PhysicsISC Class 12 · Nuclei

Mass Defect and Binding Energy: E = mc², Binding Energy per Nucleon and Nuclear Stability

Updated 15 October 2026 · Fact-checked

The mass defect of a nucleus is the difference between the total mass of its separate nucleons and the mass of the nucleus: $\Delta m = Zm_p + Nm_n - M$. Its energy equivalent, $BE = \Delta m\,c^2$ (with 1 u = 931 MeV), is the binding energy needed to pull the nucleus apart. BE per nucleon measures stability.

The nucleus is lighter than its separate nucleons; the missing mass Δm leaves as energy, and BE = Δm × 931 MeV is needed to pull it apart.

Illustration drawn for QuestionBuddy.

Understand Mass Defect and Binding Energy

Composition and units. A nucleus $_{Z}^{A} X$ contains Z protons and N = A − Z neutrons (nucleons). Atomic and nuclear masses are measured in the unified atomic mass unit: 1 u = $\frac{1}{12}$ of the mass of a $^{12}$ C atom = $1.66 \times 1 0^{- 27}$ kg. Isotopes have the same Z; isobars the same A; isotones the same N.

Size and density. Nuclear radius $R = R_{0} A^{1/3}$ with $R_{0} \approx 1.2$ fm. Volume is proportional to A, so the nuclear density (about $2 \times 1 0^{17}$ kg/m³) is the same for all nuclei.

Mass-energy equivalence. Einstein's relation $E = m c^{2}$ means mass is a form of energy. For 1 u:
$E = 1.66 \times 1 0^{- 27} \times (3 \times 1 0^{8})^{2} \approx 1.49 \times 1 0^{- 10} J \approx 931 MeV .$
ISC numericals use 1 u = 931 MeV.

Mass defect. The mass of a nucleus is always less than the total mass of its nucleons taken separately:
$Δ m = [Z m_{p} + (A - Z) m_{n}] - M_{n u c l e u s} .$
Using atomic masses (as tables give), use the mass of the hydrogen atom $m_{H}$ for each proton: the electron masses then cancel:
$Δ m = Z m_{H} + N m_{n} - M_{a t o m} .$

Binding energy. When nucleons come together, energy $Δ m c^{2}$ is released (that is where the "missing" mass goes). The same energy must be supplied to separate the nucleus completely into free nucleons:
$B E = Δ m c^{2} = Δ m (u) \times 931 MeV .$

Binding energy per nucleon. $\frac{B E}{A}$ is the average energy needed to remove one nucleon; the higher it is, the more stable the nucleus. Typical values: $^{2}$ H about 1.1 MeV; $^{4}$ He about 7.1 MeV; $^{12}$ C about 7.7 MeV; $^{56}$ Fe about 8.8 MeV (near the maximum); $^{238}$ U about 7.6 MeV. The curve of BE/A against A rises steeply for light nuclei, peaks near A ≈ 56, and falls slowly for heavy nuclei: middle-order nuclei are the most stable. (Fission and fusion follow from this curve; see the fission and fusion topics.)

Other mass-energy examples.

Pair production: a γ photon of energy at least $2 m_{e} c^{2} = 2 \times 0.511 = 1.02$ MeV, passing near a heavy nucleus, can become an electron-positron pair; extra energy becomes their kinetic energy.
Annihilation: an electron and a positron meet and vanish, producing (usually) two γ photons of 0.511 MeV each, travelling in opposite directions to conserve momentum.
Chemical reactions also convert mass to energy, but the mass change is far too small to measure.

Mass Defect and Binding Energy: formulas and key facts

Mass defect: $Δ m = Z m_{H} + N m_{n} - M_{a t o m}$; Atomic masses; electrons cancel.
Binding energy: $B E = Δ m c^{2} = Δ m (u) \times 931 MeV$; ISC uses 931.
BE per nucleon: $\frac{B E}{A}$; Higher means more stable.
Mass-energy: $E = m c^{2}$; Joules with kg.
Nuclear radius: $R = R_{0} A^{1/3}, R_{0} \approx 1.2 fm$; Density independent of A.
Pair production threshold: $E_{γ} \geq 2 m_{e} c^{2} = 1.02 MeV$; Needs a nearby nucleus.

