NCERT Solutions for Class 12 Physics Chapter 13: Nuclei

These Class 12 Physics Chapter 13 solutions cover the complete NCERT “Nuclei” exercise for session 2026–27. Every exercise question is reproduced verbatim and solved step by step, with each numerical answer verified against the official NCERT answer key and carrying correct units. The chapter explores the structure of the nucleus — its size, mass, binding energy, nuclear force, radioactivity, and the energy released in fission and fusion.

Class: 12 Subject: Physics Chapter: 13 Title: Nuclei Exercises: 13.1–13.10 Session: 2026–27
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Class 12 Physics Chapter 13 Nuclei – Overview

The nucleus holds more than 99.9% of an atom’s mass in a region about 104 times smaller than the atom. Chapter 13 builds a complete picture of this tiny, dense core. It begins with atomic masses and the atomic mass unit (u), the discovery of the neutron, and the composition of nuclei in terms of atomic number Z, neutron number N, and mass number A. It then derives the nuclear radius formula R = R0A1/3, showing that nuclear density is constant. The heart of the chapter is mass–energy equivalence (E = mc2), the mass defect, and the binding energy per nucleon curve, which explains why energy is released in both nuclear fission of heavy nuclei and nuclear fusion of light nuclei — the source of the Sun’s energy. The chapter also introduces the strong nuclear force and the three types of radioactive decay (α, β, γ).

Key Concepts & Definitions

Atomic mass unit (u): 1u = 1/12 the mass of a 12C atom = 1.660539 × 10−27 kg, equivalent to 931.5 MeV/c2.

Nucleons: protons (charge +e, mp = 1.00727 u) and neutrons (neutral, mn = 1.00866 u); a free neutron is unstable (mean life ≈ 1000 s) but stable inside the nucleus.

Z, N, A: Z = number of protons (atomic number), N = number of neutrons, A = Z + N = mass number = number of nucleons. A nuclide is written AZX.

Isotopes / isobars / isotones: same Z (e.g. 21H, 31H) / same A (e.g. 31H, 32He) / same N respectively.

Mass defect (ΔM): the nuclear mass is always less than the sum of the masses of its free nucleons; ΔM = [Zmp + (A − Z)mn] − M.

Binding energy (Eb): energy released when nucleons combine, or energy needed to separate them; Eb = ΔMc2. Binding energy per nucleon Ebn = Eb/A is nearly constant (≈ 8 MeV) for 30 < A < 170, peaking near 8.75 MeV at A = 56 (56Fe).

Nuclear force: short-ranged (< few fm), much stronger than the Coulomb force, charge-independent, and attractive beyond ≈ 0.8 fm but strongly repulsive at shorter distances.

Radioactivity: spontaneous decay of unstable nuclei by α-decay (emits 42He), β-decay (emits electrons/positrons), or γ-decay (high-energy photons).

Important Formulas & Constants

Mass–energy: E = mc2  •  1u = 931.5 MeV/c2.

Mass defect: ΔM = [Zmp + (A − Z)mn] − M (using nuclear masses), or with atomic masses ΔM = [ZmH + (A − Z)mn] − matom (electron masses cancel).

Binding energy: Eb = ΔM × 931.5 MeV (when ΔM is in u); Ebn = Eb/A.

Nuclear radius: R = R0A1/3, R0 = 1.2 fm  ⇒  R1/R2 = (A1/A2)1/3; nuclear density ρ ≈ 2.3 × 1017 kg m−3 (independent of A).

Q-value: Q = (sum of initial masses − sum of final masses)c2. Q > 0 → exothermic; Q < 0 → endothermic.

Useful data: 1 MeV = 1.6 × 10−13 J • NA = 6.023 × 1023 per mole • 1 year = 3.154 × 107 s • mH = 1.007825 u • mn = 1.008665 u.

NCERT Solutions — Exercises (13.1 to 13.10)

Questions reproduced verbatim from the NCERT textbook. Each numerical answer is solved step by step and matches the official NCERT answer key.

