Class 12 Physics NCERT Solutions

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Nuclei Class 12 Physics NCERT Solutions

The complete NCERT exercise solutions for Chapter 13, Nuclei — 10 questions from 13.1 to 13.10, each worked through step by step in the CBSE marking pattern. The atomic nucleus, mass defect, binding energy and the binding energy curve, radioactivity, and mass-energy equivalence.

Class:12Subject:PhysicsChapter:13

4 Key Formulas25 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

How many questions are in NCERT Class 12 Physics Chapter 13?

Chapter 13 carries 10 exercise questions, numbered 13.1 to 13.10. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

Table of Contents11 sections

Revision & Practice

Chapter Overview

Nuclear physics is held together by the mass-energy relation: binding energy is the mass defect converted through E = mc², the Q-value decides whether a reaction is exothermic or endothermic, and the radius of any nucleus follows R = R₀A^(1/3), which makes nuclear density a constant. Radioactivity and fission-fusion energy are then just counts — Avogadro-sized — multiplied by per-atom yields. Every question below is from the NCERT Class 12 textbook (rationalised edition), solved line by line in the board pattern.

Data supplied with the chapter

e = 1.6 × 10⁻¹⁹ C; N = 6.023 × 10²³ per mole; 1/(4πε₀) = 9 × 10⁹ N m²/C²; k = 1.381 × 10⁻²³ J K⁻¹; 1 MeV = 1.6 × 10⁻¹³ J; 1 u = 931.5 MeV/c²; 1 year = 3.154 × 10⁷ s; m_H = 1.007825 u; m_n = 1.008665 u; m(⁴₂He) = 4.002603 u; m_e = 0.000548 u.

Board pattern

Binding energy = (Zm_H + Nm_n − m_nucleus)c²; using atomic masses cancels the electron bookkeeping automatically (Z electrons cancel). Q = (Σ initial masses − Σ final masses)c², positive → exothermic. The number of nuclei in m grams is (m/M)A × N_A; nuclear energy released = (number of reactions) × (energy per reaction).

$Ex 13.1 — Binding energy of the nitrogen nucleus$
$Ex 13.2 — Binding energy of iron and bismuth nuclei$
$Ex 13.3 — Energy to separate all nucleons of a 3 g copper coin$
$Ex 13.4 — Ratio of nuclear radii: gold and silver isotopes$
$Ex 13.5 — Q-values of two reactions; exothermic or endothermic$
$Ex 13.6 — Feasibility of fissioning iron into two aluminiums$
$Ex 13.7 — Energy from all fissions in 1 kg of plutonium-239$
$Ex 13.8 — How long a 100 W lamp can burn on 2 kg of deuterium$
$Ex 13.9 — Coulomb barrier height for two touching deuterons$
$Ex 13.10 — Nuclear density is independent of mass number$

NCERT Exercise 13.1 — Binding Energy of the Nitrogen Nucleus

1Exercise question

Step-by-step solution

1¹⁴₇N has Z = 7 protons and N = 7 neutrons.
2Mass defect: Δm = 7m_H + 7m_n − m(¹⁴₇N) = 7(1.007825 + 1.008665) − 14.00307 u.
3Δm = 7 × 2.016490 − 14.00307 = 14.11543 − 14.00307 = 0.11236 u.
4Binding energy: B.E. = Δm × 931.5 = 0.11236 × 931.5 = 104.7 MeV.

Final answer

B.E. = 104.7 MeV.

NCERT Exercise 13.2 — Binding Energy of Iron and Bismuth

1Exercise question

Step-by-step solution

1For ⁵⁶₂₆Fe: 26 protons, 30 neutrons. Δm = 26m_H + 30m_n − m(Fe) = 26(1.007825) + 30(1.008665) − 55.934939.
2Δm(Fe) = 26.203450 + 30.259950 − 55.934939 = 0.528461 u.
3B.E.(Fe) = 0.528461 × 931.5 = 492.3 MeV; per nucleon = 492.3/56 = 8.79 MeV.
4For ²⁰⁹₈₃Bi: 83 protons, 126 neutrons. Δm = 83(1.007825) + 126(1.008665) − 208.980388.
5Δm(Bi) = 83.649475 + 127.091790 − 208.980388 = 1.760877 u.
6B.E.(Bi) = 1.760877 × 931.5 = 1640.3 MeV; per nucleon = 1640.3/209 = 7.85 MeV.

Final answer

B.E.(Fe) = 492.3 MeV (8.79 MeV/nucleon); B.E.(Bi) = 1640.3 MeV (7.85 MeV/nucleon).

NCERT Exercise 13.3 — Energy to Separate a Copper Coin's Nucleons

1Exercise question

Step-by-step solution

1⁶³₂₉Cu has 29 protons and 34 neutrons. Mass defect per atom: Δm = 29m_H + 34m_n − m(Cu).
2Δm = 29(1.007825) + 34(1.008665) − 62.92960 = 29.226925 + 34.294610 − 62.92960 = 0.591935 u.
3Binding energy per atom = 0.591935 × 931.5 = 551.4 MeV.
4Number of Cu atoms in 3.0 g: N = (3.0/62.93) × 6.023 × 10²³ = 2.871 × 10²².
5Total energy = 2.871 × 10²² × 551.4 = 1.58 × 10²⁵ MeV.
6In joules: 1.58 × 10²⁵ × 1.6 × 10⁻¹³ = 2.53 × 10¹² J.

Final answer

E = 1.58 × 10²⁵ MeV = 2.53 × 10¹² J.

NCERT Exercise 13.4 — Ratio of Nuclear Radii

1Exercise question

Step-by-step solution

1Nuclear radius follows R = R₀A^(1/3), so the ratio depends only on the mass numbers.
2R(Au)/R(Ag) = (197/107)^(1/3) = (1.8411)^(1/3) = 1.23.

