Nuclear Binding Energy

The concept of binding energy is fundamental to understanding why nuclear reactions release energy and why certain nuclei are more stable than others. This lesson covers mass defect, Einstein's mass-energy equivalence, the binding energy per nucleon curve, and why fission and fusion release energy. This is assessed in AQA section 3.8.1 (Paper 2).

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Mass Defect

When nucleons (protons and neutrons) come together to form a nucleus, the mass of the resulting nucleus is always less than the total mass of the individual nucleons. This difference is called the mass defect (Δm):

Δm = [Z × mₚ + (A − Z) × mₙ] − M_nucleus

where:

• Z is the number of protons
• A is the nucleon number
• mₚ is the mass of a proton (1.00728 u = 1.673 × 10⁻²⁷ kg)
• mₙ is the mass of a neutron (1.00866 u = 1.675 × 10⁻²⁷ kg)
• M_nucleus is the measured mass of the nucleus
• 1 u = 1.661 × 10⁻²⁷ kg

The "missing" mass has been converted into energy — the binding energy — according to Einstein's equation.

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Einstein's Mass-Energy Equivalence

E = mc²

where c = 3.00 × 10⁸ m s⁻¹.

This equation tells us that mass and energy are interchangeable. When nucleons bind together, the energy released (the binding energy) corresponds to a decrease in mass equal to the mass defect.

Useful Conversion

1 u of mass is equivalent to:

E = 1.661 × 10⁻²⁷ × (3.00 × 10⁸)² = 1.661 × 10⁻²⁷ × 9.00 × 10¹⁶ = 1.495 × 10⁻¹⁰ J

Converting to MeV: 1.495 × 10⁻¹⁰ / 1.60 × 10⁻¹³ = 931.5 MeV

So: 1 u = 931.5 MeV/c²

This is an extremely useful conversion factor for nuclear physics calculations.

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Binding Energy

The binding energy (BE) of a nucleus is the energy required to completely separate the nucleus into its individual protons and neutrons. Equivalently, it is the energy released when the nucleus is assembled from its constituent nucleons.

BE = Δm × c²

A nucleus with a larger binding energy is more tightly bound and therefore more stable.

Worked Example 1 — Binding energy of helium-4

Helium-4 (⁴₂He) contains 2 protons and 2 neutrons.

Total mass of separate nucleons: 2 × 1.00728 + 2 × 1.00866 = 2.01456 + 2.01732 = 4.03188 u

Measured mass of He-4 nucleus: 4.00151 u

Mass defect: Δm = 4.03188 − 4.00151 = 0.03037 u

Binding energy: BE = 0.03037 × 931.5 = 28.30 MeV

Binding energy per nucleon: 28.30 / 4 = 7.07 MeV per nucleon

Worked Example 2 — Binding energy of iron-56

Iron-56 (⁵⁶₂₆Fe) contains 26 protons and 30 neutrons.

Total mass of separate nucleons: 26 × 1.00728 + 30 × 1.00866 = 26.18928 + 30.25980 = 56.44908 u

Measured mass of Fe-56 nucleus: 55.92067 u

Mass defect: Δm = 56.44908 − 55.92067 = 0.52841 u

Binding energy: BE = 0.52841 × 931.5 = 492.3 MeV

Binding energy per nucleon: 492.3 / 56 = 8.79 MeV per nucleon

Worked Example 3 — Binding energy of uranium-235

Uranium-235 (²³⁵₉₂U) contains 92 protons and 143 neutrons.

Total mass of separate nucleons: 92 × 1.00728 + 143 × 1.00866 = 92.66976 + 144.23838 = 236.90814 u

Measured mass of U-235 nucleus: 234.99346 u

Mass defect: Δm = 236.90814 − 234.99346 = 1.91468 u

Binding energy: BE = 1.91468 × 931.5 = 1784 MeV

Binding energy per nucleon: 1784 / 235 = 7.59 MeV per nucleon

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The Binding Energy per Nucleon Curve

When the binding energy per nucleon is plotted against nucleon number (A), the resulting curve is one of the most important graphs in nuclear physics.

Key Features of the Curve