Nuclei Class 12 Physics NCERT Solutions: Chapter 13 Simplified Notes & Answers

Q: 13.21 From the relation R = R0A1/3 , where R0 is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

13.21 We have the expression for nuclear radius as: R = R0A1/3 Where R0 = constant A = mass number of nucleus Let m be the average mass of the nucleus, hence mass of the nucleus = mA Nuclear matter density ρ can be written as ρ=MassofthenucleusVolumeofthenucleus = mA43πR3 = 3mA4π(R0A13)3 = 3mA4πR03A = 3 m 4 π R 0 3 Hence, the nuclear mass density is independent of A. It is nearly constant

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 C614 present with the stable carbon isotope C612 . 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 C614 , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of C614 dating used in archaeology. Suppose a specimen from Mohenjo-Daro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization.

13.8 Decay rate of living carbon-containing matter, R = 15 decay / min Half life of C614 , T1/2 = 5730 years 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 a normal carbon-containing matter. Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period. We can relate the decay constant, λ and time t as: NN' = R'R = e-λt e-λt= R'R = 915 = 35 By taking log (ln) on both sides, -λt= loge?3 - loge?5 t = 0.5108λ Since λ = 0.693T1/2 = 0.6935730 t = 5730×0.51080.693 = 4223.5 years Hence, the approximate age of the Indus-valley is 4223.5 years.

Q: 13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as H12 + H12 → H e 2 3 + n+3.27 MeV

13.19 The given fusion reaction is H12+ H12 → H e 2 3 + n+3.27 MeV Amount of deuterium, m = 2 kg 1 mole, i.e. 2 g of deuterium contains 6.023 ×1023 atoms Hence 2 kg of deuterium contains = 6.023×10232 ×2×103atoms = 6.023 ×1026 atoms It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released. Hence total energy released from 2 kg of deuterium, E = 3.272 ×6.023 ×1026 MeV = 3.272 × 6.023 ×1026×106×1.6×10-19 J = 1.5756 ×1014 J Power of the electric bulb, P = 100 W = 100 J/s Hence energy consumed by the bulb per second = 100 J Therefore, total time the electric bulb will glow = 1.5756×1014100 seconds = 1.5756 ×1012secs = 49.96 ×103 years

Q: 13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope Au47197 and the silver isotope Ag47107 .

13.11 Nuclear radius of the gold isotope, Au47197 = RAu Nuclear radius of silver isotope, Ag47107 = RAg Mass number of gold, AAu = 197 Mass number of silver, AAg = 107 The ratio of the radii of the two nuclei is related with their mass numbers as : RAuRAg = AAuAAg1/3 = 1971071/3 =1.2256 Hence, the ratio of the nuclear radii of the gold and silver isotope is 1.23

Q: 13.14 The nucleus Ne1023 decays by β- emission. Write down the β -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m ( Ne1023) = 22.994466 u m ( Na)1123 = 22.989770 u.

13.14 In β- emission, the number of protons increase by 1 and one electron and an antineutrino are emitted from the parent nucleus. β- emission of the nucleus Ne1023 : Ne1023 →Na1123 + e- + ν? + Q It is given that: Atomic mass m ( Ne1023) = 22.994466 u Atomic mass m ( Na)1123 = 22.989770 u Mass of an electron, me = 0.000548 u Q value of the given reaction is given as Q = mNe1023-{m(Na)1123+me}c2 There are 10 electrons in Ne1023 and 11 electrons in Na1123 . Hence, the mass of the electron is cancelled in the Q-value equation. Therefore Q = {22.994466 - 22.989770} c2 = 4.696 ×10-3c2 u But 1 u = 931.5 MeV/ c2 Q = 4.374 MeV The daughter nucleus is too heavy as compared to e- and ν? . 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.

Updated on Jul 7, 2025 16:16 IST

By Pallavi Pathak, Assistant Manager Content

Class 12 Physics NCERT Nuclei curriculum explores the structure, properties, and behavior of the atomic nuclei. It is an important chapter for CBSE exam preparation. It covers fundamental concepts like radioactivity, nuclear composition, and nuclear reactions. The Nuclei Class 12 NCERT solutions provide clear and step-by-step explanations for all textbook questions. The solutions are ideal for CBSE Board preparation and preparation for other competitive examinations like NEET and JEE exams. The solutions offer simplified notes, detailed answers, and practical insights for students to build a strong understanding of the concepts of nuclear physics.
For key topics and free PDFs of Class 11 and 12 Chemistry, Physics, and Maths, check here - NCERT Solutions Class 11 and 12.

Table of content

Explore Chapter 13 – Nuclei Physics Class 12 CBSE
NCERT Class 12 Ch 13 Nuclei: Key Topics and Weightage
NCERT Physics Class 12th Solution PDF - Nuclei Chapter Download
Class 12 Physics Chapter 13 Nuclei NCERT Questions and Answers
Benefits of Using Chapter 12 Atoms Class 12 Physics NCERT Solutions
Chapter 12 Atoms Class 12 Physics NCERT Solutions- FAQs

Explore Chapter 13 – Nuclei Physics Class 12 CBSE

The following is a summary of the Nuclei chapter:

According to nuclear composition, the nucleus, containing neutrons and protons, holds over 99.9% of an atom’s mass. However, the size of the nucleus is nearly 10,000 times smaller than the atom itself.
Difference between Isobars and Isotopes, Isotopes are atoms of the same element with different numbers of neutrons, but the same number of protons. Isobars differ in neutron and proton counts but share the same mass number.
Radioactivity and Nuclear Reactions states that the unstable nuclei undergo alpha, beta, or gamma decay, the heavy nuclei split by the process of fission, and the light nuclei are combined by fusion. Both these processes release huge amounts of energy, powering reactors and stars.
Short-range but strong force that binds the nucleons is called Nuclear Force. It is independent of charge and overcomes proton repulsion.
In 1932, James Chadwick discovered the neutron. He found that a neutron is a neutral particle with a mass similar to that of a proton.
Nuclear size and density: The nuclei are spherical in shape, and their radii are determined by electron and alpha-particle scattering experiments. It reveals a constant nuclear density across all nuclei, which is higher than that of ordinary matter.
Due to mass defect, the nucleus mass is less than the sum of its individual nucleons’ masses, which corresponds to the binding energy holding the nucleus together.

For Class 12 Physics Chapter-wise Questions with Answers PDF, Important Topics & Weightage, read here - NCERT Solutions for Class 12 Physics.

NCERT Class 12 Ch 13 Nuclei: Key Topics and Weightage

Class 12 Physics Chapter 13 Nuclei is important for students who want to build a solid foundation related to concepts and topics in nuclear physics. Also, to further understand advanced topics like energy generation, nuclear reactions, and medical applications of radioactivity. Students should concentrate on solving important numerical problems related to decay laws, binding energy, and mass-energy equivalence, which are frequently asked in board and competitive exams.

The following are the topics covered in the Class 12 Physics Chapter 13 Nuclei:

Exercise	Topics Covered
13.1	Introduction To Nuclei
13.2	Atomic Masses And Composition Of Nucleus
13.3	Size Of The Nucleus
13.4	Mass-Energy And Nuclear Binding Energy
13.5	Nuclear Force
13.6	Radioactivity
13.7	Nuclear Energy

These are the important topics covered in Nuclei:

Nuclear composition (protons, neutrons, nucleons)
Isotopes, isobars, and isotones
Nuclear size and constant density
Mass defect and binding energy
Nuclear force characteristics
Discovery of the neutron
Radioactivity (alpha, beta, gamma decay)
Nuclear reactions (fission and fusion)

Nuclei Weightage in JEE Main, NEET Exams

Exam Name	No. of Questions	Percentage
NEET	2-3 questions	5%
JEE Main	2 questions	8%

NCERT Physics Class 12th Solution PDF - Nuclei Chapter Download

Get access to the comprehensive Nuclei NCERT PDF. By downloading this PDF, students will be able to access the free, detailed answers and simplified explanations to master the nuclear and radioactive concepts. It is a great material for quick revision for examinations.

