NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

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The NCERT Solutions for Class 12 Physics Chapter 13 Nuclei is provided here to help students in their CBSE Class 12 exam preparations. These NCERT Solutions for Class 12 Physics are prepared by the subject experts at BYJU’S as per the latest CBSE Syllabus for 2023-24 and its guidelines. Students can prepare well and score good grades in the Class 12 board examinations by referring to these solutions.

The NCERT Solutions for Class 12 Physics of this chapter provided here has answers to the textbook questions along with Class 12 important questions, exemplary problems, worksheets and exercises, which will help students gain complete knowledge on the topics Nuclei. Students can avail free PDF of the NCERT Solutions for Class 12 Physics Chapter 13 by downloading it from the link given below.

NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

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NCERT Solutions Class 12 Physics Chapter 13 Nuclei Important Questions

Q 13.1 (a) Lithium has two stable isotopes

$\begin{array}{l}_{3}^{6}{Li}\end{array} $

and
$\begin{array}{l}_{3}^{7}{Li}\end{array} $
, which have abundances of 7.5% and 92.5%, and masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Boron has two stable isotopes,

$\begin{array}{l}_{5}^{10}{B}\end{array} $

and
$\begin{array}{l}_{5}^{11}{B}\end{array} $
. 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
$\begin{array}{l}_{5}^{10}{B}\end{array} $
and
$\begin{array}{l}_{5}^{11}{B}\end{array} $
.

A1 :

(a) Mass of

$\begin{array}{l}_{3}^{6}{Li}\end{array} $

lithium isotope , m₁ = 6.01512 u

Mass of

$\begin{array}{l}_{3}^{7}{Li}\end{array} $

lithium isotope , m₂ = 7.01600 u

Abundance of

$\begin{array}{l}_{3}^{6}{Li}\end{array} $

, n₁= 7.5%

Abundance of

$\begin{array}{l}_{3}^{7}{Li}\end{array} $

, n₂= 92.5%

The atomic mass of the lithium atom is:

$\begin{array}{l}m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }\end{array} $

$\begin{array}{l}m =\frac{ 6.01512 \times 7.5 + 7.01600 \times 92.5 }{ 7.5 + 92.5 }\end{array} $

= 6.940934 u

Therefore, The atomic mass of the lithium atom is 6.940934 u.

(b) Mass of boron isotope

$\begin{array}{l}_{5}^{10}{B}\end{array} $

m₁ = 10.01294 u

Mass of boron isotope

$\begin{array}{l}_{5}^{11}{B}\end{array} $

m₂ = 11.00931 u

Abundance of

$\begin{array}{l}_{5}^{10}{B}\end{array} $

, n₁ = x%

Abundance of

$\begin{array}{l}_{5}^{11}{B}\end{array} $

, n₂= (100 − x)%

The atomic mass of boron, m = 10.811 u

The atomic mass of the boron atom is:

$\begin{array}{l}m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }\end{array} $

$\begin{array}{l}10.811 =\frac{ 10.01294 \times x + 11.00931 \times ( 100 – x ) }{ x + 100 -x }\end{array} $

1081.11 = 10.01294x + 1100.931 – 11.00931 x

x = 19.821/0.99637 = 19.89 %

And 100 − x = 80.11%

Hence, the abundance of

$\begin{array}{l}_{5}^{10}{B}\end{array} $

is 19.89% and that of

$\begin{array}{l}_{5}^{11}{B}\end{array} $

is 80.11%.

Q 13.2: The three stable isotopes of neon:

$\begin{array}{l}_{10}^{20}{Ne}\end{array} $

,
$\begin{array}{l}_{10}^{21}{Ne}\end{array} $
and
$\begin{array}{l}_{10}^{22}{Ne}\end{array} $
have abundances of 90.51%, 0.27% and 9.22%, respectively. 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.

Answer

Atomic mass of

$\begin{array}{l}_{10}^{20}{Ne}\end{array} $

, m₁= 19.99 u

Abundance of

$\begin{array}{l}_{10}^{20}{Ne}\end{array} $

, n₁ = 90.51%

Atomic mass of

$\begin{array}{l}_{10}^{21}{Ne}\end{array} $

, m₂ = 20.99 u

Abundance of

$\begin{array}{l}_{10}^{21}{Ne}\end{array} $

, n₂ = 0.27%

Atomic mass of

$\begin{array}{l}_{10}^{22}{Ne}\end{array} $

, m₃ = 21.99 u

Abundance of

$\begin{array}{l}_{10}^{22}{Ne}\end{array} $

, n₃ = 9.22%

The average atomic mass of neon:

$\begin{array}{l}m=\frac{ m_1 n_1 + m_2 n_2 + m_3 n_3 }{ n_1 + n_2 + n_3 }\end{array} $

$\begin{array}{l}\frac{ 19.99 \times 90.51 + 20.99 \times 0.27 + 21.99 \times 9.22 }{ 90.51 + 0.27 + 9.22 }\end{array} $

= 20.1771 u

Therefore, the average atomic mass of neon is 20.1771 u.

Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus

$\begin{array}{l}_{7}^{14}{N}\end{array} $

, given m(
$\begin{array}{l}_{7}^{14}{N}\end{array} $
)=14.00307 u.

Ans:

Atomic mass of nitrogen

$\begin{array}{l}_{7}^{14}{N}\end{array} $

, m = 14.00307 u

A nucleus of

$\begin{array}{l}_{7}^{14}{N}\end{array} $

nitrogen contains 7 neutrons and 7 protons.

∆m = 7m_H + 7m_n − m is the mass defect of the nucleus

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n= 1.008665 u

∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But, 1 u = 931.5 MeV/c²

∆m = 0.11236 × 931.5 MeV/c²

E_b = ∆mc² is the binding energy of the nucleus

Where, c =Speed of light

$\begin{array}{l}E_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\end{array} $

= 104.66334 Mev

Therefore, 104.66334 MeV is the binding energy of the nitrogen nucleus.

Q 13.4: Obtain the binding energy of the nuclei

$\begin{array}{l}_{26}^{56}{Fe}\end{array} $

and
$\begin{array}{l}_{83}^{209}{Bi}\end{array} $
in units of MeV from the following data:

(a) m (

$\begin{array}{l}_{23}^{56}{Fe}\end{array} $

) = 55.934939 u

(b) m(

$\begin{array}{l}_{83}^{209}{Bi}\end{array} $

) = 208.980388 u

Ans:

(a) Atomic mass of

$\begin{array}{l}_{26}^{56}{Fe}\end{array} $

, m₁ = 55.934939 u

$\begin{array}{l}_{26}^{56}{Fe}\end{array} $

nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = 26 × m_H + 30 × m_n − m₁

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But, 1 u = 931.5 MeV/c²

∆m = 0.528461 × 931.5 MeV/c²

$\begin{array}{l}E_{b_{1}}\end{array} $

= ∆mc²is the binding energy of the nucleus.

