NCERT Solutions for Class 12 Physics Chapter 12 Atoms

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 12
Chapter Name : Atoms

Q12.1 Choose the correct alternative from the clues given at the end of the each statement:
(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater
than/no different from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a
net force. (Thomson’s model/ Rutherford’s model)
(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution
in .......... (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the
models.)

Answer. (a) The sizes of the atoms taken in Thomson's model and Rutherford's model have the same order of
magnitude.
(b) In the ground state of Thomson's model, the electrons are in stable equilibrium. However, in Rutherford's
model, the electrons always experience a net force.
(c) A classical atom based on Rutherford's model is doomed to collapse.
(d) An atom has a nearly continuous mass distribution in Thomson's model, but has a highly non-uniform mass
distribution in Rutherford's model.
(e) The positively charged part of the atom possesses most of the mass in both the models.

Page : 435 , Block Name : Exercise

Q12.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid
hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14K.) What results do you expect?

Answer. In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil,
then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 × 10 kg) is
−27

less than the mass of incident a − particles (6.64 × 10 −27
kg) . Thus, the mass of the scattering particle is more
than the target nucleus (hydrogen). As a result, the a − particles would not bounce back if solid hydrogen is used
in the a − particles scattering experiment.

Page : 435 , Block Name : Exercise

Q12.3 What is the shortest wavelength present in the Paschen series of spectral lines?

Answer. Rydberg's formula is given as:
hc −19 1 1
= 21.76 × 10 [ − ]
2 2
λ n n
1 2

−34
h = Planck's constant = 6.6 × 10 Js
8
c = Speed of light = 3 × 10 m/s

(n1 and n2 are integers )

The shortest wavelength present in the Paschen series of the spectral lines is given for
Values n = 3 and n = ∞.
1 2

hc −19 1 1
= 21.76 × 10 [ − ]
λ (3)2 (∞)2

−34 8
6.6×10 ×3×10 ×9
λ = −19
21.76×10
−7
= 8.189 × 10 m

= 818.9nm

Page : 436 , Block Name : Exercise

Q12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted
when the atom make a transition from the upper level to the lower level?

Answer. Separation of two energy levels in an atom,
E = 2.3eV
−19
= 2.3 × 1.6 × 10
−19
= 3.68 × 10 J

Let v be the frequency of radiation emitted when the atom transits from the upper level
to the lower level.
We have the relation for energy as:
E = hv

−34
h = Planck's constant = 6.62 × 10 Js
E
∴ v =
h
−19
3.68×10 14
= −32
= 5.55 × 10 Hz
6.62×10

Hence, the frequency of the radiation is 5.6 × 10 14
Hz .

Page : 436 , Block Name : Exercise

Q12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the
electron in this state?

Answer. Ground state energy of hydrogen atom, E = −13.6eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy = −E = −(−13.6) = 13.6eV
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy = −2 × (13.6) = −27.2eV

Page : 436 , Block Name : Exercise

Q12.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine
the wavelength and frequency of photon.

Answer. For ground level, n = 1 1

Let E be the energy of this level. It is known that E is related with n as:
1 1 1
−13.6
E1 = eV
2
n
1

−13.6
= = −13.6eV
2
1

The atom is excited to a higher level, = 4.
Let E be the energy of this level.
1

−13.6
E1 = eV
2
n
1

−13.6
= = −13.6eV
2
1

The amount of energy absorbed by the photon is given as:
E = E2 − E1
−13.6 13.6
= − (− )
16 1

13.6×15
= eV
16
13.6×15 −19 −18
= × 1.6 × 10 = 2.04 × 10 J
16

For a photon of wavelength λ , the expression of energy is written as:
hc
E =
λ

h = Planck's constant = 6.6 × 10 −34
Js

c = Speed of light = 3 × 10 m/s 8

hc
∴ λ =
E
−34 8
6.6×10 ×3×10
= −18
2.04×10
−8
= 9.7 × 10 m = 97nm

And, frequency of a proton is given by the relation,
c
v =
λ
8
3×10 15
= −8
≈ 3.1 × 10 Hz
9.7×10

Hence, the wavelength of the photon is 97 nm while the frequency 3.1 × 10 15
Hz .

Page : 436 , Block Name : Exercise

Q12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3
levels. (b) Calculate the orbital period in each of these levels.

