Chapter 13 Nuclei

NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

Chapter 14: Nuclei of Physics (Class 12) is about the centre of an atom which represents the core of the atom and contains the majority of the mass of the atom. The previous chapter examined the electrons and their orbits around the atom, while this chapter explores the forces that keep neutrons and protons together, like the force between them. New concepts introduced in this chapter are: Nuclear mass, binding energy, Nuclear stability and how stable is an isotope. In addition, this chapter will provide a deeper understanding of Radioactivity (the natural breakdown of unstable nuclei) and the amount of energy released during Nuclear Fusion and Nuclear Fission. The concepts studied in this chapter will provide a fundamental basis for all other subjects that relate to Medical Imaging, Nuclear Energy and the Evolution of Stars; making it a chapter that has a great potential for scoring well on the CBSE Board Exams, JEE and NEET.

NCERT Solutions for Class 12 Physics Chapter 14 Nuclei provides students with a clear pathway to understanding the Mass-Energy Equivalence Principle as well as the Kinetics of Radioactive Decay. This provides a pathway of logical progression from the basic properties of Nuclear Physics to the Advanced Nuclear Physics of Nuclear Reactors.

1.0Class 12 Physics Chapter 13 Nuclei: Key Concepts

Class 12 Physics Chapter 13, Nuclei, explains the structure of the atomic nucleus, its properties, and the energy released during nuclear processes. The chapter covers important concepts such as nuclear size, mass defect, binding energy, radioactivity, nuclear force, fission, and fusion. The main concepts covered in this chapter include:

Composition and Classification of Nuclei: Explains the composition of the nucleus in terms of protons and neutrons and introduces isotopes, isobars, and isotones. It also discusses the role of neutrons in nuclear stability.
Size and Density of the Nucleus: Explains the relationship between nuclear radius and mass number using R=R0A1/3R = R_0 A^{1/3}. The chapter also shows why nuclear matter has extremely high density.
Mass Defect and Binding Energy: Explains mass-energy equivalence using E=mc2E = mc^2, along with mass defect and binding energy. These concepts help explain the stability of a nucleus.
Nuclear Force: Covers the main properties of nuclear force, including its short range and strong attractive nature. It explains how nuclear force holds protons and neutrons together inside the nucleus.
Radioactivity and Nuclear Decay: Explains alpha, beta, and gamma radiation and the radioactive decay law. The chapter also covers decay constant, half-life, and mean life.
Nuclear Fission and Fusion: Explains how a heavy nucleus can split into lighter nuclei during fission and how light nuclei can combine during fusion. The chapter also introduces the basic working of a nuclear reactor and the release of nuclear energy.

2.0NCERT Solutions Class 12 Physics Chapter 13 Nuclei : Detailed Solutions

SOLVED EXAMPLES

Given the mass of iron nucleus as 55.85u and $A = 56$ , find the nuclear density? Sol. $m_{Fe} = 55.85, u = 9.27 \times 1 0^{- 26} kg$ Nuclear density $= \frac{mass}{volume}$

$= \frac{9.27 \times 1 0 ^{- 26}}{( \frac{4 π}{3} ) ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \times \frac{1}{56} = 2.29 \times 1 0^{17} kg m^{- 3}$

The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Calculate the energy equivalent of 1 g of substance. Sol. Energy, $E = 1 0^{- 3} \times (3 \times 1 0^{8})^{2} J$

$E = 1 0^{- 3} \times 9 \times 1 0^{16} = 9 \times 1 0^{13} J$

Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of $_{8}^{16} O$ in $\frac{MeV}{c ^{2}}$ . Sol. $1 u = 1.6605 \times 1 0^{- 27} kg$ To convert it into energy units, we multiply it by $c^{2}$ and find that energy equivalent

$= 1.6605 \times 1 0^{- 27} \times (2.9979 \times 1 0^{8})^{2} kg m^{2} / s^{2} = 1.4924 \times 1 0^{- 10} J = \frac{1.4924 \times 1 0 ^{- 10}}{1.602 \times 1 0 ^{- 19}} eV = 0.9315 \times 1 0^{9} eV = 931.5 MeV or, 1 u = 931.5 MeV / c^{2}$

For $_{8}^{16} O, Δ M = 0.13691 u$

$= 0.13691 \times 931.5 MeV / c^{2} = 127.5 MeV / c^{2}$

The energy needed to separate $_{8}^{16} O$ into its constituents is thus $127.5 MeV / c^{2}$ .

Answer the following questions: (a) Are the equation of nuclear reaction 'balanced' in the sense a chemical equation (e.g., $2 H_{2} + O_{2} \to 2 H_{2} O$ ) is? If not, in what sense are they balanced on both sides? (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction? (c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain. Sol.

(a) A chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides of the equation. A chemical reaction merely alters original combinations of atoms. In a nuclear reaction, elements may be transmuted. Thus, the number of atoms of each element is not necessarily conserved in a nuclear reaction. However, the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. [Actually, even this is not strictly true in the realm of very high energies - what is strictly conserved is the total charge and total 'baryon number'. We need not pursue this mater here.] In nuclear reactions, the number of protons and the number of neutrons are the same on the two sides of the equation.

(b) We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in these binding energies appears as energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that difference in the total mass of nuclei on the two sides get converted into energy of vice-versa. It is in these sense that a nuclear reaction is an example of mass energy interconversion. (C) From the point of view of mass-energy interconversion, a chemical reaction is similar to a nuclear reaction in principle. The energy released or absorbed in a chemical reaction can be traced to the difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction. Since, strictly speaking, chemical binding energy also gives a negative contribution (mass defect) to the total mass of an atom or molecule, we can equally well say that the difference in the total mass of atoms or molecules, on the two sides of the chemical reaction gets converted into energy of vice-versa. However, the mass defects involved in a chemical reaction are almost a million times smaller than those in a nuclear reaction. This is the reason for the general impression, (which is incorrect) that mass-energy interconversion does not take place in a chemical reaction.

EXERCISE QUESTION WITH SOLUTIONS

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

Sol. Binding energy $= Δ m . c^{2}$

$= [7 (m_{p} + m_{n}) - M_{nucleus}] \cdot c^{2} = [7 (1.00783 + 1.00867) - 14.00307] u . c^{2} = [0.11243] \times 931.5 MeV = 104.73 MeV$

Obtain the binding energy of the nuclei $_{26}^{56} Fe$ and $_{83}^{209} Bi$ in units of MeV from the following data:

$m_{H} = 1.007825 amu m_{n} = 1.008665 amu m (_{26}^{56} Fe) = 55.934939 u m (_{83}^{209} Bi) = 208.980388 u$

Which nucleus has greater binding energy per nucleon? Sol. The $_{26}^{56} Fe$ nucleus contains 26 protons and 30 neutrons.

$Mass of 26 protons = 26 \times 1.007825 = 26.203450 amu$

$Mass of 30 neutrons Total mass = 30 \times 1.008665 = 30.259950 amu = 56.463400 amu$

Mass of $_{26}^{56} Fe$ nucleus $= 55.934939 amu$

$Mass defect Δ m B.E. of_{26}^{56} Fe nucleus = 0.528461 \times 931.5 = 0.528461 amu = Δ m \times 931.5 MeV = 492.26 MeV$

B.E./nucleon of $_{26}^{56} Fe = \frac{492.26}{56} = 8.79 MeV$ . Now, the $_{83}^{209} Bi$ nucleus contains 83 protons and 126 neutrons.

