Mass Defect and Binding Energy

In the last lesson we met Einstein's relation E = mc² and used it to compute the energy released in individual nuclear decays. The same idea, applied to entire nuclei, leads directly to the concept of binding energy: the amount of energy that holds a nucleus together. Graphing the binding energy per nucleon against mass number produces one of the most famous diagrams in physics, and explains in a single picture why iron is so abundant, why heavy nuclei can release energy by fission, and why light nuclei can release energy by fusion.

This lesson covers section 6.4.3 of the OCR A-Level Physics A specification (H556) and lays the groundwork for Lesson 5 on fission and fusion.

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The Mass Defect

If you measure the mass of a helium-4 nucleus very precisely, you find:

m(⁴He nucleus) = 4.00150 u

Now add up the masses of its constituents — two protons and two neutrons:

2 m_p + 2 m_n = 2(1.00728) + 2(1.00867) = 4.03190 u

The bound nucleus is lighter than the sum of its separate parts. The difference,

Δm = 4.03190 - 4.00150 = 0.03040 u

is called the mass defect. It is a genuine, measurable mass difference: the helium nucleus weighs less on a mass spectrometer than the constituents it is made of.

Definition. The mass defect of a nucleus is the difference between the total mass of its separated constituent nucleons and the mass of the assembled nucleus:

Δm = Z m_p + (A - Z) m_n - m_nucleus

Every stable nucleus has a positive mass defect.

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The Binding Energy

By E = mc², the mass defect corresponds to an energy:

E_B = Δm × c²

This is the binding energy of the nucleus. It is the amount of energy you would need to supply to pull the nucleus apart into its constituent nucleons, or equivalently, the amount of energy released when those free nucleons assemble themselves into the nucleus.

For helium-4:

E_B = 0.03040 u × 931.5 MeV/u ≈ 28.3 MeV

That is the total binding energy of ⁴He. It takes 28.3 MeV to pull a helium nucleus apart into two free protons and two free neutrons.

This is an enormous amount of energy compared to chemical binding (which is of order a few eV per atom). The strong nuclear force is about a million times stronger than the electromagnetic force at the distances involved.

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Binding Energy per Nucleon

The total binding energy grows with nucleon number A: bigger nuclei have more nucleons and hence more bonds. The quantity that matters physically — and that is plotted in every textbook — is the binding energy per nucleon:

E_B / A

This tells you, on average, how tightly each nucleon is bound in that nucleus. A high value means tightly bound, stable, hard to dismantle. A low value means loosely bound, potentially ready to rearrange into a more tightly bound configuration (either by fission or by fusion).

For helium-4: E_B / A = 28.3 / 4 ≈ 7.1 MeV per nucleon.

Worked Example 1: Binding energy of oxygen-16

Oxygen-16 has a nuclear mass of 15.99052 u. Find its total binding energy and the binding energy per nucleon.

Solution.

Sum of nucleons: 8(1.00728) + 8(1.00867) = 16.12760 u
Δm = 16.12760 - 15.99052 = 0.13708 u
E_B = 0.13708 × 931.5 ≈ 127.7 MeV
E_B / A = 127.7 / 16 ≈ 7.98 MeV per nucleon

Oxygen-16 is one of the most tightly bound light nuclei — a consequence of its magic number structure (8 protons and 8 neutrons).

Worked Example 2: Binding energy of iron-56

Iron-56 has a nuclear mass of 55.92068 u. Find its binding energy per nucleon.

Solution.

Sum of nucleons: 26(1.00728) + 30(1.00867) = 56.44938 u
Δm = 56.44938 - 55.92068 = 0.52870 u
E_B = 0.52870 × 931.5 ≈ 492.5 MeV
E_B / A = 492.5 / 56 ≈ 8.79 MeV per nucleon

Iron-56 has the highest binding energy per nucleon of any nucleus. It sits at the peak of the binding energy curve and is therefore the most stable nucleus in existence. This is why, in the cores of massive stars, nuclear burning proceeds through successive fusion stages until it reaches iron — and then stops, because no more energy can be extracted by further fusion.

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The Binding Energy Curve

If you plot E_B / A against A for all the stable nuclei, you get a curve that rises steeply from hydrogen, passes through helium, carbon and oxygen, peaks near iron (A ≈ 56), and slopes gently downward toward uranium. The peak value is about 8.8 MeV per nucleon; typical values across the middle of the periodic table are 8.5 MeV per nucleon.

flowchart TB
    H["Hydrogen (A=1)<br/>~0 MeV/nucleon"]
    He["Helium-4 (A=4)<br/>~7.1 MeV/nucleon"]
    C["Carbon-12 (A=12)<br/>~7.7 MeV/nucleon"]
    O["Oxygen-16 (A=16)<br/>~8.0 MeV/nucleon"]
    Fe["Iron-56 (A=56)<br/>~8.8 MeV/nucleon (PEAK)"]
    U["Uranium-238 (A=238)<br/>~7.6 MeV/nucleon"]
    H --> He --> C --> O --> Fe --> U

The qualitative shape is far more important than any individual value. You should be able to sketch this curve from memory, labelling:

• Hydrogen (A = 1) at E_B/A = 0
• Helium-4 as an unusually bound "spike" near A = 4 (~7.1 MeV)
• The rising slope from A \approx 4 to A \approx 60
• Iron-56 at the peak (E_B/A \approx 8.8 MeV)
• A gentle decline to uranium-238 (~7.6 MeV)

From this single curve you can read off two profoundly important facts: