radius of curvature ‘R’, separating the two media of refractive indices ‘n1’ and ‘n2’ (n2 > n1).
Use this diagram to deduce the relation
of the object and the image formed.
n2 n1 n2 − n1
− =
, where u and v represent respectively the distance
v u
R
(b)A convex lens of focal length f1 is kept in contact with a concave lens of focal length f2. Find the focal
length of the combination.
OR
(a)Draw a ray diagram showing the image formation by an astronomical telescope when the final image is
formed at infinite.
(b) (i) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length
5.0 cm. Find the magnifying power of the telescope for viewing distant objects when the telescope is in
normal adjustment and the final image is formed at the least distance of distinct vision.
(ii) Also find the separation between the objective lens and the eyepiece in normal adjustment.
16. (a) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s
photoelectric equation?
(b) The work function of the following metals is given : Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and
Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å
from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer
and placed 50 cm away?
OR
(a) Draw a graph showing the variation of stopping potential with frequency of incident radiation for two
photosensitive materials having work functions W1 and W2(W1 > W2). Write two important conclusions that
can be drawn from the study of these plots.
(b) Draw a plot showing the variation of photoelectric current with collector plate potential for two different
frequencies, u1 > u2, of incident radiation having the same intensity. In which case will the stopping potential
be higher? Justify your answer.
1. Yes, electromagnetic waves carry energy and
momentum.
2.
1
2
mvmax
= 3 eV or eV0 = 3 eV or V0 = 3 V.
2
OR
hc hc hc(λ 2 − λ1)
∆E = −
=
(in eV)
λ1λ 2
λ1 λ 2
=
6 ⋅ 62 × 10 −34 × 3 × 108 × (5000 − 2500) × 10 −10
2500 × 5000 × 10 −20 × 1 ⋅ 6 × 10 −19
= 2.48 eV
3. The maximum and minimum intensities are given by
(a + b)2 and (a – b)2 for the superposition of two coherent
sources. a, b are amplitudes of superposing waves.
Let I = x
The amplitude a = I = x, b = 4I = 2x
\
Maximum intensity = (a + b)2 = 9x2 = 9I
Minimum intensity = (a – b)2 = x2 = I
4. Wavelength of Ha line, l =
6563 Å
= 6563 × 10–10 m
Red shift observed in star is
(l′ – l) = 15 Å = 15 × 10–10 m
147
Practice Paper - 2
OR
The volume of the nucleus is directly proportional to
the number of nucleons (mass number) constituting
the nucleus.
4 3
πR ∝ A Where R → radius
3
R ∝ A1/3 A → Mass number
R = R0 A1/3
5. The speed of em-waves in vacuum determined by
E
the electric (E0) and magnetic fields (B0) as, c = 0
B0
6. (a) : Experimentally, it is found that the average
radius of a nucleus is given by
R = R0A1/3 where R0 = 1.1 × 10–15 m = 1.1 fm
and A = mass number
4
4
The volume of a nucleus is V = p R3 = p R03A.
3
3
Now as the masses of a proton and a neutron are roughly
equal, say m, the mass of a nucleus is also roughly
proportional to the mass number A, M = mA Hence
density within a nucleus, ρ =
m
M
mA
=
=
4
V 4 3
πR 3
πR A
3 0
3 0
is independent of the mass number A.
7. (c) : In both fission and fusion large amounts of
energy is released. The reason is correct. Charge, mass,
momentum and energy, all are conserved.
8. (i) (b) : The condition for possible interference
maxima on the screen is, dsinq = nl
where d is slit separation and l is the wavelength.
As d = 2l (given)
\ 2lsinq = nl or 2sinq = n
For number of interference maxima to be maximum,
sinq = 1 \ n = 2
The interference maxima will be formed when
n = 0, ± 1, ± 2
Hence the maximum number of possible maxima is 5.
