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

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: 13.14 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).

13.8 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.

13.19 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

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

13.14 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.

13.17 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?

13.20 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.)

13.5 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).

13.29 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

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

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

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.

13.16 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.

13.7 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?

13.9 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.

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

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.

13.22 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.

13.23 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.

13.25 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?

13.30 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.

13.31 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.

13.1 (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.

13.6 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 ${}_{54}^{120}{X e}$

13.12 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.

13.13 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.

13.15 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

13.24 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

13.26 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.

13.27 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

13.28 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.)

Benefits of Using Chapter 12 Atoms Class 12 Physics NCERT Solutions

Students can check the benefits of using Class 12 Nuclei Solutions.

Step-by-step clarity.
Easy to grasp complex nuclear physics topics.
Stay aligned with the current exam pattern and ensuring relevance for board exams.
Revise key formulas, derivations, and definitions at their own pace.
Strengthen concepts such as radioactive decay, half-life calculations, and nuclear fission and fusion.