For Class 12 Physics Chapter-wise Questions with Answers PDF, Important Topics & Weightage, read here - NCERT Solutions for Class 12 Physics.

NCERT Solution Class 12 Physics Ch 13 Nuclei PDF: Download Free PDF

Class 12 Physics Chapter 13 Nuclei NCERT Questions and Answers

Q.13.1 (a) Two stable isotopes of lithium ${L i}_{3}^{6}$ and ${L i}_{3}^{7}$ 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, ${}_{5}^{10}B$ and ${}_{5}^{11}B$ . 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 ${}_{5}^{10}B$ and ${}_{5}^{11}B$ 10.

Ans.13.1 Mass of ${}_{3}^{6}{L i}$ lithium isotope, $m_{1}$ = 6.01512 u

Mass of ${}_{3}^{7}{L i}$ lithium isotope, $m_{2}$ = 7.01600 u

Abundance of ${}_{3}^{6}{L i}$ , $η_{1}$ = 7.5%

Abundance of ${}_{3}^{7}{L i}$ , $η_{2}$ = 92.5%

The atomic mass of lithium atom is given as:

m = $\frac{m_{1} η_{1} + m_{2 η_{2}}}{η_{1} + η_{2}}$ = $\frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{7.5 + 92.5}$ = 6.940934 u

Mass of ${}_{5}^{10}B$ Boron isotope, $m_{1}$ = 10.01294 u

Mass of ${}_{5}^{11}B$ Boron isotope, $m_{2}$ = 11.00931 u

Let the abundance of ${}_{5}^{10}B$ be x % and that of ${}_{5}^{11}B$ be (100-x) %

The atomic mass of Boron atom is given as :

10.8111 = $\frac{10.01294 x + 11.00931 (100 - x)}{x + (100 - x)}$

1081.11 = 1100.931 - 0.99637x

x = 19.89 %

Hence the abundance of ${}_{5}^{10}B$ is 19.89 % and that of ${}_{5}^{11}B$ is (100-19.89) = 80.11 %

Q.13.2 The three stable isotopes of neon: ${}_{10}^{20}{N e},$ ${}_{10}^{21}{N e}$ and ${}_{10}^{22}{N e}$ 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.

Ans.13.2 Atomic mass of ${}_{10}^{20}{N e}$ neon isotope, $m_{1}$ = 19.99 u ad the abundance $η_{1}$ = 90.51 %

Atomic mass of ${}_{10}^{21}{N e}$ neon isotope, $m_{2}$ = 20.99 u ad the abundance $η_{2}$ = 0.27 %

Atomic mass of ${}_{10}^{22}{N e}$ neon isotope, $m_{3}$ = 21.99 u ad the abundance $η_{3}$ = 9.22 %

The average atomic mass of neon is given as:

m = $\frac{m_{1} η_{1} + m_{2 η_{2}} + m_{3} η_{3}}{η_{1} + η_{2} + η_{3}}$ = $\frac{19.99 \times 90.51 + 20.99 \times 0.27 + 21.99 \times 9.22}{90.51 + 0.27 + 9.22}$ = $\frac{2017.71}{100}$ = 20.1771 u

Q.13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given m( ${}_{7}^{14}{N)}$ =14.00307 u

Ans.13.3 Atomic mass of ${}_{7}^{14}N$ nitrogen , m = 14.00307 u

A nucleus of ${}_{7}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7 $m_{p}$ + 7 $m_{n}$ - m, where

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Therefore, Δm = 7 $\times$ 1.007825+ 7 $\times$ 1.008665 – 14.00307 = 0.11236 u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 104.66334 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (104.66334/ $c^{2}$ ) $\times$ $c^{2}$ = 104.66334 MeV

Q.13.4 Obtain the binding energy of the nuclei ${}_{26}^{56}{F e}$ and ${}_{83}^{209}{B i}$ in units of MeV from the following data:

m ( ${}_{26}^{56}{F e})$ = 55.934939 u m ( ${}_{83}^{209}{B i}$ ) = 208.980388 u

Ans.13.4 Atomic mass of ${}_{26}^{56}{F e}$ , $m_{1}$ = 55.934939 u

${}_{26}^{56}{F e}$ has 26 protons and (56-26) 30 neutrons

Hence the mass defect of the nucleus Δm = 26 $\times m_{p}$ + 30 $\times m_{n}$ - $m_{1}$

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Δm = 26 $\times 1.007825$ + 30 $\times 1.008665$ - 55.934939

Δm = 0.528461 u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 492.2614215 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (492.2614215 / $c^{2}$ ) $\times$ $c^{2}$ = 492.2614215 MeV

Average binding energy per nucleon = $\frac{E_{b}}{56}$ = 8.79 MeV

Atomic mass of ${}_{83}^{209}{B i}$ , $m_{2}$ = 208.980388 u

${}_{83}^{209}{B i}$ have 83 protons and (209-83) 126 neutrons

Hence the mass defect of the nucleus Δm = 83 $\times m_{p}$ + 126 $\times m_{n}$ - $m_{2}$

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Δm = 83 $\times 1.007825$ + 126 $\times 1.008665$ - 208.980388

Δm = 1.760877 u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 1640.256926 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (1640.256926 / $c^{2}$ ) $\times$ $c^{2}$ = 1640.256926 MeV

Average binding energy per nucleon = $\frac{E_{b}}{209}$ = 7.848 MeV

Commonly asked questions

13.21 From the relation R = $R_{0} A^{1 / 3}$
, where $R_{0}$ is a constant and A is the mass number of a nucleus, shows that the nuclear matter density is nearly constant (i.e. independent of A).

13.21 We have the expression for nuclear radius as:

R = $R_{0} A^{1 / 3}$

Where $R_{0}$ = constant

A = mass number of nucleus

Let m be the average mass of the nucleus, hence mass of the nucleus = mA

Nuclear matter density $ρ$ can be written as

$ρ = \frac{M a s s o f t h e n u c l e u s}{V o l u m e o f t h e n u c l e u s}$ = $\frac{m A}{\frac{4}{3} π R^{3}}$ = $\frac{3 m A}{4 π {(R_{0} A^{\frac{1}{3}})}^{3}}$ = $\frac{3 m A}{4 π R_{0}^{3} A}$ = $\frac{3 m}{4 π R_{0}^{3}}$

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

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 ${}_{6}^{14}C$ present with the stable carbon isotope ${}_{6}^{12}C$ . 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 ${}_{6}^{14}C$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of ${}_{6}^{14}C$ dating used in archaeology. Suppose a specimen from Mohenjo-Daro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization.

13.8 Decay rate of living carbon-containing matter, R = 15 decay / min

Half life of ${}_{6}^{14}C$ , $T_{1 / 2}$ = 5730 years

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 a normal carbon-containing matter.

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

We can relate the decay constant, $λ$ and time t as:

$\frac{N}{N'}$ = $\frac{R'}{R}$ = $e^{- λ t}$

$e^{- λ t} =$ $\frac{R'}{R}$ = $\frac{9}{15}$ = $\frac{3}{5}$

By taking log (ln) on both sides,

$- λ t =$ $\log_{e} ? 3$ - $\log_{e} ? 5$

t = $\frac{0.5108}{λ}$

Since $λ$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{5730}$

t = $\frac{5730 \times 0.5108}{0.693}$ = 4223.5 years

Hence, the approximate age of the Indus-valley is 4223.5 years.

