PhysicsISC Class 12 · Chapter 13

Nuclei ISC Class 12: Notes, Binding Energy and Numericals

Updated 14 October 2026 · Fact-checked

Nuclei is Chapter 13 of ISC Class 12 Physics. It covers the composition and size of the nucleus, nuclear density, isotopes, isobars and isotones, the atomic mass unit, mass defect and binding energy, the binding-energy-per-nucleon curve, E = mc² with pair production and annihilation, nuclear fission, the nuclear reactor and nuclear fusion in stars.

What Nuclei is about

The nucleus holds almost all of an atom's mass in a region about $1 0^{- 15}$ m across. This chapter (sharing 6 marks with Atoms) explains what it is made of and why it can release enormous energy.

A nucleus $_{Z}^{A} X$ has Z protons and N = A − Z neutrons (nucleons). Isotopes share Z, isobars share A, isotones share N. Nuclear radius grows as $R = R_{0} A^{1/3}$ , so nuclear density is the same (about $2.3 \times 1 0^{17} k g m^{- 3}$ ) for all nuclei. Masses are measured in unified atomic mass units: 1 u = $\frac{1}{12}$ of the mass of a $^{12} C$ atom = $1.66 \times 1 0^{- 27}$ kg, equivalent to 931 MeV (the value CISCE uses).

A nucleus weighs less than its separate nucleons: the difference is the mass defect Δm, and $B E = Δ m c^{2}$ is the binding energy, the energy needed to pull the nucleus apart. The graph of binding energy per nucleon against A rises steeply for light nuclei, peaks at about 8.8 MeV near A = 56 (iron) and falls slowly to about 7.6 MeV for the heaviest. Middle nuclei are the most stable. So splitting a heavy nucleus (fission) or joining light ones (fusion) releases energy: the syllabus asks you to estimate about 200 MeV for a nucleus of A = 240 (7.6 MeV per nucleon) splitting into two of A = 120 (8.5 MeV per nucleon). Pair production (a γ photon of at least 1.02 MeV making an electron-positron pair) and annihilation illustrate E = mc².

You also need one fission equation, the chain reaction (controlled in a reactor, uncontrolled in a bomb), the parts of a nuclear reactor and their functions (fuel, moderator, control rods, coolant, casing; criticality), and fusion, $4_{1}^{1} H \to_{2}^{4} H e$ + energy, which powers the Sun and needs temperatures of millions of kelvin. Radioactivity (decay law, half-life) is no longer in the 2027 syllabus.

Weightage: Nuclei is part of the unit Atoms and Nuclei, which carries 6 of the 70 marks in the ISC Physics Paper 1 (Theory), shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Physics syllabus, examination year 2027, distribution of marks

Key concepts

Composition, isotopes, isobars and isotones: A nucleus is written $_{Z}^{A} X$ : Z = atomic number (protons), A = mass number (protons + neutrons), N = A − Z (neutrons). Protons and neutrons are together called nucleons. Isotopes: same Z, different A (e.g. $_{6}^{12} C$ , $_{6}^{14} C$ ); same chemical properties. Isobars: same A, different Z (e.g. $_{6}^{14} C$ , $_{7}^{14} N$ ). Isotones: same N, different Z and A (e.g. $_{6}^{14} C$ and $_{8}^{16} O$ , both N = 8). Nucleons are held together by the strong, short-range nuclear force, which overcomes the electrostatic repulsion between protons.; Worked example
Classify the pairs: (a) $_{1}^{3} H$ and $_{2}^{3} H e$ ; (b) $_{17}^{35} C l$ and $_{17}^{37} C l$ ; (c) $_{6}^{14} C$ and $_{8}^{16} O$ . How many neutrons does $_{92}^{238} U$ have?
Try it first, then show the solutionSolution
(a) Same A = 3, different Z: isobars.
(b) Same Z = 17: isotopes. (c) N = 14 − 6 = 8 and 16 − 8 = 8: isotones.
$_{92}^{238} U$ : N = 238 − 92 = 146.
Answer: (a) isobars (b) isotopes (c) isotones; 146 neutrons
Size of the nucleus and nuclear density: Scattering experiments show the nuclear radius $R = R_{0} A^{1/3}$ with $R_{0} \approx 1.2 \times 1 0^{- 15}$ m (1.2 fermi). So the volume $\frac{4}{3} π R^{3} \propto A$ : each nucleon occupies about the same volume. Nuclear density $ρ = \frac{A \times 1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π R _{0}^{3} A}$ is therefore independent of A: about $2.3 \times 1 0^{17} k g m^{- 3}$ , some $1 0^{14}$ times the density of water. (A teaspoon of nuclear matter would weigh about a billion tonnes.); Worked example
Find the radius of a $^{64} C u$ nucleus and the density of nuclear matter. ( $R_{0} = 1.2 \times 1 0^{- 15}$ m, 1 u = $1.66 \times 1 0^{- 27}$ kg)
Try it first, then show the solutionSolution
$R = 1.2 \times 6 4^{1/3} = 1.2 \times 4 = 4.8 f m$
$ρ = \frac{1.66 \times 1 0 ^{- 27}}{\frac{4}{3} π ( 1.2 \times 1 0 ^{- 15} ) ^{3}} = \frac{1.66 \times 1 0 ^{- 27}}{7.24 \times 1 0 ^{- 45}} = 2.3 \times 1 0^{17} k g m^{- 3}$
Answer: 4.8 fm; $2.3 \times 1 0^{17} k g m^{- 3}$ (the same for every nucleus)
Mass-energy equivalence, pair production and annihilation: Einstein's relation $E = m c^{2}$ : mass and energy are interchangeable. 1 u of mass is equivalent to 931 MeV; 1 g to $9 \times 1 0^{13}$ J. Pair production: a γ-ray photon passing close to a nucleus can turn into an electron and a positron. Each has rest energy 0.511 MeV, so the photon needs at least 1.02 MeV; any extra becomes their kinetic energy (the nucleus is needed to take up momentum). Annihilation: an electron and a positron meet and vanish, producing (usually) two γ photons of 0.511 MeV each travelling in opposite directions. Both processes show matter and energy converting into each other.; Worked example
Find the minimum energy of a photon that can produce an electron-positron pair, and its wavelength. ( $m_{e} c^{2} = 0.511$ MeV)
Try it first, then show the solutionSolution
$E_{min} = 2 m_{e} c^{2} = 2 \times 0.511 = 1.022 M e V$
$λ = \frac{h c}{E} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{1.022 \times 1.6 \times 1 0 ^{- 13}} = 1.22 \times 1 0^{- 12} m$
Answer: 1.02 MeV; $1.22 \times 1 0^{- 12}$ m (a γ-ray)
Mass defect and binding energy: The mass of a nucleus is always less than the total mass of its separate nucleons. Mass defect: $Δ m = [Z m_{p} + (A - Z) m_{n}] - M_{n u c l e u s}$ ; using atomic masses, $Δ m = [Z m_{H} + (A - Z) m_{n}] - M_{a t o m}$ (the electron masses cancel). Binding energy $B E = Δ m c^{2}$ : the energy released when the nucleus forms from its nucleons, equal to the energy needed to separate it completely. In MeV, BE = Δm (in u) × 931. Binding energy per nucleon BE/A measures stability: the larger it is, the more tightly bound the nucleus.; Worked example
Find the mass defect, binding energy and binding energy per nucleon of $_{2}^{4} H e$ . (m(¹H) = 1.007825 u, $m_{n}$ = 1.008665 u, m(⁴He) = 4.002604 u, 1 u = 931 MeV)
Try it first, then show the solutionSolution
Mass of parts $= 2 \times 1.007825 + 2 \times 1.008665 = 4.032980 u$
$Δ m = 4.032980 - 4.002604 = 0.030376 u$
$B E = 0.030376 \times 931 = 28.28 M e V$ ; $\frac{B E}{A} = \frac{28.28}{4} = 7.07 M e V$ per nucleon
Answer: 0.030376 u; 28.3 MeV; 7.07 MeV per nucleon
Binding energy per nucleon curve: Plotting BE/A against A: (1) it rises steeply for light nuclei, with peaks for $^{4} H e$ , $^{12} C$ , $^{16} O$ (extra-stable); (2) it reaches a broad maximum of about 8.8 MeV near A = 56 ( $^{56} F e$ ), the most stable nuclei; (3) it falls slowly to about 7.6 MeV for A ≈ 240. Consequences: heavy nuclei can increase their BE/A by splitting into two middle-sized nuclei (fission); very light nuclei can increase it by joining (fusion). In each case the products are more tightly bound, the total binding energy rises, and the increase is released as energy. The nearly constant BE/A over a wide range shows that the nuclear force is short-range and saturates (each nucleon interacts only with its neighbours).; Worked example
Find BE/A for $_{8}^{16} O$ (atomic mass 15.994915 u) and compare it with $^{4} H e$ (7.07 MeV). Which is more stable?
Try it first, then show the solutionSolution
Parts: $8 \times 1.007825 + 8 \times 1.008665 = 16.131920 u$ ; $Δ m = 0.137005 u$
$B E = 0.137005 \times 931 = 127.6 M e V$ ; $B E / A = 7.97 M e V$
7.97 > 7.07, so oxygen-16 is more tightly bound per nucleon.
Answer: About 7.97 MeV per nucleon; ¹⁶O is more stable than ⁴He
Nuclear fission and the chain reaction: In fission a heavy nucleus absorbs a (slow) neutron and splits into two middle-mass nuclei, releasing 2-3 neutrons and about 200 MeV. One example: $_{92}^{235} U +_{0}^{1} n \to_{56}^{141} B a +_{36}^{92} K r + 3_{0}^{1} n$ + energy. Estimate (syllabus): a nucleus with A = 240 and BE/A ≈ 7.6 MeV splits into two with A = 120 and BE/A ≈ 8.5 MeV; Q = 2 × 120 × 8.5 − 240 × 7.6 = 2040 − 1824 = 216 MeV ≈ 200 MeV. Because each fission releases more than one neutron, a chain reaction can build up. If on average exactly one neutron from each fission causes another fission, the reaction is steady (critical): this is a controlled chain reaction in a reactor. If more than one does, the reaction grows explosively (uncontrolled): a nuclear bomb.; Worked example
Each fission of uranium-235 releases about 200 MeV. Estimate the energy released by complete fission of 1 kg of U-235. ( $N_{A} = 6.02 \times 1 0^{23} m o l^{- 1}$ )
Try it first, then show the solutionSolution
Number of nuclei $= \frac{1000}{235} \times 6.02 \times 1 0^{23} = 2.56 \times 1 0^{24}$
Energy per fission $= 200 \times 1.6 \times 1 0^{- 13} = 3.2 \times 1 0^{- 11} J$
Total $= 2.56 \times 1 0^{24} \times 3.2 \times 1 0^{- 11} = 8.2 \times 1 0^{13} J$
Answer: About $8 \times 1 0^{13}$ J (as much as burning roughly 2500 tonnes of coal)
The nuclear reactor (qualitative): A reactor sustains a controlled chain reaction and converts the energy to heat (then electricity). Fuel elements: rods of enriched uranium (U-235) or plutonium. Moderator: a material of light nuclei (heavy water, graphite, ordinary water) that slows fast neutrons to thermal speeds, at which U-235 is far more likely to fission; a neutron loses most energy colliding with a nucleus of similar mass. Control rods: cadmium or boron rods that absorb neutrons; pushed in to slow or stop (shut down) the reaction, pulled out to speed it up. Coolant: water, heavy water or liquid sodium carrying heat from the core to a heat exchanger, where steam drives turbines. Casing / shielding: thick concrete and steel walls that stop radiation escaping. The reactor is kept just critical (multiplication factor k = 1).; Worked example
In a reactor, state the function of (a) graphite blocks, (b) cadmium rods and (c) heavy water used as a coolant. How would you shut the reactor down in an emergency?
Try it first, then show the solutionSolution
(a) Graphite is a moderator: it slows fast neutrons so they can cause further fission of U-235.
(b) Cadmium rods are control rods: they absorb neutrons and control the rate of fission. (c) The coolant removes heat from the core to produce steam.
Emergency shutdown: insert the control rods fully so they absorb enough neutrons to stop the chain reaction.
Answer: Moderator; control rods; coolant; drop in all control rods
Nuclear fusion and energy in stars: In fusion, light nuclei combine to form a heavier one with larger BE/A, releasing energy. In the Sun the net reaction is $4_{1}^{1} H \to_{2}^{4} H e + 2 e^{+} + 2 ν$ + about 26.7 MeV. Nuclei must come close enough for the short-range nuclear force to act despite their electrostatic repulsion, which needs extremely high temperatures (about $1 0^{6}$ to $1 0^{7}$ K): hence thermonuclear reactions. These conditions exist in the cores of the Sun and stars, so fusion goes on continuously there; on Earth they are hard to achieve and control. The hydrogen bomb is an uncontrolled fusion reaction, triggered by a fission bomb. Per nucleon, fusion releases more energy than fission.; Worked example
Find the energy released in the fusion $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e +_{0}^{1} n$ . (Masses: ²H 2.014102 u, ³H 3.016049 u, ⁴He 4.002604 u, n 1.008665 u; 1 u = 931 MeV)
Try it first, then show the solutionSolution
Mass before $= 2.014102 + 3.016049 = 5.030151 u$ ; after $= 4.002604 + 1.008665 = 5.011269 u$
$Δ m = 0.018882 u$
$Q = 0.018882 \times 931 = 17.6 M e V$
Answer: About 17.6 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.

Energy from fission in a power station

Suppose a nuclear power station produces 3000 MW of heat in its reactor core, with each fission of U-235 releasing 200 MeV.

How much U-235 undergoes fission each day?

Try it first, then show the solutionSolution

Energy per day $= 3 \times 1 0^{9} \times 86400 = 2.59 \times 1 0^{14} J$
Fissions $= \frac{2.59 \times 1 0 ^{14}}{3.2 \times 1 0 ^{- 11}} = 8.1 \times 1 0^{24}$
Mass $= \frac{8.1 \times 1 0 ^{24}}{6.02 \times 1 0 ^{23}} \times 235 g = 3.2 k g$

Answer: Only about 3.2 kg of U-235 per day, against thousands of tonnes of coal for the same heat.

Memory hook: 200 MeV a split: a few kilograms run a city.

Mass-energy equivalence in the Sun

Imagine the Sun radiating about $3.8 \times 1 0^{26}$ W, all of it from fusion in its core.

How much mass does the Sun convert into energy every second?

Try it first, then show the solutionSolution

$Δ m = \frac{E}{c ^{2}} = \frac{3.8 \times 1 0 ^{26}}{( 3 \times 1 0 ^{8} ) ^{2}}$
$Δ m = 4.2 \times 1 0^{9} k g$ per second

Answer: About 4 million tonnes every second, yet the Sun is so massive that it will shine for billions of years.

Memory hook: Starlight is mass turned into energy.

Annihilation of matter and antimatter

Suppose a hospital PET scanner detects pairs of γ photons produced when positrons from a tracer meet electrons in the body.

What is the energy and wavelength of each photon, and why are they detected in opposite directions?

Try it first, then show the solutionSolution

Each photon carries $m_{e} c^{2} = 0.511$ MeV (the pair's rest energy shared equally)
$λ = \frac{h c}{E} = \frac{6.63 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{0.511 \times 1.6 \times 1 0 ^{- 13}} = 2.43 \times 1 0^{- 12} m$ ; they move in opposite directions to conserve momentum (the pair was almost at rest).

Answer: 0.511 MeV and $2.4 \times 1 0^{- 12}$ m each, emitted back to back.

Memory hook: Electron + positron = two 0.511 MeV gammas, back to back.

Topics in this chapter

Nuclei formula sheet

Nuclear notation: $_{Z}^{A} X, N = A - Z$; Isotopes: same Z; isobars: same A; isotones: same N.
Nuclear radius: $R = R_{0} A^{1/3}, R_{0} \approx 1.2 \times 1 0^{- 15} m$; Density about $2.3 \times 1 0^{17} k g m^{- 3}$ , independent of A.
Atomic mass unit: $1 u = \frac{1}{12} m (^{12} C) = 1.66 \times 1 0^{- 27} k g \equiv 931 M e V$; CISCE uses 931 MeV.
Mass-energy relation: $E = m c^{2}$; 1 g ↔ $9 \times 1 0^{13}$ J.
Mass defect: $Δ m = [Z m_{H} + (A - Z) m_{n}] - M_{a t o m}$; Use atomic masses consistently.
Binding energy: $B E = Δ m c^{2} = Δ m (u) \times 931 M e V$; BE/A measures stability.
Q value of a reaction: $Q = (m_{r e a c t an t s} - m_{p r o d u c t s}) c^{2}$; Q > 0: energy released.
Fission estimate: $Q \approx 2 \times 120 \times 8.5 - 240 \times 7.6 \approx 200 M e V$; From the BE/A values.
Pair production threshold: $E_{γ} \geq 2 m_{e} c^{2} = 1.02 M e V$; Annihilation gives two 0.511 MeV photons.
Fusion in the Sun: $4_{1}^{1} H \to_{2}^{4} H e + 2 e^{+} + 2 ν + 26.7 M e V$; Needs about $1 0^{6}$ - $1 0^{7}$ K.

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

Explaining stability and why fission and fusion release energy.

Draw

Axes: BE/A in MeV (0 to 9) on the y-axis, mass number A (0 to 250) on the x-axis.
Steep rise from ²H (about 1 MeV) with a spike at ⁴He (7.1 MeV); broad peak of about 8.8 MeV near A = 56; gentle fall to about 7.6 MeV at A = 238.
Mark arrows: fusion region (light nuclei moving right and up), fission region (heavy nuclei splitting towards the middle).

Label

BE/A (MeV)
A
⁴He spike
⁵⁶Fe peak (≈ 8.8 MeV)
²³⁸U (≈ 7.6 MeV)
fusion →
← fission

Watch out: State that middle-mass nuclei are most stable; that sentence explains both energy sources.

Nuclear reactor

'Name the main parts of a nuclear reactor and state their functions'.

Draw

A core vessel containing vertical fuel rods surrounded by moderator.
Control rods entering from the top, partly inserted.
Coolant pipes leading out to a heat exchanger (steam generator) and back; thick shielding wall around the core.

Label

fuel rods
moderator
control rods
coolant in/out
heat exchanger
shielding (concrete)

Watch out: Beside each label write one function word: fuel - fission; moderator - slows; control - absorbs; coolant - removes heat; shield - protects.

How the board asks Nuclei

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

Question type	How it is asked	Method
Definitions and classification	Mass defect, binding energy, isotopes/isobars/isotones, pair production, 1 u.	One precise sentence each; give an example nucleus where useful.
Binding-energy numericals	Find Δm, BE and BE/A of a nucleus from given masses.	Use 931 MeV per u; keep six decimal places until the end; do not convert Δm to kg.
Q-value of fission or fusion	Energy released in a given reaction; 200 MeV estimate from BE/A.	Sum masses on each side (count every particle); Q = Δm × 931 MeV.
BE/A graph	Sketch it and explain why fission and fusion release energy; which nuclei are most stable.	Mark the peak near A = 56; link increased BE/A of products to energy release.
Reactor and stars	Functions of moderator, control rods, coolant; shutting down; where fusion occurs; why high temperature is needed.	Use the technical names (moderator, control rods) rather than just the materials.
E = mc² applications	Minimum photon energy for pair production; mass converted per second in the Sun; annihilation photons.	Use $m_{e} c^{2} = 0.511$ MeV and E = mc² in SI.

Tricks to remember Nuclei

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

Isotopes, isobars, isotones
isotoPes = same Protons; isobArs = same A; isotoNes = same Neutrons.
Use it when: Classification items.
931, not 931.5
CISCE examiners expect 1 u = 931 MeV; use what the paper gives.
Use it when: All BE and Q numericals.
Iron sits on top
BE/A peaks at iron-56 (about 8.8 MeV): everything wants to move towards iron.
Use it when: Explaining fission and fusion.
MCC: Moderator slows, Control absorbs, Coolant carries
Moderator = slow neutrons; Control rods = absorb neutrons; Coolant = carry heat.
Use it when: Reactor-part questions.
Pair needs 1.02
Two electron masses: 2 × 0.511 = 1.02 MeV minimum.
Use it when: Pair production one-markers.

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 nuclei are isotones?
(a) $_{6}^{12} C$ and $_{6}^{14} C$
(b) $_{6}^{14} C$ and $_{7}^{14} N$
(c) $_{6}^{14} C$ and $_{8}^{16} O$
(d) $_{8}^{16} O$ and $_{8}^{17} O$

Try it first, then show the solutionSolution

Neutron numbers: (c) gives 14 − 6 = 8 and 16 − 8 = 8, equal. (a) and (d) are isotopes; (b) isobars.

Answer: (c)

Example 2: Why fission releases energy · 1 mark · Assertion-reasonBoard level

Assertion: Energy is released when a heavy nucleus such as uranium-235 undergoes fission.
Reason: The binding energy per nucleon of the fission fragments is greater than that of the heavy nucleus.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Fragments (A ≈ 120) have BE/A ≈ 8.5 MeV against 7.6 MeV for the parent, so total binding energy rises and the difference is released.

Answer: (a)

Example 3: Comparing nuclear sizes · 2 marks · Short answerHarder board level

Find the ratio of the radii of nuclei with mass numbers 125 and 8. What is the ratio of their densities? [2]

Try it first, then show the solutionSolution

$\frac{R _{1}}{R _{2}} = (\frac{125}{8})^{1/3} = \frac{5}{2} = 2.5$
Volume ∝ A and mass ∝ A, so density is the same: ratio 1 : 1.

Answer: 5 : 2; densities equal

Example 4: Binding energy of helium-4 · 3 marks · Short answerHarder board level

Calculate the mass defect, binding energy and binding energy per nucleon of $_{2}^{4} H e$ . (m(¹H) = 1.007825 u, $m_{n}$ = 1.008665 u, m(⁴He) = 4.002604 u, 1 u = 931 MeV) [3]

Try it first, then show the solutionSolution

$Δ m = 2 (1.007825) + 2 (1.008665) - 4.002604 = 0.030376 u$
$B E = 0.030376 \times 931 = 28.28 M e V$
$B E / A = 28.28/4 = 7.07 M e V$ per nucleon

Answer: 0.030376 u; 28.3 MeV; 7.07 MeV

Example 5: Estimating fission energy from BE/A · 3 marks · Short answerHarder board level

A nucleus of mass number 240 with binding energy 7.6 MeV per nucleon splits into two equal nuclei of mass number 120, each with 8.5 MeV per nucleon. Estimate the energy released, and explain where it comes from. [3]

Try it first, then show the solutionSolution

Initial total BE $= 240 \times 7.6 = 1824 M e V$ ; final $= 2 \times 120 \times 8.5 = 2040 M e V$ .
Q = 2040 − 1824 = 216 MeV ≈ 200 MeV.
The products are more tightly bound; the extra binding energy corresponds to a decrease in mass, which appears as kinetic energy of the fragments and neutrons (and γ radiation).

Answer: About 200 MeV (216 MeV)

Example 6: The BE/A curve, fission and fusion · 5 marks · Long answerHarder board level

(a) Sketch the variation of binding energy per nucleon with mass number and state two features of the curve. [2]
(b) Use the curve to explain why both fission of heavy nuclei and fusion of light nuclei release energy. [1]
(c) Calculate the energy released in $_{1}^{2} H +_{1}^{3} H \to_{2}^{4} H e +_{0}^{1} n$ . (²H 2.014102 u, ³H 3.016049 u, ⁴He 4.002604 u, n 1.008665 u) [2]

Try it first, then show the solutionSolution

(a) Steep rise for light nuclei (spike at ⁴He), broad maximum about 8.8 MeV near A = 56, slow decline to about 7.6 MeV at A ≈ 240.
Features: middle-mass nuclei are most stable; BE/A is nearly constant (about 8 MeV) for A = 30-170, showing the short range of nuclear forces.
(b) Fission moves heavy nuclei and fusion moves light nuclei towards the peak; the products have greater BE/A, so energy is released.
(c) $Δ m = (2.014102 + 3.016049) - (4.002604 + 1.008665) = 0.018882 u$
Q = 0.018882 × 931 = 17.6 MeV

Answer: (c) 17.6 MeV

Example 7: Fusion versus fission per nucleon · 3 marks · Short answerHOTS / competency

Fission of a nucleus of A = 240 releases about 216 MeV; the D-T fusion reaction releases 17.6 MeV. Compare the energy released per nucleon taking part, and explain why fusion is hard to achieve on Earth. [3]

Try it first, then show the solutionSolution

Fission: 216/240 = 0.9 MeV per nucleon. Fusion: 17.6/5 = 3.5 MeV per nucleon, about four times more.
Fusion needs the positively charged nuclei to come within about $1 0^{- 15}$ m, overcoming strong electrostatic repulsion.
That requires temperatures of millions of kelvin and confinement of the hot plasma, which is very difficult to sustain and control on Earth.