Where, c = Speed of light

$\begin{array}{l}E_{b_{1}} = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\end{array} $

= 492.26 MeV

Therefore, the average binding energy per nucleon =

$\begin{array}{l}\frac{492.26}{56} = 8.79 MeV\end{array} $

(b) Atomic mass of

$\begin{array}{l}_{83}^{209}{Bi}\end{array} $

, m₂ = 208.980388 u

$\begin{array}{l}_{83}^{209}{Bi}\end{array} $

nucleus has 83 protons and (209 − 83) 126 neutrons.

The mass defect of the nucleus is given as:

∆m’ = 83 × m_H + 126 × m_n − m₂

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But, 1 u = 931.5 MeV/c²

∆m’ = 1.760877 × 931.5 MeV/c²

E_b2= ∆m’c² is the binding energy of the nucleus.

= 1.760877 × 931.5

$\begin{array}{l} \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}\end{array} $

= 1640.26 MeV

Therefore, the average binding energy per nucleon = 1640.26/209 = 7.848 MeV.

Q 13.5: 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 entirely made of

$\begin{array}{l}_{29}^{63}{Cu}\end{array} $

with mass = 62.92960 u.

Ans :

Mass of a copper coin, m’ = 3 g

Atomic mass of

$\begin{array}{l}_{29}^{63}{Cu}\end{array} $

atom, m = 62.92960 u

The total number of

$\begin{array}{l}_{29}^{63}{Cu}\end{array} $

atoms in the coin,

$\begin{array}{l}N = \frac{N_A \times m’}{Mass \; number}\end{array} $

Where,

N_A = Avogadro’s number = 6.023 × 10²³ atoms /g

Mass number = 63 g

$\begin{array}{l}N = \frac{6.023 \times 10^{23}\times 3}{63}\end{array} $

= 2.868 x 10²² atoms

$\begin{array}{l}_{29}^{63}{Cu}\end{array} $

nucleus has 29 protons and (63 − 29) 34 neutrons

Mass defect of this nucleus, ∆m’ = 29 × m_H + 34 × m_n − m

Where,

Mass of a proton, m_H = 1.007825 u

Mass of a neutron, m_n = 1.008665 u

∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 10²²

= 1.69766958 × 10²² u

But, 1 u = 931.5 MeV/c²

∆m = 1.69766958 × 10²² × 931.5 MeV/c²

E_b = ∆mc² is the binding energy of the nuclei of the coin

= 1.69766958 × 10²² × 931.5 MeV/c²× c²

= 1.581 × 10²⁵ MeV

But 1 MeV = 1.6 × 10⁻¹³ J

E_b = 1.581 × 10²⁵ × 1.6 × 10⁻¹³

= 2.5296 × 10¹² J

Therefore, the energy required to separate all the neutrons and protons from the given coin is 2.5296 × 10¹² J.

Q 13.6: Write nuclear reaction equations for the following:

(i) α-decay of

$\begin{array}{l}_{88}^{226}{Ra}\end{array} $

(ii) α-decay of
$\begin{array}{l}_{94}^{242}{Pu}\end{array} $

(iii) β⁻-decay of

$\begin{array}{l}_{15}^{32}{P}\end{array} $

(iv) β^– -decay of
$\begin{array}{l}_{83}^{210}{Bi}\end{array} $

(v) β⁺-decay of

$\begin{array}{l}_{6}^{11}{C}\end{array} $

(vi) β⁺ -decay of
$\begin{array}{l}_{43}^{97}{Tc}\end{array} $

(vii) Electron capture of

$\begin{array}{l}_{54}^{120}{Xe}\end{array} $

Ans:

In helium, α is a nucleus

$\begin{array}{l}_{2}^{4}{He}\end{array} $

, and β is an electron (e⁻ for β⁻ and e⁺ for β⁺). 2 protons and 4 neutrons are lost in every α decay. Whereas 1 proton and a neutrino are emitted from the nucleus in every β ⁺ decay. In every β ^– decay, there is a gain of 1 proton, and an anti-neutrino is emitted from the nucleus.

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

(i)

$\begin{array}{l}_{88}^{226}{Ra} \rightarrow\; _{86}^{222}{Rn} \; + \; _{2}^{4}{He}\end{array} $

(ii)

$\begin{array}{l}_{ 94 }^{ 242 }{ Pu } \rightarrow\; _{ 92 }^{ 238 }{ U } \; + \; _{ 2 }^{ 4 }{ He }\end{array} $

(iii)

$\begin{array}{l}_{15}^{ 32 }{P} \rightarrow\; _{ 16 }^{ 32 }{ S } \; + \; e^- +\bar{v}\end{array} $

(iv)

$\begin{array}{l}_{83}^{ 210 }{B} \rightarrow\; _{ 84 }^{ 210 }{ PO } \; + \; e^- +\bar{v}\end{array} $

(v)

$\begin{array}{l}_{6}^{ 11 }{C} \rightarrow\; _{ 5 }^{ 11 }{ B } \; + \; e^+ + \; v\end{array} $

(vi)

$\begin{array}{l}_{43}^{ 97 }{Tc} \rightarrow\; _{ 42 }^{ 97 }{ MO } \; + \; e^+ + \; v\end{array} $

(vii)

$\begin{array}{l}_{54}^{ 120 }{Xe}+ \; e^+ \rightarrow\; _{ 53 }^{ 120 }{ I } \; + \; v\end{array} $

Q 13.7: A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125% and (b) 1% of its original value?

Ans:

The half-life of the radioactive isotope = T years

N₀is the actual amount of radioactive isotope.

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N₀ remains after decay. Hence, we can write:

$\begin{array}{l}\frac{N}{N_0} = 3.125 \% = \frac{3.125}{100}=\frac{1}{32}\end{array} $

But,

$\begin{array}{l}\frac{N}{N_0} = e^{-\lambda t}\end{array} $

Where, λ = Decay constant

t = Time

$\begin{array}{l}e^{-\lambda t} = \frac{1}{32}\end{array} $

$\begin{array}{l}-\lambda t = ln 1 – ln 32\end{array} $

$\begin{array}{l}-\lambda t = 0 – 3.4657\end{array} $

Since,

$\begin{array}{l}-\lambda = \frac{0.693}{T}\end{array} $

t =

$\begin{array}{l}\frac{3.466}{\frac{0.693}{T}} \approx \; 5 T years\end{array} $

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N₀ remains after decay. Hence, we can write:

$\begin{array}{l}\frac{N}{N_0} = 1 \% = \frac{1}{100}\end{array} $

$\begin{array}{l}\frac{N}{N_0} = e^{-\lambda t}\end{array} $

$\begin{array}{l}e^{-\lambda t} = \frac{1}{100}\end{array} $

$\begin{array}{l}-\lambda t = ln1 – ln100\end{array} $

$\begin{array}{l}e^{-\lambda t} =\end{array} $

0 – 4.6052

t = 4.6052/

$\begin{array}{l}\lambda\end{array} $

Since, λ = 0.693/T

$\begin{array}{l}t = \frac{4.6052}{\frac{0.693}{T}}\end{array} $

= 6.645 T years

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

Q 13.8: The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive

$\begin{array}{l}_{6}^{14}{C}\end{array} $

present with the stable carbon isotope
$\begin{array}{l}_{6}^{12}{C}\end{array} $
. When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases, and its activity begins to drop. From the known half-life (5730 years) of
$\begin{array}{l}_{6}^{14}{C}\end{array} $
and the measured activity, the age of the specimen can be approximately estimated. This is the principle of
$\begin{array}{l}_{6}^{14}{C}\end{array} $
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.