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

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

13.19 The given fusion reaction is

${}_{1}^{2}H$ + ${}_{1}^{2}H$ $\to$ ${}_{2}^{3}{H e}$ + n+3.27 MeV

Amount of deuterium, m = 2 kg

1 mole, i.e. 2 g of deuterium contains 6.023 $\times 10^{23}$
atoms

Hence 2 kg of deuterium contains =
$\frac{6.023 \times 10^{23}}{2}$ $\times 2 \times 10^{3}$ atoms = 6.023 $\times 10^{26}$ atoms

It can be inferred from the fission reaction that when 2 atoms of deuterium fuse, 3.27 MeV of energy is released.

Hence total energy released from 2 kg of deuterium, E =
$\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26}$ MeV

= $\frac{3.27}{2}$ $\times$ 6.023 $\times 10^{26} \times 10^{6} \times 1.6 \times 10^{- 19}$ J = 1.5756 $\times 10^{14}$ J

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

Hence energy consumed by the bulb per second = 100 J

Therefore, total time the electric bulb will glow = $\frac{1.5756 \times 10^{14}}{100}$ seconds = 1.5756 $\times 10^{12}$ secs

= 49.96 $\times 10^{3}$ years

13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{47}^{197}{A u}$ and the silver isotope ${}_{47}^{107}{A g}$ .

13.11 Nuclear radius of the gold isotope, ${}_{47}^{197}{A u}$ = $R_{A u}$

Nuclear radius of silver isotope, ${}_{47}^{107}{A g}$ = $R_{A g}$

Mass number of gold, $A_{A u}$ = 197

Mass number of silver, $A_{A g}$ = 107

The ratio of the radii of the two nuclei is related with their mass numbers as :

$\frac{R_{A u}}{R_{A g}}$ = ${[\frac{A_{A u}}{A_{A g}}]}^{1 / 3}$ = ${[\frac{197}{107}]}^{1 / 3}$ =1.2256

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

13.14 The nucleus ${}_{10}^{23}{N e}$ decays by $β^{-}$ emission. Write down the $β$ -decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

m ( ${}_{10}^{23}{N e})$ = 22.994466 u

m ( ${}_{11}^{23}{N a)}$ = 22.989770 u.

13.14 In $β^{-}$ emission, the number of protons increase by 1 and one electron and an antineutrino are emitted from the parent nucleus.

$β^{-}$ emission of the nucleus ${}_{10}^{23}{N e}$ :

${}_{10}^{23}{N e}$ $\to {}_{11}^{23}{N a}$ + $e^{-}$ + $\overset{?}{ν}$ + Q

It is given that:

Atomic mass m ( ${}_{10}^{23}{N e})$ = 22.994466 u

Atomic mass m ( ${}_{11}^{23}{N a)}$ = 22.989770 u

Mass of an electron, $m_{e}$ = 0.000548 u

Q value of the given reaction is given as Q = $[m ({}_{10}^{23}{N e}) - {m ({}_{11}^{23}{N a)} + m_{e}}] c^{2}$

There are 10 electrons in ${}_{10}^{23}{N e}$ and 11 electrons in ${}_{11}^{23}{N a}$ . Hence, the mass of the electron is cancelled in the Q-value equation.

Therefore Q = {22.994466 - 22.989770} $c^{2}$ = 4.696 $\times 10^{- 3} c^{2}$ u

But 1 u = 931.5 MeV/ $c^{2}$

Q = 4.374 MeV

The daughter nucleus is too heavy as compared to $e^{-}$ and $\overset{?}{ν}$ . 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.

13.17 The fission properties of ${}_{94}^{239}{P u}$ are very similar to those of ${}_{92}^{235}U$ . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ${}_{94}^{239}{P u}$ undergo fission?

13.17 The average energy released per fission of ${}_{94}^{239}{P u}$ $E_{a v g}$ = 180 MeV

Amount of pure ${}_{94}^{239}u$ , m = 1 kg = 1000 g

Avogadro's number,
$N_{A}$ = 6.023 $* 10^{23}$

Mass number of ${}_{94}^{239}{P u}$ = 239 gm

Hence, number of atoms in 1000 g ${}_{94}^{239}{P u}$ , N = $\frac{6.023 * 10^{23}}{239}$ $* 1000$ = 2.52 $* 10^{24}$

Total energy released during the fission of 1 kg of ${}_{94}^{239}{P u}$ , E = $E_{a v g}$ $*$ N

= 180 $*$ 2.52 $* 10^{24}$ MeV = 4.536 $* 10^{26}$ MeV

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.)

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

d = Radius of 1st deuteron + Radius of 2nd deuteron

Radius of the deuteron nucleus = 2 fm = 2 $\times 10^{- 15}$ m

Hence d = 2 $\times 10^{- 15} + 2 \times 10^{- 15}$ = 4 $\times 10^{- 15}$ m

Charge on a deuteron nucleus = Charge on an electron = 1.6 $\times 10^{- 19}$ C

Potential energy of two-deuteron system: $V$ = $\frac{e^{2}}{4 π ?_{0} d}$

Where $?_{0}$ = permittivity of free space

$\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 1}$

V = $\frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times 10^{- 15}}$ J = $\frac{{(1.6 \times 10^{- 19})}^{2} \times 9 \times 10^{9}}{4 \times 10^{- 15} \times 1.6 \times 10^{- 19}}$ eV = 360 $\times 10^{3}$
eV = 360 keV

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

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 ${}_{29}^{63}{C u}$ atoms (of mass 62.92960 u).

13.5 Mass of the copper coin, m’ = 3.0 g

Atomic mass of ${}_{29}^{63}{C u}$ , m = 62.92960 u

The total number of ${}_{29}^{63}{C u}$ atoms in the coin, N = $\frac{N_{A} \times m'}{M a s s n u m b e r}$ , where

$N_{A}$ = Avogadro’s number = 6.023 $\times 10^{23}$ atoms / g

Mass number = 63 g

Therefore, N = $\frac{6.023 \times 10^{23} \times 3}{63}$ = 2.868 $\times 10^{22}$ atoms

${}_{29}^{63}{C u}$ has 29 protons and (63 – 29) 34 neutrons

Hence the mass defect of the nucleus Δm = 29 $\times m_{p}$ + 34 $\times m_{n}$ - $m$

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Δm = 29 $\times 1.007825$ + 34 $\times 1.008665$ - 62.92960

Δm = 0.591935 u

Mass defect of all the atoms present in the coin, Δm = 0.591935 $\times N$

= 0.591935 $\times$ 2.868 $\times 10^{22}$ = 1.69766958 $\times 10^{22}$ u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 1.581 $\times 10^{25}$ MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (1.581 $\times 10^{25}$ / $c^{2}$ ) $\times$ $c^{2}$ = 1.581 $\times 10^{25}$ MeV

1 MeV = 1.6 $\times 10^{- 13}$ J

Hence, $E_{b} =$ 2.530 $\times 10^{12}$ J

This much of energy required to separate all the neutrons and protons from the given coin.