λD
(ii) (c) : Fringe width, β =
d
\ If we replace yellow light with blue light, i.e., longer
wavelength with shorter one, therefore the fringe width
decreases.
d
and D ′ = 2D
2
λD
Fringe width, β =
d
 2D 
= 4β
New fringe width β′ = λ 
 d / 2 
(iii) (d) : d ′ =
(iv) (c) : When Young’s double slit experiment is
λ
repeated in water, instead of air. λ ′ = , i.e., wavelength
µ
λ ′D
decreases. β =
, i.e., fringe width decreases.
d
\ The fringe become narrower.
(v) (a) : Using white light, we get white fringe at the
centre i.e., white fringe is the central maximum. When
the screen is moved, its position is not changed.
9.
Let velocity of light in rarer medium be v.
4v
Then velocity of light in glass is
.
5
\
Refractive index of glass with respect to given rarer
v
5
medium is µ =
=
4v / 5 4
sin i
i
From Snell’s law, µ =
≈
sin r r
(Q for small angles, sinq ≈ q)
\
i 5
=
r 4
\
Angle of deviation = i – r = i –
or
r=
4i
5
4i i
=
5 5
OR
The relation between the angle of incidence i, angle of
prism A, and the angle of minimum deviation dm, for a
triangular prism is given by, i =
A + dm = 100°
dm = 100° – 60° = 40°
10.
Potential energy (MeV)
and speed of light, c = 3 × 108 m s–1
Let the velocity of the star with which it is receding away
from the earth be v
\ Red shift relation
v
λ′ − λ = λ
c
c
3 × 108 × 15 × 10 −10
∴ v = × (λ′ − λ) =
= 6.86 × 105 m s −1
λ
6563 × 10 −10
A + δm
2
+100
0
–100
r0 1
2
3
r(fm)
CBSE Board Term-II Physics Class-12
148
From the above plot, following conclusions can be
drawn.
(i) Nuclear forces are short range forces.
(ii) For a separation greater than r0, the nuclear forces
are attractive and for separation less than r0, the nuclear
forces are strongly repulsive.
11. Forward bias resistance,
∆V
0. 8 − 0. 7
0. 1
R1 =
=
=
= 10
−
3
∆I for (20 − 10) × 10
10 × 10 −3
Reverse bias resistance, R2 =
10
−6
= 107
1 × 10
then, the ratio of forward to reverse bias resistance,
R1 10
=
= 10 −6
R2 107
or v0 = + 12 cm
For eye lens : fe = + 10 cm and ve = – 25 cm
1
1
1
1
1
1
1
1 −7
= −
or
= − =− − =
f e v e ue
ue v e f e
25 10 50
or ue = –7.14 cm
Length of microscope tube is,
L = |v0| + |ue| = 12 + 7.14 = 19.14 cm
(b) Magnifying power of microscope is,
L 
D
M = − 1 + 
f0 
fe 
19.14
4
or
M=−
or
M = – 16.75
 25 
1 + 10  = − 4.785 × 3.5
12. Given, l = 600 nm = 6 × 10–7 m
14. Here, input Vrms = 20 V
Peak value of input voltage
Vo = 2 Vrms = 2 × 20 = 28.28 V
Since the transformer is a step up transformer having
transformer ratio 1 : 2, the maximum value of output
voltage of the transformer applied to the diode will be
V0′ = 2 × Vo = 2 × 28.28 V
E=
\
OR
From graph,
Resistance of the germanium junction diode,
2V
∆V 2.3 V − 0.3 V
=
= 0. 2 k Ω
R=
=
10 mA − 0
∆I
10 × 10 −3 A
−34
8
hc 6.6 × 10 × 3 × 10
eV = 2.06 eV
=
λ 6 × 10 −7 × 1.6 × 10 −19
As, energy gaps of diodes D1 and D3 are greater than
the given energy of the incident radiation. Hence diodes
D1 and D3 will not be able to detect light of wavelength
600 nm.
