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NCERT Exercises and Solutions: Nuclei

🎓 Class 12 Physics CBSE Theory Ch 13 – Nuclei ⏱ ~8 min
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NCERT Exercises and Solutions: Nuclei

Chapter 13 Summary — Key Ideas at a Glance

Atomic mass unit

1 u = (1/12) × m(¹²C) = 1.6605 × 10⁻²⁷ kg ≡ 931.5 MeV/c².

Nuclear notation

\(^{A}_{Z}\mathrm{X}\): Z = protons, N = neutrons, A = Z + N = nucleons. Isotopes (same Z), isobars (same A), isotones (same N).

Nuclear radius

R = R₀ A^(1/3) with R₀ = 1.2 fm. Density ≈ 2.3 × 10¹⁷ kg/m³, independent of A.

Mass defect

ΔM = [Z m_p + (A−Z) m_n] − M_nucleus > 0.

Binding energy

E_b = ΔM·c². E_bn = E_b/A peaks at ⁵⁶Fe (≈8.79 MeV/nucleon).

Strong nuclear force

Short-ranged (~ a few fm), much stronger than Coulomb at small r, charge-independent, saturating.

Decay law

N(t) = N₀ e^(−λt). T₁/₂ = 0.693/λ. Mean life τ = 1/λ.

Activity

R = λN; SI unit becquerel (Bq); 1 Ci = 3.7 × 10¹⁰ Bq.

α, β, γ decay

α: A→A−4, Z→Z−2. β⁻: Z→Z+1. β⁺: Z→Z−1. γ: no change in Z, A.

Fission

Heavy nucleus splits into mid-mass fragments + neutrons. ²³⁵U fission ≈ 200 MeV.

Fusion

Light nuclei merge. Powers the Sun via the p-p cycle: 4¹H → ⁴He + 26.7 MeV.

Q-value

Q = (Σm_initial − Σm_final)c². Q>0 means energy released (exothermic).

Master Formula Sheet

QuantityFormulaUnits
Nuclear radiusR = R₀ A^(1/3)m (R₀ = 1.2 fm)
Nuclear densityρ = m_n / [(4/3)π R₀³] ≈ constkg/m³
Mass defectΔM = Z m_p + (A−Z) m_n − Mu
Binding energyE_b = ΔM × 931.5 MeV (if ΔM in u)MeV
BE per nucleonE_bn = E_b / AMeV/nucleon
Decay lawN = N₀ e^(−λt) ; R = R₀ e^(−λt)—; Bq
Half-lifeT₁/₂ = (ln 2)/λ = 0.693/λs
Mean lifeτ = 1/λ ; T₁/₂ = 0.693 τs
Q-value (reaction)Q = (Σm_i − Σm_f) c²MeV

NCERT Exercises — Worked Solutions

Useful constants:

e = 1.6 × 10⁻¹⁹ C 1/(4πε₀) = 9 × 10⁹ N·m²/C² 1 u = 931.5 MeV/c² m_H = 1.007825 u m_n = 1.008665 u m(⁴He) = 4.002603 u N_A = 6.022 × 10²³ /mol 1 MeV = 1.6 × 10⁻¹³ J 1 year = 3.154 × 10⁷ s
Q 13.1 — Obtain the binding energy (in MeV) of a nitrogen nucleus \(^{14}_{7}\mathrm{N}\), given m(N-14) = 14.00307 u.

ΔM = 7 m_H + 7 m_n − M(N-14)

\[ = 7(1.007825) + 7(1.008665) - 14.00307 = 0.11243\ \text{u} \] \[ E_b = 0.11243 \times 931.5 \approx \mathbf{104.7\ MeV} \]

Per nucleon: 104.7 / 14 ≈ 7.48 MeV/nucleon.

Q 13.2 — Obtain the binding energies of \(^{56}_{26}\mathrm{Fe}\) and \(^{209}_{83}\mathrm{Bi}\). Masses: m(Fe-56) = 55.934939 u, m(Bi-209) = 208.980388 u.