Ans :

The 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.

Half-life of

$\begin{array}{l}_{6}^{14}{C},T_{\frac{1}{2}}\end{array} $

= 5730 years

The decay rate of the specimen obtained from the Mohenjo-Daro site:

R’ = 9 decays/min

Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

Therefore, the relation between the decay constant, λ and time, t, is:

$\begin{array}{l}\frac{N}{N’}=\frac{R}{R’}= e^{-\lambda t}\end{array} $

$\begin{array}{l}e^{-\lambda t} = \frac{9}{15}=\frac{3}{5}\end{array} $

$\begin{array}{l}-\lambda t = log_e\frac{3}{5} = -0.5108\end{array} $

$\begin{array}{l}t = \frac{0.5108}{\lambda}\end{array} $

But

$\begin{array}{l}{\lambda} = \frac{0.693}{T_\frac{1}{2}}=\frac{0.693}{5730}\end{array} $

$\begin{array}{l}t = \frac{0.5108}{\frac{0.693}{5730}}= 4223.5 years\end{array} $

So, the approximate age of the Indus-Valley civilisation is 4223.5 years.

Q 13.9: Obtain the amount of

$\begin{array}{l}_{27}^{60}{Co}\end{array} $

necessary to provide a radioactive source of 8.0 mCi strength. The half-life of $\begin{array}{l}_{27}^{60}{Co}\end{array} $
is 5.3 years.

Ans:

The strength of the radioactive source is given as:

$\begin{array}{l}\frac{dN}{dt} = 8.0 mCi\end{array} $

= 8 x 10^-3 x 3.7 x 10¹⁰

= 29.6 x 10⁷ decay/s

Where,

N = Required number of atoms

Half-life of

$\begin{array}{l}_{ 27 }^{ 60 }{ Co }\end{array} $

T 1/2 = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 10⁸ s

The rate of decay for decay constant λ is:

$\begin{array}{l}\frac{dN}{dt} = \lambda N\end{array} $

Where,

$\begin{array}{l}\lambda = \frac{0.693}{T_\frac{1}{2}} = \frac{0.693}{1.67 \times 10^{8}}s^{-1}\end{array} $

$\begin{array}{l}N = \frac{1}{\lambda} \frac{dN}{dt}\end{array} $

$\begin{array}{l}= \frac{29.6 \times 10^{ 7}}{\frac{ 0.693 }{ 1.67 \times 10^{ 8 }}}\end{array} $

= 7.133 x 10¹⁶atoms

For

$\begin{array}{l}_{27}^{60}{Co}\end{array} $

Mass of 6.023 × 10²³ (Avogadro’s number) atoms = 60 g

Mass of 7.133 x 10¹⁶ atoms =

$\begin{array}{l}= \frac{60\times7.133\times10^{16}}{6.023\times10^{23}}\end{array} $

= 7.106 x 10^-6 g

Hence, the amount of

$\begin{array}{l}_{27}^{60}{Co}\end{array} $

necessary for the purpose is 7.106 × 10⁻⁶ g.

Q 13.10: The half-life of

$\begin{array}{l}_{ 38 }^{ 90 }{ Sr }\end{array} $

is 28 years. What is the disintegration rate of 15 mg of this isotope?

Ans:

Half-life of

$\begin{array}{l}_{ 38 }^{ 90 }{ Sr }\end{array} $

$\begin{array}{l}t_{\frac{1}{2}}\end{array} $

= 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 10⁸ s

Mass of the isotope, m = 15 mg

90 g of

$\begin{array}{l}_{ 38 }^{ 90 }{ Sr }\end{array} $

atom contains 6.023 × 10²³ (Avogadro’s number) atoms.

Therefore, 15 mg of

$\begin{array}{l}_{ 38 }^{ 90 }{ Sr }\end{array} $

contains atoms:

$\begin{array}{l}\frac{6.023 \times 10^{23} \times 15 \times 10^{-3}}{90}\end{array} $

i.e., 1.0038 x 10²⁰number of atoms

Rate of disintegration,

$\begin{array}{l}\frac{dN}{dt}= \lambda N\end{array} $

Where,

$\begin{array}{l}\lambda\end{array} $

= decay constant =

$\begin{array}{l}\frac{0.693}{8.83 \times 10^{8}}s^{-1}\end{array} $

$\begin{array}{l}\frac{dN}{dt}=\frac{0.693\times 1.0038 \times 10^{20}}{8.83 \times 10^{8}}\end{array} $

= 7.878 x 10¹⁰ atoms/s

Hence, the disintegration rate of 15 mg of

$\begin{array}{l}_{ 38 }^{ 90 }{ Sr }\end{array} $

is 7.878 × 10¹⁰ atoms/s.

Q 13.11: Obtain approximately the ratio of the nuclear radii of the gold isotope

$\begin{array}{l}_{ 79 }^{ 197 }{Au}\end{array} $

and the $\begin{array}{l}_{ 47 }^{ 107 }{Ag}\end{array} $
silver isotope.

Ans:

Nuclear radius of the gold isotope

$\begin{array}{l}_{ 79 }^{ 197 }{Au}\end{array} $

= R_Au

Nuclear radius of the silver isotope

$\begin{array}{l}_{ 47 }^{ 107 }{Ag}\end{array} $

= R_Ag

The mass number of gold, A_Au = 197

The mass number of silver, A_Ag = 107

Following is the relationship between the radii of the two nuclei and their mass number:

$\begin{array}{l}\frac{ R_{ Au }}{ R_{ Ag }}=\left (\frac{ R_{ Au }}{ R_{ Ag }} \right )^{\frac{ 1 }{ 3 } }\end{array} $

$\begin{array}{l}=\left (\frac{ 197 }{ 107 } \right )^{\frac{ 1 }{ 3 } }\end{array} $

= 1.2256

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

Q 13.12: Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of

(a)

$\begin{array}{l}_{88}^{226}{Ra}\end{array} $

(b)

$\begin{array}{l}_{86}^{220}{Rn}\end{array} $

Given,

$\begin{array}{l}m ( _{ 88 }^{ 226 }{ Ra })\end{array} $

= 226.02540 u, $\begin{array}{l}m ( _{ 86 }^{ 222 }{ Rn })\end{array} $
= 222.01750 u,

$\begin{array}{l}m ( _{ 86 }^{ 220 }{ Rn })\end{array} $

= 220.01137 u, $\begin{array}{l}m ( _{ 84 }^{ 216 }{ Po })\end{array} $
= 216.00189 u.

