Mass Defect and Nuclear Binding Energy Calculations

Weigh a helium-4 nucleus, then weigh two protons and two neutrons separately, and you’ll find something surprising: the nucleus is lighter than the particles it’s made of. The missing mass, the mass defect, has been converted into the energy that holds the nucleus together, the binding energy, according to Einstein’s famous equation E = mc². This article works through the calculations step by step, from a single nucleus to the energy released by fusion and fission.

For the concepts behind this, see inside the atomic nucleus.

Key definitions

• Mass defect (Δm): the total mass of the separate protons and neutrons minus the actual mass of the nucleus.
• Binding energy (BE): the energy needed to separate a nucleus completely into its individual protons and neutrons; equivalently, the energy released when the nucleus forms.
• Binding energy per nucleon: BE ÷ mass number (A). It measures how tightly each nucleon is held, and it’s the best guide to nuclear stability.

Data and conversion factors

Quantity	Value
proton mass, mₚ	1.007276 u
neutron mass, mₙ	1.008665 u
electron mass, mₑ	0.000549 u
hydrogen-1 atom	1.007825 u
1 u	1.660539 × 10⁻²⁷ kg
speed of light, c	2.998 × 10⁸ m/s
energy equivalent of 1 u	931.5 MeV
1 MeV	1.602 × 10⁻¹³ J

Tip: If you’re given atomic masses (which include electrons), use the mass of hydrogen-1 atoms instead of bare protons. The electrons then cancel automatically. If you’re given nuclear masses, use bare proton masses.

Worked example 1: helium-4

The mass of a helium-4 nucleus is 4.001506 u. Calculate its mass defect and binding energy.

Step 1: mass of separate nucleons 2 protons: 2 × 1.007276 = 2.014552 u 2 neutrons: 2 × 1.008665 = 2.017330 u Total = 4.031882 u

Step 2: mass defect Δm = 4.031882 − 4.001506 = 0.030376 u

Step 3: binding energy (MeV) BE = 0.030376 × 931.5 = 28.30 MeV

Step 4: binding energy per nucleon 28.30 ÷ 4 = 7.07 MeV per nucleon

Worked example 2: the same calculation in joules

Using E = mc² directly:

Δm = 0.030376 u × 1.660539 × 10⁻²⁷ kg/u = 5.0441 × 10⁻²⁹ kg

E = 5.0441 × 10⁻²⁹ × (2.998 × 10⁸)² = 5.0441 × 10⁻²⁹ × 8.988 × 10¹⁶ = 4.534 × 10⁻¹² J

Check: 4.534 × 10⁻¹² J ÷ 1.602 × 10⁻¹³ J/MeV = 28.30 MeV ✓

Per mole of helium-4 nuclei: 4.534 × 10⁻¹² × 6.022 × 10²³ = 2.73 × 10¹² J/mol

That’s about 2.7 million megajoules per mole. Compare with chemical bonds: breaking a mole of C–H bonds takes only about 0.4 MJ. Nuclear binding energies are millions of times larger than chemical bond energies.

Worked example 3: carbon-12 using atomic masses

The atomic mass of carbon-12 is exactly 12.000000 u. Calculate the binding energy per nucleon.

Using atomic masses, use hydrogen-1 atoms (which include the electrons) instead of bare protons:

• 6 hydrogen-1 atoms: 6 × 1.007825 = 6.046950 u
• 6 neutrons: 6 × 1.008665 = 6.051990 u
• Total = 12.098940 u

Δm = 12.098940 − 12.000000 = 0.098940 u

BE = 0.098940 × 931.5 = 92.16 MeV

BE per nucleon = 92.16 ÷ 12 = 7.68 MeV

Worked example 4: iron-56

The atomic mass of iron-56 is 55.934936 u (26 protons, 30 neutrons).

• 26 hydrogen-1 atoms: 26 × 1.007825 = 26.203450 u
• 30 neutrons: 30 × 1.008665 = 30.259950 u
• Total = 56.463400 u

Δm = 56.463400 − 55.934936 = 0.528464 u

BE = 0.528464 × 931.5 = 492.3 MeV

BE per nucleon = 492.3 ÷ 56 = 8.79 MeV

Iron-56 is among the most tightly bound of all nuclei (nickel-62 is marginally higher). That’s why it sits at the peak of the binding energy curve.