How to solve Mass Defect and Binding Energy questions

Work in u until the last step, then multiply by 931 MeV; keep six decimal places throughout.

1Count Z and N = A − Z.
2Total mass of parts: $Z m_{H} + N m_{n}$ (with atomic masses).
3Δm = parts − atom; it must come out positive.
4BE = Δm × 931 MeV; BE/A = BE ÷ A.
5To compare stability, compare BE/A, not total BE.
6Energy to split a nucleus into given fragments = (mass of fragments − mass of nucleus) × 931 MeV.

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 loss in an ordinary fire

Suppose burning 1 kg of coal releases about $3 \times 1 0^{7}$ J of heat.

How much mass is converted into this energy, and why is it never noticed?

Try it first, then show the solutionSolution

$Δ m = \frac{E}{c ^{2}} = \frac{3 \times 1 0 ^{7}}{9 \times 1 0 ^{16}} \approx 3.33 \times 1 0^{- 10}$ kg.
That is a third of a microgram from a kilogram: about 3 parts in $1 0^{10}$ , far below what any balance can detect.

Answer: ≈ $3.3 \times 1 0^{- 10}$ kg; too small to measure, unlike in nuclear reactions.

Memory hook: Every energy change is a mass change; only nuclear ones are big enough to see.

Gamma rays in lead shielding

Imagine a 2.5 MeV gamma photon from a hospital source entering the lead wall of a radiotherapy room.

Can it produce an electron-positron pair? If so, what total kinetic energy do the particles share?

Try it first, then show the solutionSolution

Threshold $= 2 \times 0.511 = 1.022$ MeV, less than 2.5 MeV, so pair production can occur near a lead nucleus.
Kinetic energy shared $= 2.5 - 1.022 \approx 1.48$ MeV.

Answer: Yes; about 1.48 MeV of kinetic energy.

Memory hook: One point zero two to make a pair.

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
Separate nucleons > nucleus; the difference is binding.
Use it when: Defining mass defect.
Nine-three-one
1 u = 931 MeV (ISC value).
Use it when: Every binding numerical.
Iron is the summit
BE/A peaks near A = 56 (about 8.8 MeV).
Use it when: Curve questions.
Per nucleon for stability
Compare BE/A, not total BE.
Use it when: Which is more stable?

Mass Defect and Binding Energy: solved examples

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

Example 1: The most stable nuclei · 1 mark · MCQBoard level

Binding energy per nucleon is greatest for nuclei with mass number about:
(a) 4
(b) 56
(c) 120
(d) 238

Try it first, then show the solutionSolution

The BE/A curve peaks near iron, A ≈ 56 (about 8.8 MeV).

Answer: (b).

Example 2: Binding energy of the deuteron · 2 marks · Short answerBoard level

Find the mass defect, binding energy and binding energy per nucleon of the deuteron. (m(²H) = 2.014102 u, $m_{H}$ = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV)

Try it first, then show the solutionSolution

Δm = 1.007825 + 1.008665 − 2.014102 = 0.002388 u. (1)
BE = 0.002388 × 931 ≈ 2.22 MeV; BE/A ≈ 1.11 MeV. (1)

Answer: 0.002388 u; ≈ 2.22 MeV; ≈ 1.11 MeV per nucleon.

Example 3: Why 1 u is 931 MeV · 3 marks · Long answerHarder board level

Using 1 u = $1.6605 \times 1 0^{- 27}$ kg, c = $2.998 \times 1 0^{8}$ m/s and 1 MeV = $1.602 \times 1 0^{- 13}$ J, show that 1 u is equivalent to about 931 MeV. Then explain why the mass of a nucleus is less than the sum of the masses of its nucleons.

Try it first, then show the solutionSolution

$E = m c^{2} = 1.6605 \times 1 0^{- 27} \times (2.998 \times 1 0^{8})^{2} \approx 1.492 \times 1 0^{- 10}$ J. (1)
$\frac{1.492 \times 1 0 ^{- 10}}{1.602 \times 1 0 ^{- 13}} \approx 931$ MeV. (1)
When nucleons bind, energy (the binding energy) is released; by $E = m c^{2}$ this energy carries away mass Δm, so the nucleus is lighter than its parts. (1)

Answer: ≈ 931 MeV; the missing mass is the binding energy released.