Answer

(a) Alpha particle decay of

$\begin{array}{l}_{88}^{226}{Ra}\end{array} $

emits a helium nucleus. As a result, its mass number reduces to 222 (226 − 4), and its atomic number reduces to 86 (88 − 2). This is shown in the following nuclear reaction:

$\begin{array}{l}_{ 88 }^{ 226 }{ Ra }\rightarrow _{ 86 }^{ 222 }{ Ra }+_{ 2 }^{ 4 }{ He }\end{array} $

Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass) c²

Where, c = Speed of light

It is given that:

$\begin{array}{l}m ( _{ 88 }^{ 226 }{ Ra })\end{array} $

= 226.02540 u

$\begin{array}{l}m ( _{ 86 }^{ 222 }{ Rn })\end{array} $

= 222.01750 u

$\begin{array}{l}m ( _{ 2 }^{ 4 }{ He })\end{array} $

= 4.002603 u

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 MeV

Kinetic energy of the α-particle =

$\begin{array}{l}\left ( \frac{Mass \; number \; after \; decay }{Mass \; number \; before \; decay } \right ) \times Q\end{array} $

$\begin{array}{l}= \frac{222}{226} \times 4.94\end{array} $

= 4.85 MeV.

(b) Alpha particle decay of

$\begin{array}{l}_{86}^{220}{Rn}\end{array} $

$\begin{array}{l}_{86}^{220}Ra\rightarrow _{84}^{216}Po+_{e}^{4}He\end{array} $

It is given that:

Mass of

$\begin{array}{l}_{86}^{220}{Rn}\end{array} $

= 220.01137 u

Mass of

$\begin{array}{l}_{84}^{216}{Po}\end{array} $

= 216.00189 u

Q-value = [ 220.01137 – ( 216.00189 + 4.00260 ) ] x 931.5

≈ 641 MeV.

Kinetic energy of the α-particle =

$\begin{array}{l}= \left( \frac{ 220 – 4 }{220} \right ) \times 6.41\end{array} $

= 6.29 MeV.

Q 13.13: The radionuclide ¹¹C decays according to

$\begin{array}{l}{ }_{6}^{11} \mathrm{C} \rightarrow_{5}^{11} \mathrm{~B}+e^{+}+v: \quad T_{1 / 2}=20.3 \mathrm{~min}\end{array} $

The maximum energy of the emitted positron is 0.960 MeV.
Given the mass values:
m (¹¹₆ C) = 11.011434 u and

m (¹¹₆B ) = 11.009305 u,
calculate Q and compare it with the maximum energy of the positron emitted.

Ans:

$\begin{array}{l}{ }_{6}^{11} \mathrm{C} \rightarrow_{5}^{11} \mathrm{~B}+e^{+}+v: \quad T_{1 / 2}=20.3 \mathrm{~min}\end{array} $

Mass defect in the reaction is Δm = m (₆C¹¹) – m(₅B¹¹) – m_e

This is given in terms of atomic masses. To express in terms of nuclear mass, we should subtract 6m_e from carbon, and 5m_e from boron.

Δm = m (₆C¹¹) – 6m_e – (m(₅B¹¹) – 5m_e )- m_e

= m (₆C¹¹) – 6m_e – m(₅B¹¹) + 5m_e – m_e

= m (₆C¹¹) – m(₅B¹¹) – 2 m_e

Δm = [11.011434 – 11.009305 – 2 x 0.000548] u

= 0.002129 – 0.001096 = 0.001033

Q = Δm x 931 MeV

= 0.001033 x 931

Q= 0.9617 MeV

Therefore, the Q-factor of the reaction is equal to the maximum energy of the emitted positron.

Q 13.14: The nucleus

$\begin{array}{l}_{10}^{23}{Ne}\end{array} $

decays
$\begin{array}{l}\beta ^-\end{array} $
by emission. Write down the
$\begin{array}{l}\beta \end{array} $
decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

$\begin{array}{l}m(_{10}^{23}{Ne})\end{array} $

= 22.994466 u

$\begin{array}{l}m(_{11}^{23}{Na})\end{array} $

= 22.989770 u.

Ans:

$\begin{array}{l}\beta ^-\end{array} $

emission, the number of protons increases by 1, and one electron and an antineutrino is emitted from the parent nucleus.

$\begin{array}{l}\beta ^-\end{array} $

emission from the nucleus

$\begin{array}{l}_{10}^{23}{Ne}\end{array} $

$\begin{array}{l}_{ 10 }^{ 23 }{ Ne }\rightarrow _{ 11 }^{ 23 }{ Na } + e^{ – } + \bar{ v } + Q\end{array} $

It is given that:

Atomic mass of

$\begin{array}{l}m(_{10}^{23}{Ne})\end{array} $

= 22.994466 u

Atomic mass of

$\begin{array}{l}m(_{11}^{23}{Na})\end{array} $

= 22.989770 u

Mass of an electron, m_e = 0.000548 u

The Q-value of the given reaction is given as:

$\begin{array}{l}Q = [m(_{ 10 }^{ 23 }{Ne}) – [m(_{ 11 }^{ 23 }{Na}) + m_e]]c^{ 2 }\end{array} $

There are 10 electrons and 11 electrons in

$\begin{array}{l}_{ 10 }^{ 23 }{Ne}\end{array} $

and

$\begin{array}{l}_{ 11 }^{ 23 }{Na}\end{array} $

, respectively. Hence, the mass of the electron is cancelled in the Q-value equation.

Q = [ 22.994466 – 22.989770 ] c²

= (0.004696 c²) u

But, 1 u = 931.5 MeV/c²

Q = 0.004696 x 931.5 = 4.374 MeV

The daughter nucleus is too heavy as compared to e^– and

$\begin{array}{l}\bar{v}\end{array} $

. 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.

Q 13.15: The Q value of a nuclear reaction A + b → C + d is defined by

Q = [ m_A+ m_b− m_C− m_d]c^2, 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)

$\begin{array}{l}_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}\end{array} $

(ii)

$\begin{array}{l}_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}\end{array} $

Atomic masses are given to be

$\begin{array}{l}m( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u\end{array} $

$\begin{array}{l}m( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u\end{array} $

$\begin{array}{l}m( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u\end{array} $

$\begin{array}{l}m( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u\end{array} $

Ans:

(i) The given nuclear reaction is:

$\begin{array}{l}_{ 1 }^{ 1 }{H} + _{ 1 }^{ 3 }{H}\rightarrow _{ 1 }^{ 2 }{ H } + _{ 1 }^{ 2 }{H}\end{array} $

It is given that:

Atomic mass

$\begin{array}{l}m( _{ 1 }^{ 1 }{H} ) = \; 1.007825 \; u\end{array} $

Atomic mass

$\begin{array}{l}m( _{ 1 }^{ 3 }{H} ) = \; 3.016049 \; u\end{array} $

Atomic mass

$\begin{array}{l}m( _{ 1 }^{ 2 }{H} ) = \; 2.014102 \; u\end{array} $

According to the question, the Q-value of the reaction can be written as:

Q = [

$\begin{array}{l}m( _{ 1 }^{ 1 }{H} ) \end{array} $

$\begin{array}{l}m( _{ 1 }^{ 3 }{H} ) \end{array} $

– 2m

$\begin{array}{l}m( _{ 1 }^{ 2 }{H} )\end{array} $

] c²

= [ 1.007825 + 3.016049 – 2 x 2.014102] c²

Q = ( – 0.00433 c²) u

But 1 u = 931.5 MeV/c²

Q = -0.00433 x 931.5 = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:

$\begin{array}{l}_{ 6 }^{ 12 }{C} + _{ 6 }^{ 12 }{C}\rightarrow _{ 10 }^{ 20 }{ Ne } + _{ 2 }^{ 4 }{He}\end{array} $

It is given that:

Atomic mass of

$\begin{array}{l}m( _{ 6 }^{ 12 }{C} ) = \; 12.000000 \; u\end{array} $

Atomic mass of

$\begin{array}{l}m( _{ 10 }^{ 20 }{Ne} ) = \; 19.992439 \; u\end{array} $

Atomic mass of

$\begin{array}{l}m( _{ 2 }^{ 4 }{He} ) = \; 4.002603 \; u\end{array} $

The Q-value of this reaction is given as:

Q = [

$\begin{array}{l}2m( _{ 6 }^{ 12 }{C} ) \end{array} $

–

$\begin{array}{l}m( _{ 10 }^{ 20 }{Ne} )\end{array} $

–

$\begin{array}{l}m( _{ 2 }^{ 4 }{He} )\end{array} $

] c²

= [ 2 x 12.000000 – 19.992439 – 4.002603 ] c²

= [ 0.004958 c²] u

= 0.004958 x 931.5 = 4.618377 MeV

Since we obtained a positive Q-value, it can be concluded that the reaction is exothermic.

Q 13.16: Suppose, we think of fission of a

$\begin{array}{l}_{ 26 }^{ 56 }{Fe}\end{array} $

nucleus into two equal fragments, $\begin{array}{l}_{ 13 }^{ 28 }{Al}\end{array} $
. Is fission energetically possible? Argue by working out the Q of the process. Given,

$\begin{array}{l}m( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u\end{array} $

$\begin{array}{l}m( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u\end{array} $

Ans:

The fission of

$\begin{array}{l}_{ 26 }^{ 56 }{Fe}\end{array} $

can be given as :

$\begin{array}{l}_{ 26 }^{ 56 }{Fe}\end{array} $

-> 2

$\begin{array}{l}_{ 13 }^{ 28 }{Al}\end{array} $

It is given that:

Atomic mass of

$\begin{array}{l}m( _{ 26 }^{ 56 }{Fe} ) = \; 55.93494 \; u\end{array} $

Atomic mass of

$\begin{array}{l}m( _{ 13 }^{ 28 }{Al} ) = \; 27.98191 \; u\end{array} $

The Q-value of this nuclear reaction is given as:

Q = [

$\begin{array}{l}m( _{ 26 }^{ 56 }{Fe} ) – 2m( _{ 13 }^{ 28 }{Al} ) \end{array} $

] c²

= [ 55.93494 – 2 x 27.98191 ]c²

= ( -0.02888 c²) u

But, 1 u = 931.5 MeV/c²

Q = – 0.02888 x 931.5 = -26.902 MeV

Since the Q-value is negative for fission, it is energetically not possible.

Q 13.17: The fission properties

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

of are very similar to those of
$\begin{array}{l}_{ 92 }^{ 235 }{ U }\end{array} $
. The average energy released per fission is 180 MeV. How much energy is released if all the atoms in 1 kg of pure
$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $
undergo fission?

Ans:

Average energy released per fission of

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

, E_av = 180 MeV

Amount of pure

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

, m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

= 239 g

1 mole of

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

contains N_A atoms.

Therefore, mg of

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

contains

$\begin{array}{l}\left ( \frac{N_A}{Mass \; Number}\times m \right )\end{array} $

atoms

$\begin{array}{l}\frac{6.023 \times 10^{23}}{239}\times 1000 = 2.52 \times 10^{ 24 }\end{array} $

atoms

The total energy released during the fission of 1 kg of

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

is calculated as:

E = E_avx 2.52 x 10²⁴

= 180 x 2.52 x 10²⁴

= 4.536 x 10²⁶ MeV

Hence, 4.536 x 10²⁶ MeV is released if all the atoms in 1 kg of pure

$\begin{array}{l}_{ 94 }^{ 239 }{ Pu }\end{array} $

undergo fission.

Q 13.18: A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much ₉₂²³⁵ U did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of ₉₂ ²³⁵U and that this nuclide is consumed only by the fission process.

Ans:

The reactor consumes half of its fuel in 5 years. Therefore, the half-life of the fuel of the fission reactor, t_1/2 = 5 x 365 x 24 x 60 x 60 s

200 MeV is released during the fission of 1 g of ₉₂ ²³⁵U.

1 mole, i.e., 235 g of ₉₂ ²³⁵U, contains 6.023 x 10²³ atoms.

The number of atoms in 1 g of ₉₂ ²³⁵U is (6.023 x 10²³/235) atoms.

The total energy generated per gram of ₉₂ ²³⁵U is (6.023 x 10²³/235) x 200 MeV/g

$\begin{array}{l}= \frac{200 \times 6.023 \times 10^{23}\times 1.6\times 10^{-19}\times 10^{6}}{235}\end{array} $

= 8.20 x 10¹⁰ J/g

The reactor operates only 80% of the time.

Therefore, the amount of ₉₂ ²³⁵U consumed in 5 years by the 1000 MW fission reactor is

$\begin{array}{l}= \frac{5\times 365\times24\times 60\times 60 \times 1000\times 10^{6} }{8.20\times 10^{10}}\times \frac{80}{100}\end{array} $

= 12614400/8.20

= 1538341 g

= 1538 Kg

Therefore, the initial amount of ₉₂ ²³⁵U is 2 x 1538 = 3076 Kg.

Q 13.19: How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as

$\begin{array}{l}{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\end{array} $

Ans:

The fusion reaction given is

$\begin{array}{l}{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \rightarrow{ }_{2}^{3} \mathrm{He}+\mathrm{n}+3.27 \mathrm{MeV}\end{array} $

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium, contains 6.023 x 10²³ atoms.

Therefore, 2 Kg of deuterium contains (6.023 x 10²³/2) x 2000 = 6.023 x 10²⁶ atoms

From the reaction given, it can be understood that 2 atoms of deuterium fuse, 3.27 MeV energy are released. The total energy per nucleus released during the fusion reaction is

E = (3.27/2) x 6.023 x 10²⁶MeV

= (3.27/2) x 6.023 x 10²⁶x 1.6 x 10^-19 x 10⁶

= 1.576 x 10¹⁴ J

Power of the electric lamp, P = 100 W = 100 J/s

The energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is 1.576 x 10¹⁴ /100 = 1.576 x 10¹²s

= (1.576 x 10¹²)/ (365 x 24 x 60 x 60)

= (1.576 x 10¹²)/3.1536 x10⁷

= 4.99 x 10⁴years.

Q 13.20: 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.)

