PhysicsCBSE Class 12 · Chapter 13

Nuclei Class 12: Notes, Binding Energy and Solved Numericals

Q: Is radioactivity in the 2026-27 syllabus?

No. The law of radioactive decay, half-life and mean life, and alpha, beta and gamma decay are not in the 2026-27 syllabus. Neither is the nuclear reactor.

Q: Why is the binding energy per nucleon nearly constant for middle-sized nuclei?

The nuclear force is short-ranged, so each nucleon interacts only with its nearest neighbours. Once a nucleus is large enough, adding more nucleons does not change how many neighbours each one has, so the binding per nucleon saturates at about 8 MeV.

Q: Why do both fission and fusion release energy?

Both move nuclei towards the peak of the binding energy per nucleon curve (near A = 56). The products are more tightly bound than the reactants, so their total mass is smaller and the mass difference is released as energy.

Q: Is the mass of a nucleus equal to the sum of its nucleons' masses?

No. It is always less; the difference is the mass defect, equivalent to the binding energy.

Q: What questions come most from this chapter?

In our question bank: binding energy per nucleon (curve and calculations) and nuclear density, mostly as 1-mark items and 3-mark questions.

Q: Why is a neutron used to start fission rather than a proton?

A neutron has no charge, so it is not repelled by the nucleus and can enter it easily even at low speed.

Updated 14 October 2026 · Fact-checked

Nuclei is Chapter 13 of CBSE Class 12 Physics. It covers the composition and size of the nucleus, the atomic mass unit, the nuclear force, the mass-energy relation $E = mc^2$, mass defect and binding energy, the binding energy per nucleon curve and its variation with mass number, and nuclear fission and fusion.

What Nuclei is about

Rutherford showed that an atom's positive charge and almost all its mass sit in a tiny nucleus. This chapter looks inside it. A nucleus contains Z protons and N neutrons (together, nucleons), with mass number A = Z + N. Its radius grows as $A^{1/3}$ , which means every nucleus has about the same enormous density.

The mass of a nucleus is always less than the total mass of its separate protons and neutrons. This mass defect ΔM has been converted, by Einstein's $E = m c^{2}$ , into the binding energy that holds the nucleus together. Plotting binding energy per nucleon against A gives a curve that rises steeply, peaks near A = 56 (iron) and falls slowly for heavy nuclei. The curve explains why splitting a heavy nucleus (fission) or joining two light nuclei (fusion) releases energy. The force that holds nucleons together, the nuclear force, is far stronger than the electric repulsion between protons but acts only over a few femtometres.

Radioactivity (decay law, half-life, alpha, beta and gamma decay) and the nuclear reactor are not in the 2026-27 syllabus. In our question bank this chapter appears in all eight board papers from 2023 to 2026, mostly as 1-mark items and 3-mark questions on binding energy per nucleon and nuclear density.

Weightage: This chapter is part of Units VII and VIII, Dual Nature of Radiation and Matter; Atoms and Nuclei, which carry 12 of the 70 theory marks together with Dual Nature of Radiation and Matter and Atoms.Source: CBSE Physics syllabus 2026-27, Units VII and VIII Dual Nature of Radiation and Matter; Atoms and Nuclei

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 3.4 marks per paper. See the questions

Key concepts

Composition of the nucleus and the atomic mass unit

NCERT section 13.2

A nucleus contains protons (charge +e) and neutrons (no charge), together called nucleons. A nuclide is written $_{Z}^{A} X$ , where Z is the atomic number (number of protons), A the mass number (number of nucleons) and N = A − Z the number of neutrons. The neutron, discovered by Chadwick in 1932, explained why nuclear masses are larger than Z proton masses.

Isotopes: same Z, different A (for example $_{1}^{1} H$ , $_{1}^{2} H$ , $_{1}^{3} H$ ).
Isobars: same A, different Z (for example $_{1}^{3} H$ and $_{2}^{3} H e$ ).
Isotones: same N, different Z (for example $_{80}^{198} H g$ and $_{79}^{197} A u$ , each with 118 neutrons).

Atomic and nuclear masses are measured in atomic mass units: 1 u is one-twelfth of the mass of a $^{12} C$ atom, $1 u = 1.660539 \times 1 0^{- 27} k g$ . Proton mass ≈ 1.00727 u; neutron mass ≈ 1.00866 u; mass of a hydrogen atom ≈ 1.00783 u.

Worked example

For the nuclide $_{26}^{56} F e$ , find the number of protons, neutrons and nucleons. Name one nuclide that is an isotone of $_{1}^{3} H$ . [1 mark]

Try it first, then show the solutionSolution

Protons Z = 26; nucleons A = 56; neutrons N = 56 − 26 = 30
$_{1}^{3} H$ has N = 2; $_{2}^{4} H e$ also has N = 2, so it is an isotone of tritium.

Answer: 26 protons, 30 neutrons, 56 nucleons; $_{2}^{4} H e$ is an isotone of $_{1}^{3} H$ .

Size of the nucleus and nuclear density

NCERT section 13.3

Scattering experiments show that the radius of a nucleus of mass number A is

$R = R_{0} A^{1/3}, R_{0} = 1.2 \times 1 0^{- 15} m = 1.2 f m$

So the volume $\frac{4}{3} π R^{3} \propto A$ : the volume is proportional to the number of nucleons. Since the mass is also proportional to A, the density of nuclear matter is the same for all nuclei, about $2.3 \times 1 0^{17} k g m^{- 3}$ , roughly $1 0^{14}$ times the density of ordinary matter such as water. Atoms are mostly empty space; nuclei are packed.

Worked example

Find (a) the ratio of the radii of $_{26}^{56} F e$ and $_{2}^{4} H e$ nuclei, (b) the approximate density of nuclear matter. ( $R_{0} = 1.2 f m$ , $1 u = 1.66 \times 1 0^{- 27} k g$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $\frac{R _{F e}}{R _{H e}} = (\frac{56}{4})^{1/3} = 1 4^{1/3} \approx 2.41$
(b) $ρ = \frac{A \times 1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π R _{0}^{3} A} = \frac{1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}}$
$\frac{4}{3} π (1.2 \times 1 0^{- 15})^{3} \approx 7.24 \times 1 0^{- 45} m^{3}$ , so $ρ \approx 2.3 \times 1 0^{17} k g m^{- 3}$ , independent of A

Answer: (a) about 2.4 (b) about $2.3 \times 1 0^{17} k g m^{- 3}$

Mass-energy relation

NCERT section 13.4.1

Einstein's special theory of relativity shows that mass is a form of energy:

$E = m c^{2}$

where $c = 3 \times 1 0^{8} m s^{- 1}$ . Mass and energy can be converted into each other, and in any reaction the total of mass-energy is conserved. Because $c^{2}$ is enormous, a tiny mass corresponds to a huge energy.

