NCERT Solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 13, Nuclei, provide step-by-step answers to all six questions in the NCERT exercise. The chapter deals with the composition and size of the nucleus, mass-energy equivalence, mass defect, nuclear binding energy, nuclear force, radioactivity, fission and fusion.

The solutions below follow the questions from the current NCERT Class 12 Physics textbook and show the calculations clearly, including the formulas, substitutions and final answers. They are useful for checking textbook exercises, understanding numerical methods and revising the main ideas from the chapter.

NCERT Class 12 Physics Chapter 13: Structure and Key Topics

Chapter 13 begins with the basic structure of the nucleus and then moves into the relationship between mass and energy. Students next study mass defect and binding energy, followed by nuclear reactions such as fission and fusion and the basics of radioactivity.

The chapter includes both conceptual questions and numerical problems. Numericals mainly involve mass defect, binding energy, nuclear radius, Q-value, and nuclear reactions. Questions on nuclear composition, stability, fission, fusion, and radioactivity require students to understand the underlying concepts as well.

The chapter also leads into the next part of Class 12 Physics. Chapter 14, Semiconductor Electronics, shifts from the nucleus to semiconductor materials, electronic devices, and their applications, so the two chapters cover very different areas of modern physics and electronics.

Download NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei

Use these step-by-step solutions alongside the NCERT textbook to check your calculations and understand the method used for each exercise.

For the original exercise questions, always refer to the current NCERT textbook. The questions on this page reproduce the Chapter 13 exercise exactly as given in the 2026–27 reprint.

Important Formulas

1. Nuclear radius

R = R₀A^1/3

where R₀ = 1.2 × 10⁻¹⁵ m.

2. Mass defect

When atomic masses are used, the mass defect can be calculated as:

ΔM = Zm_H + (A−Z)m_n − M

where:

• Z = number of protons
• A − Z = number of neutrons
• m_(H) = mass of a hydrogen atom
• m_(n) = mass of a neutron
• M = atomic mass of the nucleus

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Fig. 2 — The bound nucleus has slightly less mass than its separated nucleons; the missing mass ΔM is released as binding energy.

3. Binding energy

Eb = ΔM c²

Using 1 u = 931.5 MeV/c², we get:

E_b = ΔM × 931.5 MeV

4. Q-value

Q = (sum of initial masses − sum of final masses) c²

A positive Q-value indicates an exothermic reaction, while a negative Q-value indicates an endothermic reaction.

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Fig. 3 — Binding energy per nucleon rises sharply for light nuclei, peaks near A ≈ 56–62, then declines slowly — this shape is why fusion of light nuclei and fission of heavy nuclei can both release energy.

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Fig. 4 — Fission splits a heavy nucleus into lighter fragments; fusion combines light nuclei into a heavier, more tightly bound nucleus.

NCERT Questions and Solutions for Class 12 Nuclei

Question 1: Obtain the binding energy (in MeV) of a nitrogen nucleus ¹⁴₇N, given m(¹⁴₇N) = 14.00307 u.

Given

A = 14, Z = 7 ⇒ N = A − Z = 7

m_H = 1.007825 u, mn = 1.008665 u

Mass of the separated nucleons

M = Zm_H + (A−Z)m_n = 7(1.007825) + 7(1.008665)

M = 7.054775 + 7.060655 = 14.115430 u

Mass defect

ΔM = M − m(¹⁴₇N) = 14.115430 − 14.00307 = 0.112360 u

Binding energy

Eb = ΔM × 931.5 = 0.112360 × 931.5 ≈ 104.66 MeV

Answer

Eb ≈ 104.66 MeV

Question 2: Obtain the binding energy of the nuclei ⁵⁶₂₆Fe and ²⁰⁹₈₃Bi in units of MeV from the following data: m(⁵⁶₂₆Fe) = 55.934939 u and m(²⁰⁹₈₃Bi) = 208.980388 u.

For ⁵⁶₂₆Fe

Given

A = 56, Z = 26 ⇒ N = 30

M = 26(1.007825) + 30(1.008665) = 26.20345 + 30.25995 = 56.46340 u

ΔM = 56.46340 − 55.934939 = 0.528461 u

E_b = 0.528461 × 931.5 ≈ 492.26 MeV

For ²⁰⁹₈₃Bi

Given

A = 209, Z = 83 ⇒ N = 126

M = 83(1.007825) + 126(1.008665) = 83.649475 + 127.091790 = 210.741265 u

ΔM = 210.741265 − 208.980388 = 1.760877 u

E_b = 1.760877 × 931.5 ≈ 1640.26 MeV

IMAGE

Fig. 5 — Fe-56 sits almost exactly at the peak of the binding-energy curve; Bi-209 lies further along the slowly falling right-hand side.

Answer

Eb(⁵⁶Fe) ≈ 492.26 MeV

Eb(²⁰⁹Bi) ≈ 1640.26 MeV

Question 3: 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 ⁶³₂₉Cu atoms (of mass 62.92960 u).