13. Clearly, equivalent focal length of equiconvex lens
and water lens f = a
Focal length of equiconvex lens, f1 = b
Focal length f2 of water lens is given by
ab
1
1 1 1 1 b−a
= − = − =
or f 2 =
f2
f f1 a b
ab
b−a
The water lens formed between the plane mirror and the
equiconvex lens is a planoconcave lens. For this lens,
R1 = –R and R2 = ∞
Using lens maker’s formula,
1
1
1
= (µ − 1)  − 
f2
R
R
 1
2
or
b−a
1
(a − b)R
 1
= (µ − 1) 
−  or µ = 1 +
ab
ab
 −R ∞ 
OR
(a) For object lens : f0 = + 4 cm and u0 = – 6 cm
1 1
1
1
1
1 1 1 1
= −
or
= +
= − =
f v 0 u0
v 0 f 0 u0 4 6 12
2V0′ 2 × 2 × 28.28
=
= 36 V
π
22 / 7
15. (a) Refraction at convex spherical surface
When object is in rarer medium and image formed is
real.
dc voltage =
n1
O
N
A
i
u PN
n2
r
C
I
R
v
n1 < n2
In DOAC, i = a + g
and in DAIC, g = r + b or r = g – b
sin i i α + γ
∴ By Snell’s law, 1n2 =
≈ =
sin r r γ − β
n2 α + γ
=
or n2 γ − n2β = n1α + n1γ
n1 γ − β
or (n2 – n1)g = n1a + n2b...(i)
As a, b and g are small and P and N lie close to each
other,
AN AN
So, α ≈ tan α =
≈
NO PO
or
β ≈ tan β =
AN AN
≈
NI
PI
γ ≈ tan γ =
AN AN
≈
NC PC
On using them in equation (i), we get
149
Practice Paper - 2
(n2 − n1)
AN
AN
AN
= n1
+ n2
PC
PO
PI
n2 − n1 n1 n2
=
+
PC
PO PI
or
where, PC = + R, radius of curvature
PO = – u, object distance
PI = + v, image distance
n −n
n −n n n
n
n
So, 2 1 = 1 + 2 or 2 1 = 2 − 1
R
−u v
R
v u
This gives formula for refraction at spherical surface
when object is in rarer medium.
(b)
v
f1
(ii) Separation between objective and eyepiece when
final image is formed at infinity,
v
L = f 0 + fe
I
O
u
L1 L2
I
f2
For convex lens
1 1 1
= − ...(i)
f1 v′ u
For concave lens ( f2 = –ve)
−
1 1 1
= − ...(ii)
f 2 v v′
Adding equations (i) and (ii)
1
1
1 1
− =− +
f1 f 2
u v
Also,
1 1 1
− =
v u f
1
1 1
− =
f1 f 2 f
16. (a) The Einstein’s photoelectric equation is given as
Kmax = hu – f0
Since K max must be non-negative implies that
photoelectric emission is possible only if hu > f0
φ0
,
h
This shows that the greater the work function f 0 ,
higher the threshold frequency u 0 needed to emit
photoelectrons. Thus, there exists a threshold frequency
φ
u 0 = 0 for the metal surface, below which no
h
photoelectric emission is possible.
or
uf > u0 where u0 =
or
hc
> f0
λ
for l = 3300 Å
hc 1.989 × 10 −25 6.03 × 10 −19
=
= 3.77 eV
=
λ 3300 × 10 −10 1.6 × 10 −19
f f
So, f = 1 2
f 2 − f1
OR
(a)
L = 140 cm + 5.0 cm = 145 cm
(b) Condition for photo electric emission, hu > f0
where f = focal length of combination
∴
(b) (i) Given f0 = 140 cm, fe = 5 cm
When final image is at infinity, magnifying power,
−f
140
m= 0 =−
fe
5. 0
m = –28
Negative sign shows that the image is inverted.
When final image is at the least distance of distinct
vision,
−f 
f 
magnifying power, m = 0 1 + e 
fe  D 
−140  5.0 
=
1+
= −33.6
5.0  25 
\ Mo and Ni will not cause photoelectric emission.
If the laser source is brought nearer and placed 50
cm away, then photoelectric emission will not effect,
since it depends upon the work function and threshold
frequency.
OR
(a) The graph showing the variation of stopping
potential (V0) with the frequency of incident radiation
(u ) for two different photosensitive materials having
work functions W 1 and W 2 (W 1 > W 2) is shown in
figure.
CBSE Board Term-II Physics Class-12
150
(b) The stopping potential is more negative for higher
frequencies of incident radiation. Therefore, stopping
potential is higher for u1.
∆V h
= [ eDV = hDu]
∆υ e
\ Slope of the line = h i.e., it is a constant quantity
e
and does not depend on nature of metal surface.
(ii) Intercept of graph 1 on the stopping potential axis
(i)
Slope of the line =
work function(W )
hυ
=− 0
e
e
\ Intercept of the line depends upon the stopping
function of the metal surface.
=
(KE)max = eV0 = hu – f
h
φ
⇒ V0 = υ −
e
e
From this equation we can conclude that V0 will increase
if u increases.

PRACTICE PAPER 3
Time allowed : 2 hours
Maximum marks : 35
General Instructions :
(i) All questions are compulsory. There are 16 questions in all.
(ii) This question paper has five sections: Section A, Section B, Section C, Section D and Section E.
(iii)Section A contains five very short answer questions and two assertion reasoning MCQs of 1 mark each. Section
B has one case based question of 4 marks, Section C contains four short answer questions of 2 marks each,
Section D contains two short answer questions of 3 marks each and Section E contains two long answer questions
of 5 marks each.
(iv)There is no overall choice. However internal choice is provided. You have to attempt only one of the choices in
such questions.
SECTION - A
All questions are compulsory. In case of internal choices, attempt any one of them.
1.
Write the relation between the refractive index and critical angle for a given pair of optical media.
OR
State the condition for total internal reflection.
2.
Arrange the following materials in increasing order of their resistivity.
Nichrome, Copper, Germanium, Silicon
3.
In the circuit shown if drift current for the diode is 20 mA, find the potential difference across the diode.
4V
15 4.
5.
OR
Show the variation of resistivity of Si with temperature in a graph.
Is the ratio of number of holes and number of conduction electrons in a p-type semiconductor more than, less
than or equal to 1?
The circuit shown in the figure has two oppositely connected ideal diodes connected in parallel. Find the current
flowing through each diode in the circuit.
12 V
For question numbers 6-7 two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false and R is also false
6.
Assertion (A) : Velocity of light is different in different media.
Reason (R) : Light is an electromagnetic wave whose velocity in a medium depends on refractive index of the
medium.
*The paper is for practice purpose. CBSE has yet not released the official sample paper.
So, the pattern is suggestive only. For latest information visit www.cbse.gov.in.
CBSE Board Term-II Physics Class-12
152
7.
Assertion (A) : V - I characteristic of p-n diode is same as that of any other conductor.
Reason (R) : p-n diode behave as conductor at room temperature.
SECTION - B
Question 8 is a Case Study based question and it is compulsory. Attempt any 4 sub parts from this question. Each
part carries 1 mark.
Photoelectric Emission
8.
When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic
surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons
depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted
photoelectrons depends on intensity. (hu – f) is the maximum kinetic energy of emitted photoelectrons (where
f is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not
deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference
across the tube. Wavelength of characteristic X-ray depends on the atomic number.
(i)
Einstein’s photoelectric equation is
1
(d) E = mv 2
2
(ii) Light of wavelength l which is less than threshold wavelength is incident on a photosensitive material. If incident
wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential
will
(a) Emax = hu – f
(b) E = mc2
(c) E2 = p2c2 + m02c4
(a) increase
(b) decrease
(c) be zero
(d) become exactly half
(iii) When ultraviolet rays incident on metal plate then photoelectric effect does not occur, it occur by incident of
(a) Infrared rays
(b) X-rays
(c) Radio wave
(d) Micro wave
(iv) If frequency (u > u0) of incident light becomes n times the initial frequency (u), then K.E. of the emitted
photoelectrons becomes (u0 threshold frequency).
(a)
n times of the initial kinetic energy
(b) More than n times of the initial kinetic energy
(c)
Less than n times of the initial kinetic energy
(d) Kinetic energy of the emitted photoelectrons remains unchanged.