In nuclear physics it is convenient to use $1 u \equiv 931.5 M e V$ ( $1 u \times c^{2} = 931.5 M e V$ ), with $1 M e V = 1.6 \times 1 0^{- 13} J$ .

Worked example

Find the energy equivalent of 1 g of matter in joules. For how long could this run a 100 W lamp? [2 marks]

Try it first, then show the solutionSolution

$E = m c^{2} = 1 0^{- 3} \times (3 \times 1 0^{8})^{2} = 9 \times 1 0^{13} J$
$t = \frac{E}{P} = \frac{9 \times 1 0 ^{13}}{100} = 9 \times 1 0^{11} s$
That is about $\frac{9 \times 1 0 ^{11}}{3.15 \times 1 0 ^{7}} \approx 2.9 \times 1 0^{4}$ years.

Answer: $9 \times 1 0^{13} J$ ; about $9 \times 1 0^{11} s$ (nearly 30 000 years)

Mass defect and binding energy

NCERT section 13.4.2

The mass of a nucleus is less than the sum of the masses of its free protons and neutrons. The difference is the mass defect:

$Δ M = [Z m_{p} + (A - Z) m_{n}] - M_{n u c l e u s}$

When the nucleons come together, this mass is converted to energy and released. The same energy must be supplied to pull the nucleus apart into separate nucleons: this is the binding energy

$E_{b} = Δ M c^{2}$ , or $E_{b} (MeV) = Δ M (u) \times 931.5$

When atomic masses are given, use the mass of the hydrogen atom in place of $m_{p}$ ; the electron masses then cancel. The binding energy is a measure of how strongly a nucleus is held together.

Worked example

Find the binding energy of a helium-4 nucleus. Atomic masses: $^{4} H e$ = 4.002603 u, $^{1} H$ = 1.007825 u; neutron mass 1.008665 u. Take 1 u = 931.5 MeV. [3 marks]

Try it first, then show the solutionSolution

Mass of 2 hydrogen atoms + 2 neutrons $= 2 (1.007825) + 2 (1.008665) = 2.015650 + 2.017330 = 4.032980 u$
$Δ M = 4.032980 - 4.002603 = 0.030377 u$
$E_{b} = 0.030377 \times 931.5 \approx 28.3 M e V$
Binding energy per nucleon $= \frac{28.3}{4} \approx 7.07 M e V$

Answer: About 28.3 MeV (about 7.07 MeV per nucleon)

Binding energy per nucleon and its variation with mass number

NCERT section 13.4.2

The binding energy per nucleon $E_{bn} = \frac{E _{b}}{A}$ is the average energy needed to remove one nucleon. Its graph against A shows:

1. It is roughly constant, about 8 MeV, for a wide middle range (about 30 < A < 170). This is because the nuclear force is short-ranged and saturates: each nucleon is bound only to its near neighbours, so adding more nucleons does not increase the binding of each.
2. It reaches a maximum of about 8.75 MeV near A = 56 (iron): these nuclei are the most stable.
3. It is lower for light nuclei (A < 30), where many nucleons are at the surface with fewer neighbours.
4. It falls slowly for heavy nuclei (A > 170, down to about 7.6 MeV for A = 238), because the long-range Coulomb repulsion between the many protons grows.

Consequences: a heavy nucleus splitting into two middle-sized ones (fission) ends up with nucleons more tightly bound, so energy is released. Two very light nuclei joining (fusion) also move up the curve and release energy.

Worked example

Using the binding energy curve, explain why energy is released when a nucleus with A = 240 splits into two nuclei with A = 120 each. Estimate the energy released if $E_{bn}$ is about 7.6 MeV for A = 240 and 8.5 MeV for A = 120. [2 marks]

Try it first, then show the solutionSolution

Energy is released because the product nucleons are more tightly bound: total binding energy increases.
Initial total $= 240 \times 7.6 = 1824 M e V$ ; final $= 240 \times 8.5 = 2040 M e V$
Energy released $= 2040 - 1824 = 216 M e V$

Answer: Products are more tightly bound; about 216 MeV is released.

The nuclear force

NCERT section 13.5

Inside the nucleus, protons repel one another strongly, yet the nucleus holds together. The binding comes from the nuclear force, with these properties:

It is much stronger than the Coulomb force (and the gravitational force), strong enough to overcome the repulsion between protons.
It is short-ranged: it falls rapidly to zero beyond a few femtometres. This is why it saturates and why $E_{bn}$ is nearly constant for middle-sized nuclei.
The potential energy between two nucleons has a minimum at about 0.8 fm: the force is attractive for separations greater than 0.8 fm and repulsive for smaller separations, which keeps nucleons from collapsing together.
It is (nearly) charge independent: the n-n, p-p and n-p nuclear forces are about the same.

It does not follow any simple formula like Coulomb's law.

Worked example

Two protons in a nucleus are about 2 fm apart. Estimate the Coulomb repulsion between them and explain why the nucleus does not fly apart. [2 marks]

Try it first, then show the solutionSolution

$F = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 2 \times 1 0 ^{- 15} ) ^{2}} = \frac{2.304 \times 1 0 ^{- 28}}{4 \times 1 0 ^{- 30}} \approx 58 N$
This is an enormous force on a particle of mass about $1 0^{- 27} k g$ .
The attractive nuclear force at this separation is much stronger still, so it holds the nucleons together.

Answer: About 58 N; the much stronger, short-range nuclear attraction overcomes it.

Nuclear fission

NCERT section 13.7.1

In fission, a heavy nucleus splits into two middle-sized nuclei. When uranium-235 absorbs a slow neutron, it forms excited uranium-236, which splits, for example:

$_{0}^{1} n +_{92}^{235} U \to_{92}^{236} U \to_{56}^{144} B a +_{36}^{89} K r + 3_{0}^{1} n$

Each fission releases about 200 MeV, because the products are more tightly bound (higher $E_{bn}$ ) than uranium. Most of this energy appears as the kinetic energy of the fragments and neutrons. The extra neutrons released can cause more fissions, giving a chain reaction. (The nuclear reactor is not in the 2026-27 syllabus.)