For Fe-56 (Z = 26, A−Z = 30):

\[ \Delta M = 26(1.007825) + 30(1.008665) - 55.934939 = 0.5285\ \text{u} \] \[ E_b = 0.5285 \times 931.5 \approx \mathbf{492.3\ MeV};\ \ E_{bn} \approx 8.79\ \text{MeV} \]

For Bi-209 (Z = 83, A−Z = 126):

\[ \Delta M = 83(1.007825) + 126(1.008665) - 208.980388 = 1.7625\ \text{u} \] \[ E_b = 1.7625 \times 931.5 \approx \mathbf{1641.9\ MeV};\ \ E_{bn} \approx 7.85\ \text{MeV} \]

Notice E_bn(Fe) > E_bn(Bi) — Fe is more tightly bound, consistent with the BE/A peak near A = 56.

Q 13.3 — A coin has mass 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. Assume the coin is entirely \(^{63}_{29}\mathrm{Cu}\) (m = 62.92960 u).

Number of Cu atoms in 3.0 g: N = (3.0 × 6.022 × 10²³)/63 ≈ 2.868 × 10²² atoms.

For one Cu-63 nucleus (Z = 29, A − Z = 34):

\[ \Delta M = 29(1.007825) + 34(1.008665) - 62.92960 = 0.59225\ \text{u} \] \[ E_b\ \text{(per atom)} = 0.59225 \times 931.5 \approx 551.6\ \text{MeV} \]

Total nuclear binding energy:

\[ E_{\text{total}} = 2.868 \times 10^{22} \times 551.6\ \text{MeV} \approx \mathbf{1.58 \times 10^{25}\ MeV} \approx 2.53 \times 10^{12}\ \text{J} \]
Q 13.4 — Obtain approximately the ratio of nuclear radii of \(^{197}_{79}\mathrm{Au}\) and \(^{107}_{47}\mathrm{Ag}\).
\[ \frac{R_{\text{Au}}}{R_{\text{Ag}}} = \left(\frac{197}{107}\right)^{1/3} = (1.841)^{1/3} \approx \mathbf{1.226} \]

Au radius is about 23% larger than Ag radius.

Q 13.5 — Determine the Q-value of the following reactions and state whether each is exothermic or endothermic. Given m(²H) = 2.014102 u, m(³H) = 3.016049 u, m(¹²C) = 12.000000 u, m(²⁰Ne) = 19.992439 u.
(i) ¹H + ³H → ²H + ²H
(ii) ¹²C + ¹²C → ²⁰Ne + ⁴He

(i) ¹H + ³H → ²H + ²H:

\[ \Delta m = (1.007825 + 3.016049) - 2(2.014102) = -0.00433\ \text{u} \] \[ Q = -0.00433 \times 931.5 \approx \mathbf{-4.03\ MeV}\ \text{(endothermic)} \]

(ii) ¹²C + ¹²C → ²⁰Ne + ⁴He:

\[ \Delta m = 2(12.000000) - (19.992439 + 4.002603) = 0.004958\ \text{u} \] \[ Q = 0.004958 \times 931.5 \approx \mathbf{+4.62\ MeV}\ \text{(exothermic)} \]
Q 13.6 — Suppose we think of fission of \(^{56}_{26}\mathrm{Fe}\) into two equal fragments \(^{28}_{13}\mathrm{Al}\). Is the fission energetically possible? Given m(Fe-56) = 55.93494 u, m(Al-28) = 27.98191 u.
\[ Q = [m(\text{Fe-56}) - 2 m(\text{Al-28})]c^2 \] \[ = (55.93494 - 2 \times 27.98191)\times 931.5\ \text{MeV} = (-0.02888)\times 931.5 \] \[ Q \approx \mathbf{-26.9\ MeV} \]

Q is negative, so this fission is energetically not possible. Iron-56 sits near the peak of the BE/A curve — it would cost energy to split it. (Stars stop fusing when the core reaches Fe.)

Q 13.7 — The fission properties of \(^{239}_{94}\mathrm{Pu}\) are very similar to those of ²³⁵U. Average energy per fission = 180 MeV. How much energy is released if all atoms in 1 kg of pure ²³⁹Pu undergo fission?