The binding energy per nucleon curve

Nucleus	BE per nucleon (MeV, approx.)
hydrogen-2 (deuterium)	1.1
helium-4	7.1
carbon-12	7.7
oxygen-16	8.0
iron-56	8.8
uranium-235	7.6

• It rises steeply for light nuclei, with helium-4 unusually high.
• It peaks near iron-56 and nickel-62.
• It falls slowly for heavy nuclei.

Energy is released whenever nuclei move towards the peak, because the products are more tightly bound than the starting nuclei.

Worked example 5: energy from fusion

Calculate the energy released when deuterium and tritium fuse:

²H + ³H → ⁴He + ¹n

Atomic masses: ²H = 2.014102 u; ³H = 3.016049 u; ⁴He = 4.002602 u; n = 1.008665 u

Mass before = 2.014102 + 3.016049 = 5.030151 u Mass after = 4.002602 + 1.008665 = 5.011267 u

Δm = 5.030151 − 5.011267 = 0.018884 u

Energy = 0.018884 × 931.5 = 17.6 MeV per reaction

This D–T reaction is the one most fusion reactor designs aim to use. See nuclear fission vs fusion.

Worked example 6: energy from fission (using BE per nucleon)

A uranium-235 nucleus absorbs a neutron and splits into two fragments with a total of about 234 nucleons (after releasing two neutrons). Estimate the energy released, given BE per nucleon of about 7.6 MeV for uranium and about 8.5 MeV for the fragments.

Energy released ≈ (8.5 − 7.6) × 235 ≈ 0.9 × 235 ≈ 210 MeV per fission

The accepted value is about 200 MeV per fission, around 10⁸ times the energy released per molecule in burning fuels such as methane. That’s why a few kilograms of uranium can power a city for a day.

Why the curve has this shape

• Small nuclei: each nucleon has few neighbours, and many nucleons sit at the surface, so the strong force binds them less effectively.
• Medium nuclei (around iron): the attractive strong force is best balanced against electrical repulsion between protons.
• Large nuclei: the growing number of protons increases electrical repulsion (every proton repels every other), which reduces binding per nucleon and eventually makes nuclei unstable. See radioactive elements.

Does chemistry involve a mass defect too?

In principle, yes. When a chemical reaction releases energy, the products are very slightly lighter than the reactants, by exactly E/c². But the effect is tiny. Burning one mole of methane releases about 890 kJ, which corresponds to a mass change of only about 10⁻¹¹ kg, far too small to detect with a balance. That’s why the law of conservation of mass works perfectly for chemistry, while nuclear reactions, which release millions of times more energy per atom, show measurable mass changes.

Common mistakes

1. Subtracting the wrong way round. Mass defect = (separate nucleons) − (nucleus), always positive.
2. Mixing atomic and nuclear masses. Use hydrogen-1 atoms with atomic masses, bare protons with nuclear masses.
3. Forgetting to square c.
4. Unit slips: convert u to kg before using E = mc² in joules, or use 931.5 MeV/u directly.
5. Confusing total BE with BE per nucleon. Uranium has a much larger total binding energy than iron, but a smaller binding energy per nucleon.

Practice questions

1. Calculate the binding energy of deuterium (atomic mass 2.014102 u) in MeV.
2. The atomic mass of oxygen-16 is 15.994915 u. Calculate its binding energy per nucleon.

Answers:

1. 1.007825 + 1.008665 − 2.014102 = 0.002388 u → × 931.5 = 2.22 MeV
2. 8(1.007825) + 8(1.008665) = 16.131920 u; Δm = 0.137005 u; BE = 127.6 MeV; per nucleon = 7.98 MeV

Key takeaways

• Mass defect = mass of separate nucleons − mass of the nucleus; it’s converted into binding energy by E = mc².
• 1 u of mass is equivalent to 931.5 MeV.
• Binding energy per nucleon peaks near iron-56 (about 8.8 MeV).
• Fusion of light nuclei and fission of heavy nuclei both release energy by moving towards the peak.
• Nuclear energies are millions of times larger than chemical bond energies. See also the atomic mass unit.