Compared with chemical reactions, which release a few eV per molecule, fission releases about a hundred million times more energy per reacting particle.

Worked example

Each fission of a U-235 nucleus releases about 200 MeV. Find the energy released by the complete fission of 1 kg of U-235. ( $N_{A} = 6.02 \times 1 0^{23} m o l^{- 1}$ , $1 M e V = 1.6 \times 1 0^{- 13} J$ ) [2 marks]

Try it first, then show the solutionSolution

Number of nuclei $= \frac{1000}{235} \times 6.02 \times 1 0^{23} \approx 2.56 \times 1 0^{24}$
Energy $= 2.56 \times 1 0^{24} \times 200 M e V = 5.12 \times 1 0^{26} M e V$
$= 5.12 \times 1 0^{26} \times 1.6 \times 1 0^{- 13} \approx 8.2 \times 1 0^{13} J$

Answer: About $8.2 \times 1 0^{13} J$

Nuclear fusion

NCERT section 13.7.2

In fusion, two light nuclei join to form a heavier nucleus. Because light nuclei lie on the steep rising part of the $E_{bn}$ curve, the product is more tightly bound and energy is released, for example

$_{1}^{2} H +_{1}^{2} H \to_{2}^{3} H e + n + 3.27 M e V$

The nuclei must come close enough (a few fm) for the nuclear force to act, which means overcoming their Coulomb repulsion. This needs extremely high temperatures (of the order of $1 0^{7} K$ or more), so fusion is also called a thermonuclear reaction.

Energy in stars: in the sun, hydrogen is fused into helium through the proton-proton cycle; the net result is that four protons (with two electrons) become one helium-4 nucleus, releasing 26.7 MeV. Fusion releases more energy per kilogram of fuel than fission. Controlled fusion on earth is still being developed, because the hot fuel (a plasma) is very hard to confine.

Worked example

Using atomic masses $^{1} H$ = 1.007825 u and $^{4} H e$ = 4.002603 u, estimate the energy released when four hydrogen atoms fuse (overall) into one helium atom. (1 u = 931.5 MeV) [2 marks]

Try it first, then show the solutionSolution

$Δ m = 4 \times 1.007825 - 4.002603 = 4.031300 - 4.002603 = 0.028697 u$
$E = 0.028697 \times 931.5 \approx 26.7 M e V$

Answer: About 26.7 MeV

Physics in real life: Nuclei

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Mass-energy relation

Suppose a power station could convert the mass of a single paper clip of 1 g completely into electrical energy.

How much energy would that be, and for how many days could it supply a town using 50 MW?

Try it first, then show the solutionSolution

$E = m c^{2} = 1 0^{- 3} \times 9 \times 1 0^{16} = 9 \times 1 0^{13} J$
Time $= \frac{9 \times 1 0 ^{13}}{5 \times 1 0 ^{7}} = 1.8 \times 1 0^{6} s$
$\frac{1.8 \times 1 0 ^{6}}{86 400} \approx 21$ days

Answer: $9 \times 1 0^{13} J$ , enough for about 21 days.

Memory hook: A gram of mass is a mountain of energy: c² does the multiplying.

Energy from fission

Imagine comparing burning 1 kg of coal (releasing about $3 \times 1 0^{7} J$ ) with fission of 1 kg of uranium-235.

Suppose each fission releases 200 MeV. How many times more energy does the uranium give than the coal?

Try it first, then show the solutionSolution

Uranium: about $8.2 \times 1 0^{13} J$ per kg (from $2.56 \times 1 0^{24}$ nuclei × 200 MeV)
Ratio $= \frac{8.2 \times 1 0 ^{13}}{3 \times 1 0 ^{7}} \approx 2.7 \times 1 0^{6}$

Answer: Nearly three million times more energy.

Memory hook: Chemistry trades eV per atom; nuclear physics trades MeV per nucleus.

Nuclear density

Suppose you could fill a teaspoon (volume $5 \times 1 0^{- 6} m^{3}$ ) with nuclear matter.

Using a nuclear density of $2.3 \times 1 0^{17} k g m^{- 3}$ , find the mass of that teaspoonful.

Try it first, then show the solutionSolution

$m = ρ V = 2.3 \times 1 0^{17} \times 5 \times 1 0^{- 6}$
$m \approx 1.2 \times 1 0^{12} k g$
That is about a billion tonnes in a teaspoon.

Answer: About $1.2 \times 1 0^{12} k g$ .

Memory hook: Atoms are empty; nuclei are packed solid.

Topics in this chapter

Nuclei formula sheet

Mass number: $A = Z + N$
Atomic mass unit: $1 u = 1.660539 \times 1 0^{- 27} k g \equiv 931.5 M e V / c^{2}$
Nuclear radius: $R = R_{0} A^{1/3}, R_{0} = 1.2 f m$
Nuclear density: $ρ = \frac{m}{\frac{4}{3} π R _{0}^{3}} \approx 2.3 \times 1 0^{17} k g m^{- 3}$; Independent of A.
Mass-energy relation: $E = m c^{2}$
Mass defect: $Δ M = [Z m_{p} + (A - Z) m_{n}] - M$
Binding energy: $E_{b} = Δ M c^{2} = Δ M (u) \times 931.5 M e V$
Binding energy per nucleon: $E_{bn} = \frac{E _{b}}{A}$
Energy from a reaction (Q value): $Q = (mass of reactants - mass of products) c^{2}$; Q > 0 means energy is released.
Unit conversion: $1 M e V = 1.6 \times 1 0^{- 13} J$

Nuclei: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Binding energy per nucleon against mass number

"Draw the curve and explain" (3-5 marks).

Draw

Draw axes: binding energy per nucleon (MeV, 0 to 10) against mass number A (0 to about 240).
Draw a steep rise from A = 1, reaching about 8 MeV by A ≈ 30, with a small peak at helium-4.
Continue to a broad maximum of about 8.75 MeV near A = 56.
Draw a gentle fall to about 7.6 MeV at A = 238.

Label

$E_{bn}$ (MeV)
Mass number A
Peak near A = 56 (Fe)
Fusion region (light nuclei)
Fission region (heavy nuclei)

Watch out: Mark arrows showing light nuclei moving up the curve by fusion and heavy nuclei moving up by fission.