Number of atoms in 1 kg = (1000 × 6.022 × 10²³)/239 ≈ 2.52 × 10²⁴ atoms.

\[ E = 2.52 \times 10^{24} \times 180 \approx \mathbf{4.54 \times 10^{26}\ MeV} \]

In joules: 4.54 × 10²⁶ × 1.6 × 10⁻¹³ ≈ 7.26 × 10¹³ J.

Q 13.8 — How long can a 100 W lamp be kept glowing by fusion of 2.0 kg of deuterium? Reaction: ²H + ²H → ³He + n + 3.27 MeV.

Atoms in 2.0 kg of ²H: N_atoms = (2000 × 6.022 × 10²³)/2 = 6.022 × 10²⁶.

Two deuterons fuse per reaction → number of reactions = 3.011 × 10²⁶.

Energy: E = 3.011 × 10²⁶ × 3.27 × 1.6 × 10⁻¹³ J ≈ 1.576 × 10¹⁴ J.

\[ t = \frac{E}{P} = \frac{1.576 \times 10^{14}}{100} \approx 1.58 \times 10^{12}\ \text{s} \approx \mathbf{5 \times 10^{4}\ \text{years}} \]
Q 13.9 — Calculate the height of the potential barrier for a head-on collision of two deuterons. Treat them as hard spheres of radius 2.0 fm.

Centre-to-centre at touch: r = 2 × 2.0 fm = 4.0 × 10⁻¹⁵ m.

\[ U = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r} = \frac{(9 \times 10^{9})(1.6 \times 10^{-19})^2}{4 \times 10^{-15}} \approx 5.76 \times 10^{-14}\ \text{J} \]

≈ 5.76 × 10⁻¹⁴ / 1.6 × 10⁻¹³ ≈ 0.36 MeV (≈ 360 keV).

Q 13.10 — From R = R₀ A^(1/3), show that nuclear matter density is nearly independent of A.

The mass of a nucleus M ≈ A × m_avg, where m_avg ≈ 1 u (since m_p ≈ m_n ≈ 1 u).

The volume:

\[ V = \tfrac{4}{3}\pi R^3 = \tfrac{4}{3}\pi (R_0 A^{1/3})^3 = \tfrac{4}{3}\pi R_0^3 \cdot A \]

Therefore density:

\[ \rho = \frac{M}{V} = \frac{A \cdot m_{\text{avg}}}{\tfrac{4}{3}\pi R_0^3 \cdot A} = \frac{m_{\text{avg}}}{\tfrac{4}{3}\pi R_0^3} \]

The A cancels, leaving ρ independent of A. Plugging in numbers:

\[ \rho = \frac{1.66 \times 10^{-27}}{\tfrac{4}{3}\pi (1.2\times 10^{-15})^3} \approx 2.3 \times 10^{17}\ \text{kg/m}^3 \]

Interactive Practice — Decay Calculator

Half-life ↔ activity practice

Set initial activity, half-life, and observation time; see how many nuclei and what activity remain.

3.0 (3.0 T½)
R(t): 100 Bq N(t)/N₀: 0.125 Decay constant λ: 6.42×10⁻⁵ s⁻¹

Quick Self-Check

Final-Round Practice Questions

Q1 (MCQ). The most tightly bound nucleus per nucleon is:

  • (a) ²³⁸U
  • (b) ²³⁵U
  • (c) ⁵⁶Fe
  • (d) ⁴He
(c) ⁵⁶Fe sits at the peak of the BE/A curve at ~8.79 MeV/nucleon.

Q2 (MCQ). After 3 mean lives, the fraction of nuclei left is:

  • (a) 1/2
  • (b) 1/8
  • (c) 1/e³ ≈ 0.0498
  • (d) 1/3
(c) N/N₀ = e^(−t/τ) = e^(−3) ≈ 0.0498. (Note: 3 half-lives gives 1/8.)

Q3 (Short Answer). Differentiate clearly between fission and fusion in terms of: (i) the type of nuclei involved, (ii) energy released per nucleon.

Fission: a heavy nucleus (A > 200) splits into two intermediate-mass fragments. Energy released per nucleon ~ 0.9 MeV (e.g. 200 MeV / 235 nucleons). Fusion: two light nuclei (A < 10) merge into a heavier one. Energy released per nucleon is much higher — e.g. d-t fusion gives 17.6 MeV / 5 nucleons ≈ 3.5 MeV/nucleon.