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

m( ${}^{198}{A u}$ ) = 197.968233 u

m( ${}^{198}{H g}$ ) =197.966760 u

13.29 It can be observed from the given $γ -$ decay diagram that $γ_{1}$ decays from 1.088 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to decay is given as:

= 1.088 – 0 = 1.088 MeV = 1.088 $\times 10^{6}$ eV = 1.088 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 1.7408 $\times 10^{- 13}$ J

We know, $E_{1} = h ν_{1}$ , where

$ν_{1}$ = Frequency of radiation radiated by $γ_{1}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{1}$ = $\frac{E_{1}}{h}$ = $\frac{1.7408 \times 10^{- 13}}{6.6 \times 10^{- 34}}$ = 2.637 $\times 10^{20}$ Hz

It can be observed from the given $γ -$ decay diagram that $γ_{2}$ decays from 0.412 MeV energy level to the 0 MeV energy level.

Hence the energy corresponding to $γ_{2}$ decay is given as:

$E_{2}$ = 0.412 – 0 = 0.412 MeV = 0.412 $\times 10^{6}$ eV = 0.412 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 6.592 $\times 10^{- 14}$ J

We know, $E_{2} = h ν_{2}$ , where

$ν_{2}$ = Frequency of radiation radiated by $γ_{2}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{2}$ = $\frac{E_{2}}{h}$ = $\frac{6.592 \times 10^{- 14}}{6.6 \times 10^{- 34}}$ = 9.987 $\times 10^{19}$ Hz

It can be observed from the given $γ -$ decay diagram that $γ_{3}$ decays from 1.008 MeV energy level to the 0.412 MeV energy level.

Hence the energy corresponding to decay is given as:

= 1.088 - 0.412 = 0.676 MeV = 0.676 $\times 10^{6}$ eV = 0.676 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

= 1.0816 $\times 10^{- 13}$ J

We know, $E_{3} = h ν_{3}$ , where

$ν_{3}$ = Frequency of radiation radiated by $γ_{3}$ decay

$h = P l a n c k^{'} s c o n s t a n t$ = 6.6 $\times 10^{- 34}$ Js

Hence, $ν_{3}$ = $\frac{E_{3}}{h}$ = $\frac{1.0816 \times 10^{- 13}}{6.6 \times 10^{- 34}}$ = 1.639 $\times 10^{20}$ Hz

Mass of ( ${}_{79}^{198}{A u}$ ) = 197.968233 u

Mass of ( ${}_{79}^{198}{A u}$ ) =197.966760 u

Energy of the highest level is given as:

E = $[m ({}_{79}^{198}{A u}) - m ({}_{80}^{198}{H g})] c^{2}$

= $[197.968233 - 197.966760] c^{2}$ u

= 1.473 $\times 10^{- 3} c^{2}$ u

= 1.473 $\times 10^{- 3} \times 931.5$ MeV

= 1.3720995 MeV

$β_{1}^{-} d e c a y s$ from 1.3720995 MeV level to 1.088 MeV level

Hence maximum kinetic energy of the $β_{1}^{-}$ particles = 1.3720995 - 1.088 = 0.2840995 MeV

$β_{2}^{-}$ from 1.3720995 MeV level to 0.412 MeV level

Hence maximum kinetic energy of the $β_{2}^{-}$ particles = 1.3720995 – 0.412 = 0.9600995 MeV

13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus ( ${}_{7}^{14}{N)}$ , given

m( ${}_{7}^{14}{N)}$ =14.00307 u

13.3 Atomic mass of ${}_{7}^{14}N$ nitrogen , m = 14.00307 u

A nucleus of ${}_{7}^{14}N$ nitrogen contains 7 protons and 7 neutrons.

Hence, the mass defect of this nucleus, Δm = 7 $m_{p}$ + 7 $m_{n}$ - m, where

Mass of a proton, $m_{p}$ = 1.007825 u

Mass of a neutron, $m_{n}$ = 1.008665 u

Therefore, Δm = 7 $\times$ 1.007825+ 7 $\times$ 1.008665 – 14.00307 = 0.11236 u

But 1 u = 931.5 MeV/ $c^{2}$

Δm = 104.66334 MeV/ $c^{2}$

The binding energy of the nucleus, $E_{b}$ = Δm $c^{2}$ , where c = speed of light

$E_{b} =$ (104.66334/ $c^{2}$ ) $\times$ $c^{2}$ = 104.66334 MeV

13.2 The three stable isotopes of neon: ${}_{10}^{20}{N e},$ ${}_{10}^{21}{N e}$ and ${}_{10}^{22}{N e}$ 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.

13.2 Atomic mass of ${}_{10}^{20}{N e}$ neon isotope, $m_{1}$ = 19.99 u ad the abundance $η_{1}$ = 90.51 %

Atomic mass of ${}_{10}^{21}{N e}$ neon isotope, $m_{2}$ = 20.99 u ad the abundance $η_{2}$ = 0.27 %

Atomic mass of ${}_{10}^{22}{N e}$ neon isotope, $m_{3}$ = 21.99 u ad the abundance $η_{3}$ = 9.22 %

The average atomic mass of neon is given as:

13.16 Suppose, we think of fission of a ${}_{26}^{56}{F e}$ nucleus into two equal fragments, ${}_{13}^{28}{A l}$ . Is the fission energetically possible? Argue by working out Q of the process. Given

m ( ${}_{26}^{56}{F e})$ = 55.93494 u and m ( ${}_{13}^{28}{A l})$ = 27.98191 u.

13.16 The fission can be shown as:

${}_{26}^{56}{F e} \to 2 {}_{13}^{28}{A l}$

It is given that atomic mass

m ( ${}_{26}^{56}{F e})$ = 55.93494 u

m ( ${}_{13}^{28}{A l})$ = 27.98191 u

The Q-value of this reaction is given as:

Q = $[m ({}_{26}^{56}{F e}) - 2 m ({}_{13}^{28}{A l})] c^{2}$

= $[55.93494 - 2 \times 27.98191] c^{2}$

= -0.02888 $c^{2}$

= -0.02888 $\times 931.5$ MeV

= - 26.902 MeV

The Q value of the fission is negative, therefore the fission is not possible energetically. Q value needs to be positive for a fission.

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%, b) 1% of its original value?

13.7 Half life of the radioactive isotope = T years

Original amount of the radioactive isotope = $N_{o}$

After decay, the amount of radioactive isotope = N

It is given that only 3.125% of $N_{o}$ remains after decay. Hence, we can write,

$\frac{N}{N_{o}}$ = 3.125% = $\frac{3.125}{100}$ = $\frac{1}{32}$

But $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{32}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{32}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 32$

$- λ t$ = 0 – 3.465

$t$ = $\frac{3.465}{λ}$

Since $λ$ = $\frac{0.693}{T}$

t = $\frac{3.465}{\frac{0.693}{T}}$ = 5T years

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

After decay, the amount of radioactive isotope = N

It is given that only 1.0% of $N_{o}$ remains after decay. Hence, we can write,

$\frac{N}{N_{o}}$ = 1% = $\frac{1}{100}$

But $\frac{N}{N_{o}}$ = $e^{- λ t}$ , where $λ$ = decay constant, t = time

Therefore,

$e^{- λ t} = \frac{1}{100}$

By taking log on both sides

$\log_{e} ? e^{- λ t}$ = $\log_{e} ? \frac{1}{100}$

$- λ t =$ $\log_{e} ? 1$ - $\log_{e} ? 100$

$- λ t$ = 0 – 4.605

$t$ = $\frac{4.605}{λ}$

Since $λ$ = $\frac{0.693}{T}$

t = $\frac{4.605}{\frac{0.693}{T}}$ = 6.645T years

Hence, all the isotopes will take about 6.645T years to reduce 1.0% of its original value.

