PhysicsCBSE Class 12 · Nuclei

Mass Defect and Binding Energy Class 12: BE per Nucleon and Nuclear Force

Q: Where does the missing mass go when a nucleus forms?

It is released as energy (the binding energy), usually as gamma radiation or kinetic energy, in line with E = mc².

Q: Why is binding energy per nucleon a better measure of stability than total binding energy?

Total binding energy grows with the number of nucleons. Per-nucleon binding shows how tightly each nucleon is held, so it compares nuclei of different sizes fairly.

Q: Why do both fission and fusion release energy?

Both move nuclei towards the peak of the BE/A curve near A = 56. The products are more tightly bound than the starting nuclei, and the extra binding energy is released.

Q: Why is the nuclear force called short-range?

It acts only over a few femtometres. Beyond that it becomes negligible, unlike the Coulomb force, which falls off as 1/r² but never stops.

Q: What is meant by charge independence of nuclear force?

The nuclear force between two protons, two neutrons, or a proton and a neutron is essentially the same, regardless of electric charge.

Updated 15 October 2026 · Fact-checked

A nucleus has less mass than its separate protons and neutrons. This mass defect ΔM, times c², is the binding energy needed to pull the nucleus apart (1 u ≡ 931.5 MeV). Binding energy per nucleon is about 8 MeV for most nuclei, peaks near A = 56, and comes from the short-range nuclear force.

Notice that the free nucleons weigh more than the nucleus they form; the missing mass ΔM appears as binding energy, ΔM × 931.5 MeV per u.

Illustration drawn for QuestionBuddy.

Understand Mass Defect and Binding Energy

Weigh a helium-4 nucleus and then weigh two protons and two neutrons separately: the separate particles are heavier. Mass seems to disappear when a nucleus forms. Einstein's mass-energy relation explains where it goes.

Mass-energy relation. Mass and energy are interchangeable: $E = m c^{2}$ . A mass of 1 u corresponds to $E = 1.6605 \times 1 0^{- 27} \times (3 \times 1 0^{8})^{2} \approx 1.49 \times 1 0^{- 10}$ J $\approx 931.5$ MeV. So in nuclear calculations, 1 u ≡ 931.5 MeV.

Mass defect. For a nucleus with Z protons and N neutrons and mass M, the mass defect is $Δ M = [Z m_{p} + N m_{n}] - M$ . In practice we use atomic masses (with $m_{H}$ for hydrogen atoms) so that electron masses cancel: $Δ M = Z m_{H} + N m_{n} - M_{a t o m}$ . ΔM is always positive for a stable nucleus.

Binding energy. When the nucleons come together, energy $Δ M c^{2}$ is released; equally, the same energy must be supplied to separate the nucleus completely into free nucleons. This is the binding energy: $E_{b} = Δ M c^{2} = Δ M (in u) \times 931.5$ MeV. The binding energy per nucleon, $E_{bn} = \frac{E _{b}}{A}$ , measures how tightly each nucleon is held, i.e. how stable the nucleus is.

The binding energy curve. Plotting $E_{bn}$ against A shows:

It rises steeply for light nuclei (deuterium about 1.1 MeV, helium-4 about 7.1 MeV).
It is nearly constant, about 8 MeV, for a wide middle range (roughly 30 < A < 170), reaching a maximum of about 8.75 MeV near A = 56 (iron), the most stable nuclei.
It falls slowly for heavy nuclei, to about 7.6 MeV for uranium-238.

Consequences: (1) the near-constant value shows that the nuclear force is short-range and saturates: each nucleon interacts only with its close neighbours, so adding nucleons does not increase the binding per nucleon. (2) A heavy nucleus splitting into two middle-mass nuclei (fission) or two light nuclei joining (fusion) moves towards higher binding per nucleon, so energy is released.

Nuclear force. The force that binds nucleons:

is much stronger than the electrostatic force (it must overcome the huge repulsion between protons only a few fermi apart) and than gravity;
is short-range: it falls rapidly to zero beyond a few fermi;
is attractive for separations above about 0.8 fm, but becomes strongly repulsive at smaller distances (so nucleons do not collapse into each other); its potential energy is a minimum near 0.8 fm;
is nearly independent of charge: proton-proton, neutron-neutron and proton-neutron forces are about the same;
saturates, as shown by the binding energy curve.