Ans :

When two deuterons collide head-on, the distance between their centres, d, is given as:

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2 f m = 2 × 10⁻¹⁵ m

d = 2 × 10⁻¹⁵ + 2 × 10⁻¹⁵ = 4 × 10⁻¹⁵ m

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

The potential energy of the two-deuteron system:

$\begin{array}{l}v = \frac{ e^{ 2 } }{ 4 \pi \epsilon_0 d }\end{array} $

Where,

$\begin{array}{l}\epsilon_0\end{array} $

= Permittivity of free space

$\begin{array}{l}\frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} Nm^2C^{-2}\end{array} $

Therefore,

$\begin{array}{l}V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} } J\end{array} $

$\begin{array}{l}V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} \times (1.6 \times 10^{-19})}eV\end{array} $

= 360 k eV

Hence, the height of the potential barrier of the two-deuteron system is 360 k eV.

Q 13.21: From the relation R = R₀A^1/3, 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).

Ans:

The nuclear radius is given as

R = R₀A^1/3

Here,

R₀ is Constant

A is the mass number of the nucleus

Nuclear matter density, ρ = Mass of the nucleus/Volume of the nucleus

Mass of the nucleus = mA

Density of the nucleus = (4/3)πR³

= (4/3)π(R₀A^1/3)³

= (4/3)πR₀³A

ρ = mA/[(4/3)πR₀³A]

= 3mA/(4πR₀³A)

ρ = 3m/(4πR₀³)

Hence, nuclear matter density is nearly a constant and is independent of A.

Q 13.22: For the

$\begin{array}{l}\beta^{+}\end{array} $

(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).

$\begin{array}{l}e^+ + _{Z}^{A}{X} \rightarrow _{ Z-1 }^{A}{Y}+v\end{array} $

Show that if

$\begin{array}{l}\beta^{+}\end{array} $

emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

Ans :

The chemical equations of the two competing processes are written as follows:

$\begin{array}{l}_{A}^{Z}\textrm{X}\rightarrow _{A}^{Z-1}\textrm{Y}+e+\nu+Q_1\end{array} $

[Positron Emission]

$\begin{array}{l}e^{-1}+_{A}^{Z}\textrm{X}\rightarrow _{A}^{Z-1}\textrm{Y}+v+Q_2\end{array} $

[Electron Capture]

The Q-value of the positron emission reaction is determined as follows :

$\begin{array}{l}\\Q_{1}=[m_{N}(_{Z}^{A}X)-m_{N}(_{Z-1}^{A}Y)-m_{e}]c^{2} \\ \\Q_{1}=[m(_{Z}^{A}X)-Zm_{e}-m(_{Z-1}^{A}Y)+(Z-1)m_{e}-m_{e}]c^{2} \\ \\Q_{1}=[m(_{Z}^{A}X)-m(_{Z-1}^{A}Y)-2m_{e}]c^{2}\end{array} $

The Q-value of the electron capture reaction is determined as follows:

$\begin{array}{l}\\Q_{2}=[m_{N}(_{Z}^{A}X)+m_{e}-m_{N}(_{Z-1}^{A}Y)]c^{2} \\ \\Q_{2}=[m(_{Z}^{A}X)-m(_{Z-1}^{A}Y)]c^{2}\end{array} $

In the equation, m_Nis the mass of the nucleus, and m denotes the mass of the atom.

From equations (1) and (2), we understand that,

Q₁ = Q₂ – 2m_ec²

From the above relation, we can infer that if Q1 > 0, then Q2 > 0; Also, if Q2> 0, it does not necessarily mean that Q1 > 0.

In other words, this means that if $β^{+}$ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically allowed nuclear reaction.

Q 13.23: 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 masses are ²⁴ ₁₂Mg (23.98504u), ²⁵₁₂Mg(24.98584u) and ²⁶₁₂Mg (25.98259u). The natural abundance of ²⁴ ₁₂Mg is 78.99% by mass. Calculate the abundances of the other two isotopes.

Ans:

Let the abundance of ²⁵₁₂Mg be x%

The abundance of ²⁶₁₂Mg = (100 – 78.99 -x)%

= (21.01 – x)%

The average atomic mass of magnetic

$\begin{array}{l}24.312 = \frac{23.98504\times 78.99+24.98584x+25.98259(21.01-x))}{100}\end{array} $

24.312 x 100 = 1894.57 + 24.98584x + 545. 894 – 25.98259x

2431.2 = 2440.46 – 0.99675x

0.99675x = 2440.46 – 2431.2

0.99675x = 9.26

x = 9.26/0.99675

x = 9.290%

The abundance of ²⁵₁₂Mg is 9.290%

The abundance of ²⁶₁₂Mg = (21.01 – x)%

= (21.01 -9.290 )% = 11.71%.

Q 13.24: The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei ⁴¹₂₀Ca and ²⁷₁₃Al from the following data:
m(⁴⁰₂₀Ca ) = 39.962591 u
m(⁴¹₂₀Ca ) = 40.962278 u
m(²⁶₁₃Al ) = 25.986895 u
m(²⁷₁₃Al ) = 26.981541 u

Ans:

When the nucleons are separated from₂₀Ca⁴¹ → ₂₀ Ca⁴⁰ + ₀n¹

Mass defect, Δm = m (₂₀ Ca⁴⁰) + m_n – m(₂₀ Ca⁴¹)

= 39.962591 + 1.008665 – 40.962278

= 0.008978 amu

Neutron separation energy = 0.008978 x 931 MeV = 8.362 MeV

Similarly, ₁₃Al²⁷ → ₁₃ Al²⁶ + ₀n¹

Mass defect, Δm = m (₁₃ Al²⁶) + m_n – m(₁₃Al²⁷)

= 25.986895 + 1.008665 – 26.981541

= 26.99556 – 26.981541 = 0.014019 amu

Neutron separation energy = 0.014019 x 931 MeV

= 13.051 MeV.

Q 13.25: A source contains two phosphorous radionuclides ³²₁₅P (T_1/2 = 14.3d) and ³³₁₅P(T_1/2 = 25.3d). Initially, 10% of the decays come from ³³₁₅ P. How long one must wait until 90% do so?

Ans:

The rate of disintegration is

– (dN/dt) ∝ N

Initially, 10% of the decay is due to ³³₁₅ P, and 90% of the decay comes from ³²₁₅P.

We have to find the time at which 90% of the decay is due to ³³₁₅ P and 10% of the decay comes from ³²₁₅P.

Initially, if the amount of ³³₁₅ P is N, then the amount of ³²₁₅P is 9N.

Finally, if the amount of ³³₁₅ P is 9N’, then the amount of ³²₁₅P is N’.

For ³²₁₅P,

$\begin{array}{l}\frac{N’}{9N}=\left ( \frac{1}{2} \right )^{t/T_{1/2}}\end{array} $

$\begin{array}{l}N’=9N \left ( \frac{1}{2} \right )^{t/T_{1/2}}\end{array} $

$\begin{array}{l}N’=9N \left ( 2\right )^{-t/14.3}\end{array} $

—–(1)

For ³³₁₅ P

$\begin{array}{l}\frac{9N’}{N}=\left ( \frac{1}{2} \right )^{t/T_{1/2}}\end{array} $

$\begin{array}{l}9N’= N\left ( \frac{1}{2} \right )^{t/T_{1/2}}\end{array} $

$\begin{array}{l}9N’= N\left ( 2 \right )^{-t/25.3}\end{array} $

——(2)

On dividing (1) by (2),

$\begin{array}{l}\\\frac{1}{9}=9\times 2^{\frac{t}{25.3}-\frac{t}{14.3}} \\ \\\frac{1}{81}=2^{-\frac{11t}{25.3\times 14.3}} \\ \\log1-log81=-\frac{11t}{25.3\times 14.3}log2 \\ \\-\frac{11t}{25.3\times 14.3}=\frac{0-1.908}{0.301} \\ \\t=\frac{25.3\times 14.3\times 1.908}{11\times 0.301}\approx 208.5 \: days\end{array} $

Hence, it will take about 208.5 days for a 90% decay of ₁₅P³³.