Potential energy between two nucleons

Nuclear force questions (2-3 marks).

Draw

Draw axes: potential energy against separation r (fm).
Draw a curve that is steeply positive (repulsive) at very small r, falls to a minimum near r = 0.8 fm, then rises towards zero by about 2-3 fm.

Label

Potential energy
r (fm)
Minimum at about 0.8 fm
Repulsive region (r < 0.8 fm)
Attractive region (r > 0.8 fm)

Watch out: State that the force is attractive where the slope is positive (r > 0.8 fm) and repulsive where it is negative.

How the board asks Nuclei

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Nuclear size and density	Ratio of radii of two nuclei; show that density is independent of A; calculate density.	$R \propto A^{1/3}$ ; density = A u ÷ $(\frac{4}{3} π R_{0}^{3} A)$ , so A cancels.
Mass defect and binding energy	Calculate ΔM, $E_{b}$ and $E_{bn}$ from given masses; compare the stability of two nuclei.	Use atomic masses of H and the nucleus (electrons cancel); multiply ΔM in u by 931.5; divide by A; larger $E_{bn}$ means more stable.
Binding energy curve	Draw the curve, state its features and explain fission and fusion from it.	Sketch: rise, peak about 8.75 MeV at A ≈ 56, slow fall; link the flat part to saturation, the fall to Coulomb repulsion.
Nuclear force	Properties; potential energy against separation graph; why nuclei are stable.	Strong, short-range, charge independent, attractive beyond 0.8 fm and repulsive below; draw the PE curve with its minimum.
Fission and fusion	Energy per fission or fusion; energy from a mass of fuel; why fusion needs high temperatures; energy in stars.	Count nuclei from moles, multiply by energy per event, convert MeV to J; explain the Coulomb barrier.

Tricks to remember Nuclei

Short memory aids. Say each one aloud once, then use it on a question the same day.

Iron sits on top
Iron (A ≈ 56) is at the peak of the binding curve: lighter nuclei gain by fusing up, heavier ones by splitting down. Everything wants to become iron.
Use it when: Explaining fission and fusion.
931.5
Mass defect in u × 931.5 = binding energy in MeV.
Use it when: Every binding-energy numerical.
Volume ∝ A, so density is constant
R ∝ A^{1/3} ⇒ R³ ∝ A ⇒ volume ∝ mass ⇒ same density for all nuclei.
Use it when: Density questions.
Nuclear force: Strong, Short, Same, Saturating
Four S's: strong, short-ranged, the same for n-n, p-p, n-p, and saturating.
Use it when: Properties of the nuclear force.
Isotopes, isobars, isotones
isotoPes: same Protons; isobArs: same A; isotoNes: same Neutrons.
Use it when: Definitions.

Nuclei: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Isotones · 1 mark · MCQBoard level

Which pair of nuclides are isotones?
(a) $_{6}^{12} C$ and $_{6}^{14} C$
(b) $_{1}^{3} H$ and $_{2}^{3} H e$
(c) $_{6}^{13} C$ and $_{7}^{14} N$
(d) $_{8}^{16} O$ and $_{8}^{17} O$

Try it first, then show the solutionSolution

Isotones have the same N. $^{13} C$ : N = 7; $^{14} N$ : N = 7.

Answer: (c)

Example 2: Nuclear force · 1 mark · Assertion-reasonBoard level

Assertion (A): The binding energy per nucleon is nearly constant for nuclei with 30 < A < 170.
Reason (R): The nuclear force is short-ranged.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

A short-range force means each nucleon interacts only with its neighbours, so the binding per nucleon saturates. R explains A.

Answer: (a)

Example 3: Deuteron binding energy · 2 marks · Very short answerHarder board level

Find the binding energy of the deuteron in MeV and in joules. Masses: $^{2} H$ = 2.014102 u, $^{1} H$ = 1.007825 u, neutron 1.008665 u; 1 u = 931.5 MeV. [2 marks]

Try it first, then show the solutionSolution

$Δ M = 1.007825 + 1.008665 - 2.014102 = 0.002388 u$
$E_{b} = 0.002388 \times 931.5 \approx 2.22 M e V = 2.22 \times 1.6 \times 1 0^{- 13} \approx 3.56 \times 1 0^{- 13} J$

Answer: About 2.22 MeV ( $3.6 \times 1 0^{- 13} J$ )

Example 4: Size and density of a nucleus · 3 marks · Short answerHarder board level

(a) Find the radius of the $_{13}^{27} A l$ nucleus. (b) Show that nuclear density is independent of A and estimate it. ( $R_{0} = 1.2 f m$ , $1 u = 1.66 \times 1 0^{- 27} k g$ ) [3 marks]

Try it first, then show the solutionSolution

(a) $R = 1.2 \times 2 7^{1/3} = 1.2 \times 3 = 3.6 f m$
(b) $ρ = \frac{A \times 1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π R _{0}^{3} A}$ : A cancels.
$ρ = \frac{1.66 \times 1 0 ^{- 27}}{7.24 \times 1 0 ^{- 45}} \approx 2.3 \times 1 0^{17} k g m^{- 3}$

Answer: (a) 3.6 fm (b) about $2.3 \times 1 0^{17} k g m^{- 3}$ , independent of A

Example 5: Binding energy curve: full answer · 5 marks · Long answerHarder board level

(a) Define mass defect and binding energy. (b) Draw the binding energy per nucleon curve and explain its main features. (c) Calculate the binding energy per nucleon of helium-4 (atomic mass 4.002603 u; $m_{H}$ = 1.007825 u; $m_{n}$ = 1.008665 u) and say where it lies on the curve. [5 marks]

Try it first, then show the solutionSolution

(a) Mass defect: $Δ M = Z m_{p} + N m_{n} - M$ . Binding energy: $Δ M c^{2}$ , the energy needed to separate the nucleus into free nucleons.
(b) Curve rises steeply for light nuclei, is nearly flat at about 8 MeV for 30 < A < 170, peaks at about 8.75 MeV near A = 56, and falls to about 7.6 MeV for A = 238.
Flat region: short-range nuclear force saturates. Fall for heavy nuclei: Coulomb repulsion among many protons. Low values for light nuclei: many nucleons on the surface.
(c) $Δ M = 0.030377 u$ ; $E_{b} \approx 28.3 M e V$ ; $E_{bn} \approx 7.07 M e V$ .
Helium-4 stands out as a small peak above its light neighbours on the rising part of the curve: it is unusually stable for its size.