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

SOLUTION For 147N: Z = 7 protons, N = A − Z = 14 − 7 = 7 neutrons. Using atomic masses (electron masses cancel), mH = 1.007825 u, mn = 1.008665 u. Sum of nucleon masses = 7 × 1.007825 + 7 × 1.008665 = 7.054775 + 7.060655 = 14.115430 u. Mass defect ΔM = 14.115430 − 14.00307 = 0.112360 u. Binding energy Eb = ΔM × 931.5 MeV = 0.112360 × 931.5 ≈ 104.7 MeV.

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

SOLUTION For 5626Fe: Z = 26, N = 56 − 26 = 30. Sum of nucleon masses = 26 × 1.007825 + 30 × 1.008665 = 26.20345 + 30.25995 = 56.46340 u. ΔM = 56.46340 − 55.934939 = 0.528461 u ⇒ Eb = 0.528461 × 931.5 ≈ 492.26 MeV. Binding energy per nucleon = 492.26 / 56 ≈ 8.79 MeV. For 20983Bi: Z = 83, N = 209 − 83 = 126. Sum of nucleon masses = 83 × 1.007825 + 126 × 1.008665 = 83.649475 + 127.091790 = 210.741265 u. ΔM = 210.741265 − 208.980388 = 1.760877 u ⇒ Eb = 1.760877 × 931.5 ≈ 1640.26 MeV. Binding energy per nucleon = 1640.26 / 209 ≈ 7.84 MeV. (Fe is more tightly bound than Bi, consistent with the binding-energy curve.)

13.3 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).

SOLUTION Number of atoms N = (mass / atomic mass) × NA = (3.0 / 63) × 6.023 × 1023 = 2.868 × 1022 atoms. Binding energy of one 6329Cu nucleus: Z = 29, N = 63 − 29 = 34. Sum of nucleon masses = 29 × 1.007825 + 34 × 1.008665 = 29.226925 + 34.29461 = 63.521535 u. ΔM per nucleus = 63.521535 − 62.92960 = 0.591935 u ⇒ Eb = 0.591935 × 931.5 ≈ 551.38 MeV. Total energy = N × Eb = 2.868 × 1022 × 551.38 ≈ 1.584 × 1025 MeV. In joules: 1.584 × 1025 × 1.6 × 10−13 J ≈ 2.535 × 1012 J.

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

SOLUTION Since R = R0A1/3, the ratio depends only on mass numbers: RAu / RAg = (AAu / AAg)1/3 = (197 / 107)1/3 = (1.8411)1/3. (1.8411)1/3 ≈ 1.23. The gold nucleus has a radius about 1.23 times that of the silver nucleus.

13.5 The Q value of a nuclear reaction A + b → C + d is defined by Q = [mA + mb − mC − md]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 + 42HeAtomic masses are given to be m(21H) = 2.014102 u, m(31H) = 3.016049 u, m(126C) = 12.000000 u, m(2010Ne) = 19.992439 u.

SOLUTION Atomic masses may be used directly because the number of electrons is the same on both sides, so electron masses cancel. Take m(11H) = 1.007825 u and m(42He) = 4.002603 u. (i) Q = [m(11H) + m(31H) − 2m(21H)]c2. Δm = (1.007825 + 3.016049) − (2 × 2.014102) = 4.023874 − 4.028204 = −0.004330 u. Q = −0.004330 × 931.5 ≈ −4.03 MeV. Q is negative → the reaction is endothermic. (ii) Q = [2m(126C) − m(2010Ne) − m(42He)]c2. Δm = (2 × 12.000000) − (19.992439 + 4.002603) = 24.000000 − 23.995042 = 0.004958 u. Q = 0.004958 × 931.5 ≈ 4.62 MeV. Q is positive → the reaction is exothermic.

13.6 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.

SOLUTION The reaction is 5626Fe → 2 × 2813Al. (Note 26 = 2 × 13 and 56 = 2 × 28, so charge and nucleon numbers balance.) Q = [m(5626Fe) − 2m(2813Al)]c2. Δm = 55.93494 − (2 × 27.98191) = 55.93494 − 55.96382 = −0.02888 u. Q = −0.02888 × 931.5 ≈ −26.90 MeV. Q is negative, so energy must be supplied. The fission of 56Fe into two 28Al fragments is not energetically possible — expected, because 56Fe sits near the peak of the binding-energy curve, so it cannot split spontaneously.