(v)
A monochromatic light is used in a photoelectric experiment. The stopping potential
(a)
is related to the mean wavelength
(b) is related to the shortest wavelength
(c)
is not related to the minimum kinetic energy of emitted photoelectrons
(d) intensity of incident light.
SECTION - C
All questions are compulsory. In case of internal choices, attempt anyone.
9.
In a certain double slit experimental arrangement, interference fringes of width 1 mm each are observed when
light of wavelength 5000 Å is used. Keeping the setup unaltered, if the source is replaced by another of wavelength
6000 Å, then find the fringe width.
OR
A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a
screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen.
Calculate the width of the slit.
153
Practice Paper - 3
10.
11.
Name the parts of the electromagnetic spectrum which is
(a) suitable for radar systems used in aircraft navigation.
(b) used to treat muscular strain.
Draw the output waveform across the resistor.
A
V0
+1 V
0
–1 V
Input waveform at A
OR
Three photo diodes D1, D2 and D3 are made of semiconductors having band gaps of 2.5 eV, 2 eV and 2.75 eV,
respectively. Which ones will be able to detect light of wavelength 5500 Å?
12. State the process responsible for energy generation in Sun. Why such process is difficult to achieve an Earth.
SECTION - D
All questions are compulsory. In case of internal choices, attempt any one.
13. (a) Two slits in Young’s double slit experiment are illuminated by two different lamps emitting light of the same
wavelength. Will you observe the interference pattern? Justify your answer.
(b) In Young’s double slit experiment using monochromatic light of wavelength l, the intensity of light at a
point on the screen where path difference is l, is K units. Find out the intensity of light at a point where path
difference is l/3.
14. Use the lens equation to deduce algebraically:
(a) An object placed within the focus of a convex lens produces a virtual and enlarged image.
(b) A concave lens produces a virtual and diminished image independent of the location of the object.
OR
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The
two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram
to show the formation of the image by the combination. Determine the nature and position of the image formed.
SECTION - E
All questions are compulsory. In case of internal choices, attempt any one.
15. Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen
atom undergoes transition from higher energy state (quantum number ni) to the lower state, (nf). When electron
in hydrogen atom jumps from energy state ni = 4 to nf = 3, 2, 1. Identify the spectral series to which the emission
lines belong.
OR
(i)In hydrogen atom, an electron undergoes transition from 2nd excited state to the first excited state and then
to the ground state. Identify the spectral series to which these transitions belong.
(ii) Find out the ratio of the wavelengths of the emitted radiations in the two cases.
1 1 1
16. Define power of a lens. Write its units. Deduce the relation = +
for two thin lenses kept in contact
f f1 f2
coaxially.
OR
A screen is placed 80 cm from an object. The image of the object on the screen is formed by a convex lens placed
between them at two different locations separated by a distance 20 cm. Determine the focal length of the lens.
1
, where a and b are the rarer and denser
sin C
media respectively and C is the critical angle for the given
pair of optical media.
1.
a
µb =
OR
(a) Conditions for total internal reflection are
(i) Light must travel from denser to rarer medium.
(ii) Angle of incidence must be greater than critical
angle.
CBSE Board Term-II Physics Class-12
154
2. Materials
Copper (Cu) Nichrome
(alloy of Ni, Fe, Cr)
Germanium
Silicon
Resistivity (r) (W m at 0°C)
1.7 × 10–8
100 × 10–8
0.46
2300
3. (c) : Since the diode is reversed biased, only drift
current exists in circuit which is 20 mA.
Potential drop across 15 W resistor
= 15 W × 20 mA = 300 mV = 0.0003 V
Potential difference across the diode
= 4 – 0.0003 = 3.99 ; 4V
OR
Resistivity ()
The variation of resistivity of Silicon (Si) with temperature
is shown as
Temperature (T)
4.The ratio of number of holes and number of
conduction electrons in a p-type semiconductor is more
than 1.
5. Diode D1 is reverse biased, so it offers an infinite
resistance. So no current flows in the branch of diode
D 1.