Answer: $E_{bn}$ of He-4 ≈ 7.07 MeV, a local peak on the rising part of the curve.

Example 6: Energy from fission · 3 marks · Short answerHarder board level

Each fission of U-235 releases about 200 MeV. A small fission source uses up 1 g of U-235 per day. Find the average power it produces. ( $N_{A} = 6.02 \times 1 0^{23}$ , $1 M e V = 1.6 \times 1 0^{- 13} J$ ) [3 marks]

Try it first, then show the solutionSolution

Nuclei in 1 g $= \frac{1}{235} \times 6.02 \times 1 0^{23} \approx 2.56 \times 1 0^{21}$
Energy $= 2.56 \times 1 0^{21} \times 200 \times 1.6 \times 1 0^{- 13} \approx 8.2 \times 1 0^{10} J$
Power $= \frac{8.2 \times 1 0 ^{10}}{86 400} \approx 9.5 \times 1 0^{5} W$ (about 1 MW)

Answer: About $9.5 \times 1 0^{5} W$ (roughly 1 MW)

Example 7: Is this fission possible? · 2 marks · Very short answerHOTS / competency

Suppose a nucleus of mass number 56 (binding energy per nucleon about 8.8 MeV) were to split into two equal nuclei of mass number 28 (binding energy per nucleon about 8.4 MeV). Would energy be released or absorbed? How much? [2 marks]

Try it first, then show the solutionSolution

Initial total binding energy $= 56 \times 8.8 = 492.8 M e V$ ; final $= 56 \times 8.4 = 470.4 M e V$
The products are less tightly bound, so 22.4 MeV must be supplied: energy is absorbed, and this fission cannot happen on its own.

Answer: Energy would have to be supplied (about 22.4 MeV); such a fission is not energetically possible.

Example 8: Why fusion needs heat · 3 marks · Short answerHOTS / competency

Two deuterons must come within about 2 fm of each other to fuse. Estimate the Coulomb potential energy at that separation in MeV, and explain why fusion needs temperatures of millions of kelvin. ( $\frac{e ^{2}}{4 π ε _{0}} = 1.44 M e V f m$ ) [3 marks]

Try it first, then show the solutionSolution

$U = \frac{e ^{2}}{4 π ε _{0} r} = \frac{1.44 M e V f m}{2 f m} = 0.72 M e V$
The deuterons must bring this energy as kinetic energy to get that close (the Coulomb barrier).
Average thermal energies of about $k T$ are only about 0.026 eV at room temperature; even at $1 0^{7} K$ they are about 1 keV. Only at such extreme temperatures do enough nuclei in the fast tail of the energy distribution get close enough, so fusion is called thermonuclear.

Answer: About 0.72 MeV; huge temperatures are needed to give nuclei enough energy to overcome this barrier.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Energy of the sun

Suppose a student reads that the sun produces energy by fusing hydrogen into helium. Overall, four hydrogen nuclei (with two electrons) become one helium-4 nucleus, releasing 26.7 MeV. Atomic masses: $^{1} H$ = 1.007825 u, $^{4} H e$ = 4.002603 u; 1 u = 931.5 MeV.

(i) · 1 mark
Name the process.
Answer
Answer: Nuclear fusion (thermonuclear)
Light nuclei join to form a heavier one.
(ii) · 1 mark
Verify the energy released from the masses.
Answer
Answer: About 26.7 MeV
$(4 \times 1.007825 - 4.002603) \times 931.5 = 0.028697 \times 931.5$
(iii) · 1 mark
Why can this happen in the core of the sun but not at room temperature?
Answer
Answer: Only at the sun's extremely high core temperature do nuclei have enough energy to overcome their Coulomb repulsion.
Fusion needs nuclei to come within the range of the nuclear force.
(iv) · 1 mark
Why does fusion of light nuclei release energy?
Answer
Answer: The product (helium) has a higher binding energy per nucleon than the reactants.
The total mass decreases and the mass difference appears as energy.

Comparing two nuclei

Suppose a table gives binding energies per nucleon: lithium-7, 5.6 MeV; iron-56, 8.8 MeV; uranium-238, 7.6 MeV.

(i) · 1 mark
Which of the three is the most stable?
Answer
Answer: Iron-56
Highest binding energy per nucleon.
(ii) · 1 mark
Find the total binding energy of iron-56.
Answer
Answer: About 493 MeV
$56 \times 8.8 = 492.8 M e V$
(iii) · 1 mark
Which of lithium-7 and uranium-238 could release energy by fusion, and which by fission?
Answer
Answer: Lithium-7 by fusion; uranium-238 by fission.
Both processes move towards the peak near A = 56.
(iv) · 1 mark
Why is $E_{bn}$ lower for uranium than for iron?
Answer
Answer: The Coulomb repulsion between its 92 protons is large, while each nucleon is still bound only to its near neighbours.
The long-range repulsion grows with Z while the short-range attraction per nucleon does not.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Show that the density of nuclear matter is the same for all nuclei. [2 marks] (2 marks)

Step	Marks
Mass ≈ A u; volume $= \frac{4}{3} π R^{3} = \frac{4}{3} π R_{0}^{3} A$ using $R = R_{0} A^{1/3}$	1
$ρ = \frac{A u}{\frac{4}{3} π R _{0}^{3} A} = \frac{u}{\frac{4}{3} π R _{0}^{3}}$ : no A, so the same for all nuclei	1

Calculate the binding energy per nucleon of the deuteron. Masses: $^{2} H$ = 2.014102 u, $^{1} H$ = 1.007825 u, neutron 1.008665 u; 1 u = 931.5 MeV. [3 marks] (3 marks)

Step	Marks
Mass of constituents $= 1.007825 + 1.008665 = 2.016490 u$	1
$Δ M = 2.016490 - 2.014102 = 0.002388 u$ ; $E_{b} = 0.002388 \times 931.5 \approx 2.22 M e V$	1
Per nucleon $= 2.22/2 \approx 1.11 M e V$	1

(a) Draw the curve of binding energy per nucleon against mass number and state three features. (b) Use it to explain why energy is released in fission and in fusion. (c) Estimate the energy released if a nucleus with A = 240 ( $E_{bn}$ ≈ 7.6 MeV) splits into two nuclei of A = 120 ( $E_{bn}$ ≈ 8.5 MeV). [5 marks] (5 marks)

Step	Marks
(a) Curve: steep rise for light nuclei, peak about 8.75 MeV near A = 56, slow fall to about 7.6 MeV at A = 238	1
Features: nearly constant for 30 < A < 170 (saturation); maximum near iron; lower for light and heavy nuclei	1
(b) Fission: heavy nucleus → middle nuclei with higher $E_{bn}$ , so energy released	1
Fusion: light nuclei → heavier nucleus with higher $E_{bn}$ , so energy released	1
(c) $240 \times (8.5 - 7.6) = 216 M e V$	1

Where students lose marks in Nuclei

The slips examiners see most often, and what to write instead.