$Mass of 83 protons = 83 \times 1.007825 = 83.649475 amu$

$Mass of 126 neutrons = 126 \times 1.008665 = 127.091790 amu$

Total mass $= 210.741265 amu$ Mass of $_{83}^{209}$ Bi nucleus $= 208.980388$ amu Mass defect, $Δ m = 1.760877 amu$ B.E. of M nucleus = $1.760877 \times 931.5$

$= 1640.3 MeV$

$B.E./nucleon of_{83}^{209} Bi = \frac{1640.3}{209} = 7.85 MeV$

Clearly, $_{26}^{56} Fe$ has a greater B.E. per nucleon. In fact, it is the maximum value.

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

Sol. Binding energy of

$_{29}^{63} Cu = [(29 m_{p} + 34 m_{n}) - M_{nucleus}] \cdot c^{2} = 551.681 MeV$

∵ nuclear energy required to separate all the neutrons and protons from each other

$= (\frac{N _{A}}{M _{w}} \times m) \times (B.E. of_{29}^{63} Cu) = (\frac{6.023 \times 1 0 ^{23}}{63} \times 3) (551.681 MeV) = 1.582 \times 1 0^{25} MeV = 1.582 \times 1 0^{25} \times 1 0^{6} \times 1.6 \times 1 0^{- 19} = 2.53 \times 1 0^{12} J$

Obtain approximately the ratio of the nuclear radii of the gold isotope $_{79}^{197} Au$ and silver isotope $_{47}^{107} Ag$ ?

Sol. As

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

where

$∴ R_{0} = 1.1 \times 1 0^{- 15} m \frac{R ( ^{197} Au )}{R ( ^{107} Ag )} = (\frac{197}{107})^{1/3} ≃ 1.23$

Find the Q-value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of (a) $_{88}^{226} Ra$ and (b) $_{86}^{220} Rn$ . Given $m (_{88}^{226} Ra) = 226.02540 u$ ,

$m (_{86}^{222} Rn) = 222.01750 u, m (_{86}^{220} Rn) = 220.01137 u m (_{84}^{216} Po) = 216.00189 u .$

Sol. (a) $_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He + Q$

$Q = = K_{α} = [m (_{88}^{226} Ra) - m (_{86}^{222} Rn) - m (_{2}^{4} He)] c^{2} [226.02540 - 222.01750 - 4.00260] \times 931.5 MeV 0.0053 \times 931.5 = 4.937 MeV = \frac{A - 4}{A} Q = \frac{226 - 4}{226} \times 4.937 = 4.85 MeV$

(b)

$_{86}^{220} Rn \to_{84}^{216} Po +_{2}^{4} He + Q Q = [(_{86}^{220} Rn) - m (_{84}^{216} Po) - m (_{2}^{4} He)] c^{2} = [220.01137 - 216.00189 - 4.00260] \times 931.5 MeV = 0.00688 \times 931.5 = 6.41 MeV K_{α} = \frac{A - 4}{A} Q = \frac{220 - 4}{220} \times 6.41 = 6.29 MeV$

Suppose, we think of fission of a $_{26}^{56} Fe$ nucleus into two equal fragments, $_{13}^{28} Al$ . Is the fission energetically possible? Argue by working out Q of the process. Given $m (_{26}^{56} Fe) = 55.93494 u$ and m $(_{13}^{28} Al) = 27.98191 u$ .

Sol. $_{26}^{56} Fe ⟶_{13}^{28} Al +_{13}^{28} Al + Q$

$Q = [m (_{26}^{56} Fe) - 2 m (_{13}^{28} Al)] c^{2} = [55.93494 - 2 \times 27.98191] \times 931.5 MeV = - 0.02888 \times 931.5 = - 26.90 MeV$

As the Q-value is negative, the fission is not possible energetically.