Q 13.26: Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:

$\begin{array}{l}{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\end{array} $

$\begin{array}{l}{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}\end{array} $

Calculate the Q-values for these decays and determine that both are energetically allowed.

Ans:

$\begin{array}{l}{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{82}^{209} \mathrm{~Pb}+{ }_{6}^{14} \mathrm{C}\end{array} $

Δm = m (₈₈Ra²²³) – m (₈₂Pb²⁰⁹) – m(₆C¹⁴)

= 223.01850 – 208.9817 – 14.00324

= 0.03419 u

The amount of energy released is given as

Q = Δm x 931 MeV

= 0.03419 x 931 MeV = 31.83 MeV

$\begin{array}{l}{ }_{88}^{223} \mathrm{Ra} \rightarrow{ }_{86}^{219} \mathrm{Rn}+{ }_{2}^{4} \mathrm{He}\end{array} $

Δm = m (₈₈Ra²²³) – m (₈₆Rn²¹⁹) – m(₂He⁴)

= 223.01850 – 219.00948 – 4.00260

= 0.00642 u

Q = 0.00642 x 931 MeV = 5.98 MeV

As both the Q-factor values are positive, the reaction is energetically allowed.

Q 13.27: Consider the fission of ²³⁸₉₂U by fast neutrons. In one fission event, no neutrons are emitted, and the final end products, after the beta decay of the primary fragments, are ¹⁴⁰₅₈Ce and ⁹⁹₄₄Ru. Calculate Q-value for this fission process. The relevant atomic and particle masses are as follows:

m(²³⁸ ₉₂U ) = 238.05079 u
m(¹⁴⁰₅₈Ce ) = 139.90543 u
m(⁹⁹₄₄Ru ) = 98.90594 u

Ans:

In the fission of ²³⁸₉₂U, 10 β^– particles decay from the parent nucleus. The nuclear reaction can be written as:

$\begin{array}{l}{ }_{92}^{238} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99} \mathrm{Ru}+10{ }_{-1}^{0} e\end{array} $

Given:

m₁ = (²³⁸ ₉₂U ) =238.05079 u
m₂ = (¹⁴⁰₅₈Ce ) =139.90543 u
m₃ = (⁹⁹₄₄Ru ) = 98.90594 u

m₄ = ¹₀n = 1.008665 u

The Q value is given as

Q = [ m'(²³⁸ ₉₂U ) + m (¹₀n) – m'(¹⁴⁰₅₈Ce ) – m'(⁹⁹₄₄Ru ) – 10m_e ] c²

Here,

m’ is the atomic masses of the nuclei

m'(²³⁸ ₉₂U ) = m₁ – 92 m_e

m'(¹⁴⁰₅₈Ce ) = m₂ – 58m_e

m'(⁹⁹₄₄Ru ) = m₃ – 44m_e

Therefore, Q = [m₁ – 92 m_e+ m₄ – m₂ + 58m_e– m₃ + 44m_e– 10m_e ] c²

= [m₁ + m₄ – m₂ – m₃]c²

= [238.0507 + 1.008665 – 139.90543 – 98.90594]c²

= 0.247995 c² u

[1 u = 931.5 MeV/c²]

Q = 0.247995 x 931.5 = 231.007 MeV

Hence, the Q-value of the fission process is 231.007 MeV.

Q 13.28: Consider the D-T reaction (deuterium-tritium fusion)

$\begin{array}{l}{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}\end{array} $

(a) Calculate the energy released in MeV in this reaction from the following data:
m(²₁H ) = 2.014102 u
m(³₁H ) = 3.016049 u
(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.)

Ans:

$\begin{array}{l}{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}\end{array} $

The Q-value is given as

Q = Δm x 931 MeV

= (m (₁H²) + m(₁H³) – m(₂He⁴) – m_n) x 931

= (2.014102 + 3.016049- 4.002603-1.00867) x 931

Q = 0.0188 x 931 = 17.58 MeV

Therefore, the energy released in the given reaction is 17.58 MeV.

(b) Repulsive potential energy of two nuclei when they touch each other is

$\begin{array}{l}=\frac{e^{2}}{4\pi \epsilon _{0}(2r)}\end{array} $

Where ε₀ = Permittivity of free space

$\begin{array}{l}\frac{1}{4\Pi \epsilon _{0}}\end{array} $

=9 x 10⁹ Nm²C^-2

$\begin{array}{l}= \frac{9\times 10^{9}\times (1.6\times 10^{-19})^{2}}{2\times 2\times 10^{-15}} joule\end{array} $

= (23.04 x 10^-29)/(4 x 10^-15)

= 5.76 x 10^-14 J

The kinetic energy needed to overcome the coulomb repulsion between the two nuclei is

K.E = 5. 76 x 10^-14 J

K.E= 2 x (3/2) KT

K is the Boltzmann constant = 1.38 x 10^-23

T = K.E/3K = 5. 76 x 10^-14/(3 x 1.38 x 10^-23)

= 5. 76 x 10^-14/(4.14 x 10^-23)

= 1.3913 x 10⁹ K

Therefore, the kinetic energy needed to overcome the coulomb repulsion between the two nuclei is 1.3913 x 10⁹ K.

Q 13.29: Obtain the maximum kinetic energy of β-particles and the radiation frequencies of γ decay in the decay scheme shown in Fig. 13.6. You are given that

m(¹⁹⁸Au) = 197.968233 u
m(¹⁹⁸Hg) =197.966760 u

Chapter 13- Class 12 - NCERT solutions - Nuclei

Ans:

From the γ decay diagram, it can be seen that γ₁ decays from the 1.088 MeV energy level to the 0 MeV energy level.
The energy corresponding to γ₁ decay is given as
E₁ = 1.088−0=1.088MeV
hv₁ = 1.088×1.6×10⁻¹⁹ ×10⁶J
Where,
h = Planck’s constant = 6.6 x 10^-34 Js

v₁ = frequency of radiation radiated by γ₁ decay.

$\begin{array}{l}v_{1}=\frac{E_{1}}{h}=\frac{1.088\times 1.6\times 10^{-19}\times 10^{6}}{6.6\times 10^{-34}}= 2.637 \times 10^{20} Hz\end{array} $

From the diagram, it can be observed that γ₂ decays from the 0.412 MeV energy level to the 0 MeV energy level.
The energy corresponding to γ₂ decay is given as
E₂ = 0.412−0=0.412MeV
hv₂ = 0.412×1.6×10⁻¹⁹ × 10⁶ J
Where v₂ is the frequency of the radiation radiated by γ₂ decay.