Diode D2 is forward biased, and offers negligible
resistance in the circuit. So current in the branch
12 V
V
I=
=
= 2A
Req 2 Ω + 4 Ω
In the given graph knee voltage is a voltage at which
forward bias becomes greater than the potential barrier,
the forward current increases almost linearly.
From this graph we can verify that p-n diode
characteristics are very different from that of conductor
which obey’s Ohm’s law.
8. (i) (a) Einstein’s photo ele c t r ic e quat ion is
Emax = hu- f
(ii) (a) : According to Einstein’s photoelectric
equation,
hc hc
−
eV0 =
λ λ0
As l0 is constant, so when l is decreased, stopping
potential (V0) increases.
(iii) (b) : It indicates that threshold frequency is
greater than that of ultraviolet light. As X-rays have
greater frequency than ultraviolet rays, so they can
cause photoelectric effect.
(iv) (b) : K.E.1 = hu – f
K.E.2 = nhu – f = n(hu – f) + (n – 1)f
K.E.2 = nKE1 + (n – 1)f
K.E.2 > nKE1
(v) (b) : S topping potential is the measurement of
maximum kinetic energy of emitted photoelectrons
and kinetic energy of emitted photoelectrons is
linearly related with the frequency of incident light
(i.e., corresponding to shortest wavelength, K.E. is
maximum).
Stopping potential is independent of intensity.
λD
d
Since D and d are unaltered, b ∝ l
λ′
6000
β′ λ′
or β′ = β × = 1 ×
= 1.2 mm
∴
=
λ
5000
β λ
9.Fringe width, β =
+
12 V
6. (a) : Velocity of light has different values in different
media.
It depends on the refractive index of the medium. Related
by formula
velocity in vacuum
vmedium =
refractive index of medium
7. (d) : The V-I characteristic of p-n diode depends
whether the junction is forward biased or reverse
biased. This can be showed by graph between voltage
and current.
OR
Position of first minimum in diffraction pattern
Dλ
y=
a
Dλ 1 × 500 × 10 −9
So, slit width a =
=
= 2 × 10 −4 m
−
3
y
2.5 × 10
10. (a) Microwaves are suitable for radar systems
used in aircraft navigation.
155
Practice Paper - 3
These waves are produced by special vacuum tubes,
namely klystrons, magnetrons and Gunn diodes.
(b) Infra-red waves are used to treat muscular pain.
These waves are produced by hot bodies and molecules.
11.
The diode acts as half wave rectifier, it offers low resistance
when forward biased and high resistance when reverse
biased.
OR
We know that, energy of incident photon
hc
E=
λ
l = 6000 Å = 600 nm (given)
1242 eVnm
E=
= 2.25 eV
550 nm
or v = –ve
m = v / u, As
1
1
<
or | v | > | u |
|v | |u|
So m > 1
and hence image formed is enlarged.
(b) For concave lens : f is –ve and u is –ve
For u < 0 (object is on left)
| f | + | u |
 1
1 1 1
1 
= + =−
+
=−

v f u
| f | | u |
 | f |⋅| u | 
or v = –ve
As v is –ve, so image lies on the left of lens and is virtual
in nature.
v
1
1
m = , As
>
or | v | < | u |
u
|v | |u |
So m < 1, and hence image formed is diminished.
OR
L
M
D2 will detect these radiations because energy of incident
radiation is greater than the band gap.
12. Nuclear fusion is responsible for energy generation
in the Sun. For fusion, temperature of 107 K is required,
such a temperature is difficult to achieve.
13. (a) Two different lamps emit light waves which
are not coherent, as they are not in same phase or not
have stable phase difference. Due to this, no sustained
interference pattern can be obtained on screen.
 φ
(b) Intensity at a point, I = 4I0 cos2  
 2
2π
Phase difference =
× Path difference
λ
At path difference l,
2π
× λ = 2π
Phase difference, φ =
λ
 2π 
∴ Intensity, K = 4I0 cos2  
 2
[Q Given I = K at path difference l]
K = 4I0
...(i)
λ
At path difference
3
φ′ =
2π λ 2π
× =
λ 3 3
K
 2π 
Intensity, I ′ = 4I0 cos2   =
(Using (i))
 6
4
14. (a) For convex lens : f is +ve and u is – ve.