The fission properties of $_{94}^{239} Pu$ 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 $_{94}^{239} Pu$ undergo fission? Sol. The number of atoms in 1 kg of

$_{94}^{239} Pu = \frac{6.023 \times 1 0 ^{23} \times 1000}{239}$

Energy released per fission $= 180 MeV$ Energy released by 1 kg of $_{94}^{239} Pu$

$= \frac{6.023 \times 1 0 ^{23} \times 1000}{239} \times 180 MeV = 4.536 \times 1 0^{26} MeV$

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

$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n + 3.27 MeV$

Sol. The number of atoms in 2 kg of deuterium

$= \frac{6.023 \times 1 0 ^{23}}{2} \times 2 \times 1 0^{3} = 6.023 \times 1 0^{26}$

Energy released by combination of

$6.023 \times 1 0^{26} atoms E = \frac{3.27}{2} \times 6.023 \times 1 0^{26} MeV = \frac{3.27}{2} \times 6.023 \times 1 0^{26} \times 1.6 \times 1 0^{- 13} J = 1.57 \times 1 0^{14} J$

$Using power = \frac{Energy}{time} = \frac{E}{t}$

$t t = \frac{1.57 \times 1 0 ^{14}}{100} = 1.57 \times 1 0^{12} s = \frac{1.57 \times 1 0 ^{12}}{365 \times 24 \times 60 \times 60} years = 4.98 \times 1 0^{4} years$

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.) Sol. Charge on each deuteron,

$e = 1.6 \times 1 0^{- 19} C$

Radius of deuteron,

$R = 2.0 fm = 2.0 \times 1 0^{- 15} m U = \frac{1}{4 π ε _{0}} \cdot \frac{e ^{2}}{2 R}$

The Coulomb barrier is given by

$K = \frac{1}{4 π ε _{0}} \cdot \frac{e ^{2}}{2 R} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{2 \times 2 \times 1 0 ^{- 15}} = \frac{23.04 \times 1 0 ^{- 29}}{4 \times 1 0 ^{- 15}} = 5.76 \times 1 0^{- 14} J = \frac{5.76 \times 1 0 ^{- 14}}{1.6 \times 1 0 ^{- 19}} eV = 3.6 \times 1 0^{5} eV = 360 keV$

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

$Volume of nucleus = \frac{4}{3} π R^{3} = \frac{4}{3} π (R_{o} A^{1/3})^{3} = \frac{4}{3} π R_{0}^{3} A$

So, Mass density $= \frac{mass}{volume} = \frac{mA}{\frac{4}{3} π R _{0}^{3} A}$

$ρ = \frac{3 m}{4 π R _{0}^{3}}$

So density is independent of mass number.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics for every chapter, with exercise answers, formulas, and step-by-step solutions to help students study each topic and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 13

Binding Energy Curve: The solutions explain the binding energy per nucleon curve and show why energy is released during both nuclear fission and nuclear fusion.
Step-by-Step Radioactive Decay Problems: Numerical questions based on half-life, decay constant, and activity are solved step by step using the required exponential and logarithmic formulas.
Clear Explanation of Alpha, Beta, and Gamma Decay: The solutions explain the differences between α\alpha, β\beta, and γ\gamma decay, including the changes in atomic number and mass number during each type of decay.
Easy Mass Defect Calculations: Problems involving mass defect and nuclear energy are explained step by step using atomic mass units (u) and the mass-energy relation. This helps students calculate the energy released or absorbed in nuclear reactions.
Clear Understanding of Nuclear Reactors: The roles of moderators, control rods, and coolants are explained clearly to help students understand how a controlled nuclear chain reaction works in a nuclear reactor.

