NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 13
Chapter Name : Nuclei

6 7
Q13.1 (a) Two stable isotopes of lithium 3 Li and 3 Li have respective abundances of 7.5% and 92.5%.
These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of
lithium.
(b) Boron has two stable isotopes, 10 11
5 B and 5 B. 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 10 11
5 B and 5 B.

6
Answer. (a) Mass of lithium isotope 3 Li, m 1 = 6.01512 u
7
Mass of lithium isotope 3 Li, m 2 = 7.01600 u
Abundance of 63 Li, m 1, η 1 = 7.5%
7
Abundance of 3 Li, m 1, η 2, 92.5%
The atomic mass of lithium atom is given as:
m 1η 1 + m 2η 2
m= η1 + n2
6.01512 × 7.5 + 7.01600 × 92.5
= 92.5 + 7.5
= 6.940934 u

10 11
(b) Mass of boron isotope 5 B, m 1 = 10.01294 u Mass of boron isotope 5 B, m 2 = 11.00931 u
Abundance of 10
5 B, η 1 = x%
Abundance of , (100 — x)%
Atomic mass of boron, m = 10.811 u
The atomic mass of boron atom is given as:
m 1η 1 + m 2η 2
m= η1 + η2
10.01294 × x + 11.00931 × ( 100 − x )
10.811 = x + 100 − x
1081.11 = 10.01294x + 1100.931 − 11.00931x
19.821
∴ x = 0.99637 = 19.89%
And 100 - x = 80.11%
10 11
Hence, the abundance of 5 B is 19.89% and that of 5 B is 80.11%.

Page : 462 , Block Name : Exercise

20 22
Q13.2 The three stable isotopes of neon: 10Ne and 10Ne 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.

20
Answer. Atomic mass of 10Ne, m 1 = 19.99 u
20
Abundance of 10Ne, η 1 = 90.51%
21
Atomic mass of 10Ne , m = 20.99 U
2

Abundance of 21
10Ne, η 2 = 0.27% Ne
22
Atomic mass of 10 Ne, m 3 = 21.99u
22
Abundance of 10 Ne, η 3 = 9.22%
The average atomic mass of neon is given as:
m 1η 1 + m 2η 2 + m 3η 3
m= η1 + η2 + η3
19.99 × 90.51 + 20.99 × 0.27 + 21.99 × 9.22
= 90.51 + 0.27 + 9.22
= 20.1771 u

Page : 462 , Block Name : Exercise

14 14
Q13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( 7 N) given m ( 7 N) = 14.00307 u

( )
Answer. Atomic mass of nitrogen 7N 14 , m 14.00307 u
A nucleus of nitrogen 7N 14 contains 7 protons and 7 neutrons.
Hence, the mass defect of this nucleus, Δm = 7m H + 7m n − m
Where,
Mass of a proton, m H = 1.007825 u
Mass of a neutron, m n = 1.008665 u
∴ Δm = 7 x 1.007825 + 7 x 1.008665 - 14.00307
= 7.054775 + 7.06055 — 14.00307
= 0.11236 u
But 1 u = 931.5MeV / c 2
∴ Δm = 0.11236 x 931.5 MeV/c2
Hence, the binding energy of the nucleus is given as:
E b = Δmc 2
Where,
c = Speed of light

∴ E b = 0.11236 × 931.5
( )
MeV
c2
× c2

= 104.66334 MeV
Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

Page : 462 , Block Name : Exercise

56 209
Q13.4 Obtain the binding energy of the nuclei Fe and 83 Bi in units of MeV from the following
26
data:

( )
m 56
26Fe = 55.934939 u

m ( Bi )= 208.980388 u
209
83

Answer. Atomic mass of 36Fc, = 55.934939 u
26
36Fc nucleus has 26 protons and (56 - 26) = 30 neutrons
26
Hence, the mass defect of the nucleus, Δm = 26 × m H + 30 × m n − m 1
Where,
Mass of a proton, m H = 1.007825 u
Mass of a neutron, m n = 1.008665 u
∴ Δm 26 x 1.007825 + 30 x 1.008665 — 55.934939
= 26.20345 + 30.25995 = 55.934939
= 0.528461 u
But 1 u = 931.5 MeV / c 2
∴ Δm = 0.528461 x 931.5 MeV/c2
The binding energy of this nucleus is given as:
E b1 = Δmc 2
Where,
c = Speed of light

∴ E b1 = 0.528461 × 931.5
( )MeV
c2
× c2

492.26
Average binding energy per nucleon = 56
= 8.79MeV
Atomic mass of 20Bi, m 2 = 208.980388 u
83
209Bi nucleus has 83 protons and (209 — 83) 126 neutrons.
83
Hence, the mass defect of this nucleus is given as:
Δm ′ = 83 × m H + 126 × m n − m 2
Where,

Mass of a proton, m H = 1.007825 u
Mass of a neutron, m n = 1.008665 u
∴ Δm ′ = 83 × 1.007825 + 126 × 1.008665 − 208.980388
83.649475 + 127.091790 - 208.980388
= 1.760877 u
But 1 u 931.5 MeV / c 2
∴ Δm ′ = 1.760877 x 931.5 MeV / c 2
Hence, the binding energy of this nucleus is given as:
E b2 = Δm ′c 2

= 1.760877 x 931.5
( )
MeV
c2
× c2

1640.26 MeV
1640.26
Average binding energy per nucleon = 209
= 7.848 Mev

Page : 462 , Block Name : Exercise

Q13.5 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to
separate all the neutrons and protons from each other. For simplicity assume that the coin is

( )
56
entirely made of m 26Fe atoms (of mass 62.92960 u).