13.7 The fission properties of 23994Pu are very similar to those of 23592U. 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 23994Pu undergo fission?

SOLUTION Number of atoms in 1 kg (1000 g) of 239Pu = (mass / molar mass) × NA = (1000 / 239) × 6.023 × 1023. N = 4.1841 × 6.023 × 1023 = 2.520 × 1024 atoms. Total energy = N × 180 MeV = 2.520 × 1024 × 180 ≈ 4.536 × 1026 MeV. (In joules: 4.536 × 1026 × 1.6 × 10−13 J ≈ 7.26 × 1013 J — an enormous amount from just 1 kg of fuel.)

13.8 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as 21H + 21H → 32He + n + 3.27 MeV.

SOLUTION Number of deuterium atoms in 2.0 kg (2000 g): N = (2000 / 2) × 6.023 × 1023 = 6.023 × 1026 atoms. Each fusion reaction consumes 2 deuterons and releases 3.27 MeV, so number of reactions = N/2 = 3.0115 × 1026. Total energy E = 3.0115 × 1026 × 3.27 MeV = 9.848 × 1026 MeV. Convert to joules: E = 9.848 × 1026 × 1.6 × 10−13 J = 1.576 × 1014 J. Time t = E / Power = 1.576 × 1014 / 100 = 1.576 × 1012 s. In years: t = 1.576 × 1012 / (3.154 × 107) ≈ 4.9 × 104 years.

13.9 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.)

SOLUTION When two deuterons (each charge +e, radius r = 2.0 fm) just touch, their centres are separated by d = 2r = 4.0 fm = 4.0 × 10−15 m. The barrier height equals the Coulomb potential energy: U = (1/4πε0) × e² / d, with 1/(4πε0) = 9 × 109 N m² C−2. U = (9 × 109 × (1.6 × 10−19)2) / (4.0 × 10−15) = (9 × 109 × 2.56 × 10−38) / (4.0 × 10−15). U = (2.304 × 10−28) / (4.0 × 10−15) = 5.76 × 10−14 J. Convert to keV: U = (5.76 × 10−14) / (1.6 × 10−19) eV = 3.6 × 105 eV = 360 keV. (This 360 keV is shared equally, so each deuteron needs about 180 keV of kinetic energy.)

13.10 From the relation R = R0A1/3, where R0 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).

SOLUTION Nuclear density ρ = mass of nucleus / volume of nucleus. Mass of a nucleus ≈ A × m, where m is the average mass of one nucleon (≈ 1.66 × 10−27 kg). Volume of nucleus (assumed spherical) = (4/3)πR3 = (4/3)π(R0A1/3)3 = (4/3)πR03A. Therefore ρ = (A × m) / [(4/3)πR03A] = m / [(4/3)πR03] = 3m / (4πR03). The mass number A cancels completely, so ρ depends only on constants. Hence nuclear matter density is independent of A — it is the same (≈ 2.3 × 1017 kg m−3) for all nuclei. Different nuclei behave like drops of a liquid of constant density.

Extra Practice Questions

Short Answer Type Questions

Q1. Define isotopes, isobars and isotones with one example of each.

ANSWERIsotopes have the same atomic number Z but different mass number A (e.g. 11H, 21H, 31H). Isobars have the same A but different Z (e.g. 31H and 32He). Isotones have the same neutron number N but different Z (e.g. 19880Hg and 19779Au).

Q2. Why is the mass of a nucleus always less than the sum of the masses of its constituent nucleons?

ANSWERWhen nucleons bind to form a nucleus, energy (the binding energy) is released. By Einstein’s relation E = mc2, this released energy corresponds to a loss of mass called the mass defect, ΔM. Hence the bound nucleus is lighter than its separated nucleons by ΔM = Eb/c2.

Q3. State two key properties of the nuclear force.