13.9 Obtain the amount of ${}_{27}^{60}{C o}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of ${}_{27}^{60}{C o}$ is 5.3 years.

13.9 The strength of the radioactive source is given as:

$\frac{d N}{d t}$ = 8.0 mCi = 8 $\times 10^{- 3}$ $\times 3.7 \times 10^{10}$ decay/s = 296 $\times 10^{6}$ decay/s, where

N = Required number of atoms

Given, half life of ${}_{27}^{60}{C o}$ , $T_{1 / 2}$ = 5.3 years = 5.3 $\times 365 \times 24 \times 60 \times 60$ secs = 167 $\times 10^{6}$ s

For decay constant $λ$ , we have rate of decay as:

$\frac{d N}{d t}$ = $λ N$ or

N = $\frac{1}{λ} \frac{d N}{d t}$ , where $λ$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{167 \times 10^{6}}$ /s = 4.1497 $\times 10^{- 9}$ $s^{- 1}$

N = $\frac{296 \times 10^{6}}{4.1497 \times 10^{- 9}}$ = 7.133 $\times 10^{16}$ atoms

For ${}_{27}^{60}{C o}$ , mass of 6.023 $\times 10^{23}$ atoms = 60 gms

Therefore, the mass of 7.133 $\times 10^{16}$ atoms = $\frac{60}{6.023 \times 10^{23}} \times$ 7.133 $\times 10^{16}$ gms = 7.106 $\times 10^{- 6}$ g

13.10 The half-life of ${}_{38}^{90}{S r}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

13.10 Half life of ${}_{38}^{90}{S r}$ , $T_{1 / 2}$ = 28 years = 28 $\times 365 \times 24 \times 60 \times 60$ secs = 0.883 $\times 10^{9}$ s

Mass of the isotope, m = 15 mg = 15 $\times 10^{- 3}$ gms

90 g of ${}_{38}^{90}{S r}$ contains 6.023 $\times 10^{23}$ atoms

No. of atoms in 15 mg of ${}_{38}^{90}{S r}$ contains = $\frac{6.023 \times 10^{23}}{90} \times$ 15 $\times 10^{- 3}$ = 1.0038 $\times 10^{20}$

Rate of disintegration $\frac{d N}{d t}$ = $? N$ , where $?$ = $\frac{0.693}{T_{1 / 2}}$ = $\frac{0.693}{0.883 \times 10^{9}}$ /s = 7.848 $\times 10^{- 10}$ $s^{- 1}$

$\frac{d N}{d t}$ = 7.848 $\times 10^{- 10} \times$ 1.0038 $\times 10^{20}$ = 7.878 $\times 10^{10}$ atoms / second.

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

13.18 Half life of the fuel in the fission reactor, $T_{1 / 2}$
= 5 years = 5 $\times 365 \times 24 \times 60 \times 60 s$

= 157.68 $\times 10^{6} s$

We know that in the fission of 1 g of ${}_{92}^{235}U$
, the energy released = 200 MeV

1 mole i.e. 235 gm of ${}_{92}^{235}U$ contains 6.023 $\times 10^{23}$ atoms

Therefore 1 gm of ${}_{92}^{235}U$
contains = $\frac{6.023 \times 10^{23}}{235}$ = 2.563 $\times 10^{21}$ atoms

The total energy Q generated per gm of ${}_{92}^{235}U$ = 200 $\times$ 2.563 $\times 10^{21}$ MeV/g = 5.126 $\times 10^{23}$
MeV/g = 5.126 $\times 10^{23} \times 1.6 \times 10^{- 19} \times 10^{6}$ J/g = 8.20 $\times 10^{10}$
J/g

Since the reactor operates only 80% of the time, hence the amount of ${}_{92}^{235}U$
in 5 years is given by $\frac{0.8 \times 157.68 \times 10^{6} \times 1000 \times 10^{6}}{8.20 \times 10^{10}}$ = 1538 kg

Hence, initial amount of fuel = 2 $\times 1538 k g$ = 3076 kg

13.22 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).

$e^{+}$ + ${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + v}$

Show that if $β^{+ +}$ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa.

13.22 Let the amount on energy released during the electron capture process be $Q_{1}$ . The nuclear reaction can be written as:

$e^{+}$ + ${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + v} + Q_{1}$ ……………………..(1)

Let the amount of energy released during the positron capture process be $Q_{2}$ . The nuclear reaction can be written as:

${}_{Z}^{A}X$ $\to {}_{z - 1}^{A}{Y + e^{+} + v}$ + $Q_{2}$ …………………….(2)

Let us assume

$m_{N} ({}_{Z}^{A}X)$ = Nuclear mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z}^{A}X)$ = Atomic mass of ${}_{Z}^{A}X$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Nuclear mass of ${}_{Z - 1}^{A}Y$

$m_{N} ({}_{Z - 1}^{A}Y)$ = Atomic mass of ${}_{Z - 1}^{A}Y$

$m_{e}$ = mass of the electron

c = speed of light

Q-value of the electron capture reaction is given as:

$Q_{1}$ = [ $m_{N} ({}_{Z}^{A}X) + m_{e} - m_{N} ({}_{Z - 1}^{A}Y)] c^{2}$

= [ $m ({}_{Z}^{A}X) - Z m_{e} + m_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y)] c^{2}$ ………………….(3)

Q-value of the positron capture reaction is given as:

$Q_{2}$ = [ $m_{N} ({}_{Z}^{A}X) - m_{N} ({}_{Z - 1}^{A}Y) - m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - Z m_{e} - m ({}_{Z - 1}^{A}Y) + (Z - 1) m_{e} - m_{e}] c^{2}$

= [ $m ({}_{Z}^{A}X) - m ({}_{Z - 1}^{A}Y) - 2 m_{e}] c^{2}$ ……………(4)

It can be inferred that if $Q_{2} > 0$ , then $Q_{1} > 0$ , also if $Q_{1} > 0, i t d o e s n o t n e c e s s a r i l y$ means that $Q_{2} > 0$ .

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 Q-value must be positive for an energetically-allowed nuclear reaction.

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 ${}_{12}^{24}{M g}$ (23.98504u), ${}_{12}^{25}{M g}$

(24.98584u) and ${}_{12}^{26}{M g}$ (25.98259u). The natural abundance of ${}_{12}^{24}{M g}$ is 78.99% by mass. Calculate the abundances of other two isotopes.

13.23 Average atomic mass of magnesium, m = 24.312 u

Mass of magnesium isotope ${}_{12}^{24}{M g}$ , $m_{1}$ = 23.98504u

Mass of magnesium isotope ${}_{12}^{25}{M g}$ , $m_{2}$ = 24.98584u

Mass of magnesium isotope ${}_{12}^{26}{M g}$ , $m_{3}$ = 25.98259u

Let the abundance of magnesium isotope ${}_{12}^{24}{M g}$ be $η_{1}$ = 78.99 %

Let the abundance of magnesium isotope ${}_{12}^{25}{M g}$ be $η_{2}$ = x %

Therefore, the abundance of magnesium isotope ${}_{12}^{26}{M g}$ be $η_{3}$ = (100 - 78.99 - x) %

= (21.01 – x)%

The average atomic mass can be expressed as:

m = $\frac{m_{1} η_{1} + m_{2} η_{2} + m_{3} η_{3}}{η_{1} + η_{2} + η_{3}}$ = $\frac{23.98504 \times 78.99 + 24.98584 x + 25.98259 (21.01 - x)}{100}$ = $\frac{1894.578 + 24.98584 x + 545.894 - 25.98259 x}{100}$ = $\frac{2440.472 - 0.99675 x}{100}$

24.312 = $\frac{2440.472 - 0.99675 x}{100}$

x = 9.3%

Therefore the abundance of ${}_{12}^{25}{M g}$ is 9.3% and the abundance of ${}_{12}^{26}{M g}$ is (21.01 – 9.3)11.71%

13.25 A source contains two phosphorous radio nuclides ${}_{15}^{32}P {, T}_{1 / 2}$ = 14.3d and ${}_{15}^{33}P, T_{1 / 2}$ = 25.3d. Initially, 10% of the decays come from ${}_{15}^{33}P$ . How long one must wait until 90% do so?