Answer. Mass of a copper coin, m ′ 3 g
Atomic mass of 29Cu 63 atom, m - 62.92960 u
NA × m′
The total number of 29 Cu 63 atoms in the coin, N = Mass number
Where,
N A = Avogadro's number = 6.023 × 10 23 atoms/g
Mass number = 63 g
6.023 × 10 23 × 3
∴N 63 = 2.868 × 10 22 atoms
63
29Cu nucleus has 29 protons and (63 — 29) 34 neutrons
Therefore, mass defect of this nucleus, Δm r = 29 × m H + 34 × m n − m
Where,
Mass of a proton, m H = 1.007825 u
Mass of a neutron, m H = 1.008665 u
∴ Δm ′ = 29 × 1.007825 + 34 × 1.008665 − 62.9296
= 0.591935 u
Mass defect or all the atoms present In the coin, Δm = 0.591935 × 2.868 × 10 22
= 1.69766958 u x 10 22u
But 1 u = 931.5 MeV / c 2
∴ Δm = 1.69766958 x 10 22 × 931.5MeV / c 2

Hence, the binding energy of the nuclei of the coin is given as:
E b = Δmc 2

1.69766950 10 22 × 931.5
( )
MeV
c 2 × c2

1,581 x 10 25 MeV
But 1 MeV = 1.6 x 10 − 13J
E b = 1.581 × 10 25 × 1.6 × 10 − 13
2.5296 × 10 12J
This much energy required to separate all the neutrons and protons from the given coin.

Page : 462 , Block Name : Exercise

Q13.6 Write nuclear reaction equations for
226
(i) α-decay of 88 Ra
242
(ii) α-decay of 94 Pu
32
(iii) β − -decay of 15 P
210
(iv) β + -decay of 83 Bi
6
(v) β + -decay of 11 P
43
(vi) β + -decay of 97 Tc
120
(vii) Electron capture of 54 Xe

Answer.

( ) ( )
α is a nucleus of helium 2Hc 4 and β is an electron e − for β − and e + for β + . In every α − decay,
there Is a loss of 2 protons and 4 neutrons. In every β + − decay there is a loss of 1 proton and
neutrino is emitted from the nucleus. In every β − − decay, there is a gain of 1 proton and an
antineutrino Is emitted from the nucleus.

For the given cases, the various nuclear reactions can be written as:

(i) 88Ra 226 86Ra 222 + 2He 4

(ii) 94Pu 242 92Ra 238 + 2He 4

¯
(iii) 15P 13 16S 32 + e − + v

¯
210 210 +
(iv) 83B 84PO +e +v

(v) 6C 11 5B 11 + e + + v

(vi) 43Tc 97 42MO 42 + e + + v

120
(vii) 54 Xe + e + 53I 120 + v

Page : 462 , Block Name : Exercise

Q13.7 A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to
a) 3.125%, b) 1% of its original value?

Answer. Half-life of the radioactive isotope = T years
Original amount of the radioactive isotope = No

(a) After decay, the amount of the radioactive isotope N
It is given that only 3.125% of No remains after decay. Hence, we can write:
N 3.125 1
N0
= 3.125% = 100 = 32
N
But N = e − it
0
Where,
λ = Decay constant
t = Time
1
∴ − λt = 32
− λt = ln1 − ln32
− λt = 0 − 3.4657
3.4657
t= λ
0.693
Since λ = T
3.466
0.693
∴t= T
≈ 5T
Hence, the isotope will take about ST years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N
N 1
It is given that only 1% of No remains after decay. Hence. we can write: N = 1% = 100
0
N
But N = e − / lambdat
0
1
∴ e − λ′ = 100
− λt = 0 − 4.6052
4.6052
t= λ
Since, λ = 0.693 / T

4.6052
∴t= 0.693 = 6.645T years
T
Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

Page : 462 , Block Name : Exercise

Q13.8 The normal activity of living carbon-containing matter is found to be about 15 decays per
14
minute for every gram of carbon. This activity arises from the small proportion of radioactive 6 C
14
present with the stable carbon isotope 6 c. When the organism is dead, its interaction with the
atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to
14
drop. From the known half-life (5730 years) of 6 c, and the measured activity, the age of the
14
specimen can be approximately estimated. This is the principle of 6 c dating used in archaeology.
Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of
carbon. Estimate the approximate age of the Indus-Valley civilisation.

Answer. Decay rate Of living carbon-containing matter, R = 15 decay/min
Let N be the number of radioactive atoms present in a normal carbon- containing matter.
14
Half life of 6 C, T 1 / 2 = 5730 years
The decay rate of the specimen Obtained from the Mohenjodaro Site:
R' = 9 decays/min
Let N' be the number of radioactive atoms present in the specimen during the Mohenjodaro
period.
Therefore, we can relate the decay constant, λ and time, t as:
N R
′ = = e − 2t
N R′
9 3
e − it = 15 = 5
3
− λt = log e 5 = − 0.5108
0.5108
∴t= λ
0.693 0.693
λ= T = 5730
1/2
0.5108
∴t= 0.693 = 4223.5
5730
Hence, the approximate age Of the Indus-Valley civilisation is 4223.5 years.