Mass Defect and Binding Energy: formulas and key facts

Mass-energy relation: $E = m c^{2}, 1 u \equiv 931.5 MeV$; Use the MeV equivalent to save effort.
Mass defect: $Δ M = Z m_{H} + N m_{n} - M_{a t o m}$; Atomic masses make electron masses cancel.
Binding energy: $E_{b} = Δ M c^{2} = Δ M (u) \times 931.5 MeV$; Energy needed to separate all nucleons.
Binding energy per nucleon: $E_{bn} = \frac{E _{b}}{A}$; Peak ≈ 8.75 MeV near A = 56.
Useful masses: $m_{H} = 1.007825 u, m_{n} = 1.008665 u, m_{p} = 1.007276 u$; Use the values given in the question if different.

How to solve Mass Defect and Binding Energy questions

Use this method for mass defect and binding energy numericals.

1Find Z and N = A − Z.
2Write ΔM = Z m_H + N m_n − M (atomic masses), keeping 6 decimal places.
3Convert: E_b = ΔM × 931.5 MeV (or ΔM in kg × c² in joule).
4Divide by A for binding energy per nucleon.
5Compare E_bn values to say which nucleus is more stable.
6For energy released in a reaction, use (total mass before − total mass after) × 931.5 MeV, or the change in total binding energy.
7State units clearly: u, MeV, J.

Mass Defect and Binding Energy in real life

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 changes in everyday energy

Suppose a 2 kW kettle heats water for 5 minutes.

(a) How much energy does the water gain? (b) Using E = mc², by how much does the mass of the water increase? (c) Why do we notice mass changes in nuclear reactions but never in kettles?

Try it first, then show the solutionSolution

E = 2000 × 300 = 6 × 10⁵ J.
Δm = E/c² = 6 × 10⁵/(9 × 10¹⁶) ≈ 6.7 × 10⁻¹² kg, far too small to measure.
Nuclear binding energies are millions of electronvolts per nucleus, a measurable fraction (around 0.1% to 1%) of the mass, whereas chemical and thermal energies are only a few eV per atom.

Answer: (a) 6 × 10⁵ J; (b) about 6.7 × 10⁻¹² kg; (c) nuclear energies per particle are millions of times larger.

Memory hook: E = mc² works for kettles too; the mass change is just too tiny to see.

Binding energy on a large scale

Imagine taking 1 mole of helium-4 nuclei (6.022 × 10²³ nuclei) and pulling every one apart into free protons and neutrons. Each needs about 28.3 MeV.

(a) Find the total energy needed in joule. (b) Roughly how much mass would the products gain?

Try it first, then show the solutionSolution

E = 28.3 × 6.022 × 10²³ MeV ≈ 1.70 × 10²⁵ MeV ≈ 1.70 × 10²⁵ × 1.6 × 10⁻¹³ J ≈ 2.7 × 10¹² J.
Δm = E/c² ≈ 2.7 × 10¹²/9 × 10¹⁶ ≈ 3 × 10⁻⁵ kg = 0.03 g, out of 4 g: about 0.75%.

Answer: (a) about 2.7 × 10¹² J; (b) about 0.03 g (0.75% of the mass).

Memory hook: Nuclear glue is worth almost 1% of the mass.

Tricks to remember Mass Defect and Binding Energy

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

Parts weigh more than the whole
Free nucleons are heavier than the nucleus; the difference is the binding energy.
Use it when: Defining mass defect.
931.5 per u
Multiply mass defect in u by 931.5 to get MeV.
Use it when: Quick conversion.
Iron at the top
BE/A peaks near A = 56: light nuclei fuse up to it, heavy nuclei split down to it.
Use it when: Explaining fission and fusion.
SAS-C for the nuclear force
Strong, Attractive (repulsive very close), Short-range, Charge-independent (and saturating).
Use it when: Listing properties.

Mass Defect and Binding Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Binding energy of tritium · 2 marks · Very short answerBoard level

Find the mass defect and binding energy of tritium (³H, atomic mass 3.016049 u). Take m_H = 1.007825 u and m_n = 1.008665 u. Also find the binding energy per nucleon.

Try it first, then show the solutionSolution

ΔM = 1.007825 + 2(1.008665) − 3.016049 = 3.025155 − 3.016049 = 0.009106 u. (1)
E_b = 0.009106 × 931.5 ≈ 8.48 MeV; per nucleon ≈ 2.83 MeV. (1)

Answer: 0.009106 u; about 8.48 MeV; about 2.83 MeV per nucleon.

Example 2: The binding energy curve · 3 marks · Short answerBoard level

Sketch the variation of binding energy per nucleon with mass number. State three features of the curve and explain how it shows that the nuclear force is short-range.