For 0 < |u| < f,
1 1 1 1
1 |u| − f
= + = −
=
= − ve
f |u|
v f u f |u |
O
I
I′
15 cm
60 cm
15 cm
30 cm
For the convex lens,
u = –60 cm, f = +20 cm
1 1 1
− = gives v = +30 cm
v u f
For the convex mirror,
u = +(30 – 15) cm = 15 cm, f = +
20
cm = 10 cm
2
1 1 1
+ = gives v = +30 cm
v u f
Final image is formed at the distance of 30 cm from the
convex mirror (or 45 cm from the convex lens) to the
right of the convex mirror.
The final image formed is a virtual.
mv 2
1
=
rn
4πε0
nh
and mvrn =
2π
From eqn. (i) and (ii)
ε h2n2
∴ rn = 0 2
πme
Total energy
15.
Q
1
1
En = mvn2 −
2
4πε0
e2
…(i)
rn2
e2
1
=
rn 8πε0
…(ii)
e2
1
−
rn 4πε0
e2
rn
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CBSE Board Term-II Physics Class-12
156
A
−Rhc
1 e2
1 me 4
= − 2 2 2 , En = 2
n
8πε0 rn
8ε0 h n
me 4
where Rydberg constant R = 2 3
8ε0h c
B
En = −
I
I1
v


1
1
Energy emitted DE = Ei – Ef , ∆E = Rhc  2 − 2 
n
ni 
 f

But DE = hu


1
1
1
me 4 1
υ = Rc  2 − 2  or υ =


−
8ε20h3  n2f ni2 
 n f ni 
When electron in hydrogen atom jumps from energy
state ni = 4 to nf = 3, 2, 1, the Paschen, Balmer and Lyman
spectral series are found.
OR
(i) An electron undergoes transition from 2nd excited
state to the first excited state is Balmer series and then
to the ground state is Lyman series.
(ii) The wavelength of the emitted radiations in the two
cases.
v1
u
An object is placed at point O. The lens A produces an
image at I1 which serves as a virtual object for lens B
which produces final image at I.
Given, the lenses are thin. The optical centres (P) of the
lenses A and B coincide with each other.
For lens A, we have
1 1 1
− =
…(i)
v1 u f1
1 1
1
For lens B, we have − =
…(ii)
v v1 f2
Adding equations (i) and (ii),
1 1 1 1
− = +
…(iii)
v u f1 f2
If two lenses are considered as equivalent to a single lens
of focal length f, then
1 1 1
…(iv)
− =
v u f
From equation (iii) and equation (iv), we can write
1 1 1
= +
f f1 f2
OR
Case I : u = –x, v = 80 – x, f = f
For n2 → n1
DE = (–3.40 + 13.6) = 10.20 eV
λ2 =
P
O
x
80 – x
80 cm
6.626 × 10 −34 × 3 × 108
\
10.2 × 1.6 × 10 −19
19.878 × 10 −7
= 1.218 × 10 −7 m = 1218 Å
λ2 =
10.2 × 1.6
1 1 1
1
1
= − =
+
f v u 80 − x x
...(i)
Case II : u = –(x + 20), v = 60 – x
For n3 → n2
DE = (–1.5 + 3.4) = 1.9 eV
19.878 × 10 −7
= 6.538 × 10 −7 m = 6538 Å
1. 9 × 1. 6
λ
6538
= 5.36
The ratio 1 =
λ 2 1218
16. Power of lens : It is the reciprocal of focal length
of a lens.
1
P = (f is in metre)
f
Unit of power of lens : Dioptre
λ1 =
x + 20
60 – x
80 cm
1
1
1
=
+
f 60 − x x + 20
From (i) and (ii)
1
1
1
1
+ =
+
⇒ x = 30 cm
80 − x x 60 − x x + 20
1
1
1
=
+ ⇒ f = 18.75 cm
\
f 80 − 30 30

...(ii)