Page : 462 , Block Name : Exercise

Q13.9 Obtain the amount of 60Co necessary to provide a radioactive source of 8.0 mCi strength.
27
The half-life of 2 Co is 5.3 years.
22

Answer. The strength of the radioactive source is given as:
dN
dt
= 8.0 mCi
= 8 × 10 − 3 × 3.7 × 10 10
= 29.6 × 10 7 decay/s
Where,
N — Required number of atoms
Half-life of 60Co = T 1 / 2 5.3 years
27
= 5.3 x 365 x 24 x 60 x 60
= 1.67 x 10 8 s
For decay constant λ, we have the rate of decay as:
dN
dt
= λN
0.693 0.693
Where, λ = T = s −1
1/2 1.67 × 10 8
1 dN
∴ N = λ dt
29.6 × 10 7
= 0.693
= 7.133 × 10 16 atoms
For 60Co
27
Mass Of 6.023 × 10 23 (Avogadro's number) atoms 60 g
60 × 7.133 × 10 16
16
7.133 × 10 atoms = = 7.106 × 10 − 6g
6.023 × 10 23
Hence, the amount of 27Co 60 necessary for the purpose is 7.106 x 10 − 6 g.

Page : 463 , Block Name : Exercise

Q13.10 The half-life of 90Sr is 28 years. What is the disintegration rate of 15 mg of this isotope?
38

Answer. Half life of 90Sr = 28 years
38
= 28 x 365 x 24 x 60 x 60
= 8.83 x 10 8 s
Mass of the isotope, m = 15 mg
90 g of 90Sr atom contains 6.023 x 10 23)(Avogadro ′ snumber)atoms.
38

Page : 463 , Block Name : Exercise

Q13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope 197
79 Au and the silver
107
isotope 47 Ag.

Answer. Nuclear radius of the gold isotope 197Au = R Au
79

Nuclear radius of the silver isotope 107Ag = R Ag
47
Mass number of gold, A Au = 197
Mass number of silver A Ag = 107
The ratio of the radii of the two nuclei is related with their mass numbers as:

( )
1
R Au R Au 3

R Ag
= R Ag

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Page : 463 , Block Name : Exercise

Q13.12 Find the q-value and the kinetic energy of the emitted α-particle in the α-decay of
(a) 226Ra and (b) 226Rn.
88 86

( )
Given m
226
88
Ra = 226.02540 u,

m
( ) 222
86
Rn = 222.01750 u,

m ( ) 222
86
Rn = 220.01137 u,

m
( ) 216
84
Pon = 216.00189 u.

Answer. (a) Alpha particle decay of 226Ra emits a helium nucleus. As a result, its mass number
88
reduces to (226 - 4) 222 and its atomic number reduces to (88 - 2) 86. This is shown in the
following nuclear reaction.
86
26Ra ⟶ 222 Ra 42 He
ss
q-value of
emitted α-particle = (Sum of initial mass — Sum Of nal mass) c²
Where, c = Speed of light
It is given that:

( ) 266
m 88 Ra = 226.02540u

( ) 222
m 86 Rn = 222.01750u

( )
m 42 He = 4.002603u
q-value = [226.02540 - (222.01750 + 4.002603)] u c²
0.005297 u c²

But 1 u = 931.5 Mev/c²
∴ Q = 0.005297 × 931.5 ≈ 4.94

Kinetic energy of the α-particle = =
222
( Mass number after decay
Mass number before decay ) ×Q

= 226 × 4.94 = 4.85MeV

(b) Alpha particle decay of

220
( 220
86
Rn) is shown by the following nuclear reaction.

86
Rn ⟶ 216Po + 42 He
It is given that:

Mass of
( )
200
86
Rn = 220.01137 u

Mass of
( )
216
84
Rn = 216.00189 u

Therefore, Q-value = [220.01137 - (216.00189 + 4.00260)] x 931.5
= 641 Mev

Kinetic energy of the α-particle = ( )
220 − 4
220
× 6.41

6.29 Mev

Page : 463 , Block Name : Exercise

Q13.13 The radionuclide 11C decays according to
11 11 +
6 C → 5 B + e + v : T 1 / 2 = 20.3min
The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:

m ( )
11
6 ( )
11
C = 11.011434 u and m m 6 B = 11.009305 u.

Calculate q and compare it with the maximum energy of the positron emitted.

Answer. The given nuclear reaction is:
11
Half life of 6 C nuclei, T 1 / 2 = 20.3 min
11
Atomic mass of m 6 C = - 11.011434 u
11
Atomic mass of m 6 B = 11.009305 u
Maximum energy possessed by the emitted positron = 0.960 MeV
11
The change in the Q-value (δQ) of the nuclear masses of the 6 C nucleus is given as:

[( )[ ]]
11
ΔQ = m ′ 6C 11 − m ′( s B) + m e c 2 ………….(i)

Where,
mₑ = Mass of an electron or positron = 0.000548 u
c = Speed of light
m' Respective nuclear masses
If atomic masses are used instead of nuclear masses, then we have to add 6 mₑ in the case of 11C
and 5 mₑ in the case of 11B.
Hence, equation (1) reduces to:

[( ) ( )
ΔQ = m 0C 11 − m 11
s B − 2m e c
2
]
Here, m 11 11
6 C and m 5 B are tthe atomic masses.
∴ ΔQ = [11.011434 − 11.009305 − 2 × 0.000548]c 2
= (0.001033 c²) u
But 1 u = 931.5 Mev/c²
Therefore, δQ 0.001033 x 931.5 0.962 Mev
The value of Q is almost comparable to the maximum energy of the emitted positron.

Page : 463 , Block Name : Exercise

23
Q13.14 The nucleus 10 Ne decays by β − emission. Write down the β-decay equation and determine
the maximum kinetic energy of the electrons emitted.
Given that:

( )
23
10 Ne = 22.994466 u

( )
23
11 Na = 22.089770 u.