Try it first, then show the solutionSolution

Sketch: steep rise for small A, broad maximum of about 8.75 MeV near A = 56, slow fall to about 7.6 MeV at A ≈ 238. (1)
Features: BE/A is small for light nuclei; nearly constant (about 8 MeV) for 30 < A < 170; decreases for heavy nuclei. (1)
Because BE/A stays almost constant as A grows, each nucleon interacts only with a few close neighbours (saturation), which means the force is short-range. (1)

Answer: Steep rise, broad peak near A = 56, slow decline; the near-constant middle region shows saturation and short range.

Example 3: Helium-4 versus the deuteron · 5 marks · Long answerHarder board level

(a) Find the binding energy and binding energy per nucleon of ⁴He (atomic mass 4.002603 u). (3)
(b) Compare with the deuteron (1.11 MeV per nucleon) and comment on stability. (2)

Try it first, then show the solutionSolution

(a) ΔM = 2(1.007825) + 2(1.008665) − 4.002603. (1)
= 4.032980 − 4.002603 = 0.030377 u. (1)
E_b = 0.030377 × 931.5 ≈ 28.3 MeV; per nucleon ≈ 7.07 MeV. (1)
(b) ⁴He binds each nucleon about six times more strongly than the deuteron. (1)
So ⁴He is far more stable; this is why fusing hydrogen isotopes into helium releases a lot of energy. (1)

Answer: (a) about 28.3 MeV and 7.07 MeV per nucleon; (b) helium-4 is much more tightly bound and stable.

Example 4: Binding energy of oxygen-16 · 3 marks · Short answerHarder board level

Find the binding energy per nucleon of ¹⁶₈O (atomic mass 15.994915 u).

Try it first, then show the solutionSolution

ΔM = 8(1.007825) + 8(1.008665) − 15.994915 = 16.131920 − 15.994915. (1)
ΔM = 0.137005 u; E_b ≈ 0.137005 × 931.5 ≈ 127.6 MeV. (1)
E_bn ≈ 127.6/16 ≈ 7.98 MeV. (1)

Answer: About 7.98 MeV per nucleon.

Example 5: Mass-energy equivalence · 5 marks · Long answerHOTS / competency

(a) Show that 1 u is equivalent to about 931.5 MeV (1 u = 1.6605 × 10⁻²⁷ kg). (2)
(b) Find the energy equivalent of 1 g of matter and express it in kWh. (2)
(c) Why is the mass of a nucleus less than the sum of the masses of its nucleons? (1)

Try it first, then show the solutionSolution

(a) E = mc² = 1.6605 × 10⁻²⁷ × (2.998 × 10⁸)² ≈ 1.492 × 10⁻¹⁰ J. (1)
In MeV: 1.492 × 10⁻¹⁰/1.602 × 10⁻¹³ ≈ 931.5 MeV. (1)
(b) E = 10⁻³ × (3 × 10⁸)² = 9 × 10¹³ J. (1)
= 9 × 10¹³/3.6 × 10⁶ ≈ 2.5 × 10⁷ kWh. (1)
(c) When nucleons bind, energy (the binding energy) is released; by E = mc² this energy carries away mass, so the nucleus is lighter. (1)

Answer: (a) about 931.5 MeV; (b) 9 × 10¹³ J ≈ 2.5 × 10⁷ kWh; (c) the released binding energy carries away mass.

Example 6: Properties of the nuclear force · 4 marks · Case-basedHarder board level

Read the passage and answer.

Two protons in a nucleus are about 1 fm apart.
(a) Estimate the electrostatic repulsion between them. (1)
(b) What must be true of the nuclear force for the nucleus to be stable? (1)
(c) Why does the nuclear force not pull nucleons together into a point? (1)
(d) What does charge independence of the nuclear force mean? (1)

Try it first, then show the solutionSolution

(a) F = 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/(10⁻¹⁵)² ≈ 230 N, enormous for such tiny particles. (1)
(b) It must be attractive and much stronger than this repulsion at nuclear distances. (1)
(c) It becomes strongly repulsive at separations below about 0.8 fm, so nucleons keep a minimum spacing. (1)
(d) The force between two protons, two neutrons, or a proton and a neutron is about the same. (1)

Answer: (a) about 230 N; (b) much stronger and attractive; (c) repulsive at very short distances; (d) same for pp, nn and np pairs.

Where students lose marks in Mass Defect and Binding Energy

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

Mistake: Mixing nuclear and atomic masses.
Using m_p with an atomic mass (or m_H with a nuclear mass).
Do this: Use m_H with atomic masses (electrons cancel) or m_p with nuclear masses.
Usually costs: The answer mark.
Mistake: Rounding masses too early.
Mass defects are small differences of large numbers.
Do this: Keep all six decimal places until the subtraction is done.
Usually costs: 1 mark.
Mistake: Saying a larger total binding energy always means a more stable nucleus.
Ignoring per-nucleon values.
Do this: Compare binding energy per nucleon for stability.
Usually costs: 1 mark.
Mistake: Describing the nuclear force as always attractive.
Forgetting the repulsive core.
Do this: Attractive beyond about 0.8 fm, strongly repulsive at closer range.
Usually costs: ½ to 1 mark.