Therefore, v₂ =E₂/h

$\begin{array}{l}v_{2}=\frac{E_{2}}{h}=\frac{0.412\times 1.6\times 10^{-19}\times 10^{6}}{6.6\times 10^{-34}}\end{array} $

= 9.988×10¹⁹ Hz
From the gamma decay diagram, it can be γ₃ decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

The energy corresponding to γ₃ is given as
E₃ = 1.088−0.412=0.676 MeV
hv₃ = 0.676×10⁻¹⁹ ×10⁶ J
Where v₃ = frequency of the radiation radiated by γ₃ decay

Therefore, v₃ =E₃/h

$\begin{array}{l}v_{3}=\frac{E_{3}}{h}=\frac{0.676\times 1.6\times 10^{-19}\times 10^{6}}{6.6\times 10^{-34}}\end{array} $

=1.639×10²⁰ Hz
The mass of Au is 197.968233 u

Mass of Hg = 197.966760 u

[1u=931.5MeV/c²]

The energy of the highest level is given as:
E=[m(¹⁹⁸₇₈Au−m(¹⁹⁰₈₀Hg)]

=197.968233−197.96676=0.001473u

=0.001473×931.5=1.3720995MeV

β₁ decays from the 1.3720995 MeV level to the 1.088 MeV level

Therefore, the maximum kinetic energy of the β₁ particle =1.3720995−1.088=0.2840995MeV

β₂ decays from the 1.3720995 MeV level to the 0.412 MeV level
Therefore, the maximum kinetic energy of the β₂ particle = 1.3720995−0.412 = 0.9600995.

Q 13.30. Calculate and compare the energy released by a) the fusion of 1.0 kg of hydrogen deep within the sun and b) the fission of 1.0 kg of ²³⁵U in a fission reactor.

Ans:

Amount of hydrogen, m = 1 kg = 1000 g

(a) 1 mole, i.e., 1 g of hydrogen, contains 6.023 x 10²³ atoms. Therefore, 1000 g of hydrogen contains 6.023 x 10²³x 1000 atoms.

In the sun, four hydrogen nuclei fuse to form a helium nucleus and will release the energy of 26 MeV

The energy released by the fusion of 1 kg of hydrogen,

E₁ = (6.023 x 10²³x 1000 x 26)/4

E₁= 39.16 x 10²⁶ MeV.

(b) Amount of ²³⁵U = 1 kg = 1000 g

1 mole, i.e., 235 g of uranium, contains 6.023 x 10²³ atoms. Therefore, 1000 g of uranium contains (6.023 x 10²³x 1000)/235 atoms.

The energy released during the fission of one atom of ²³⁵U = 200 MeV

The energy released by the fission of 1 Kg of ²³⁵U,

E₂= (6.023 x 10²³x 1000 x 200)/235

E₂ = 5.1 x 10²⁶ MeV

(E₁/E₂) = (39.16 x 10²⁶/5.1 x 10²⁶)

= 7.67

The energy released in the fusion of hydrogen is 7.65 times more than the energy released during the fission of Uranium.

Q 13.31. Suppose India had a target of producing, by 2020 AD, 200,000 MW of electric power, ten per cent of which was to be obtained from nuclear power plants. Suppose we are given that, on average, the efficiency of utilisation (i.e., conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of ²³⁵U to be about 200MeV.

Ans:

Electric power to be generated = 2 x 10⁵ MW = 2 x 10⁵x 10⁶ J/s = 2 x 10¹¹ J/s

10 % of the amount is obtained from the nuclear power plant

P₁ = (10/100) x 2 x 10¹¹x 60 x 60 x 24 x 365 J/year

The heat energy released during per fission of a ²³⁵U nucleus, E = 200 MeV

Efficiency of the reactor = 25%

The amount of energy converted to electrical energy in fission is (25/100) x 200 = 50 MeV

= 50 x 1.6 x 10^-19 x 10⁶ = 80 x 10^-13 J

Required number of atoms for fission per year

= [(10/100) x 2 x 10¹¹x 60 x 60 x 24 x 365]/80 x 10^-13

= 788400 x 10²³atoms

1 mole, i.e., 235 g of ²³⁵U, contains 6.023 x 10²³ atoms

Mass of 6.023 x 10²³ atoms of ²³⁵U = 235 x 10^-3 kg

Mass of 788400 x 10²³atoms of ²³⁵U

=[ (235 x 10^-3)/(6.023 x 10²³) ] x 78840 x 10²⁴

= 3.076 x 10⁴ Kg

Hence, the mass of uranium needed per year is 3.076 x 10⁴ Kg.

Class 12 Physics NCERT Solutions for Chapter 13 Nuclei

Chapter 13 Nuclei of Class 12 Physics is well-structured as per the latest CBSE Syllabus 2023-24. One of the best methods to prepare for Physics is by taking notes and writing down all the important formulas and points in a notebook using the NCERT Solutions for Class 12. The notes will also help students during the time of revision. Once you get an idea about the type of questions asked in the examination, it is easy for you to collect important points and prepare Class 12 Physics Chapter 13 notes.

Subtopics of Class 12 Physics Chapter 13 Nuclei

Introduction
Atomic Masses and Composition of Nucleus
Size of the Nucleus
Mass energy and Nuclear Binding Energy
1. Mass energy
2. Nuclear binding energy
Nuclear Force
Radioactivity
1. Law of radioactive decay
2. Alpha decay
3. Beta-decay
4. Gamma decay
Nuclear Energy
1. Fission
2. Nuclear reactor
3. Nuclear fusion – energy generation in stars
4. Controlled thermonuclear fusion.

Also Access

NCERT Exemplar for Class 12 Physics Chapter 13

CBSE Notes for Class 12 Physics Chapter 13

Physics is one of the most interesting and scoring subjects in the Class 12 board examination. But, students must learn the concepts in-depth to score well in the board exams. Students can practise these NCERT Solutions to prepare well for the board exams. Class 12 Physics Chapter 13 Nuclei are one of the most interesting topics in Physics Class 12. Students must prepare this topic effectively to score well in Physics subject, in the board examination.

Students should have a detailed knowledge of all the subtopics and important points of the chapter during their preparation. In order to make students thorough with the concepts covered in chapter Nuclei, apart from NCERT Solutions, BYJU’S introduces an interactive method of learning with videos and animations.

BYJU’S interactive teaching methods will assist students in getting complete knowledge of the topic. Students also can download study materials, sample papers, exercises, previous years’ question papers, prepared notes and textbooks.

Disclaimer –

Dropped Topics –

13.6.1 Law of Radioactive Decay
13.6.2 Alpha Decay
13.6.3 Beta Decay
13.6.4 Gamma Decay
13.7.2 Nuclear Reactor
Exercises 13.1, 13.2, 13.6–13.10, 13.12–13.14, 13.18, 13.22–13.31

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NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

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Class 12 Physics NCERT Solutions for Chapter 13 Nuclei

Subtopics of Class 12 Physics Chapter 13 Nuclei

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