13.25 Half life of ${}_{15}^{32}P {, T}_{1 / 2}$ = 14.3 days

Half life of ${}_{15}^{33}P, T_{1 / 2}$ = 25.3days and ${}_{15}^{33}P$ decay is 10% of the total amount of decay

The source has initially 10 % of ${}_{15}^{33}P$ nucleus and 90% of ${}_{15}^{32}P$ nucleus

Suppose after t days, the source has 10% of ${}_{15}^{32}P$ nucleus and 90% of ${}_{15}^{33}P$ nucleus

Initially:

Number of ${}_{15}^{33}P$ nucleus = N

Number of ${}_{15}^{32}P$ nucleus = 9N

Finally:

Number of ${}_{15}^{33}P$ nucleus = 9N’

Number of ${}_{15}^{32}P$ nucleus = N’

For ${}_{15}^{32}P$ nucleus, the number ratio, $\frac{N'}{9 N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

N’ = 9N ${(2)}^{- t / 14.3}$ …………….(1)

For ${}_{15}^{33}P$ nucleus, the number ratio, $\frac{9 N'}{N}$ = ${(\frac{1}{2})}^{t / T_{1 / 2}}$

9N’ = N ${(2)}^{- t / 25.3}$ …………….(2)

On dividing equation (1) by equation (2), we get:

$\frac{1}{9}$ = 9 $\times 2^{(\frac{t}{25.3} - \frac{t}{14.3})}$

$\frac{1}{81}$ = $2^{(\frac{- 11 t}{25.3 \times 14.3})}$

Taking log on both sides

log 1 – log 81 = $\frac{- 11 t}{25.3 \times 14.3}$ log2

0 – 1.908 = ( $\frac{- 11 t}{361.79}$ ) $\times 0.301$

t = 208.5 days

Hence, it will take about 208.5 days for 90% decay of $P_{15}^{33}$

13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of ${}_{92}^{235}U$ in a fission reactor.

13.30 Amount of hydrogen, m = 1 kg = 1000 g

1 mole of hydrogen, i.e. 1 g of hydrogen ( ${}_{1}^{1}{H)}$ contains 6.023 $\times 10^{23}$ atoms

1 kg of hydrogen contains = 1000 $\times$ 6.023 $\times 10^{23}$ atoms = 6.023 $\times 10^{26}$ atoms

Within Sun, four ( ${}_{1}^{1}{H)}$ nuclei combine and forms 1 ${}_{2}^{4}{H e}$ nucleus. In this process 26 MeV of energy is released.

Hence the energy released from fusion of 1 kg of ${}_{1}^{1}H$ is:

$E_{1}$ = $\frac{6.023 \times 10^{26} \times 26}{4}$ MeV = 3.91495 $\times 10^{27}$ MeV

Amount of ${}_{92}^{235}U$ , m = 1 kg = 1000 g

1 mole of ${}_{92}^{235}U$ , i.e. 235 g of ${}_{92}^{235}U$ contains 6.023 $\times 10^{23}$ atoms

1 kg of ${}_{92}^{235}U$ contains = $\frac{1000}{235}$ $\times$ 6.023 $\times 10^{23}$ atoms = 2.563 $\times 10^{24}$ atoms

It is known that the amount of energy released in the fission of 1 atom of ${}_{92}^{235}U$ is 200 MeV.

Hence the energy released from fusion of 1 kg of ${}_{92}^{235}U$ is:

= 2.563 $\times 10^{24} \times 200$ MeV = 5.126 $\times 10^{26}$ MeV

$\frac{E_{1}}{E_{2}}$ = $\frac{3.91495 \times 10^{27}}{5.126 \times 10^{26}}$ = 7.64

Therefore, the energy released in the fusion of 1 kg of hydrogen is nearly 8 times of the energy release during the fission of 1 kg of ${}_{92}^{235}U$ .

13.31 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 efficiency of utilization (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 ${}^{235}U$ to be about 200MeV.

13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $\times 10^{5}$ MW = 2 $\times 10^{4}$ MW = 2 $\times 10^{4} \times 10^{6}$ J/s

= 2 $\times 10^{4} \times 10^{6} \times 60 \times 60 \times 24 \times 365$ J/y = 6.3072 $\times 10^{17}$ J/y

Heat energy release per fission of a ${}^{235}U$ nucleus, E = 200 MeV

Efficiency of a reactor = 25%

So the amount of electrical energy converted from heat energy per fission = 25% of 200 MeV = 50 MeV

= 50 $\times 10^{6}$ eV = 50 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J = 8 $\times 10^{- 12}$ J

Therefore, number of atoms required per year = $\frac{6.3072 \times 10^{17}}{8 \times 10^{- 12}}$ = 7.884 $\times 10^{28}$

1 mole of ${}^{235}U$ = 235 gm of ${}^{235}U$ contains 6.023 $\times 10^{23}$ atoms

Hence the mass of 7.884 $\times 10^{28} a t o m s$ = $\frac{235 \times 10^{-3}}{6.023 \times 10^{23}}$ $\times$ 7.884 $\times 10^{28}$ = 30.76 $\times 10^{3}$ kg

Hence, the Uranium needed per year is 30.76 $\times 10^{3}$ kg

13.1 (a) Two stable isotopes of lithium ${}_{3}^{6}{L i}$ and ${}_{3}^{7}{L i}$ 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, ${}_{5}^{10}B$ and ${}_{5}^{11}B$ . 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 ${}_{5}^{10}B$ and ${}_{5}^{11}B$ 10.

13.1 Mass of ${}_{3}^{6}{L i}$ lithium isotope, $m_{1}$ = 6.01512 u

Mass of ${}_{3}^{7}{L i}$ lithium isotope, $m_{2}$ = 7.01600 u

Abundance of ${}_{3}^{6}{L i}$ , $η_{1}$ = 7.5%

Abundance of ${}_{3}^{7}{L i}$ , $η_{2}$ = 92.5%

The atomic mass of lithium atom is given as:

m = $\frac{m_{1} η_{1} + m_{2 η_{2}}}{η_{1} + η_{2}}$ = $\frac{6.01512 \times 7.5 + 7.01600 \times 92.5}{7.5 + 92.5}$ = 6.940934 u

Mass of ${}_{5}^{10}B$ Boron isotope, $m_{1}$ = 10.01294 u

Mass of ${}_{5}^{11}B$ Boron isotope, $m_{2}$ = 11.00931 u

Let the abundance of ${}_{5}^{10}B$ be x % and that of ${}_{5}^{11}B$ be (100-x) %

The atomic mass of Boron atom is given as :

10.8111 = $\frac{10.01294 x + 11.00931 (100 - x)}{x + (100 - x)}$

1081.11 = 1100.931 - 0.99637x

x = 19.89 %

Hence the abundance of ${}_{5}^{10}B$ is 19.89 % and that of ${}_{5}^{11}B$ is (100-19.89) = 80.11 %