Answer. In β − emission, the number of protons increases by 1, and one electron and an
antineutrino are emitted from the parent nucleus.
23
β − emission of the nucleus 10 Ne is given as:
23 ¯
23 −
10 Ne → 11 Na + e + v + Q
It is given that:

Atomic mass of m ( )
23
10
Ne = 22.994466 u

Atomic mass of m ( )
23
11
Ne = 22.989770 u

Mass of an electron, mₑ = 0.000548 u

q-value of the given reaction is given as:

[ ( ) [ ( ) ]]
23
q = m 10 Ne − m
23
11
N + me
23
c2
23
There are 10 electrons in 10N e and 11 electrons in 11 N C.
Hence, the mass of the electron is cancelled in the q-value equation.
∴ Q = [22.994466 − 22.989770]c 2

( )
= 0.004696c 2 u
But 1 u = 931.5 MeV/c².
∴ q = 0.004696 × 931.5 = 4.374MeV
The daughter nucleus is too heavy as compared to e⁻ and ⊽ . Hence, it carries negligible energy.
The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the
emitted electrons is almost equal to the q-value, i.e., 4.374 MeV.

Page : 463 , Block Name : Exercise

Q13.15 The q value of a nuclear reaction A + b →C + d is de ned by

[
q = mA + mb − mC − md c2 ]
where the masses refer to the respective nuclei. Determine from the given data the q-value of the
following reactions and state whether the reactions are exothermic or endothermic.
(i) 1H + 31 H → 2l H + 21 H
(ii) 12C + 12 20 4
6 C → 10Ne + 2 He
Atomic masses are given to be

( )
2
m 1 H = 2.014102u

m ( H ) = 3.016049u
3
1

m ( C ) = 12.000000u
12
6

m ( Ne ) = 19.992439u
20
10

Answer. (i) The given nuclear reaction is:
3 2 2
H1 +1H →1H +1H
It is given that:
1
Atomic mass m(1H) = 1.007825 u
1
Atomic mass m(3H) = 3.016049 u
1
Atomic mass m(2H) = 2.014102 u
According to the question, the Q-value of the reaction can be written as:

[ ( )]
1 2
Q = m(1H) + m 1 H − 2m ( )
3
1
c2

= [1.007825 + 3.016049 − 2 × 2.014102]c 2

Q= ( − 0.00433c )u 2

But 1u = 931.5MeV / c 2
∴ q = − 0.00433 × 931.5 = − 4.0334MeV
The negative q-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:
12
12 20 4
C + 6 C → 10Ne + 2 He
It is given that:

( )
Atomic mass of m 6( 12 ) C = 12.0 u

Atomic mass of m ( Ne ) = 19.992439 u
(2)
10

Atomic mass of m ( He ) = 4.002603 u
(4)
2

The q-value of this reaction is given as:

q = 2m
[ () ( ) ) ( )
n
6
−m
20
10
4
Ne − m 2 He ]c 2

[2 x 12.0 - 19.992439 - 4.002603]c²
= (0.004958 c²) u
= 0.004958 931.5 4.618377 MeV
The positive q-value Of the reaction shows that the reaction is exothermic.

Page : 463 , Block Name : Exercise

56 28
Q13.16 Suppose, we think of ssion of a Fe nucleus into two equal fragments, 13Al. Is the ssion
26

( )
56
energetically possible? Argue by working out q of the process. Given m 26 Fe = 55.93494 u and

m 28 ( )
13Al = 27.98191u

Answer. The ssion of 26Fe can be given as:
26
56 28
13Fe ⟶ 2 13Al
It is given that:
Atomic mass of m 56Fe = 55.93494 u
26
Atomic mass of m 28Al = 27.98191 u
13

The q-value of this nuclear reaction is given as:
Q = [ m 56Fe -2m m 28Al]c²
26 13
=[55.93494 - 2 x 27.98191]c²
= ( -0.02888 c² ) u
But 1u = 1u = 931.5MeV / c 2
∴ q = − 0.02888 × 931.5 = − 26.902MeV
The Q-value of the ssion is negative. Therefore, the ssion is not possible energetically. For an
energetically-possible ssion reaction, the q-value must be positive.

Page : 463 , Block Name : Exercise

( )
92
Q13.17 The ssion properties of 239Pu are very similar to those of 235 U = 27.98191u. The average
94
energy released per ssion is 180 MeV. How much energy, in MeV, is released if all the atoms in 1
kg of pure 239Pu undergo ssion?
94

Answer. Average energy released per ssion of 239Pu, E av = 180 MeV
94
Amount of pure 239Pu, E av, m = 1 kg = 1000 g
94
N A Avogadro number = 6.023 x 10 23
Mass number of 23994Pu = 239 g
1 mole of contains 239Pu, E av contains N A atoms.
94

( NA
Therefore, m g of 23994Pu contains Mass number × m

6.023 × 10 23
)
= 239 × 1000 = 2.52 × 10 24
239
Therefore, Total energy released during the ssion of 1 kg of 94 Pu is calculated as:
E = E ∞ × 2.52 × 10 24
= 180 × 2.52 × 10 24 = 4.536 × 10 26MeV
239
Hence, = 4.536 × 10 26 MeV is released if all the atoms in 1 kg of pure 94 Pu undergo.

Page : 463 , Block Name : Exercise

235
Q13.18 A 1000 MW ssion reactor consumes half of its fuel in 5.00 y. How much 92 U did it contain
initially? Assume that the reactor operates 80% of the time, that all the energy generated arises
from the ssion of 235
92 U and that this nuclide is consumed only by the ssion process.