Frequently Asked Questions

NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

1.0Class 12 Physics Chapter 13 Nuclei: Key Concepts

Composition and Classification of Nuclei: Explains the composition of the nucleus in terms of protons and neutrons and introduces isotopes, isobars, and isotones. It also discusses the role of neutrons in nuclear stability.
Size and Density of the Nucleus: Explains the relationship between nuclear radius and mass number using R=R0A1/3R = R_0 A^{1/3}. The chapter also shows why nuclear matter has extremely high density.
Mass Defect and Binding Energy: Explains mass-energy equivalence using E=mc2E = mc^2, along with mass defect and binding energy. These concepts help explain the stability of a nucleus.
Nuclear Force: Covers the main properties of nuclear force, including its short range and strong attractive nature. It explains how nuclear force holds protons and neutrons together inside the nucleus.
Radioactivity and Nuclear Decay: Explains alpha, beta, and gamma radiation and the radioactive decay law. The chapter also covers decay constant, half-life, and mean life.
Nuclear Fission and Fusion: Explains how a heavy nucleus can split into lighter nuclei during fission and how light nuclei can combine during fusion. The chapter also introduces the basic working of a nuclear reactor and the release of nuclear energy.

2.0NCERT Solutions Class 12 Physics Chapter 13 Nuclei : Detailed Solutions

SOLVED EXAMPLES

Given the mass of iron nucleus as 55.85u and $A = 56$ , find the nuclear density? Sol. $m_{Fe} = 55.85, u = 9.27 \times 1 0^{- 26} kg$ Nuclear density $= \frac{mass}{volume}$

$= \frac{9.27 \times 1 0 ^{- 26}}{( \frac{4 π}{3} ) ( 1.2 \times 1 0 ^{- 15} ) ^{3}} \times \frac{1}{56} = 2.29 \times 1 0^{17} kg m^{- 3}$

Calculate the energy equivalent of 1 g of substance. Sol. Energy, $E = 1 0^{- 3} \times (3 \times 1 0^{8})^{2} J$

$E = 1 0^{- 3} \times 9 \times 1 0^{16} = 9 \times 1 0^{13} J$

Thus, if one gram of matter is converted to energy, there is a release of enormous amount of energy.

Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of $_{8}^{16} O$ in $\frac{MeV}{c ^{2}}$ . Sol. $1 u = 1.6605 \times 1 0^{- 27} kg$ To convert it into energy units, we multiply it by $c^{2}$ and find that energy equivalent

For $_{8}^{16} O, Δ M = 0.13691 u$

$= 0.13691 \times 931.5 MeV / c^{2} = 127.5 MeV / c^{2}$

The energy needed to separate $_{8}^{16} O$ into its constituents is thus $127.5 MeV / c^{2}$ .

Answer the following questions: (a) Are the equation of nuclear reaction 'balanced' in the sense a chemical equation (e.g., $2 H_{2} + O_{2} \to 2 H_{2} O$ ) is? If not, in what sense are they balanced on both sides? (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction? (c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. This is strictly speaking, incorrect. Explain. Sol.

EXERCISE QUESTION WITH SOLUTIONS

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

Sol. Binding energy $= Δ m . c^{2}$

$= [7 (m_{p} + m_{n}) - M_{nucleus}] \cdot c^{2} = [7 (1.00783 + 1.00867) - 14.00307] u . c^{2} = [0.11243] \times 931.5 MeV = 104.73 MeV$

Obtain the binding energy of the nuclei $_{26}^{56} Fe$ and $_{83}^{209} Bi$ in units of MeV from the following data:

$m_{H} = 1.007825 amu m_{n} = 1.008665 amu m (_{26}^{56} Fe) = 55.934939 u m (_{83}^{209} Bi) = 208.980388 u$

Which nucleus has greater binding energy per nucleon? Sol. The $_{26}^{56} Fe$ nucleus contains 26 protons and 30 neutrons.

$Mass of 26 protons = 26 \times 1.007825 = 26.203450 amu$

$Mass of 30 neutrons Total mass = 30 \times 1.008665 = 30.259950 amu = 56.463400 amu$

Mass of $_{26}^{56} Fe$ nucleus $= 55.934939 amu$

$Mass defect Δ m B.E. of_{26}^{56} Fe nucleus = 0.528461 \times 931.5 = 0.528461 amu = Δ m \times 931.5 MeV = 492.26 MeV$

B.E./nucleon of $_{26}^{56} Fe = \frac{492.26}{56} = 8.79 MeV$ . Now, the $_{83}^{209} Bi$ nucleus contains 83 protons and 126 neutrons.