Final answer

R(Au)/R(Ag) = 1.23.

NCERT Exercise 13.5 — Q-Values of Two Reactions

1Exercise question

Step-by-step solution

1(i) Q = [m(¹H) + m(³H) − 2m(²H)]c² with m(¹H) = 1.007825 u.
2Q = [1.007825 + 3.016049 − 2 × 2.014102] × 931.5 = (4.023874 − 4.028204) × 931.5 MeV.
3Q = −0.004330 × 931.5 = −4.03 MeV — negative, so the reaction is endothermic.
4(ii) Q = [2m(¹²C) − m(²⁰Ne) − m(⁴He)]c² = [2 × 12.000000 − 19.992439 − 4.002603] × 931.5 MeV.
5Q = (24.000000 − 23.995042) × 931.5 = 0.004958 × 931.5 = +4.62 MeV — positive, so the reaction is exothermic.

Final answer

(i) Q = −4.03 MeV, endothermic. (ii) Q = +4.62 MeV, exothermic.

NCERT Exercise 13.6 — Feasibility of Iron Fission

1Exercise question

Step-by-step solution

1Q = [m(⁵⁶Fe) − 2m(²⁸Al)]c² = [55.93494 − 2 × 27.98191] × 931.5 MeV.
2Q = (55.93494 − 55.96382) × 931.5 = −0.02888 × 931.5 = −26.9 MeV.
3The negative Q means energy must be supplied; the fission is endothermic and clearly not energetically possible.

Final answer

Q = −26.9 MeV — the fission of ⁵⁶Fe into two ²⁸Al nuclei is not energetically possible.

NCERT Exercise 13.7 — Energy from 1 kg of Plutonium-239

1Exercise question

Step-by-step solution

1Number of Pu atoms in 1 kg: N = (1000/239) × 6.023 × 10²³ = 2.52 × 10²⁴.
2Each fission releases 180 MeV, so total energy = 2.52 × 10²⁴ × 180.
3E = 4.54 × 10²⁶ MeV.

Final answer

E ≈ 4.5 × 10²⁶ MeV (4.54 × 10²⁶ MeV).

NCERT Exercise 13.8 — Lamp Lifetime from 2 kg of Deuterium

1Exercise question

Step-by-step solution

1Number of deuterium atoms in 2.0 kg: N = (2000/2) × 6.023 × 10²³ = 6.023 × 10²⁶.
2Each reaction burns two deuterons, so number of reactions = 6.023 × 10²⁶/2 = 3.01 × 10²⁶.
3Energy released = 3.01 × 10²⁶ × 3.27 = 9.85 × 10²⁶ MeV = 9.85 × 10²⁶ × 1.6 × 10⁻¹³ J = 1.58 × 10¹⁴ J.
4Time = E/P = (1.58 × 10¹⁴)/100 = 1.58 × 10¹² s.
5In years: t = (1.58 × 10¹²)/(3.154 × 10⁷) = 5.0 × 10⁴ years.

Final answer

t ≈ 5.0 × 10⁴ years (1.58 × 10¹² s).

NCERT Exercise 13.9 — Coulomb Barrier Height for Two Deutrons

1Exercise question

Step-by-step solution

1When the two deuterons just touch, their centres are separated by 2r = 2 × 2.0 = 4.0 fm = 4.0 × 10⁻¹⁵ m.
2Barrier height: U = (1/4πε₀)(e·e)/2r = (9 × 10⁹)(1.6 × 10⁻¹⁹)²/(4.0 × 10⁻¹⁵).
3U = (2.304 × 10⁻²⁸)/(4.0 × 10⁻¹⁵) = 5.76 × 10⁻¹⁴ J.
4In keV: U = (5.76 × 10⁻¹⁴)/(1.6 × 10⁻¹⁶) = 360 keV.

Final answer

Barrier height = 360 keV.

NCERT Exercise 13.10 — Nuclear Density is Independent of A

1Exercise question

Step-by-step solution

1Mass of a nucleus of mass number A is roughly m = A × m_p.
2Its volume is V = (4/3)πR³ = (4/3)πR₀³A.
3Density ρ = m/V = (A m_p)/((4/3)πR₀³A) = (3m_p)/(4πR₀³).
4The factor A cancels, so ρ is the same for every nucleus — nuclear matter density is constant.

Final answer

ρ = 3m_p/(4πR₀³), independent of A — nuclear density is a constant.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Mass-energy relation

E = m c^{2}

Mass defect

Δ m = (Z m_{p} + (A - Z) m_{n}) - m_{n u c l e u s}

Binding energy

E_{b} = Δ m c^{2}, \frac{E _{b}}{A}

Radioactive decay

N = N_{0} e^{- λ t}, T_{1/2} = \frac{ln 2}{λ}

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

The binding energy per nucleon curve peaks near iron-56, so fusion above it and fission below it both release energy — that single fact is the reason both processes matter.
In decay problems find the decay constant from the half-life first, then use the exponential law; the number of half-lives elapsed is often the quickest route.

FAQ

Frequently asked questions

How many questions are in NCERT Class 12 Physics Chapter 13 (Nuclei)?

There are 10 exercise questions in this chapter, numbered 13.1 to 13.10. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Nuclei Class 12 Physics?

The formulas this chapter's questions actually turn on are: Mass-energy relation, Mass defect, Binding energy, Radioactive decay. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Nuclei important for JEE Main and NEET?

Very important — mass defect, binding energy per nucleon and half-life problems are standard in every board paper and in JEE Main and NEET, and the curve is a frequent one-mark question.

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