Example 4: Carbon-12 · 3 marks · Short answerHarder board level

Find the binding energy and binding energy per nucleon of $_{6}^{12}$ C (atomic mass exactly 12.000000 u). ( $m_{H}$ = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV)

Try it first, then show the solutionSolution

Parts: 6 × 1.007825 + 6 × 1.008665 = 12.098940 u. (1)
Δm = 12.098940 − 12.000000 = 0.098940 u; BE = 0.098940 × 931 ≈ 92.1 MeV. (1)
BE/A = $\frac{92.1}{12}$ ≈ 7.68 MeV. (1)

Answer: ≈ 92.1 MeV; ≈ 7.68 MeV per nucleon.

Example 5: Splitting carbon into alpha particles · 3 marks · Short answerHOTS / competency

How much energy is needed to split a $^{12}$ C nucleus into three $^{4}$ He nuclei? (m(⁴He) = 4.002604 u, m(¹²C) = 12.000000 u, 1 u = 931 MeV). Check your answer using binding energies.

Try it first, then show the solutionSolution

Mass of products − mass of $^{12}$ C = 3 × 4.002604 − 12.000000 = 0.007812 u (products heavier: energy must be supplied). (1)
Energy = 0.007812 × 931 ≈ 7.27 MeV. (1)
Check: BE(¹²C) ≈ 92.11 MeV; BE(⁴He) = 0.030376 × 931 ≈ 28.28 MeV; 92.11 − 3 × 28.28 ≈ 7.27 MeV. (1)

Answer: ≈ 7.27 MeV must be supplied.

Example 6: Iron-56 on the curve · 4 marks · Case-basedHOTS / competency

Suppose you are asked to place $_{26}^{56}$ Fe (atomic mass 55.934937 u) on the BE/A curve. ( $m_{H}$ = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV)
(i) Find the mass defect. (1 mark)
(ii) Find the binding energy. (1 mark)
(iii) Find BE/A and say where it lies on the curve. (1 mark)
(iv) Find its nuclear radius ( $R_{0}$ = 1.2 fm). (1 mark)

Try it first, then show the solutionSolution

(i) 26 × 1.007825 + 30 × 1.008665 − 55.934937 = 0.528463 u.
(ii) BE = 0.528463 × 931 ≈ 492.0 MeV.
(iii) BE/A ≈ 8.79 MeV, at the peak of the curve: among the most stable nuclei.
(iv) R = 1.2 × $5 6^{1/3}$ ≈ 4.59 fm.

Answer: (i) 0.528463 u; (ii) ≈ 492 MeV; (iii) ≈ 8.79 MeV, near the peak; (iv) ≈ 4.6 fm.

Where students lose marks in Mass Defect and Binding Energy

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

Mistake: Rounding too early
The ISC 2024 examiners (Q19(ii)(a), neon-20) saw intermediate values rounded, giving deviated answers.
Do this: Keep six decimal places in u until the final multiplication.
Usually costs: 1 mark.
Mistake: Converting Δm to kg needlessly or using 931.5
Also 2024: unnecessary kg conversion, or 931.5 used when 931 was given.
Do this: Use the value the paper gives: Δm (u) × 931 MeV.
Usually costs: ½ to 1 mark.
Mistake: Wrong definition of mass defect
In 2025 Q19(i)(b) candidates confused the mass of the nucleus with the mass of nucleons.
Do this: Mass defect = (sum of masses of free nucleons) − (mass of nucleus).
Usually costs: 1 mark.
Mistake: Pair production threshold as 1.02 eV or 931 MeV
The 2025 examiners (Q1(B)(vi)) saw wrong units or values.
Do this: $2 m_{e} c^{2} = 1.02$ MeV.
Usually costs: 1 mark.

Exam tips

Write Z and N explicitly before the mass-defect line.
Give BE in MeV and BE/A in MeV per nucleon.
Sketch the BE/A curve with its peak near A = 56 when explaining stability.
Use atomic masses with $m_{H}$ , not $m_{p}$ , unless the question gives nuclear masses.