$Mass of 83 protons = 83 \times 1.007825 = 83.649475 amu$

$Mass of 126 neutrons = 126 \times 1.008665 = 127.091790 amu$

Total mass $= 210.741265 amu$ Mass of $_{83}^{209}$ Bi nucleus $= 208.980388$ amu Mass defect, $Δ m = 1.760877 amu$ B.E. of M nucleus = $1.760877 \times 931.5$

$= 1640.3 MeV$

$B.E./nucleon of_{83}^{209} Bi = \frac{1640.3}{209} = 7.85 MeV$

Clearly, $_{26}^{56} Fe$ has a greater B.E. per nucleon. In fact, it is the maximum value.

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

Sol. Binding energy of

$_{29}^{63} Cu = [(29 m_{p} + 34 m_{n}) - M_{nucleus}] \cdot c^{2} = 551.681 MeV$

∵ nuclear energy required to separate all the neutrons and protons from each other

Obtain approximately the ratio of the nuclear radii of the gold isotope $_{79}^{197} Au$ and silver isotope $_{47}^{107} Ag$ ?

Sol. As

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

where

$∴ R_{0} = 1.1 \times 1 0^{- 15} m \frac{R ( ^{197} Au )}{R ( ^{107} Ag )} = (\frac{197}{107})^{1/3} ≃ 1.23$

Find the Q-value and the kinetic energy of the emitted $α$ -particle in the $α$ -decay of (a) $_{88}^{226} Ra$ and (b) $_{86}^{220} Rn$ . Given $m (_{88}^{226} Ra) = 226.02540 u$ ,

$m (_{86}^{222} Rn) = 222.01750 u, m (_{86}^{220} Rn) = 220.01137 u m (_{84}^{216} Po) = 216.00189 u .$

Sol. (a) $_{88}^{226} Ra \to_{86}^{222} Rn +_{2}^{4} He + Q$

(b)

Suppose, we think of fission of a $_{26}^{56} Fe$ nucleus into two equal fragments, $_{13}^{28} Al$ . Is the fission energetically possible? Argue by working out Q of the process. Given $m (_{26}^{56} Fe) = 55.93494 u$ and m $(_{13}^{28} Al) = 27.98191 u$ .

Sol. $_{26}^{56} Fe ⟶_{13}^{28} Al +_{13}^{28} Al + Q$

$Q = [m (_{26}^{56} Fe) - 2 m (_{13}^{28} Al)] c^{2} = [55.93494 - 2 \times 27.98191] \times 931.5 MeV = - 0.02888 \times 931.5 = - 26.90 MeV$

As the Q-value is negative, the fission is not possible energetically.

The fission properties of $_{94}^{239} Pu$ 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 $_{94}^{239} Pu$ undergo fission? Sol. The number of atoms in 1 kg of

$_{94}^{239} Pu = \frac{6.023 \times 1 0 ^{23} \times 1000}{239}$

Energy released per fission $= 180 MeV$ Energy released by 1 kg of $_{94}^{239} Pu$

$= \frac{6.023 \times 1 0 ^{23} \times 1000}{239} \times 180 MeV = 4.536 \times 1 0^{26} MeV$

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

$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He +_{0}^{1} n + 3.27 MeV$

Sol. The number of atoms in 2 kg of deuterium

$= \frac{6.023 \times 1 0 ^{23}}{2} \times 2 \times 1 0^{3} = 6.023 \times 1 0^{26}$

Energy released by combination of

$6.023 \times 1 0^{26} atoms E = \frac{3.27}{2} \times 6.023 \times 1 0^{26} MeV = \frac{3.27}{2} \times 6.023 \times 1 0^{26} \times 1.6 \times 1 0^{- 13} J = 1.57 \times 1 0^{14} J$