Mistake: Forgetting to convert u to MeV or MeV to J
Mixing units in one line.
Do this: Keep ΔM in u, multiply by 931.5 to get MeV, and only then convert to J (× 1.6 × 10⁻¹³) if asked.
Usually costs: 1 mark.
Mistake: Using proton mass with atomic mass of the nucleus
Mixing nuclear and atomic masses.
Do this: If atomic masses are given, use the hydrogen atom mass (1.007825 u) for each proton; the electrons then cancel.
Usually costs: 1 mark (wrong ΔM).
Mistake: Saying binding energy is the energy stored in the nucleus and released when it breaks apart
Misreading the sign.
Do this: Binding energy is the energy that must be SUPPLIED to separate the nucleus into free nucleons (and that was released when it formed).
Usually costs: 1 mark.
Mistake: Drawing the binding curve peaking at uranium
Thinking big means stable.
Do this: The peak (about 8.75 MeV) is near A = 56; the curve falls slowly towards uranium.
Usually costs: 1 mark in graph questions.
Mistake: Writing about half-life or radioactive decay
Old syllabi included radioactivity.
Do this: Radioactivity and the nuclear reactor are not in the 2026-27 syllabus.
Usually costs: Time.

NCERT exercise guide: Nuclei

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 13.1 and 13.2 (binding energy)

What it tests: Binding energy of nitrogen-14, and comparing binding energies per nucleon of iron-56 and bismuth-209.

How to approach it: ΔM from atomic masses (hydrogen atom for each proton), × 931.5 MeV; divide by A to compare.

Same kind as Exercise 13.1

Find the binding energy and binding energy per nucleon of carbon-12. Atomic masses: $^{12} C$ = 12.000000 u, $^{1} H$ = 1.007825 u; neutron 1.008665 u; 1 u = 931.5 MeV.

Show the solutionSolution

$Δ M = 6 (1.007825) + 6 (1.008665) - 12.000000 = 6.046950 + 6.051990 - 12.000000 = 0.098940 u$
$E_{b} = 0.098940 \times 931.5 \approx 92.2 M e V$
$E_{bn} = 92.2/12 \approx 7.68 M e V$

Answer: 92.2 MeV; 7.68 MeV per nucleon

Exercises 13.3 and 13.4 (energy and size)

What it tests: Nuclear energy needed to separate all nucleons in a coin, and the ratio of radii of two nuclei.

How to approach it: Number of nuclei × binding energy per nucleus; $R \propto A^{1/3}$ .

Same kind as Exercise 13.4

Find the ratio of the nuclear radii of $^{216} X$ and $^{27} Y$ .

Show the solutionSolution

$\frac{R _{X}}{R _{Y}} = (\frac{216}{27})^{1/3} = 8^{1/3}$
$= 2$

Answer: 2 : 1

Exercises 13.5 to 13.8 (Q values, fission and fusion)

What it tests: Q value of reactions from masses, whether a fission is energetically possible, energy from fission of a given mass, and how long a lamp could run on fusion energy.

How to approach it: Q = (mass of reactants − mass of products) × 931.5 MeV; positive Q means energy released. Count nuclei with moles.

Same kind as Exercise 13.8

Each D-D fusion ( $_{1}^{2} H +_{1}^{2} H \to_{2}^{3} H e + n$ ) releases 3.27 MeV. How much energy is released when 1 g of deuterium undergoes this fusion completely? ( $N_{A} = 6.02 \times 1 0^{23}$ )

Show the solutionSolution

Number of deuterium nuclei $= \frac{1}{2} \times 6.02 \times 1 0^{23} = 3.01 \times 1 0^{23}$
Each reaction uses 2 nuclei: $1.505 \times 1 0^{23}$ reactions
$E = 1.505 \times 1 0^{23} \times 3.27 M e V \approx 4.92 \times 1 0^{23} M e V \approx 7.9 \times 1 0^{10} J$

Answer: About $7.9 \times 1 0^{10} J$

Nuclei: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 3.4 marks per paper; 16 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Nuclei

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 55/1/1, Q15 · 1 mark · Assertion-reasonCompetency-based

Reason (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons.
Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Both Assertion (A) and Reason (R) are false.

Try it first, then show our solutionSolution

The assertion is true: the mass defect exists.
The reason is false: energy is released, not absorbed, when nucleons bind (this is the binding energy).
1 mark

Answer: Option (C)

Where the marks are

A true and R false because binding releases energy (1 mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/1/1, Q21 · 2 marks · Very short answer

Draw the plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions that can be drawn from this plot.

Try it first, then show our solutionSolution

Graph: Plot potential energy $U$ (MeV) on the y-axis against separation $r$ (fm) on the x-axis.
- For small $r$ , $U$ is large and positive.
- $U$ falls steeply and crosses zero.
- It reaches a minimum (about $- 100$ MeV) at $r_{0} \approx 0.8$ fm.
- It then rises and approaches zero asymptotically for $r$ beyond about 2–3 fm.
1 mark
Conclusion 1: The nuclear force is attractive for separations greater than $r_{0} \approx 0.8$ fm, where the slope of $U$ is positive. It is strongly repulsive for separations less than $r_{0}$ , which keeps nucleons from collapsing into one another.
½ mark
Conclusion 2: The nuclear force is short-ranged. It becomes negligible beyond a few femtometres, since $U \to 0$ . It is maximally binding (minimum potential energy) near $r_{0} \approx 0.8$ fm.
½ mark

Answer: U–r curve with a minimum at about 0.8 fm. The force is repulsive below 0.8 fm, attractive above it, and short-ranged (vanishes beyond a few fm).