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
One unified atomic mass unit is defined as:
(a) the mass of a proton
(b) the mass of a hydrogen atom
(c) $\frac{1}{12}$ of the mass of a carbon-12 atom
(d) $\frac{1}{16}$ of the mass of an oxygen atom
Hint
Carbon standard.
Answer
(c).
Q2 · 1 mark · Short answerBoard level
Find the energy equivalent of 1 mg of mass.
Hint
$E = m c^{2}$ with m in kg.
Answer
$9 \times 1 0^{10}$ J.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mass of a nucleus is less than the sum of the masses of its free nucleons.
Reason (R): Energy equal to the binding energy is released when the nucleons combine.
(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
E = mc².
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Find the binding energy per nucleon of $_{3}^{7}$ Li (atomic mass 7.016004 u).
Hint
Z = 3, N = 4.
Answer
Δm = 0.042131 u; BE ≈ 39.2 MeV; BE/A ≈ 5.60 MeV.
Q5 · 2 marks · Short answerHarder board level
Find the ratio of the nuclear radii of $^{27}$ Al and $^{64}$ Cu, and the ratio of their densities.
Hint
R ∝ $A^{1/3}$ .
Answer
3 : 4; densities equal (1 : 1).
Q6 · 2 marks · Short answerHOTS / competency
Find the energy needed to remove one neutron from $^{13}$ C (13.003355 u), leaving $^{12}$ C (12.000000 u). ( $m_{n}$ = 1.008665 u)
Hint
Products − reactant.
Answer
(12.000000 + 1.008665 − 13.003355) × 931 ≈ 4.94 MeV.
Q7 · 2 marks · Short answerHOTS / competency
Explain, with the BE/A curve, why $^{4}$ He (7.1 MeV) is more stable than $^{7}$ Li (5.6 MeV) although $^{7}$ Li has the larger total binding energy.
Hint
Per nucleon.
Answer
Stability depends on the energy per nucleon: each nucleon in ⁴He is bound by about 7.1 MeV compared with 5.6 MeV in ⁷Li, so ⁴He is harder to break apart.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose you analyse $_{20}^{40}$ Ca (atomic mass 39.962591 u). ( $m_{H}$ = 1.007825 u, $m_{n}$ = 1.008665 u, 1 u = 931 MeV)
(i) Find the total mass of its separate nucleons. (1 mark)
(ii) Find the mass defect. (1 mark)
(iii) Find the binding energy. (1 mark)
(iv) Find BE/A and compare its stability with $^{12}$ C (7.68 MeV). (1 mark)
Hint
Z = N = 20.
Answer
(i) 40.329800 u. (ii) 0.367209 u. (iii) ≈ 341.9 MeV. (iv) ≈ 8.55 MeV per nucleon: more stable than ¹²C.

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.Show answer
The energy needed to separate a nucleus completely into its free nucleons (equal to $Δ m c^{2}$ ).
2.1 u in MeV?Show answer
931 MeV.
3.Peak of the BE/A curve?Show answer
Near A = 56, about 8.8 MeV.
4.Minimum energy for pair production?Show answer
1.02 MeV.
5.How does nuclear density depend on A?Show answer
It is practically independent of A.

Mass Defect and Binding Energy: frequently asked questions

Where does the missing mass go?

It is released as energy (radiation and kinetic energy) when the nucleus forms; Δm × c² equals that energy.

Why use the hydrogen atom mass instead of the proton mass?

Tables give atomic masses, which include electrons. Using m_H for each proton adds exactly Z electrons, which cancel the Z electrons in the atomic mass.

Does a bigger total binding energy mean a more stable nucleus?

No. Larger nuclei have more nucleons; stability is judged by binding energy per nucleon.

What does the ISC 2027 syllabus need here?

Composition and size of the nucleus, nuclear density, isotopes, isobars and isotones, the atomic mass unit, mass defect and BE = Δm c², the BE/A graph and its features, E = mc² calculations including annihilation and pair production.

How was this asked recently?

ISC 2024 Q19(ii)(a) (2 marks) asked the mass defect and binding energy of neon-20; ISC 2025 Q19(i)(b) defined mass defect, and Q1(B)(vi) asked the pair-production threshold.