$Using power = \frac{Energy}{time} = \frac{E}{t}$

$t t = \frac{1.57 \times 1 0 ^{14}}{100} = 1.57 \times 1 0^{12} s = \frac{1.57 \times 1 0 ^{12}}{365 \times 24 \times 60 \times 60} years = 4.98 \times 1 0^{4} years$

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.) Sol. Charge on each deuteron,

$e = 1.6 \times 1 0^{- 19} C$

Radius of deuteron,

$R = 2.0 fm = 2.0 \times 1 0^{- 15} m U = \frac{1}{4 π ε _{0}} \cdot \frac{e ^{2}}{2 R}$

The Coulomb barrier is given by

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

$Volume of nucleus = \frac{4}{3} π R^{3} = \frac{4}{3} π (R_{o} A^{1/3})^{3} = \frac{4}{3} π R_{0}^{3} A$

So, Mass density $= \frac{mass}{volume} = \frac{mA}{\frac{4}{3} π R _{0}^{3} A}$

$ρ = \frac{3 m}{4 π R _{0}^{3}}$

So density is independent of mass number.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics for every chapter, with exercise answers, formulas, and step-by-step solutions to help students study each topic and solve questions.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 4	Moving Charges and Magnetism
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 13

Binding Energy Curve: The solutions explain the binding energy per nucleon curve and show why energy is released during both nuclear fission and nuclear fusion.
Step-by-Step Radioactive Decay Problems: Numerical questions based on half-life, decay constant, and activity are solved step by step using the required exponential and logarithmic formulas.
Clear Explanation of Alpha, Beta, and Gamma Decay: The solutions explain the differences between α\alpha, β\beta, and γ\gamma decay, including the changes in atomic number and mass number during each type of decay.
Easy Mass Defect Calculations: Problems involving mass defect and nuclear energy are explained step by step using atomic mass units (u) and the mass-energy relation. This helps students calculate the energy released or absorbed in nuclear reactions.
Clear Understanding of Nuclear Reactors: The roles of moderators, control rods, and coolants are explained clearly to help students understand how a controlled nuclear chain reaction works in a nuclear reactor.

Frequently Asked Questions

Where can I find NCERT Solutions Class 12 Physics Chapter 13 Nuclei?

What do NCERT Solutions Class 12 Physics Chapter 13 cover?

How can I solve radioactive decay questions using NCERT Solutions Class 12 Physics Chapter 13?

What is mass defect and binding energy in NCERT Solutions Class 12 Physics Chapter 13?

How are alpha, beta and gamma decay questions solved in NCERT Solutions Class 12 Physics Chapter 13?

What is the difference between nuclear fission and fusion according to NCERT Solutions Class 12 Physics Chapter 13?

Are NCERT Solutions Class 12 Physics Chapter 13 useful for JEE and NEET?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

1.0Class 12 Physics Chapter 13 Nuclei: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 13 Nuclei : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 13

Table of Contents

Frequently Asked Questions

Where can I find NCERT Solutions Class 12 Physics Chapter 13 Nuclei?

What do NCERT Solutions Class 12 Physics Chapter 13 cover?

How can I solve radioactive decay questions using NCERT Solutions Class 12 Physics Chapter 13?

What is mass defect and binding energy in NCERT Solutions Class 12 Physics Chapter 13?

How are alpha, beta and gamma decay questions solved in NCERT Solutions Class 12 Physics Chapter 13?

What is the difference between nuclear fission and fusion according to NCERT Solutions Class 12 Physics Chapter 13?

Are NCERT Solutions Class 12 Physics Chapter 13 useful for JEE and NEET?

Join ALLEN!

NCERT Solutions Class 12 Physics Chapter 13 - Nuclei

1.0Class 12 Physics Chapter 13 Nuclei: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 13 Nuclei : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTION WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 13

Table of Contents