Answer. (a) Let v be the orbital speed of the electron in a hydrogen atom in the ground state level, n
1 1 = 1 . For
charge (e) of an electron, v is given by the relation, 1
2 2
e e
v1 = =
n1 4πϵ0 (h/2π) 2ϵ0 h

−19
e = 1.6 × 10 C
−12 −1 −1 −2
ϵ0 = Permittivity of free space = 8.85 × 10 N C m

−34
h = Planck's constant = 6.62 × 10 Js
2
−19
(1.6×10 )
∴ v1 =
−12 −34
2×8.85×10 ×6.62×10
8 6
= 0.0218 × 10 = 2.18 × 10 m/s

For level n = 2, we can write the relation for the corresponding orbital speed as:
2
2
e
v2 =
n2 2ϵ0 h
2
−19
(1.6×10 )
=
−12 −34
2×2×8.85×10 ×6.62×10
6
= 1.09 × 10 m/s

And, for n 3 = 3, we can write the relation for the corresponding orbital speed as:
2
e
v3 =
n3 2ϵ0 h
2
−19
(1.6×10 )
=
−12 −34
3×2×8.85×10 ×6.62×10
5
= 7.27 × 10 m/s

(b) Let T be the orbital period of the electron when it is in level n
1 1 = 1

Orbital period is related to orbital speed as:
2πr1
T1 =
v1

r = Radius of the orbit
1
2 2
n h ϵ0
1
= 2
πme

h = Planck's constant = 6.62 × 10 Js −34

e = Charge on an electron = 1.6 × 10 C −19

ϵ = Permittivity of free space = 8.85 × 10
−12 −1 2 −2
0 N C m

m = Mass of an electron = 9.1 × 10 kg −31

2πr1
∴ T1 =
v1
2
2 −34 −12
2π×(1) ×(6.62×10 ) ×8.85×10
=
2
6 −31 −19
2.18×10 ×π×9.1×10 ×(1.6×10 )

−17 −16
= 15.27 × 10 = 1.527 × 10 s

For level n 2 = 2 , we can write the period as:
2πr2
T2 =
v2

r = Radius of the electron in n
2 2 = 2
2 2
(n2 ) h ϵ0
=
πme2
2πr2
∴ T2 =
v2
2
2 −34 −12
2π×(2) ×(6.62×10 ) ×8.85×10
= 2
6 6 −31 −19
1.09×10 ×10 ×9.1×10 ×(1.6×10 )

−15
= 1.22 × 10 s

And, for level n 3 = 3 , we can write the period as:
2πr3
T3 =
v3

r = Radius of the electron in n
3 3 = 3
2 2
(n3 ) h ϵ0
=
πme2
2πr3
∴ T3 =
v3
2
2 −34 −12
2π×(3) ×(6.62×10 ) ×8.85×10
=
2
5 −31 −19
7.27×10 ×π×9.1×10 ×(1.6×10 )

−15
= 4.12 × 10 s

Hence, the orbital period in each of these levels is 1.52 × 10 −16
s, 1.22 × 10
−15
s and 4.12 × 10 −15
s respectively.

Page : 436 , Block Name : Exercise

Q12.8 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?

Answer. The radius of the innermost orbit of a hydrogen atom, r = 5.3 × 10 m 1
−11

Let r be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
2
2
r2 = (n) r1
−11 −10
= 4 × 5.3 × 10 = 2.12 × 10 m

For n = 3, we can write the corresponding electron radius as:
2
r3 = (n) r1
−11 −10
= 9 × 5.3 × 10 = 4.77 × 10 m

Hence, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10 −10
m and 4.77× 10 −10
m respectively.

Page : 436 , Block Name : Exercise

Q12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of

wavelengths will be emitted?

Answer. It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is
12.5eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes
−13.6 + 12.5 eV i.e., − 1.1eV.

Orbital energy is related to orbit level (n) as:
−13.6
E = 2
eV
(n)

−13.6
E = = −1.5eV
9

For n = 3
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has
jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series
of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
1 1 1
= Ry ( 2
− 2
)
λ 1 n

R = Rydberg constant = 1.097 × 10 m
7 −1
y

λ = Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
1 7 1 1
= 1.097 × 10 ( − )
2 2
λ 1 3

7 1 7 8
= 1.097 × 10 (1 − ) = 1.097 × 10 ×
9 9

9
λ = = 102.55nm
7
8×1.097×10

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1 7 1 1
= 1.097 × 10 ( − )
2 2
λ 2 3

7 1 1 7 5
= 1.097 × 10 ( − ) = 1.097 × 10 ×
4 9 36

36
λ = = 656.33m
7
5×1.097×10

This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one
wavelength i.e., 656.33 nm is emitted.

