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Nuclear radiation · Subtopic 1 of 2

Mass Defect and Binding Energy

Mass and energy are equivalent: E=mc2E=mc^2. A nucleus has less mass than its separate protons and neutrons; this mass defect Δm\Delta m corresponds to the binding energy c2Δmc^2\Delta m, the energy needed to separate the nucleus into its nucleons. Binding energy per nucleon peaks near iron-56. Splitting heavy nuclei (fission) or joining light ones (fusion) moves towards the peak, releasing energy.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 2≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use E=mc2E=mc^2 and the unified atomic mass unit.
  • Define and calculate mass defect and binding energy.
  • Sketch and interpret the binding energy per nucleon curve.
  • Calculate energy released in fission and fusion.
Pause and think

A helium-4 nucleus is lighter than two protons and two neutrons weighed separately. Where did the missing mass go?

Section 2 of 14

Core idea

Start hereThe whole weighs less than its parts

Weigh two protons and two neutrons separately, then a helium nucleus made of them: the nucleus is lighter. The missing mass became energy when they bound together. That's where nuclear power and starlight come from.

The big idea

Mass defect Δm\Delta m; binding energy Δmc2\Delta mc^2. Binding energy per nucleon peaks near iron: fusion (light) and fission (heavy) both release energy.

1. Mass defect

E=Δmc2,1 u≡931.5 MeVE=\Delta mc^2,\qquad1\text{ u}\equiv931.5\text{ MeV}
where
Δm\Delta m
mass of separate nucleons − mass of nucleus.
Worked exampleHelium-4

Nucleus 4.001506 u; mp=1.007276m_p=1.007276 u, mn=1.008665m_n=1.008665 u.

Goal
Binding energy per nucleon.
Strategy
Defect × 931.5.
  1. 2(1.007276)+2(1.008665)=4.031882 u2(1.007276)+2(1.008665)=4.031882\text{ u}
    Separate nucleons.
  2. Δm=0.030376 u\Delta m=0.030376\text{ u}
    Subtract.
  3. 0.030376×931.5=28.3 MeV0.030376\times931.5=28.3\text{ MeV}
    Binding energy.
  4. 28.3÷4=7.07 MeV28.3\div4=7.07\text{ MeV}
    Per nucleon.
Answer7.077.07 MeV per nucleon.
✓ CheckHelium sits high on the curve for a light nucleus ✓.

2. The curve

Binding energy per nucleon against nucleon number
501001502002468100nucleon number Abinding energy per nucleon (MeV)He-4Fe-56 (most stable)U-235fusion releases energy →← fission releases energy
Moving towards the peak from either side increases the binding energy per nucleon, releasing energy: fusion of light nuclei, fission of heavy ones.
Worked exampleEnergy from a fission

U-235 (~7.6 MeV/nucleon) → products (~8.5 MeV/nucleon).

Goal
Energy released.
Strategy
Gain per nucleon × number of nucleons.
  1. 8.5−7.6=0.9 MeV8.5-7.6=0.9\text{ MeV}
    Gain per nucleon.
  2. 235×0.9≈210 MeV235\times0.9\approx210\text{ MeV}
    Total.
AnswerAbout 200 MeV.
✓ CheckMatches the accepted ~200 MeV per fission ✓.
Common misconceptionMore binding energy = less stable? Higher binding energy per nucleon means more stable.
Pause and think

Why don't stars release energy by fusing iron?

Summary

  • ΔE=Δmc2\Delta E=\Delta mc^2.
  • Peak near Fe-56.
Key words
Mass defect
Mass lost when nucleons bind.
HistoryMeitner on a winter walk. Over Christmas 19381938, Lise Meitner and her nephew Otto Frisch, walking in the Swedish snow, realised that Hahn's uranium experiments meant the nucleus was splitting in two. Using E=mc2E=mc^2, they calculated about 200200 MeV per fission, and named it after cell division.
Fun factThe Sun loses weight. The Sun converts about 44 million tonnes of mass into energy every second by fusing hydrogen into helium. Even so, in its whole 1010-billion-year life it will lose less than 0.1%0.1\% of its mass this way.
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