13.6 Write nuclear reaction equations for

(i) $α$ -decay of ${}_{88}^{226}{R a}$

(ii) $α$ -decay of ${}_{94}^{242}{P u}$

(iii) $β$ –-decay of ${}_{15}^{32}P$

(iv) $β$ –-decay of ${}_{83}^{210}{B i}$

(v) $β$ +decay of ${}_{6}^{11}C$

(vi) $β$ –-decay of ${}_{43}^{97}{T c}$

(vii) Electron capture of ${}_{54}^{120}{X e}$

13.6 In $α$ - decay, there is a loss of 2 protons and 4 neutrons. In every $β +$ decay, there is a loss of 1 proton and a 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.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

${}_{94}^{242}{P u}$ $\to$ ${}_{92}^{238}U$ + ${}_{2}^{4}{H e}$

${}_{15}^{32}P$ $\to$ ${}_{16}^{32}S$ + $e^{-}$ + $\overset{?}{ν}$

${}_{83}^{210}{B i}$ $\to$ ${}_{84}^{210}{P o}$ + $e^{-}$ + $\overset{?}{ν}$

${}_{6}^{11}C$ $\to$ ${}_{5}^{11}B$ + $e^{+}$ + $ν$

${}_{43}^{97}{T c}$ $\to$ ${}_{42}^{97}{M o}$ + $e^{+}$ + $ν$

${}_{54}^{120}{X e} \to$ ${}_{53}^{120}I$ + $ν$

13.12 Find the Q-value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of (a) ${}_{88}^{226}{R a}$ and (b) ${}_{86}^{220}{R n}$ . Given m ( ${}_{88}^{226}{R a})$ = 226.02540 u, m ( ${}_{86}^{222}{R n})$ = 222.01750 u, m ( ${}_{86}^{222}{R n})$ = 220.01137 u, m ( ${}_{84}^{216}{P o)}$ = 216.00189 u.

13.12 $α -$ particle decay of ${}_{88}^{226}{R a}$ emits a helium nucleus. As a result, its mass number reduces to (226-4) 222 and its atomic number reduces to (88-2) 86.

${}_{88}^{226}{R a}$ $\to$ ${}_{86}^{222}{R n}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $\times c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{88}^{226}{R a})$ = 226.02540 u

m ( ${}_{86}^{222}{R n})$ = 222.01750 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(226.02540) – (222.01750 + 4.002603)] $c^{2}$

= 5.297 $\times 10^{- 3} {u c}^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Hence Q = 4.934 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $\times Q$ = $\frac{222}{226}$ $\times$ 4.934= 4.85 MeV

$α -$ particle decay of ${}_{86}^{220}{R n}$ emits a helium nucleus. As a result, its mass number reduces to (220-4) 216 and its atomic number reduces to (86-2) 84.

${}_{86}^{220}{R n}$ $\to$ ${}_{84}^{216}{P o}$ + ${}_{2}^{4}{H e}$

Q value of emitted $α -$ particle = (Sum of initial mass – Sum of final mass) $\times c^{2}$ , where

c = Speed of light.

It is given that

m ( ${}_{86}^{222}{R n})$ = 220.01137 u

m ( ${}_{84}^{216}{P o)}$ = 216.00189 u

m ( ${}_{2}^{4}{H e})$ = 4.002603 u

Q value = [(220.01137) – (216.00189 + 4.002603)] $c^{2}$

= 6.877 $\times 10^{- 3} {u c}^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Hence Q = 6.406 MeV

Kinetic energy of the $α -$ particle = $\frac{M a s s n u m b e r a f t e r d e c a y}{M a s s n u m b e r b e f o r e d e c a y}$ $\times Q$ = $\frac{216}{222}$ $\times$ 6.406= 6.23 MeV

13.13 The radionuclide ${}_{6}^{11}C$ decays according to ${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v :}} T_{1 / 2}$ =20.3 min. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

13.13 The given values are

m ( ${}_{6}^{11}{C)}$ = 11.011434 u and m ( ${}_{5}^{11}B$ ) = 11.009305 u

The given nuclear reaction:

${}_{6}^{11}{C \to {}_{5}^{11}{B + e^{+} + v}}$

Half life of ${}_{6}^{11}C$ nuclei, $T_{1 / 2}$ =20.3 min

The maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (ΔQ) of the nuclear masses of the ${}_{6}^{11}C$

ΔQ = $[m' ({}_{6}^{11}{C)} - {m' ({}_{5}^{11}B) + m_{e}}] c^{2} \dots \dots \dots \dots \dots (1)$

where

$m_{e}$ = Mass of an electron or positron = 0.000548 u

c = speed of the light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 $m_{e}$ in the case of ${}_{6}^{11}C$ and 5 $m_{e}$ in the case of ${}_{5}^{11}B$ .

Hence the equation (1) reduces to

ΔQ = $[m ({}_{6}^{11}{C)} - m ({}_{5}^{11}B) - {2 m}_{e}] c^{2}$

= $[11.011434 - 11.009305 - 2 \times 0.000548] c^{2}$ u

=1.033 $\times 10^{- 3} c^{2}$ u

But 1 u = 931.5 MeV/ $c^{2}$

ΔQ = 0.962 MeV

The value of $Δ Q$ is almost comparable to the maximum energy of the emitted positron.

13.15 The Q value of a nuclear reaction A + b $\to$ 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) ${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

(ii) ${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic masses are given to be

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

13.15 The given nuclear reaction is

${}_{1}^{1}H$ + ${}_{1}^{3}H$ $\to$ ${}_{1}^{2}H$ + ${}_{1}^{2}H$

Atomic mass

m ( ${}_{1}^{1}H$ ) = 1.007825 u

m ( ${}_{1}^{2}H$ ) = 2.014102 u

m ( ${}_{1}^{3}H$ ) = 3.016049 u

The Q-value of the reaction can be written as:

Q = $[m ({}_{1}^{1}H) + m ({}_{1}^{3}H) - 2 m ({}_{1}^{2}H)] c^{2}$

= $[1.007825 + 3.016049 - 2 \times 2.014102] c^{2}$

= (-4.33 $\times 10^{- 3}$ ) $c^{2}$

But 1 u = 931.5 MeV/ $c^{2}$

Q = -4.0334 MeV

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

The given nuclear reaction is

${}_{6}^{12}C$ + ${}_{6}^{12}C$ $\to$ ${}_{10}^{20}{N e}$ + ${}_{2}^{4}{H e}$

Atomic mass

m ( ${}_{6}^{12}C$ ) = 12.000000 u

m ( ${}_{10}^{20}{N e}$ ) = 19.992439 u

m ( ${}_{2}^{4}{H e}$ ) = 4.002603 u

The Q-value of this reaction is given as:

Q = $[2 m ({}_{6}^{12}C) - m ({}_{10}^{20}{N e}) - m ({}_{2}^{4}{H e})] c^{2}$

= $[2 \times 12.0 - 19.992439 - 4.002603] c^{2}$

=4.958 $\times 10^{- 3} c^{2}$ u

=4.958 $\times 10^{- 3} \times 931.5$

=4.6183 MeV

The positive sign shows that the reaction is exothermic.