Page : 436 , Block Name : Exercise

Q12.10 In accordance with the Bohr’s model, nd the quantum number that characterises the earth’s revolution
around the sun in an orbit of radius 1.5 × 10 m with orbital speed 3 ×10 m/s. (Mass of earth = 6.0 × 10 kg.)
11 4 24

Answer. Radius of the orbit of the Earth around the Sun, r = 1.5 × 10 11
m

Orbital speed of the Earth, v = 3 × 10 m/s 4

Mass of the Earth, m = 6.0 × 10 kg 24

According to Bohr's model, angular momentum is quantized and given as:
nh
mvr =
2π

h = Planck's constant = 6.62 × 10
−34
Js

n = Quantum number
mvr2π
∴ n =
h
24 4 11
2π×6×10 ×3×10 ×1.5×10
= −34
6.62×10

73 74
= 25.61 × 10 = 2.6 × 10

Hence, the quanta number that characterizes the Earth' revolution is 2.6 × 10 74

Page : 436 , Block Name : Exercise

Q12.11 Answer the following questions, which help you understand the difference between Thomson’s model and
Rutherford’s model better.
(a) Is the average angle of de ection of α-particles by a thin gold foil predicted by Thomson’s model much less,
about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by
Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors xed, it is found experimentally that for small thickness t, the number of α-particles
scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of
scattering of α-particles by a thin foil?

Answer. (a) about the same
The average angle of de ection of a-particles by a thin gold foil predicted by Thomson's model is about the same
size as predicted by Rutherford's model. This is because the average angle was taken in both models.
(b) much less the probability of scattering of a-particles at angles greater than 90 predicted by Thomson's model
∘

is much less than that predicted by Rutherford's model.
(c) Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number
of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability
depends linearly on the thickness of the target.
(d) Thomson's model It is wrong to ignore multiple scattering in Thomson's model for the calculation of average
angle of scattering of Q particles by a thin foil. This is because a single collision causes very little de ection in this
model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Page : 436 , Block Name : Additional Exercise

Q12.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb
attraction by a factor of about 10 . An alternative way of looking at this fact is to estimate the radius of the rst
−40

Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will nd the
answer interesting.

Answer. Radius of the rst Bohr orbit is given by the relation,
2
h
4πϵ0 ( )

…(1)
2π

r1 = 2
me e

ϵ = Permittivity of free space
0

h = Planck’s constant = 6.63 × 10 Js −34

m = Mass of an electron = 9.1 × 10
−31
e kg

e = Charge of an electron = 1.9 × 10 C −19

m = Mass of a proton = 1.67 × 10
−27
p kg

r = Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
2

FC =
e

4πϵ0 r
2
… (2)
Gravitational force of attraction between an electron and a proton is given as:
Gmp me
FG =
2
… (3)
r

G = Gravitational constant = 6.67 × 10 −11 2
Nm /kg
2

If the electrostatic (Coulomb) force and the gravitational force between an electron and
a proton are equal, then we can write:
∴ F6 = Fc
Gmp me 2
e
2
= 2
r 4πϵ0 r
2

∴
4πϵ0
e
= Gmp mc …. (4)
Putting the value of equation (4) in equation (1), we get:
2
h
( )
2π

r1 =
2
Gmp me
2
−34
6.63×10
( )
2×3.14
29
= ≈ 1.21 × 10 m
2
−11 −27 −31
6.67×10 ×1.67×10 ×(9.1×10 )

It is known that the universe is 156 billion light years wide or 1.5 × 10 m wide. Hence, we can conclude that the 27

radius of the rst Bohr orbit is much greater than the estimated size of the whole universe.

Page : 436 , Block Name : Additional Exercise

Q12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n
to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in
the orbit.

Answer. It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n - 1).
We have the relation for energy (E ) of radiation at level n as: 1

4

E1 = hv1 =
hme
3
× (
1

n
2
) … (i)
2 h
(4π)3 ϵ ( )
0 2π

v = Frequency of radiation at level n
1

h = Planck’s constant

m = Mass of hydrogen atom

e = Charge on an electron

ϵ = Permittivity of free space
0

Now, the relation for energy (E ) of radiation at level (n - 1) is given as: 2
4

… (ii)
hme 1
E2 = hv2 = 3
× 2
2 h (n−1)
(4π)3 ϵ ( )
0 2π

v2 = Frequency of radiation at level (n − 1)

Energy (E) released as a result of de-excitation:
E = E2 − E1

hv = E2 − E1 …)(iii)W here, (v = Frequency of radiation emitted
Putting values from equation (i) and )ii) in equation (iii), we get:
4
me 1 1
v = 3
[ 2
− 2
]
2 h (n−1) n
3
(4π) ϵ ( )
0 2π

4
me (2n−1)
= 3
2 h
(4π)3 ϵ ( ) n2 (n−1)2
0 2π

For large n, we can write (2n − 1) ≃ 2n and (n − 1) ≃ n.
4
me v
∴ v = 3
…)(iv)Classicalrelationof f requencyof revolutionof anelectronisgivenas : (vc =
h 2πr
3 2 3
32π ϵ ( ) n
0 2π

… (v)
Velocity of the electron in the n th
orbit is given as:

… (vi)
2
e

h
4πϵ0 ( )n
2π

And, radius of the n th
orbit is given as:
2
h
4πϵ0 ( )

…. (vii)
2π
2
r = n
2
me

Putting the values of equations (vi) and (vii) in equation (v), we get:
4

vc =
me
3
… (viii)
2 h
3 3
32π ϵ ( ) n
0 2π

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

Page : 436 , Block Name : Additional Exercise

Q12.14 Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the
typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly
puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what
he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we
can get a quantity with the dimensions of length that is roughly equal to the known size of an atom ( ∼ 10 m ). −10

(a) Construct a quantity with the dimensions of length from the fundamental constants e, m , and c. Determine e

its numerical value.
(b) You will nd that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions.
Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play
any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic
size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in
recognising that h, m , and e will yield the right atomic size. Construct a quantity with the dimension of length
e

from h, me, and e and con rm that its numerical value has indeed the correct order of magnitude.

Answer. (a) Charge on an electron, e = 1.6 × 10 −19
C

Mass of an electron, m = 9.1 × 10 kg e
−31

Speed of light, c = 3 × 10 m/s 8

Let us take a quantity involving the given quantities as (
2
e
)
4πϵ0 me c2

ϵ = Permittivity of free space
0

And, 1
= 9 × 10 Nm C
4πϵ0
9 2 −2

The numerical value of the taken quantity will be:
2
1 e
×
4πϵ0 me c2
2
−19
(1.6×10 )
9
= 9 × 10 × 2
−31 8
9.1×10 ×(3×10 )

−15
= 2.81 × 10 m

Hence, the numerical value of the taken quantity is much smaller than the typical size of
an atom.
(b) Charge on an electron, e = 1.6 × 10 C −19

Mass of an electron, m = 9.1 × 10 kg e
−31

Planck’s constant, h = 6.63 × 10 Js −34

2
h
4πϵ0 ( )

Let us take a quantity involving the given quantities as
2π

2
mc e

ϵ = Permittivity of the free space
0

And, 1
= 9 × 10 Nm C
9 2 −2

4πϵ0

The numerical value of the taken quantity will be:

2
h
( )
2π

4πϵ0 ×
me e2
2
−34
6.63×10
( )
2×3.14
1
= ×
9 2
9×10 −31 −19
9.1×10 ×(1.6×10 )

−10
= 0.53 × 10 m

Hence, the value of the quantity taken is of the order of the atomic size.

Page : 437 , Block Name : Additional Exercise

Q12.15 The total energy of an electron in the rst excited state of the hydrogen atom is about –3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Answer. (a) Total energy of the electron, E = −3.4eV
Kinetic energy of the electron is equal to the negative of the total energy.
⇒ K = −E

= −(−3.4) = +3.4eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

(b) Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.
⇒ U = −2K

= −2 × 3.4 = −6.8eV

Hence, the potential energy of the electron in the given state is 6.8 ev.

(c) The potential energy of a system depends on the reference point taken. Here, the potential energy of the
reference point is taken as zero. If the reference point is
changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic
and potential energies, total energy of the system will also change.

Page : 437 , Block Name : Additional Exercise

Q12.16 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?

Answer. We never speak of quantization of orbits of planets around the Sun because the angular momentum
associated with planetary motion is largely relative to the value of Planck's constant (h). The angular momentum
of the Earth in its orbit is of the order of 10 h. This leads to a very high value of quantum levels n of the order of
70

10 . For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum
70

levels for planetary motion are considered continuous.

Page : 437 , Block Name : Additional Exercise

Q12.17 Obtain the rst Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in
which a negatively charged muon (μ ) of mass about 207m orbits around a proton].
−
e

Answer. Mass of a negatively charged muon, m μ = 207me

According to Bohr’s model,
Bohr radius, r e ∝ (
1

me
)

And, energy of a ground state electronic hydrogen atom, E ∝ m
e e

Also, energy of a ground state muonic hydrogen atom, E ∝ m .
μ μ

We have the value of the rst Bohr orbit, r = 0.53A = 0.53 × 10
−10
e m

Let r be the radius of muonic hydrogen atom.
μ

At equilibrium, we can write the relation as:
mμ r μ = me r e

207me × rμ = me re

−10
0.53×10 −13
∴ rμ = = 2.56 × 10 m
207

Hence, the value of the rst Bohr radius of a muonic hydrogen atom is
−13
2.56 × 10 m

We have,
Ee = −13.6eV

Tke the ratio of these energies as:
Ee me me
= =
Eμ mμ 207me

Eμ = 207Ee

= 207 × (−13.6) = −2.81keV

Hence, the ground state energy of a muonic hydrogen atom is −2.81keV

Page : 437 , Block Name : Additional Exercise