Mass defect and binding energy

The mass defect of a nucleus is the difference between the total mass of its separate protons and neutrons and the mass of the nucleus itself. A nucleus always weighs less than its parts. That missing mass is the binding energy, found with E = mc²: the energy needed to pull the nucleus apart into separate nucleons, which is also the energy released when it forms.

What is mass defect?

Weigh 3 protons and 4 neutrons separately and add up the masses. Then weigh a lithium-7 nucleus, which is made of exactly those seven particles. The nucleus is lighter by about 0.6%. Nothing has been lost: the mass has gone into the energy that holds the nucleus together.

For a nucleus with proton number Z and nucleon number A, there are Z protons and A − Z neutrons, so:

Here m_p is the proton mass, m_n is the neutron mass and m_nucleus is the measured mass of the nucleus. At A-level, masses are usually given in atomic mass units, u. The AQA data booklet gives m_p = 1.00728 u, m_n = 1.00867 u and 1 u = 1.661 × 10⁻²⁷ kg.

What is binding energy?

Binding energy is the energy needed to break a nucleus up into its separate nucleons. The same amount is released when those nucleons come together to form the nucleus. Einstein's equation turns the mass defect into this energy:

With Δm in kg and c = 3.00 × 10⁸ m/s, E comes out in joules. It is usually quicker to work in u and MeV, because 1 u of mass is equivalent to 931.5 MeV of energy:

Try E = mc² on its own first. This question from the Nuclear Energy course turns a reactor's lost mass into energy.

How to calculate binding energy

1. Count the protons (Z) and neutrons (A − Z).
2. Add up the masses of the separate nucleons: Zm_p + (A − Z)m_n.
3. Subtract the mass of the nucleus to get the mass defect, Δm.
4. Multiply Δm in u by 931.5 to get the binding energy in MeV, or change Δm to kg and use E = Δmc² for joules.
5. For binding energy per nucleon, divide by A.

Example 1: lithium-7 in MeV

A lithium-7 nucleus has a mass of 7.01436 u. Find its mass defect and binding energy.

Lithium has proton number 3, so the nucleus has 3 protons and 7 − 3 = 4 neutrons:

The mass defect is 0.04216 u and the binding energy is 39.3 MeV, or 39.3 ÷ 7 = 5.61 MeV per nucleon.

Example 2: the same nucleus in joules

Change the mass defect to kg, then use E = Δmc²:

That is 6.30 × 10⁻¹² J. Dividing by 1.60 × 10⁻¹⁹ J per eV gives 39.4 MeV. The small difference from example 1 comes from the rounded constants, so use whichever route your question asks for.

Example 3: iron-56 per nucleon

An iron-56 nucleus has 26 protons, 30 neutrons and a mass of 55.92067 u. Find its binding energy per nucleon.

Iron-56 holds each nucleon far more tightly than lithium-7 does. Check these in the binding energy calculator.

Binding energy per nucleon

Total binding energy grows with the size of the nucleus, so to compare stability you divide by the number of nucleons. The higher the binding energy per nucleon, the more tightly bound and stable the nucleus.

These values come from measured atomic masses. They differ from the worked examples in the last decimal place because the examples use the data booklet's rounded nucleon masses.

Here x is the nucleon number A and y is the binding energy per nucleon in MeV. This smooth curve comes from a model of the nucleus called the semi-empirical mass formula, so it is close to the real values but not exact: it peaks at about 8.7 MeV, and light nuclei such as helium-4 really sit well above it. The shape is what matters. The curve rises steeply, peaks near iron, then falls slowly for heavy nuclei.

Why fission and fusion release energy

Any reaction that moves nucleons up the curve, towards iron, releases energy, because the products are more tightly bound than what you started with.

• Fission: a heavy nucleus such as uranium-235 (7.59 MeV per nucleon) splits into two middle-sized nuclei with about 8.4 MeV per nucleon. A gain of about 0.8 MeV for each of 235 nucleons is why one fission releases around 200 MeV.
• Fusion: light nuclei join to make a heavier one. The steep left side of the curve means fusion releases even more energy per nucleon than fission.

You can find the energy released from the masses directly. In deuterium and tritium fusion, a deuteron (2.013553 u) and a triton (3.015501 u) make a helium-4 nucleus (4.001506 u) and a neutron (1.00867 u):

Each fusion releases 17.6 MeV. The Nuclear Energy course covers fission, chain reactions and fusion step by step.

Common mistakes

• Mixing atomic and nuclear masses. An atomic mass includes the electrons. If you are given the mass of an atom, subtract Z electron masses (5.5 × 10⁻⁴ u each) before comparing with the nucleons.
• Rounding too early. The mass defect is a small difference between two large numbers, so keep at least five decimal places in u until the end.
• Using u in E = mc². E = Δmc² needs Δm in kg. With u, multiply by 931.5 to get MeV instead.
• Thinking binding energy is stored in the nucleus. It is the energy you would have to put in to pull the nucleus apart. A bound nucleus has less energy than its separate parts.
• Comparing total binding energy. Uranium-235 has far more total binding energy than iron-56, but less per nucleon, so it is less stable.

Practise nuclear physics questions

Nuclear physics is in the second year of A-level Physics, so it comes up in the daily A-level Physics: Year 2 paper, a short timed paper of 8 questions with a new one every day.

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Physics · Atoms and particlesBinding energy calculator: from the mass defect, in MeVTurn a nucleus's mass defect into its binding energy in MeV or joules, with the working, plus how to find the mass defect and binding energy per nucleon.