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 ${}_{20}^{41}{C a}$ and from the following data:

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

13.24 If a neutron ${}_{0}^{1}n)$ is removed from ${}_{20}^{41}{C a}$ , the corresponding reaction can be written as:

${}_{20}^{41}{C a} \to$ ${}_{20}^{40}{C a}$ + ${}_{0}^{1}n$

The separation energies are

For ${}_{20}^{41}{C a}$ : Separation energy = 8.363007 MeV

For ${}_{13}^{27}{A l}$ : Separation energy = 13.059 MeV

It is given that

m( ${}_{20}^{40}{C a)}$ = 39.962591 u

m( ${}_{20}^{41}{C a)}$ = 40.962278 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m ${}_{20}^{40}{C a)} +$ ( m( ${}_{0}^{1}n) -$ m( ${}_{20}^{41}{C a)}$

= 39.962591 + 1.008665 - 40.962278

= 8.978 $\times 10^{3}$ u

=8.363007 MeV

For ${}_{13}^{27}{A l}$ , the neutron removal reaction can be written as

${}_{13}^{27}{A l} \to$ ${}_{13}^{26}{A l}$ + ${}_{0}^{1}n$

It is given that

m( ${}_{13}^{26}{A l)}$ = 25.986895 u

m( ${}_{13}^{27}{A l)}$ = 26.981541 u

m( ${}_{0}^{1}n)$ = 1.008665 u

The mass defect of the reaction is given as:

Δm = m( ${}_{13}^{26}{A l)} +$ m( ${}_{0}^{1}n) -$ m( ${}_{13}^{27}{A l)}$

= 25.986895 + 1.008665 - 26.981541

= 0.014019 u

=13.0586985 MeV

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

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

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

13.26 For the emission of ${}_{6}^{14}C$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{82}^{209}{P b + {}_{6}^{14}C}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{82}^{209}{P b}$ , $m_{2}$ = 208.98107 u

Mass of ${}_{6}^{14}C$ , $m_{3}$ = 14.00324 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 208.98107 - 14.00324) $c^{2}$ u

= 0.03419 $c^{2}$ u

= 0.03419 MeV = 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.

For the emission of ${}_{2}^{4}{H e}$ , the nuclear reaction is:

${}_{88}^{223}{R a \to {}_{86}^{219}{R n + {}_{2}^{4}{H e}}}$

We know that:

Mass of ${}_{88}^{223}{R a}$ , $m_{1}$ = 223.01850 u

Mass of ${}_{86}^{219}{R n}$ , $m_{2}$ = 219.00948 u

Mass of ${}_{2}^{4}{H e}$ , $m_{3}$ = 4.00260 u

Hence, the Q-value of the reaction is given as:

Q = ( $m_{1}$ - $m_{2}$ - $m_{3}$ ) $c^{2}$

= (223.01850 - 219.00948 - 4.00260) $c^{2}$ u

= 6.42 $\times 10^{- 3} c^{2}$ u

= 6.42 $\times 10^{- 3} \times 931.5$ MeV = 5.980 MeV

Hence, the Q-value of the nuclear reaction is 5.98 MeV, since the value is positive, the reaction is energetically allowed.

13.27 Consider the fission of ${}_{92}^{238}U$ 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 ${}_{58}^{140}{C e}$ and ${}_{44}^{99}{R u})$ . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${}_{92}^{238}U)$ =238.05079 u

m( ${}_{58}^{140}{C e}$ ) =139.90543 u

m( ${}_{44}^{99}{R u})$ = 98.90594 u

13.27 In the fission of ${}_{92}^{238}U$ , $β^{-}$ 10 particles decay from the parent nucleus. The nuclear reaction can be written as:

${}_{92}^{238}U + {}_{0}^{1}{n \to {}_{58}^{140}{C e +}} {}_{44}^{99}{R u + 10 {}_{- 1}^{0}e}$

It is given that:

Mass of a ${}_{92}^{238}U n u c l e u s, m_{1}$ = 238.05079 u

Mass of a ${}_{58}^{140}{C e} n u c l e u s, m_{2}$ =139.90543 u

Mass of a ${}_{44}^{99}{R u} n u c l e u s, m_{3}$ = 98.90594 u

Mass of a neutron ${}_{0}^{1}n$ , $m_{4}$ = 1.00865 u

Q value of the above equation,

Q = $[m^{'} ({}_{92}^{238}U) + m' ({}_{0}^{1}{n) - m^{'} (} {}_{58}^{140}{C e) - m^{'} ({}_{44}^{99}{R u) - 10 m_{e}}}] c^{2}$

Where

m’ = represents the corresponding atomic masses of the nuclei.

$m^{'} ({}_{92}^{238}U)$ = $m_{1} - 92 m_{e}$

m’( ${}_{58}^{140}{C e)}$ = $m_{2} - 58 m_{e}$

m’( ${}_{44}^{99}{R u)}$ = $m_{3} - 44 m_{e}$

m’( ${}_{0}^{1}{n)} = m_{4}$

Substituting these values, we get

Q = $[m_{1} - 92 m_{e} + m_{4} - m_{2} + 58 m_{e} - m_{3} + 44 m_{e} - 10 m_{e}] c^{2}$

= $[m_{1} + m_{4} - m_{3} - m_{2}] c^{2}$

= $[238.05079 + 1.00865 - 98.90594 - 139.90543] c^{2}$ u

=0.24807 $c^{2}$ u

= 231.077 MeV

13.28 Consider the D–T reaction (deuterium–tritium fusion)

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

(a) Calculate the energy released in MeV in this reaction from the data:

m( ${}_{1}^{2}{H)}$ =2.014102 u

m( ${}_{1}^{3}{H)}$ =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 = Boltzmann’s constant, T = absolute temperature.)

13.28 The equation for deuterium-tritium fusion is given as:

${}_{1}^{2}{H + {}_{1}^{3}{H \to {}_{2}^{4}{H e + n}}}$

It is given that

Mass of ( ${}_{1}^{2}{H), m_{1}}$ = 2.014102 u

Mass of ( ${}_{1}^{3}{H)}$ , $m_{2} =$ 3.016049 u

Mass of ( ${}_{2}^{4}{H e), m_{3}}$ = 4.002603 u

Mass of ( ${}_{0}^{1}{n), m_{4}}$ = 1.008665 u

Q-value of the given D-T reaction is:

Q = $[m_{1} + m_{2} - m_{3} - m_{4}] c^{2}$

= $[2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} u$

= 0.018883 $c^{2} u$

= 17.59 MeV

Radius of the deuterium and tritium, r $\approx 2.0 f m$ = 2 $\times 10^{- 15}$ m

Distance between the centers of the nucleus when they touch each other,

d = r +r = 4 $\times 10^{- 15}$ m

Charge on the deuterium and tritium nucleus = e

Hence the repulsive potential energy between the two nuclei is given as:

V = $\frac{e^{2}}{4 π ?_{0} d}$

Where,

$?_{0}$ = permittivity of free space

It is given that $\frac{1}{4 π ?_{0}}$ = 9 $\times 10^{9}$ N $m^{2} C^{- 2}$

Hence, V = $\frac{{(1.6 \times 10^{- 19})}^{2}}{4 \times 10^{- 15}}$ $\times$ 9 $\times 10^{9}$ = 5.76 $\times 10^{- 14}$ J = $\frac{5.76 \times 10^{- 14}}{1.6 \times 10^{- 19}}$ eV = 360 keV

Hence, 360 keV of kinetic energy is needed to overcome the Coulomb repulsion between the two nuclei.

It is given that

KE = 2 $\times \frac{3}{2} k T$

Where

k = Boltzmann constant = 1.38 $\times 10^{- 23}$ $m^{2} k g s^{- 2} K^{- 1}$

T = Temperature required to trigger the reaction

Therefore T = $\frac{K E}{3 k}$ = $\frac{5.76 \times 10^{- 14}}{3 \times 1.38 \times 10^{- 23}}$ = 1.39 $\times 10^{9}$ K

Hence the gas must be heated to 1.39 $\times 10^{9}$ K to initiate the reaction.

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