Given

A = 63, Z = 29 ⇒ N = 34, M = 62.92960 u

Step 1: Mass defect of one copper nucleus

ΔM = Zm_H + (A−Z)m_n − M = 29(1.007825) + 34(1.008665) − 62.92960

29(1.007825) = 29.227925, 34(1.008665) = 34.294610

ΔM = 29.227925 + 34.294610 − 62.92960 = 0.592935 u

Step 2: Binding energy of one copper nucleus

E_b = 0.592935 × 931.5 ≈ 552.36 MeV

Step 3: Number of copper atoms in the coin

number of moles = 3.0/62.92960 ≈ 0.04767 mol

N = 0.04767 × 6.023 × 10²³ ≈ 2.87 × 10²² atoms

Step 4: Total nuclear energy

E_(total) = (2.87 × 10²²)(552.36 MeV) ≈ 1.58 × 10²⁵ MeV

Using 1 MeV = 1.6 × 10⁻¹³ J:

E_(total) = 1.58 × 10²⁵ × 1.6 × 10⁻¹³ ≈ 2.53 × 10¹² J

Answer

E(total) ≈ 2.53 × 10¹² J

Question 4: Obtain approximately the ratio of the nuclear radii of the gold isotope ¹⁹⁷₇₉Au and the silver isotope ¹⁰⁷₄₇Ag.

Given

A(Au) = 197, A(Ag) = 107

Formula

R = R₀A^1/3 ⇒ R(Au)/R(Ag) = [A(Au)/A(Ag)]^1/3

The R₀ terms cancel since both nuclei use the same R₀.

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Fig. 6 — The gold nucleus is larger than the silver nucleus, in proportion to A¹ᵟ³.

Calculation

197/107 ≈ 1.841

R(Au)/R(Ag) ≈ (1.841)1/3 ≈ 1.23

Answer

R(Au)/R(Ag) ≈ 1.23

Question 5: The Q value of a nuclear reaction A + b → C + d is defined by Q = [m(A) + m(b) − m(C) − m(d)]c², where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. (i) ¹₁H + ³₁H → ²₁H + ²₁H (ii) ¹²₆C + ¹²₆C → ²⁰₁₀Ne + ⁴₂He

Given

m(²H) = 2.014102 u, m(³H) = 3.016049 u

m(¹²C) = 12.000000 u, m(²⁰Ne) = 19.992439 u, m(⁴He) = 4.002603 u

IMAGE

Fig. 7 — Comparing total mass before and after each reaction shows directly whether Q is positive or negative.

(i) ¹H + ³H → ²H + ²H

Q = [m(¹H) + m(³H) − 2m(²H)]c2 = [1.007825 + 3.016049 − 2(2.014102)]c²

Q = [4.023874 − 4.028204]c2 = −0.004330 u·c²

Q = −0.004330 × 931.5 ≈ −4.03 MeV

The Q-value is negative, so the reaction is endothermic.

(ii) ¹²C + ¹²C → ²⁰Ne + ⁴He

Q = [2m(¹²C) − m(²⁰Ne) − m(⁴He)]c² = [24.000000 − 19.992439 − 4.002603]c²

Q = [24.000000 − 23.995042]c2 = 0.004958 u·c²

Q = 0.004958 × 931.5 ≈ 4.62 MeV

The Q-value is positive, so the reaction is exothermic.

Answer

(i) Q ≈ −4.03 MeV — endothermic

(ii) Q ≈ +4.62 MeV — exothermic

Question 6: Suppose, we think of fission of a ⁵⁶₂₆Fe nucleus into two equal fragments, ²⁸₁₃Al. Is the fission energetically possible? Argue by working out Q of the process. Given m(⁵⁶₂₆Fe) = 55.93494 u and m(²⁸₁₃Al) = 27.98191 u.

The proposed fission reaction is ⁵⁶Fe → 2 ²⁸Al.

Q-value

Q = [m(⁵⁶Fe) − 2m(²⁸Al)]c² = [55.93494 − 2(27.98191)]c²

Q = [55.93494 − 55.96382]c² = −0.02888 u·c²

Q = −0.02888 × 931.5 ≈ −26.90 MeV

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Fig. 8 — The two aluminium fragments together weigh more than the original iron nucleus, so the process would absorb energy rather than release it.

The Q-value is negative. Therefore, the proposed fission does not release energy — instead, approximately 26.90 MeV would have to be supplied for the process to occur.

Answer

Q ≈ −26.90 MeV

Since Q < 0, the fission is not energetically possible as an energy-releasing process.

How to Solve Nuclei Numericals

Most numerical questions in this chapter can be approached systematically.

Step 1: Identify A and Z

For a nuclide ᴬₓX: A gives the total number of nucleons, Z gives the number of protons, and A − Z gives the number of neutrons.

Step 2: Select the appropriate formula

For binding energy:

E_b = ΔM × 931.5 MeV

For nuclear radius:

R = R₀A^1/3

For a nuclear reaction:

Q = (initial mass − final mass) × 931.5 MeV

Step 3: Keep the units consistent

Masses given in atomic mass units should be used with 1 u = 931.5 MeV/c². If the final answer is required in joules, use 1 MeV = 1.6 × 10⁻¹³ J.

Step 4: Check the sign

• Q > 0: energy is released, so the reaction is exothermic.
• Q < 0: energy is absorbed, so the reaction is endothermic.

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

• Detailed Step-by-Step Solutions: The solutions take students through the reasoning and calculations so they can see how each answer is obtained.
• Know which formula to use: Each numerical begins with the relevant formula, followed by the values and working needed to reach the answer.
• Understand the binding energy curve: The explanation helps students see what the curve represents and how it relates to the energy released during fission and fusion.
• Updated for CBSE 2026–27: The solutions follow the latest CBSE syllabus and exam requirements for Class 12 Physics.
• Useful for last-minute revision: Students can quickly review the important formulas, methods, calculations, and answers before their Physics exam.

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• Learn from Experienced Physics Faculty: Get explanations from Physics faculty with experience teaching students preparing for JEE and NEET, along with board examinations.
• Make Difficult Concepts Easier: Understand topics such as radioactive decay, nuclear stability, mass defect, and binding energy through clear explanations.
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