Answer. Half life of the fuel of the ssion reactor, t 1 = 5 years
2
= 5 x 365 x 24 x 60 x 60 s

235
We know that in the ssion of 1 g of 92 U nucleus, the energy released is equal to 200 MeV.
235
1 mole, i.e., 235 g of 92 U contains 6.023 x 10²³ atoms.
235
1 g 92 U contains 6.023 x 10²³ atoms.
235
The total energy generated per gram of 92 U is calculated as:
6.023 × 10 23
E= × 200MeV / g
235
200 × 6.023 × 10 23 × 1.6 × 10 − 19 × 10 6
= = 8.20 × 10 10Jg
235
The reactor operates only 80% of the time.
235
Hence, the amount of 92 U consumed in 5 years by the 1000 MW ssion reactor is calculated as:
5 × 80 × 60 × 60 × 365 × 24 × 1000 × 10 6
= g
100 × 8.20 × 10 10
= 1538 kg
235
Therefore, initial amount of 92 U = 2 x 1538 = 3076 kg.

Page : 464 , Block Name : Exercise

Q13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?
Take the fusion reaction as
2 2 3
1 H + 1 H → 2 He + n + 3.27MeV

Answer. The given fusion reaction is:
2 2 3
1 H + 1 H ⟶ 2 He + n + 3.27MeV
Amount of deuterium, m = 2 kg
1 mole, i.e., 2 g of deuterium contains 6.023 x 10 23 atoms.
6.023 × 10 23
.•. 2.0 kg of deuterium contains = 2
× 2000 = 6.023 × 10 26 atoms
It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV
energy released. Therefore, total energy per nucleus released in the fusion reaction:
3.27
E= × 6.023 × 10 26MeV
2
3.27
= × 6.023 × 10 26 × 1.6 × 10 − 19 × 10 6
2
= 1.576 × 10 14J
Power of the electric lamp, P = 100 W = 100 J/s
Hence, the energy consumed by the lamp per second = 100 J
The total time for which the electric lamp will glow is calculated as:

1.576 × 10 14
100
s
1.576 × 10 14
100 × 60 × 60 × 24 × 365
≈ 4.9 × 10 4 years

Page : 464 , Block Name : Exercise

Q13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint:
The height of the potential barrier is given by the Coulomb repulsion between the two deuterons
when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Answer. When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1 st deuteron + Radius of 2 nd deuteron
Radius of a deuteron nucleus = 2 fm = 2 x 10⁻¹⁵ m
∴ d = 2 × 10 − 15 + 2 × 10 − 15 = 4 × 10 − 15m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 x 10 − 19 C
Potential energy of the two-deuteron system:
e2
V = 4πϵ d
0
Where,
∈ 0 = permittivity of free space
1
= 9 × 10 9Nm 2C − 2
4πϵ 0

(
9 × 10 9 × 1.6 × 10 − 19 ) 2

∴V= eV
4 × 10 − 15

(
9 × 10 9 × 1.6 × 10 − 19 ) 2

= eV
4 × 10 − 15 × ( 1.6 × 10 − 19 )
( )
= 3 × 10 9 × 1.6 × 10 − 19
2

= 3 × 10 × (1.6 × 10
9
) − 19 2

= 360keV
Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

Page : 464 , Block Name : Exercise

Q13.21 From the relation R = R 0A 1 / 3, where R 0 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).

Answer. We have the expression for nuclear radius as:
Where, Ro Constant.

A = Mass number of the nucleus
Mass of the nucleus
Nuclear matter density, ρ = Volume of the nucleus
Let m be the average mass of the nucleus.
Hence. mass of the nucleus mA
mA 3mA 3mA 3m
∴ρ= 4 = = 3 = 3

( )
πR 3 1 3 4πR 0 A 4πR 0
3
4π R 0A 3

Hence, the nuclear matter density is independent of A. It is nearly constant.

Page : 464 , Block Name : Exercise

Q13.22 For the β + (positron) emission from a nucleus, there is another competing process known
as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a
neutrino is emitted).
e t + AzX → Az− 1 Y + v
Show that if β + emission is energetically allowed, electron capture is necessarily allowed but not
vice–versa.

Answer. Let the amount of energy released during the electron capture process be q₁. The nuclear
reaction can be written as:
d
+ A
e + zX → z − 1 Y + v + q 1 ……(i)
Let the amount of energy released during the positron capture process be q₂.
The nuclear reaction can be written as:
1
4
zX → z − 1 Y + e + + v + Q 2 …….(ii)
A
m N( zX) = Nuclear mass of z X

( ) A A
m N z − 1Y Y
= Nuclear mass of z − 1

m (z Y )
A A
N −1 Y
= Nuclear mass of z − 1

m (z Y )
A A
−1 z − − 1Y
= Atomic mass of
Mₑ = Mass of an electron
C = Speed of light
q-value of the electron capture reaction is given as:

[ ]
A
q 1 = m N(ZX) + m e − m N Az− 1 Y ( )c 2

[ ]
A
= m(ZX) − Zm e + m e − m Az− 1 Y + (Z − 1)m e c 2 ( )

= m
[ ( ) ( )]4
z
− m z − 1Y
A
c 2……………. . (3)

= m[( ) (A
ZX )
− m z− 1 AY − 2m e c 2 ]
q-value of the positron capture reaction is given as:

[ ]
A
( A
q 2 = m N( zX) − m N z − 1 Y − m e c 2 )
[( ) ( ) ]
= m AzX − Zm e − m Az− 1 Y + (Z − 1)m e − m e c 2

[( ) z ]
= m X − m ( Y ) − 2m c ………..(4)
A A
z−1 c
2

It can be inferred that if q 2 > O, then q 2 > O; Also, if q 2, it does not necessarily mean that q 2 > O.