Where the marks are

Correct shape of the U versus r plot with its minimum near 0.8 fm (1 mark)
Attractive for r > r₀ and repulsive for r < r₀ (½ mark)
Short-range nature of the nuclear force (½ mark)

See it in the full 2026 board paper, set 55/1/1

2026, set 55/5/1, Q26(a) · 3 marks · Short answer

(i) Write any two features of nuclear forces.
(ii) If both the number of protons and the neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction ? Explain.

Try it first, then show our solutionSolution

(i) Features of nuclear forces (any two):
• They are very strong — much stronger than Coulomb force between protons.
• They are short-range — negligible beyond a few femtometres (attractive for separation > ~0.8 fm, strongly repulsive below it).
• They are charge independent — n–n, p–p and n–p forces are about the same.
1 mark
(ii) Although the total numbers of protons and neutrons are unchanged, the binding energies of the nuclei before and after the reaction are different. The mass of a nucleus is less than the sum of the masses of its nucleons by the mass defect $Δ m = B / c^{2}$ .
1 mark
Hence the total rest mass of the reactant nuclei differs from that of the product nuclei. If the products are more tightly bound, the difference in mass $Δ m$ appears as energy $Q = Δ m c^{2}$ (released); if less tightly bound, energy is absorbed and converted into mass. Thus mass–energy conversion takes place through the difference in binding energies.
1 mark

Answer: (i) Strong, short-range, charge-independent (any two); (ii) Difference in binding energy (mass defect) of initial and final nuclei appears as energy Q = Δm c²

Where the marks are

Any two correct features of nuclear force (½ + ½ = 1 mark)
Identifies that binding energies/mass defects of nuclei differ before and after reaction (1 mark)
Explains mass difference Δm converted to energy Q = Δmc² (or vice versa) (1 mark)

See it in the full 2026 board paper, set 55/5/1

Every Nuclei question, by year

For 2027

Sample paper 2026-27: Q7 (1 mark, solved), Q24 (3 marks, solved)

2026

Set 55/1/1: Q15 (1 mark, solved), Q21 (2 marks, solved)

Set 55/5/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q15 (1 mark, solved), Q26(a) (3 marks, solved)

Sample paper 2025-26: Q11 (1 mark, solved), Q24 (3 marks, solved)

2025

Set 55/1/1: Q11 (1 mark, solved), Q27 (3 marks, solved)

Set 55/7/1: Q12 (1 mark, solved), Q21 (2 marks, solved)

Sample paper 2024-25: Q15 (1 mark, solved), Q20 (2 marks, solved)

2024

Set 55/1/1: Q28 (3 marks, solved)

Set 55/5/1: Q11 (1 mark, solved)

Sample paper 2023-24: Q22 (3 marks, solved)

2023

Set 55/1/1: Q12 (1 mark, solved), Q30(b) (3 marks, solved)

Set 55/5/1: Q8 (1 mark, solved), Q19 (2 marks, solved)