Exam tips

Binding energy questions appeared in 2023 (55/1/1 Q30(b)), 2024 (55/5/1 Q11), 2025 (55/1/1 Q11) and 2026 (55/1/1 Q15 and 55/5/1 Q15).
Draw the BE/A curve with labelled axes and mark ²H, ⁴He, ⁵⁶Fe and ²³⁸U approximately.
Show the mass defect subtraction line clearly; examiners award marks for it.
Pair the BE curve with fission and fusion in one sentence; it is the most useful application.
Learn the potential energy versus distance graph for two nucleons (minimum near 0.8 fm, steep rise at shorter distance).

Practice questions on Mass Defect and Binding Energy

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 binding energy per nucleon is maximum for nuclei with mass number about
(a) 4 (b) 56 (c) 120 (d) 238
Hint
Iron.
Answer
(b) 56.
Q2 · 2 marks · Very short answerBoard level
Define mass defect and binding energy of a nucleus.
Hint
Difference of masses; energy to separate.
Answer
Mass defect: the difference between the sum of the masses of the nucleons and the mass of the nucleus. Binding energy: the energy needed to separate a nucleus into its free nucleons, equal to the mass defect times c².
Q3 · 1 mark · Very short answerHarder board level
Find the energy equivalent of a mass defect of 0.1 u.
Hint
× 931.5.
Answer
0.1 × 931.5 = 93.15 MeV.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Two protons about 1 fm apart inside a nucleus stay bound in spite of their electrostatic repulsion.
Reason (R): At such separations the attractive nuclear force between nucleons is much stronger than the Coulomb repulsion.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare the two forces at nuclear distances.
Answer
(a).
Q5 · 3 marks · Short answerHOTS / competency
Find the binding energy per nucleon of ⁷₃Li (atomic mass 7.016004 u).
Hint
Z = 3, N = 4.
Answer
ΔM = 3(1.007825) + 4(1.008665) − 7.016004 = 0.042131 u; E_b ≈ 39.2 MeV; per nucleon ≈ 5.61 MeV.
Q6 · 2 marks · Very short answerHOTS / competency
A gamma-ray photon is to split a deuteron (binding energy 2.22 MeV) into a proton and a neutron. Find the longest wavelength that can do it.
Hint
λ(nm) ≈ 1240/E(eV).
Answer
λ ≈ 1240/(2.224 × 10⁶) nm ≈ 5.6 × 10⁻⁴ nm (5.6 × 10⁻¹³ m).
Q7 · 2 marks · Very short answerHarder board level
State three properties of the nuclear force.
Hint
Strength, range, charge.
Answer
Any three: very strong (much stronger than Coulomb force at nuclear distances); short-range (a few fm); attractive above about 0.8 fm but repulsive below; charge independent; saturating.

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.

1.1 u in MeV?Show answer
931.5 MeV.
2.Mass defect formula?Show answer
ΔM = Zm_H + Nm_n − M.
3.Binding energy per nucleon of ⁵⁶Fe?Show answer
About 8.75 MeV (the maximum).
4.Range of the nuclear force?Show answer
A few femtometres.
5.Binding energy of the deuteron?Show answer
About 2.22 MeV.
6.What does saturation of the nuclear force mean?Show answer
Each nucleon interacts only with a limited number of close neighbours.

Mass Defect and Binding Energy: frequently asked questions

Where does the missing mass go when a nucleus forms?

It is released as energy (the binding energy), usually as gamma radiation or kinetic energy, in line with E = mc².

Why is binding energy per nucleon a better measure of stability than total binding energy?

Total binding energy grows with the number of nucleons. Per-nucleon binding shows how tightly each nucleon is held, so it compares nuclei of different sizes fairly.

Why do both fission and fusion release energy?

Both move nuclei towards the peak of the BE/A curve near A = 56. The products are more tightly bound than the starting nuclei, and the extra binding energy is released.

Why is the nuclear force called short-range?

It acts only over a few femtometres. Beyond that it becomes negligible, unlike the Coulomb force, which falls off as 1/r² but never stops.

What is meant by charge independence of nuclear force?

The nuclear force between two protons, two neutrons, or a proton and a neutron is essentially the same, regardless of electric charge.