Page : 464 , Block Name : Exercise

Q13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The
average value is based on their relative natural abundance on earth. The three isotopes and their
24
masses are 12Mg (23.98504u),
25
12Mg (24.98584u) and (25.98259u). The natural
24
abundance of 12Mg is 78.99% by mass. Calculate the abundances of other two isotopes.

Answer. Average atomic mass of magnesium. m = 24.312 u
24 25
Mass of magnesium 12Mg isotope, m 1 = 23.98504 u, mass of magnesium 12Mg isotope, m 2 =
24.98584 u
26 24
Mass of magnesium 12Mg isotope, m 3 = 25.98259 u, Abundance of 12Mg, η 1 = 78.99%
Let the abundance of 25 25
12Mg, η 1 = x% therefore, the abundance of 12Mg,
η 3 = 100 − x − 78.99% = (21.01 − x)9 / 6
m 1η 1 + m 2η 2 + m 1η 1
m= η1 + η2 + η3
23.98504 × 78.99 + 24.98584 × x + 25.98259 × ( 21.01 − x )
24.312 = 100

2431.2 = 1894.5783096 + 24.98584x + 545.8942159 − 25.98259x
0.99675x = 9.2725255
∴ x ≈ 9.3%
And 21.01 − x = 11.71%
Hence, the abundance of 26 26
12Mg is 9.3% and that of 12Mg is 11.71%.

Page : 464 , Block Name : Additional Exercise

Q13.24 The neutron separation energy is de ned as the energy required to remove a neutron from
41 27
the nucleus. Obtain the neutron separation energies of the nuclei Ca and from the following
20 13
data:

[ ] 40
m 20Ca = 39.962591u

m ( Ca ) = 40.962278u
40
20

m ( Al ) = 25.986895u
2ϵ
13

m { Al } = 26.981541u
27
13

41
Answer. For 20Ca: Separation energy = 8.363007 MeV
For 13 27Al: Separation energy = 13.059 MeV
1 41
A neutron 0 n is removed from a 20Ca nucleus. The corresponding nuclear reaction can be written
as:
41 40 1
20 Ca ⟶ 20 Ca + 0
n
It is given that:

Mass m
( )
40
20
Ca) = 39.962591 u )

Mass m
( )
41
20
Ca) = 40.962278 u

( )
Mass m 0n 1 = 1.008665 u
The mass defect Of this reaction is given as:

Δm = ( Ca ) + ( n ) − m ( Ca )
40 1 41
m 20 o 20

= 39.962591 + 1.008665 - 40.962278= 0.008978 u
But 1 u = 931.5 MeV / c 2
∴ Δm = 0.008978 x 931.5 MeV / c 2
Hence, the energy required for neutron removal is calculated as:
E = Δmc 2
= 0.008978 x 931S -8.363007 MeV
For 27
13Al, the neutron removal reaction can be written as:
It is given that:
Mass = 26.981541 u
Mass = 25.986895 u
The mass defect of this reaction is given as:
25.986895 + 1.008665 - 26.981541
= 0.01401911 u
= 0.014019 x 931.5 MeV / c 2

Hence, the energy required for neutron removal is calculated as:
vE = Δmc 2
= 0.014019 x 9315=13.059 Mev

Page : 464 , Block Name : Additional Exercise

32
(
Q13.25 A source contains two phosphorous radio nuclides 15 P T 1 / 2 = 14.3d and )
33 33
( )
15 P T 1 / 2 = 25.3d . Initially, 10% of the decays come from 15 P. How long one must wait until 90%
do so?

32
Answer. Half life of 15 P, T 1 / 2 = 14.3 days
32 ′
Half life of 15 P, T 1 / 2 = 25.3 days
32
15
P nucleus decay is 10% of the total amount of decay.
32 32
The source has initially 10% of 15 P nucleus and 90% Of 15 P nucleus.
32 32
Suppose after t days, the source has 10% of 15 P nucleus and 90% of 15 P nucleus.
Initially:
32
Number of 15 P nucleus = N
32
Number of 15 P nucleus = 9 N
Finally:
32
Number of 15 P nucleus = 9 N’
32
Number of 15 P nucleus = N’
32
For 15 P nucleus, we can write the number ratio as:
1
N′
9N
= ()
1
2
Ts / 2

−t
N ′ = 9N(2) 14.3 ……………….(1)
32
For 15 P, we can write the number ratio as:
1
9N ′ ′
N
= () 1
2
Ts / 2

−t
9N ′ = N(2) 25.3 ………………………(2)
On dividing equation (1) by equation (2), we get:
1
9
=9×2 ( 25.3
1 1
− 14.3
)
1 11t

81
= 2 − ( 25.3 × 14.3 )

− 11t
log1 − log81 = 25.3 × 14.3 log2
− 11t 0 − 1.908
25.3 × 14.3
= 0.301
25.3 × 14.3 × 1.908
t= 11 × 0.301
≈ 208.5 days
Hence, it will take about 208.5 days for 90% decay of 15P 33.

Page : 464 , Block Name : Additional Exercise

Q13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than
an α-particle. Consider the following decay processes:
223 209 14
88
Ra → 82 Pb + 6 C
223 219 4
88 Ra → 86 Rn + 2 He
Calculate the q-values for these decays and determine that both are energetically allowed.