Answer: Fusion ≈ 3.5 MeV/nucleon against ≈ 0.9 MeV/nucleon for fission

Example 8: Threshold for pair production · 2 marks · Short answerHOTS / competency

What is the minimum energy a γ photon needs to produce an electron-positron pair? What happens to any energy above this? [2]

Try it first, then show the solutionSolution

Rest energy of each particle is 0.511 MeV, so $E_{min} = 2 \times 0.511 = 1.02 M e V$ .
Any excess energy appears as kinetic energy of the electron and positron (a nearby nucleus takes up the momentum).

Answer: 1.02 MeV; the excess becomes kinetic energy of the pair

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Inside a research reactor

Suppose students visit a research reactor whose core sits at the bottom of a pool of water. Their guide explains that the water slows fast neutrons and carries away heat, that cadmium rods hang above the core, and that the fuel is enriched uranium.

(i) · 1 mark
What is the general name for a material that slows fast neutrons, and why must they be slowed?
Answer
Answer: A moderator; slow (thermal) neutrons are far more likely to cause fission of U-235.
Fast neutrons tend to escape or be captured without causing fission; slowing them sustains the chain reaction.
(ii) · 1 mark
What do the cadmium rods do, and how would the reactor be shut down in an emergency?
Answer
Answer: They absorb neutrons (control rods); lower them fully into the core.
Absorbing neutrons reduces the number causing new fissions below one per fission, so the chain reaction stops.
(iii) · 1 mark
What does 'critical' mean for a reactor?
Answer
Answer: On average exactly one neutron from each fission causes another fission.
The reaction then proceeds at a steady rate (multiplication factor 1).
(iv) · 2 marks
If the reactor produces 3000 MW of heat, estimate the mass of U-235 fissioned per day (200 MeV per fission).
Answer
Answer: About 3.2 kg
$2.59 \times 1 0^{14} J \div 3.2 \times 1 0^{- 11} J = 8.1 \times 1 0^{24}$ fissions = 13.5 mol × 235 g = 3.2 kg.

Reading the binding-energy curve

Suppose a student is given the binding energy per nucleon for three nuclei: ²H 1.1 MeV, ⁵⁶Fe 8.8 MeV and ²³⁸U 7.6 MeV.

(i) · 1 mark
Which nucleus is the most stable? Why?
Answer
Answer: ⁵⁶Fe
It has the highest binding energy per nucleon, so the most energy is needed per nucleon to break it apart.
(ii) · 1 mark
Find the total binding energy of ²³⁸U.
Answer
Answer: About 1809 MeV
238 × 7.6 = 1808.8 MeV
(iii) · 1 mark
Which process, fission or fusion, could release energy from (a) ²H and (b) ²³⁸U?
Answer
Answer: (a) fusion (b) fission
Both move the nuclei towards the high BE/A near iron.
(iv) · 2 marks
Where in the universe does fusion take place continuously, and what conditions make it possible there?
Answer
Answer: In the cores of the Sun and other stars.
Temperatures of about $1 0^{7}$ K and enormous pressures let nuclei overcome their electrostatic repulsion and fuse.

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.

In a nuclear reactor, state the use of (a) graphite and (b) cadmium rods. (2 marks)

Step	Marks
(a) Graphite: moderator, slows down fast neutrons	1
(b) Cadmium rods: control rods, absorb neutrons to control (or stop) the chain reaction	1

Find the mass defect and binding energy per nucleon of $_{8}^{16} O$ (15.994915 u), given m(¹H) = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV. (3 marks)

Step	Marks
Mass of nucleons: 8(1.007825) + 8(1.008665) = 16.131920 u; Δm = 0.137005 u	1
BE = 0.137005 × 931 = 127.6 MeV	1
BE/A = 7.97 MeV per nucleon	1

Explain, using the BE/A curve, why energy is released in fission. Estimate the energy for A = 240 (7.6 MeV per nucleon) splitting into two nuclei of A = 120 (8.5 MeV per nucleon), and write one fission equation for U-235. (5 marks)

Step	Marks
Sketch/description of curve: maximum near A = 56; lower BE/A for heavy nuclei	1
Fragments have higher BE/A, so total BE increases and energy is released	1
BE before 1824 MeV; after 2040 MeV	1
Q ≈ 216 MeV (≈ 200 MeV)	1
Balanced equation, e.g. $_{92}^{235} U +_{0}^{1} n \to_{56}^{141} B a +_{36}^{92} K r + 3_{0}^{1} n$	1