Q4 (Numerical). For half-wave radioactive decay: a sample has activity 6400 Bq initially. After 4 half-lives, what is the activity?

R = R₀ × (1/2)⁴ = 6400 / 16 = 400 Bq.

Q5 (HOTS). Why does NCERT say nuclear processes release ~10⁶ times the energy of chemical processes for the same mass of fuel?

Chemical bonds involve binding energies of a few eV per atom (rearranging outer electrons). Nuclear binding energies are MeV per nucleon (rearranging the strong-force-bound nucleons). The ratio is roughly MeV / eV = 10⁶. So per kilogram, nuclear fuels release roughly a million times more energy than chemical fuels.

Final Assertion–Reason Set

Options: (A) Both true, R correct explanation. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: Q-value of an exothermic nuclear reaction is positive.

Reason: The total mass of products is less than the total mass of reactants, and the missing mass appears as released kinetic energy.

(A) Both correct; the reason explains the assertion.

Assertion: The fission of ⁵⁶Fe is energetically forbidden.

Reason: ⁵⁶Fe lies at the maximum of the binding-energy-per-nucleon curve.

(A) Both correct and the reason explains the assertion. Splitting Fe would produce less-tightly-bound fragments — Q < 0.

Assertion: A control rod made of cadmium can stop a nuclear reactor.

Reason: Cadmium is an efficient absorber of thermal neutrons.

(A) Both correct; pushing in cadmium rods captures neutrons, reducing k below 1 and stopping the chain reaction.
Onward! You now have the toolkit to compute binding energies, predict decay rates, evaluate Q-values for reactions, and explain why fission and fusion both release energy. Next chapter (Ch 14) opens up the world of semiconductor electronics — the physics behind every modern device.

Frequently Asked Questions - NCERT Exercises and Solutions: Nuclei

What are the key NCERT exercise types in Chapter 13 Nuclei?
NCERT Class 12 Physics Chapter 13 Nuclei exercises cover conceptual MCQs, short numerical problems (2-3 marks), long numerical derivations (5 marks), and assertion-reason questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT exercise question, aligned with the CBSE board exam pattern. Students should focus on dimensional analysis, formula application, and unit consistency to score full marks.
How should students approach numerical problems in Nuclei?
For numerical problems in NCERT Class 12 Physics Chapter 13 Nuclei: (1) list all given data with units, (2) write the relevant formula(e), (3) substitute values carefully, (4) calculate with proper significant figures, (5) state the final answer with the correct unit. Always draw a free-body or schematic diagram where relevant. The MyAiSchool solutions follow this 5-step CBSE-aligned format consistently.
What are the most-asked CBSE board questions from Chapter 13?
From NCERT Class 12 Physics Chapter 13 (Nuclei), the most-asked CBSE board questions test conceptual understanding of fundamental principles, application of derived formulas to standard scenarios, and proof-style derivations. 5-mark questions usually combine derivation + application. The MyAiSchool exercise set tags each question by board frequency so students can prioritize high-yield problems before exams.
How do I check the dimensional correctness of my answer?
Dimensional correctness in NCERT Class 12 Physics is verified by ensuring the LHS and RHS of any equation have the same dimensional formula. For Chapter 13 Nuclei problems, write the dimensions of each quantity, substitute, and simplify. If the dimensions match, the equation is dimensionally valid (necessary but not sufficient). The MyAiSchool solutions include dimensional checks in every numerical answer.
What are common mistakes students make in Chapter 13 exercises?
Common mistakes in NCERT Class 12 Physics Chapter 13 Nuclei exercises include: (1) unit conversion errors (CGS vs SI), (2) sign convention mistakes for charges/currents/vectors, (3) forgetting to consider all field/force contributions, (4) algebra errors during derivations, (5) misreading the problem. The MyAiSchool solutions highlight these traps with red-flag annotations so students learn to avoid them.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Physics Chapter 13 Nuclei solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with reasoning, dimensional checks, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Each solution is verified against NCERT and CBSE marking schemes for accuracy.
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