14
Answer. Take a 6 C emission nuclear reaction:
223 209 14
88 Ra 82 Pb + 6 C
We know that:
223
Mass of 88 Ra, m 1 = 223.01850 u
209
Mass of 82 Pb, m 1 = 208.98107 u
6
Mass of 14 C, m 3 = 14.00324 u
Hence, the q-value of the reaction is given as:
Q = (m 1 - m 2 - m 3 )c²
= (223.01850 - 208.98107 = 14.00324)
= (0.03419 c²) u
But 1 u = 931.5 MéV/c²
Therefore, q = 0.03419 x 931.5
=31.848 Mev
Hence, the q-value of the nuclear reaction is 31.848 MeV. Since the value is positive, the reaction
is energetically allowed.
2
Now take a 4 He emission nuclear reaction:
223 229 4
88
Ra 86 Rn + 2 He
We know that:
223
Mass of 88 Ra, m 1 = 223.01850
219
Mass of 82 Ra, m 2 = 219.00948
4
Mass of 2 He, m 3 = 4.00260
q-value of this nuclear reaction is given as:

q = (m 1 - m 2 - m 3 )c²
= (223.01850 - 219.00948 — 4.00260)c²
= (0.00642 c²) u
= 0.00642 x 931.5
= 5.98 mev
Hence, the q value of the second nuclear reaction is 5.98 MeV. Since the value is positive, the
reaction is energetically allowed.

Page : 464 , Block Name : Additional Exercise

Q13.27 Consider the ssion of 238U by fast neutrons. In one ssion event, no neutrons are emitted
92
140 99
and the nal end products, after the beta decay of the primary fragments, are Ce and 44Ru.
58
Calculate q for this ssion process. The relevant atomic and particle masses are
m 238U =238.05079 u
92
m 140U =139.90543 u
58
m 99U = 98.90594 u
44

Answer. In the ssion of 238U, 10 β − particles decay from the parent nucleus. The nuclear reaction
92
can be written as:
1
238U + 0 n 140Ce + 44Ru + 10 0− 1 e
92 58 99
It is given that:
Mass of a nucleus 238U, m 1 = 238.05079 u
92
Mass of a nucleus 140Ce, m 2 = 139.90543 u
58
1
Mass of a nucleus 99Ru, m 3, = 98.90594 u Mass of a neutron 0 n m 4 = 1.008665 u
44
q-value of the above equation,

( ) ()
238 1
( 140
) ( )
99
q = [m ′ 92 U + m 0 n - m ′ 58 Ce - m ′ 44Ru - 10 m − e]c²

Where,
m' = Represents the corresponding atomic masses of the nuclei

( )
238
m ′ 92 U = m 1 − 92m e
140
m ′( 58 Ce) = m 2 − 58m e
44
′ ′
m (99Ru) = m 3 − 44m e

( )
m 10 n
= m4

[ ]
q = m 1 − 92m e + m 4 − m 2 + 58m e − m 3 + 44m e − 10m e c 2

[
= m1 + m4 − m2 − m3 c2 ]
= [238.0507 + 1.008665 − 139.90543 − 98.90594]c 2

[
= 0.247995c 2 u ]
But 1 u =931.5 Mev/c²
∴ q = 0.247995 × 931.5 = 231.007MeV
Hence, the q-value of the ssion process is 231.007 MeV.

Page : 465 , Block Name : Additional Exercise

Q13.28 Consider the D–T reaction (deuterium–tritium fusion)
2 3 4
1 H + 1 H → 2 He + n
(a) Calculate the energy released in MeV in this reaction from the data:

( )
2
m 1 H = 2.014102u

m ( H ) = 3.016049u
3
1

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the
kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what
temperature must the gas be heated to initiate the reaction?
(Hint: Kinetic energy required for one fusion event = average thermal kinetic energy available with
the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

3 4
Answer. (a) Take the D-T nuclear reaction: 1 2H + 1 H ⟶ 2 He + n
It is given that:
Mass of 21 H, m 1 = 2.014102 u
3
Mass of 1 H, m 2 = 3.016049 u
4
Mass of 2 He, m 3 = 4.002603 u
1
Mass of 0 n, m 4 = 1.008665 u
q-value of the given D-T reaction is:

[
q = m1 + m2 − m3 − m4 c2 ]
= [2.014102 + 3.016049 — 4.002603 - 1.008665]c²
= [0.018883 c²] u
But 1 u = 931.5 Mev/c²
Therefore, q = 0.018883 x 931.5 - 17.59 Mev

(b) Radius Of deuterium and tritium, r 2.0 fm = 2 x 10⁻¹⁵ m
Distance between the two nuclei at the moment when they touch each other, d = r + r = 10⁻¹⁵ m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e

Hence, the repulsive potential energy between the two nuclei is given as:
e2
V = 4πϵ ( d )
0
Where,
ϵ 0 Permittivity of free space
1
9 2 −2
4πϵ 0 = 9 × 10 Nm C

(
9 × 10 9 × 1.6 × 10 − 19 ) 2

∴V= = 5.76 × 10 − 14J
4 × 10 − 15
5.76 × 10 − 14
= = 3.6 × 10 5eV = 360keV
1.6 × 10 − 19
Hence, 5.76 x 10⁻¹⁴ J or 360 keV of kinetic energy (KE) is needed to overcome the Coulomb
repulsion between the two nuclei.
However, it is given that:
3
KE = 2 × 2 kT
Where,
k = Boltzmann constant = 1.38 × 10 − 23m 2kgs − 2K − 1
T = Temperature required for triggering the reaction
Therefore, T = KE/3K
5.76 × 10 − 14
= = 1.39 × 10 9K
3 × 1.38 × 10 − 23
Hence, the gas must be heated to a temperature of 1.39 x 10⁹ K to initiate the reaction.