Where students lose marks in Nuclei

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

Mistake: Leaving the answer as a mass difference or using 931.5
CISCE examiners' comments, 2024: in binding-energy and fusion numericals, candidates left the answer as a mass difference, used wrong units, used only one reactant, did not use the 931 MeV conversion, used 931.5, converted Δm to kg unnecessarily, or rounded intermediate digits.
Wrong working
Energy released = 0.0256 u; or Δm converted to kg then multiplied by $c^{2}$ with rounding.
Right working
Count both reactants; keep all decimals; Q = Δm (u) × 931 MeV.
Do this: Finish every mass-defect answer in MeV.
Usually costs: 1-2 marks.
Mistake: Defining mass defect wrongly
CISCE examiners' comments, 2025: candidates defined mass defect wrongly or confused the mass of the nucleus with the mass of its nucleons, and gave vague answers (solar system, Earth, Moon) for where fusion goes on continuously.
Wrong working
Mass defect is the mass of the nucleus lost in a reaction.
Right working
Mass defect = (total mass of the separate nucleons) − (mass of the nucleus). Fusion occurs continuously in the Sun and stars.
Do this: Write the formula alongside the definition.
Usually costs: 1 mark each.
Mistake: Naming materials instead of the parts of a reactor
CISCE examiners' comments, 2024 and 2025: candidates confused graphite rods with control rods, gave cadmium a moderating role, named water or graphite instead of 'moderator', and wrote 'turn off the reactor' or 'remove the fuel' instead of inserting control rods.
Wrong working
Cadmium slows down neutrons; shut the reactor by switching it off.
Right working
Cadmium (or boron) absorbs neutrons (control rods); emergency shutdown = insert control rods fully.
Do this: Learn part → function → typical material as a three-column table.
Usually costs: 1 mark each.
Mistake: Wrong threshold for pair production
CISCE examiners' comments, 2025: asked for the minimum γ energy to produce an electron-positron pair, candidates wrote the equation instead of the value, or answered 931 MeV or 1.02 eV.
Wrong working
1.02 eV
Right working
1.02 MeV (twice the electron's rest energy, 0.511 MeV).
Do this: Check the prefix: MeV, not eV.
Usually costs: 1 mark.

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
1 u of mass is equivalent to about:
(a) 9.31 MeV
(b) 93.1 MeV
(c) 931 MeV
(d) 9310 MeV
Hint
E = mc² for $1.66 \times 1 0^{- 27}$ kg.
Answer
(c) 931 MeV
Q2 · 1 mark · MCQBoard level
The density of nuclear matter:
(a) increases with mass number
(b) decreases with mass number
(c) is about the same for all nuclei
(d) is equal to that of water
Hint
R³ ∝ A.
Answer
(c)
Q3 · 1 mark · MCQBoard level
In a nuclear reactor, the control rods are made of:
(a) graphite
(b) heavy water
(c) cadmium or boron
(d) uranium
Hint
They absorb neutrons.
Answer
(c)
Q4 · 1 mark · Very short answerBoard level
What is meant by the mass defect of a nucleus?
Hint
Parts minus whole.
Answer
The difference between the total mass of the separate nucleons and the mass of the nucleus.
Q5 · 2 marks · Short answerBoard level
From $_{18}^{40} A r$ , $_{20}^{40} C a$ , $_{19}^{39} K$ and $_{18}^{38} A r$ , pick a pair of isotopes, a pair of isobars and a pair of isotones.
Hint
Compare Z, A and N.
Answer
Isotopes: ⁴⁰Ar and ³⁸Ar (Z = 18). Isobars: ⁴⁰Ar and ⁴⁰Ca (A = 40). Isotones: ⁴⁰Ca and ³⁹K (both N = 20; ³⁸Ar also has N = 20, while ⁴⁰Ar has N = 22).
Q6 · 2 marks · Short answerHarder board level
Find the binding energy per nucleon of $_{8}^{16} O$ (15.994915 u). (m(¹H) = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV)
Hint
Δm then × 931.
Answer
Δm = 0.137005 u; BE = 127.6 MeV; BE/A = 7.97 MeV
Q7 · 2 marks · Short answerHarder board level
An electron and a positron, nearly at rest, annihilate. Find the energy of each photon produced and explain why two photons are needed.
Hint
Momentum conservation.
Answer
Total energy 1.022 MeV shared equally: 0.511 MeV each. The pair has nearly zero momentum, so the photons must move in opposite directions with equal momentum; a single photon would carry momentum.
Q8 · 2 marks · Short answerHOTS / competency
Why are light nuclei such as hydrogen (in water) or carbon (in graphite) used as moderators rather than heavy nuclei like lead?
Hint
Collision between equal masses.
Answer
In an elastic collision a neutron loses most energy to a nucleus of similar mass (all of it in a head-on collision with a proton). A heavy nucleus just makes the neutron bounce off with almost the same speed, so it hardly slows it.
Q9 · 3 marks · Short answerHarder board level
Estimate the energy released by complete fission of 1 kg of U-235 if each fission releases 200 MeV.
Hint
Number of nuclei from N_A.
Answer
Nuclei = (1000/235) × 6.02 × 10²³ = 2.56 × 10²⁴; energy = 2.56 × 10²⁴ × 3.2 × 10⁻¹¹ J = 8.2 × 10¹³ J.
Q10 · 3 marks · Short answerHarder board level
Write the net fusion reaction in the Sun and find the energy released using atomic masses (¹H 1.007825 u, ⁴He 4.002604 u). Why is a very high temperature needed?
Hint
Δm = 4m(H) − m(He).
Answer
$4_{1}^{1} H \to_{2}^{4} H e + 2 e^{+} + 2 ν$ ; Δm = 4.031300 − 4.002604 = 0.028696 u; Q = 26.7 MeV. High temperature gives nuclei enough kinetic energy to overcome their electrostatic repulsion and come within range of the nuclear force.
Q11 · 3 marks · Short answerHarder board level
Find the radius of a $^{64} C u$ nucleus and the nuclear density, taking $R_{0} = 1.2$ fm and 1 u = $1.66 \times 1 0^{- 27}$ kg. Compare with the density of water.
Hint
Density is independent of A.
Answer
R = 4.8 fm; ρ = 2.3 × 10¹⁷ kg m⁻³, about 2 × 10¹⁴ times the density of water (1000 kg m⁻³).
Q12 · 5 marks · Long answerHarder board level
(a) Name the main parts of a nuclear reactor and state the function of each. [3]
(b) What is meant by a controlled chain reaction? [1]
(c) Why is a nuclear bomb an uncontrolled chain reaction? [1]
Hint
Fuel, moderator, control rods, coolant, shield.
Answer
(a) Fuel rods (enriched U-235: undergo fission); moderator (graphite/heavy water: slows neutrons); control rods (cadmium/boron: absorb neutrons to control rate); coolant (water/liquid sodium: removes heat to produce steam); shielding (concrete: protects from radiation). (b) A chain reaction in which, on average, exactly one neutron per fission causes another fission, so the rate stays steady (critical). (c) In a bomb, more than one neutron per fission causes further fission, so the number of fissions multiplies rapidly and the energy is released almost instantly.
Q13 · 2 marks · Short answerHarder board level
The Sun radiates $3.8 \times 1 0^{26}$ W. Find the mass it converts into energy per second.
Hint
Δm = E/c².
Answer
$Δ m = 3.8 \times 1 0^{26} /9 \times 1 0^{16} = 4.2 \times 1 0^{9}$ kg each second