ANSWER(i) It is a short-range force, effective only up to a few femtometres, becoming negligible beyond that (causing saturation). (ii) It is charge-independent — the force between p–p, n–n and p–n is essentially the same — and is far stronger than the Coulomb force at nuclear distances.

Q4. Why does a free neutron decay but a bound neutron remain stable?

ANSWERA free neutron is slightly more massive than a proton, so it decays into a proton, an electron and an antineutrino with a mean life of about 1000 s. Inside a stable nucleus, this decay is energetically forbidden because the resulting nucleus would have a higher total energy; the strong binding makes the bound neutron stable.

Q5. Calculate the energy equivalent of 1 g of matter completely converted to energy. (c = 3 × 108 m s−1)

ANSWERE = mc2 = 10−3 × (3 × 108)2 = 10−3 × 9 × 1016 = 9 × 1013 J — an enormous amount of energy from just one gram.

Long Answer Type Questions

Q1. Draw and explain the main features of the binding-energy-per-nucleon versus mass-number curve, and use it to explain why both fission and fusion release energy.

ANSWERThe curve plots binding energy per nucleon (Ebn) against mass number A. It rises steeply for light nuclei, reaches a broad maximum of about 8.75 MeV near A = 56 (56Fe), then falls slowly to about 7.6 MeV at A = 238. In the middle region (30 < A < 170), Ebn is nearly constant (≈ 8 MeV) due to the short-range, saturating nuclear force. A higher Ebn means a more tightly bound, lower-energy nucleus. Fission: when a heavy nucleus (A ≈ 240, Ebn ≈ 7.6 MeV) splits into two medium nuclei (A ≈ 120, Ebn ≈ 8.5 MeV), the fragments are more tightly bound, so energy (≈ 200 MeV per fission) is released. Fusion: when two very light nuclei (low Ebn) join to form a heavier nucleus with higher Ebn, the product is more tightly bound and energy is again released — this powers the Sun.

Q2. Describe the proton–proton cycle by which the Sun produces energy, and state the net result.

ANSWERIn the Sun’s core (T ≈ 1.5 × 107 K) hydrogen is burned to helium through the p–p cycle: (i) 11H + 11H → 21H + e+ + ν + 0.42 MeV; (ii) e+ + e− → γ + γ + 1.02 MeV; (iii) 21H + 11H → 32He + γ + 5.49 MeV; (iv) 32He + 32He → 42He + 11H + 11H + 12.86 MeV. The first three steps must occur twice for the fourth to happen. The net effect is that four hydrogen nuclei combine to form one 42He nucleus, releasing about 26.7 MeV of energy along with neutrinos and gamma photons.

Q3. Explain mass defect and binding energy, and outline how the binding energy of a nucleus is calculated, using 168O as an example.

ANSWERThe mass defect ΔM = [Zmp + (A − Z)mn] − M is the difference between the total mass of the free nucleons and the actual nuclear mass M. By E = mc2, this missing mass corresponds to the binding energy Eb = ΔMc2 — the energy needed to break the nucleus into free nucleons, or released when they combine. For 168O (8 protons, 8 neutrons): expected mass = 8(1.00727) + 8(1.00866) + 8(0.00055) electrons = 16.12744 u; the measured nuclear mass is 15.99053 u, giving ΔM = 0.13691 u. Thus Eb = 0.13691 × 931.5 ≈ 127.5 MeV, and the binding energy per nucleon is about 7.97 MeV — a highly stable nucleus.

MCQs & Assertion–Reason

1. The radius of a nucleus of mass number A is proportional to:

(a) A    (b) A2    (c) A1/3    (d) A1/2

2. The energy equivalent of 1 atomic mass unit (1u) is approximately:

(a) 931.5 MeV    (b) 1.6 MeV    (c) 511 keV    (d) 8 MeV

3. Nuclear density is:

(a) directly proportional to A    (b) inversely proportional to A    (c) independent of A    (d) proportional to A1/3

4. 31H and 32He are examples of:

(a) isotopes    (b) isobars    (c) isotones    (d) isomers

5. The binding energy per nucleon is maximum (about 8.75 MeV) for a nucleus near mass number:

(a) 2    (b) 56    (c) 120    (d) 238

6. In β−-decay, the particle emitted from the nucleus is:

(a) a helium nucleus    (b) a proton    (c) an electron    (d) a photon

7. The mass defect of a nucleus is the:

(a) mass of the nucleus    (b) total mass of nucleons minus nuclear mass    (c) mass of the electrons    (d) mass of the neutrons only

8. Energy is released in nuclear fusion because:

(a) the product nucleus is less tightly bound    (b) the product nucleus has higher binding energy per nucleon    (c) mass increases    (d) charge is destroyed

9. The nuclear force between two nucleons is:

(a) long-range and charge-dependent    (b) short-range and charge-independent    (c) weaker than gravity    (d) always repulsive

10. The average energy released per fission of 235U is of the order of:

(a) a few eV    (b) a few keV    (c) about 200 MeV    (d) about 2 GeV

Answer key: 1-(c), 2-(a), 3-(c), 4-(b), 5-(b), 6-(c), 7-(b), 8-(b), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The density of nuclear matter is the same for all nuclei.

Reason: Since R = R0A1/3, the volume of a nucleus is proportional to its mass A, so the mass-to-volume ratio is constant.

A-R 2. Assertion: Energy is released both when a heavy nucleus undergoes fission and when light nuclei fuse.

Reason: In both processes the products have a higher binding energy per nucleon than the reactants.

A-R 3. Assertion: A free neutron is stable.

Reason: A neutron carries no electric charge.

A-R 4. Assertion: The nuclear force is charge-independent.

Reason: The force between two protons, two neutrons, or a proton and a neutron is approximately the same.

A-R 5. Assertion: The fission of an 56Fe nucleus into two equal fragments is not energetically possible.

Reason: 56Fe lies near the peak of the binding-energy-per-nucleon curve, so it is one of the most stable nuclei.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Common Mistakes & Exam Tips

Common mistakes to avoid

  • Mixing up atomic mass and nuclear mass — when given atomic masses, the electron masses cancel only if electron numbers balance on both sides; otherwise account for them.
  • Forgetting to compute neutrons as N = A − Z before finding the mass defect.
  • Using the wrong conversion: 1 u = 931.5 MeV/c2, and 1 MeV = 1.6 × 10−13 J (not 10−19 J).
  • Confusing binding energy (Eb) with binding energy per nucleon (Ebn = Eb/A).
  • Getting the sign of Q wrong: Q > 0 means exothermic (energy released), Q < 0 means endothermic (energy absorbed).
  • In the deuteron-barrier problem, using the radius instead of the centre-to-centre separation d = 2r.

How to score full marks in this chapter

For binding-energy numericals, always show the mass-defect line clearly, then multiply by 931.5 to get MeV — examiners award method marks. Memorise the key constants (mH = 1.007825 u, mn = 1.008665 u, 1 u = 931.5 MeV/c2, 1 year = 3.154 × 107 s). For the binding-energy curve, learn the peak (≈ 8.75 MeV at A = 56) and use it to argue both fission and fusion. State units at every step and round only the final answer to match the NCERT key.

Frequently Asked Questions

What is Class 12 Physics Chapter 13 Nuclei about?

Chapter 13, Nuclei, studies the structure and properties of the atomic nucleus — its size, mass, the atomic mass unit, mass defect and binding energy, the strong nuclear force, radioactivity, and the energy released in nuclear fission and fusion (the Sun’s energy source).

How many exercises are in Class 12 Physics Chapter 13?

There are 10 exercise questions, numbered 13.1 to 13.10. Most are numericals on binding energy, Q-value, nuclear radius and fission/fusion energy; all are solved step by step on this page with answers matching the NCERT key.

What is the binding energy of the nitrogen nucleus in Exercise 13.1?

For 147N the mass defect is 0.112360 u, giving a binding energy of 0.112360 × 931.5 ≈ 104.7 MeV.

Are these Class 12 Physics Chapter 13 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 12 Physics are free and follow the official NCERT textbook for session 2026–27, with every numerical verified against the NCERT answer key.

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