Practice questions on Nuclei

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The nuclear radius of a nucleus with A = 64 compared with one with A = 8 is:
(a) 8 times
(b) 4 times
(c) 2 times
(d) the same
Hint
$R \propto A^{1/3}$ .
Answer
(c) 2 times, since $(64/8)^{1/3} = 2$ .
Q2 · 1 mark · MCQBoard level
The energy equivalent of 1 u is:
(a) 9.315 MeV
(b) 93.15 MeV
(c) 931.5 MeV
(d) 9315 MeV
Hint
Learn this number.
Answer
(c) 931.5 MeV
Q3 · 1 mark · Very short answerBoard level
Define isobars and give one example.
Hint
Same A.
Answer
Nuclides with the same mass number but different atomic numbers, e.g. $_{1}^{3} H$ and $_{2}^{3} H e$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mass of a nucleus is less than the sum of the masses of its nucleons.
Reason (R): Energy is released when nucleons combine to form a nucleus.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Mass-energy equivalence.
Answer
(a). The released binding energy corresponds to the missing mass, $Δ M = E_{b} / c^{2}$ .
Q5 · 2 marks · Very short answerHarder board level
Calculate the binding energy per nucleon of oxygen-16 and decide whether it is more or less stable than helium-4 ( $E_{bn} \approx 7.07$ MeV). Masses: $^{16} O$ = 15.994915 u, $^{1} H$ = 1.007825 u, $m_{n}$ = 1.008665 u; 1 u = 931.5 MeV.
Hint
Eight hydrogen atoms and eight neutrons; ΔM × 931.5, then divide by 16.
Answer
$Δ M = 8 (1.007825) + 8 (1.008665) - 15.994915 = 16.131920 - 15.994915 = 0.137005 u$ ; $E_{b} \approx 0.137005 \times 931.5 \approx 127.6 M e V$ ; $E_{bn} \approx 7.98 M e V$ . This is larger than 7.07 MeV, so oxygen-16 is more tightly bound (more stable) than helium-4.
Q6 · 2 marks · Very short answerHarder board level
The radius of a nucleus is three times the radius of a helium-4 nucleus. (a) Find its mass number. (b) Find both radii, taking $R_{0} = 1.2 f m$ . (c) What is the ratio of their nuclear densities?
Hint
$R \propto A^{1/3}$ , so $A \propto R^{3}$ .
Answer
(a) $A = 4 \times 3^{3} = 108$ . (b) $R_{H e} = 1.2 \times 4^{1/3} \approx 1.9 f m$ ; $R = 3 \times 1.9 \approx 5.7 f m$ (check: $1.2 \times 10 8^{1/3} \approx 5.7 f m$ ). (c) 1 : 1, because nuclear density does not depend on A.
Q7 · 3 marks · Short answerHarder board level
Draw the potential energy curve for a pair of nucleons and use it to state two properties of the nuclear force.
Hint
Minimum near 0.8 fm.
Answer
The curve is steeply positive at very small r, has a minimum near 0.8 fm and rises to zero by a few fm. Properties: attractive for r > 0.8 fm and repulsive for r < 0.8 fm (prevents collapse); short-ranged (negligible beyond a few fm). Also: much stronger than the Coulomb force, charge independent.
Q8 · 3 marks · Short answerHarder board level
A power station produces 300 MW of electrical power from the fission of U-235 and is 30% efficient. If each fission releases 200 MeV, how much U-235 does it use each day? ( $N_{A} = 6.02 \times 1 0^{23} m o l^{- 1}$ , $1 M e V = 1.6 \times 1 0^{- 13} J$ )
Hint
Find the fission power first (output ÷ efficiency), then the energy per day, then the number of fissions.
Answer
Fission power $= \frac{300}{0.30} = 1000 M W$ ; energy per day $= 1 0^{9} \times 86400 = 8.64 \times 1 0^{13} J$ . Energy per fission $= 200 \times 1.6 \times 1 0^{- 13} = 3.2 \times 1 0^{- 11} J$ , so fissions per day $= \frac{8.64 \times 1 0 ^{13}}{3.2 \times 1 0 ^{- 11}} = 2.7 \times 1 0^{24}$ . Mass $= \frac{2.7 \times 1 0 ^{24}}{6.02 \times 1 0 ^{23}} \times 235 g \approx 1.05 k g$ per day.
Q9 · 2 marks · Very short answerHOTS / competency
Explain, using the binding energy per nucleon curve, why heavy nuclei can undergo fission and light nuclei fusion, but iron nuclei can do neither with a release of energy.
Hint
Iron is at the peak.
Answer
Energy is released when nucleons end up more tightly bound. Heavy nuclei move up the curve by splitting; light nuclei by joining. Iron is at the peak: splitting it or fusing it would lower the binding per nucleon, so either process would absorb energy.
Q10 · 2 marks · Very short answerHOTS / competency
A nucleus of mass number 236 ( $E_{bn}$ = 7.6 MeV) splits into fragments of mass numbers 140 ( $E_{bn}$ = 8.3 MeV) and 94 ( $E_{bn}$ = 8.6 MeV), and two free neutrons. Find the energy released. Why do the free neutrons add nothing to the final binding energy?
Hint
Compare total binding energies; a free neutron is not bound to anything.
Answer
Initial: $236 \times 7.6 = 1793.6 M e V$ . Final: $140 \times 8.3 + 94 \times 8.6 = 1162.0 + 808.4 = 1970.4 M e V$ ; the two free neutrons have zero binding energy. Released: $1970.4 - 1793.6 = 176.8 M e V$ , mostly as kinetic energy of the fragments.
Q11 · 2 marks · Very short answerHarder board level
The binding energy per nucleon is about 1.11 MeV for the deuteron and about 7.07 MeV for helium-4. Find the energy released when two deuterons fuse to form one helium-4 nucleus, and explain why the deuteron is so loosely bound.
Hint
Energy released = total binding energy of products − total binding energy of reactants.
Answer
Reactants: $2 \times (2 \times 1.11) = 4.44 M e V$ . Product: $4 \times 7.07 = 28.28 M e V$ . Released: about 23.8 MeV (matching the mass-difference value $(2 \times 2.014102 - 4.002603) \times 931.5 \approx 23.8 M e V$ ). With only two nucleons, each has just one neighbour to bind to, so the deuteron is loosely bound.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define mass defect and binding energy.Show answer
Mass defect: the difference between the total mass of the separate nucleons and the mass of the nucleus. Binding energy: the energy equivalent of the mass defect, $Δ M c^{2}$ , needed to separate the nucleus into its nucleons.
2.Write the expression for nuclear radius.Show answer
$R = R_{0} A^{1/3}$ , $R_{0} = 1.2$ fm.
3.Why is nuclear density independent of mass number?Show answer
Volume ∝ R³ ∝ A and mass ∝ A, so mass/volume is constant.
4.Where does the binding energy per nucleon curve peak?Show answer
Near A = 56 (iron), at about 8.75 MeV.
5.Give three properties of the nuclear force.Show answer
Strongest force, short-ranged, charge independent (also: attractive beyond 0.8 fm, repulsive below, saturating).
6.Energy equivalent of 1 u?Show answer
931.5 MeV.
7.About how much energy is released per fission of U-235?Show answer
About 200 MeV.
8.Why does fusion need very high temperatures?Show answer
The nuclei must have enough kinetic energy to overcome their Coulomb repulsion and come within range of the nuclear force.

Best order to study it

Start with the composition of the nucleus, notation, isotopes, isobars, isotones and the atomic mass unit.
Learn nuclear size ( $R = R_{0} A^{1/3}$ ) and show that nuclear density is constant.
Study $E = m c^{2}$ and the equivalence 1 u = 931.5 MeV.
Calculate mass defect and binding energy, then binding energy per nucleon.
Draw and explain the binding energy curve, then learn the properties of the nuclear force.
Finish with fission and fusion: equations, energy per event, energy per kilogram and energy in stars.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the number of protons, neutrons and nucleons from $_{Z}^{A} X$ ?
Can you define and give examples of isotopes, isobars and isotones?
Can you find the ratio of radii of two nuclei and show density is independent of A?
Can you convert between u, MeV and J?
Can you calculate mass defect, binding energy and binding energy per nucleon?
Can you draw the $E_{bn}$ -A curve and mark its peak?
Can you explain the curve's flat region and its fall for heavy nuclei?
Can you list the properties of the nuclear force and draw its potential energy curve?
Can you write a fission equation and calculate energy from a given mass of uranium?
Can you explain fusion, the need for high temperature and energy production in the sun?

Nuclei: frequently asked questions

Is radioactivity in the 2026-27 syllabus?

No. The law of radioactive decay, half-life and mean life, and alpha, beta and gamma decay are not in the 2026-27 syllabus. Neither is the nuclear reactor.

Why is the binding energy per nucleon nearly constant for middle-sized nuclei?

The nuclear force is short-ranged, so each nucleon interacts only with its nearest neighbours. Once a nucleus is large enough, adding more nucleons does not change how many neighbours each one has, so the binding per nucleon saturates at about 8 MeV.

Why do both fission and fusion release energy?

Both move nuclei towards the peak of the binding energy per nucleon curve (near A = 56). The products are more tightly bound than the reactants, so their total mass is smaller and the mass difference is released as energy.

Is the mass of a nucleus equal to the sum of its nucleons' masses?

No. It is always less; the difference is the mass defect, equivalent to the binding energy.

What questions come most from this chapter?

In our question bank: binding energy per nucleon (curve and calculations) and nuclear density, mostly as 1-mark items and 3-mark questions.

Why is a neutron used to start fission rather than a proton?

A neutron has no charge, so it is not repelled by the nucleus and can enter it easily even at low speed.

Based on: CBSE Physics (042) syllabus 2026-27; NCERT Physics Part II, Textbook for Class XII, Chapter 13 (Reprint 2026-27)