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 binding energy of a nucleus.Show answer
The energy required to separate a nucleus completely into its nucleons, $B E = Δ m c^{2}$ .
2.1 u in kg and in MeV?Show answer
$1.66 \times 1 0^{- 27}$ kg; 931 MeV.
3.Nuclear radius formula?Show answer
$R = R_{0} A^{1/3}$ , $R_{0} \approx 1.2$ fm.
4.Which nuclei are the most stable?Show answer
Those near A = 56 (iron), with BE/A about 8.8 MeV.
5.Energy released per fission of U-235?Show answer
About 200 MeV.
6.Function of a moderator?Show answer
To slow fast neutrons to thermal energies so they cause fission.
7.What is pair production?Show answer
Conversion of a γ photon (at least 1.02 MeV) into an electron and a positron near a nucleus.
8.Why does fusion need very high temperature?Show answer
To give nuclei enough energy to overcome electrostatic repulsion and come close enough to fuse.

Best order to study it

Learn nuclear notation and classify isotopes, isobars and isotones.
Learn R = R₀A^{1/3} and show nuclear density is constant.
Learn E = mc², 1 u = 931 MeV, pair production and annihilation.
Practise mass-defect and binding-energy numericals (keep decimals; use 931).
Draw the BE/A curve and use it to explain fission and fusion; learn the 200 MeV estimate.
Learn the reactor's parts and functions, chain reactions, and fusion in stars.

Quick revision checklist

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

Can you classify isotopes, isobars and isotones?
Can you show that nuclear density does not depend on A?
Can you define mass defect and binding energy and calculate BE/A?
Can you sketch the BE/A curve and state its features?
Can you estimate the fission energy of 200 MeV from BE/A values?
Can you name each reactor part, its function and material?
Can you write the fusion reaction in the Sun and explain why high temperature is needed?
Can you give the threshold for pair production (1.02 MeV)?

Nuclei: frequently asked questions

Is radioactivity part of this chapter for ISC 2027?

No. Radioactivity (the decay law, half-life and mean life) was removed in the February 2026 revision of the ISC Physics syllabus. Nuclei now covers composition and size, mass-energy, binding energy, fission, the reactor and fusion.

Should I use 931 or 931.5 MeV per u?

Use 931 MeV, as CISCE does; examiners noted candidates who used 931.5 when 931 was given. Always use the value printed in the paper.

How much detail is needed on the nuclear reactor?

Qualitative only: the main parts (fuel elements, moderator, control rods, coolant, casing), their functions, criticality and how the energy is used. Details of the chain reaction are not required.

Why is energy released in both fission and fusion?

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

Do I need to know a fission equation?

Yes, any one balanced fission equation, for example U-235 absorbing a neutron and splitting into barium-141, krypton-92 and three neutrons.

How many marks do Atoms and Nuclei carry?

Together, 6 of the 70 theory marks.

Based on: CISCE ISC 2027 Physics syllabus (revised, uploaded February 2026); CISCE ISC 2027 Physics specimen Paper 1 (Theory) question paper and answer key; CISCE Analysis of Pupil Performance, ISC Physics, 2025; CISCE Analysis of Pupil Performance, ISC Physics, 2024