Page : 465 , Block Name : Additional Exercise

Q13.29 Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of ү-

( )
decays in the decay scheme shown in Fig. You are given that m 198Au = 197.968233u
ml 198Hg) = 197.966760u

Answer. It can be observed from the given y-decay diagram that Y₂ decays from the 0.412 MeV
energy level to the 0 MeV energy level.
Hence, the energy corresponding to VI-decay is given as:

E 2 = 0.412 − 0 = 0.412MeV
hv 2 = 0.412 × 1.6 × 10 − 19 × 10 6J
Where,
V₂ = Frequency of radiation radiated by Y₂-decay
E2
∴ v2 =
h
0.412 × 1.6 × 10 − 19 × 10 6
= = 9.988 × 10 19Hz
6.6 × 10 − 34
It can be observed from the given Y-decay diagram that Y₃ decays from the 1.088 MeV energy level
to the O MeV energy level. Hence, the energy corresponding to vz-decay is given as: = 0.412 MeV
energy level.
Hence, the energy corresponding to Y₃-decay is given as:
E 3 = 1.088 − 0.412 = 0.676MeV
hv 3 = 0.676 × 10 − 19 × 10 6J
Where,
v₃ = Frequency of radiation radiated by Y₃-decay
E3
∴ v3 = h
0.676 × 1.6 × 10 − 19 × 10 6
= = 1.639 × 10 20Hz
6.6 × 10 − 19

( ) 198
Mass of m 78 Au = 197.968233 u

Mass of m ( Hg ) = 197.966760 u
198
80

1 u = 931.5 MeV / c 2
Energy of the highest level is given as:
= 197.968233 - 197.966760 -0.001473 u
= 0.001473 x 931.5=1.3720995 MeV
β 1 decays from the 1.3720995 MeV level to the 1.088 MeV level
Therefore, Maximum kinetic energy of the β 1 particle = 1.3720995 - 1.088
= 0.2840995 MEV
β 2 decays from the 1.3720995 MeV level to the 0.412 MeV level
Therefore, Maximum kinetic energy of the β 2 particle — 1.3720995 — 0.412 = 0.9600995 Mev

Page : 465 , Block Name : Additional Exercise

Q13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within

( )
Sun and b) the ssion of 1.0 kg of 235U in a ssion reactor.

Answer. (a) Amount Of hydrogen, m = 1 kg = 1000 g

(
1 mole, i.e., 1 g of hydrogen 1 1H ) contains 6.023 x 10²³ atoms.

(
Therefore, 1000 q 1 1H ) contains 6.023 x 10²³ x 1000 atoms.

( (
Within the sun, four 1 1H ) nuclei combine and form one 4 2H ) nucleus. In this process 26 MeV of
energy is released.
6.023 × 10 23 × 26 × 10 3
E1 =
Hence, the energy released from the fusion of 1 kg 1 1H ) is: ( 4
= 39.1495 × 10 26MeV

(b) Amount of 235U = 1 kg = 1000 g
92
1 mole, i.e., 235 g of 235U contains 6.023 x 10²³ atoms.
92
6.023 × 10 23 × 1000
Therefore, 1000 g of 235U contains 235
atoms.
92
235
It is known that the amount of energy released in the ssion of one atom of 92 U is 200 MeV.
Hence, energy released from the ssion of 1 kg of 235U is:
92
6 × 10 25 × 1000 × 200
E2 =
235
= 5.106 × 10 26MeV
E1 39.1495 × 10 26
= = 7.67 ≈ 8
E1 5.106 × 10 26
Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times the energy
released in the ssion of 1 kg of uranium.

Page : 466 , Block Name : Additional Exercise

Q13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten
percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an
average, the ef ciency of utilization (i.e. conversion to electric energy) of thermal energy
produced in a reactor was 25%. How much amount of ssionable uranium would our country need
per year by 2020? Take the heat energy per ssion of 235U to be about 200 MeV.

Answer. Amount of electric power to be generated, P = 2 x 10s MW 10% of this amount has to be
obtained from nuclear power plants.
10
Therefore, Amount of nuclear power, = P 1 = 100 × 2 × 10 5
= 2 × 10 4 × 10 6J / s
= 2 × 10 10 × 60 × 60 × 24 × 365J / y
Heat energy released per ssion of a 235U nucleus, E = 200 MeV
Ef ciency of a reactor = 25%
Hence, the amount of energy converted into the electrical energy per ssion is calculated as:
25
100
× 200 = 50MeV

= 50 × 1.6 × 10 − 19 × 10 6 = 8 × 10 − 12J
Number of atoms required for ssion per year:
2 × 10 10 × 60 × 60 × 24 × 365
= 78840 × 10 24
8 × 10 − 12
1 mole, i.e., 235 g of U 235 contains 6.023 × 10 23 atoms.
Mass of 6.023 × 10 23 atoms of U 235 = 235 g = 235 x 10⁻³ kg
Mass of 78840 x 10²⁴ atoms of U 235
235 × 10 − 3
= × 78840 × 10 24
6.023 × 10 23
= 3.076 x 10⁴ kg
Hence, the mass of uranium needed per year is 3.076 x 10⁴ kg.

Page : 466 , Block Name : Additional Exercise