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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).
Spec mapping (AQA 7408): This lesson develops the concept of mass defect Δm, Einstein's mass-energy equivalence E = Δmc², the binding energy per nucleon as the appropriate measure of nuclear stability, the shape of the BE-per-nucleon curve with its peak near Fe-56 (≈ 8.8 MeV/nucleon), and the consequence that energy is released in any reaction whose products lie higher on the curve than the reactants — explaining why both fusion (light nuclei) and fission (heavy nuclei) liberate energy. It maps to AQA 7408 section 3.8.1.4 (mass and energy) and section 3.8.1.5 (fission and fusion). (Refer to the official AQA specification document for exact wording.)
Synoptic links: (1) E = mc² is the relativistic mass-energy relation from special relativity (section 3.12, Turning Points option), and the 1 u → 931.5 MeV/c² conversion is the most-used corollary in nuclear physics. (2) The binding-energy curve is the engine of stellar nucleosynthesis (the fusion lessons in this course) — stars fuse up to Fe-56 and stop, because beyond iron fusion is endothermic. (3) Mass defects matter in particle physics too (section 3.2.1.5): the masses of hadrons exceed the sum of their constituent quark masses because of the binding energy of the strong interaction, and a similar mass-defect argument applies.
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:
The "missing" mass has been converted into energy — the binding energy — according to Einstein's equation.
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.
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.
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.
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
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
Deuterium (²₁H) contains 1 proton and 1 neutron — the simplest bound nuclear system.
Total mass of separate nucleons: 1.00728 + 1.00866 = 2.01594 u.
Measured mass of deuterium nucleus: 2.01355 u.
Mass defect: Δm = 2.01594 − 2.01355 = 0.00239 u.
Binding energy: BE = 0.00239 × 931.5 = 2.22 MeV.
Binding energy per nucleon: 2.22 / 2 = 1.11 MeV per nucleon.
Compare with He-4 (BE/A = 7.07 MeV/nucleon). Adding two more nucleons to deuterium to form He-4 increases the BE/A by a factor of more than six — a dramatic stability gain that reflects the closed-shell structure of He-4 (a doubly magic α-particle). This is why α emission is energetically favoured over individual nucleon emission for heavy unstable nuclei: ejecting a tightly bound He-4 unit releases ~28 MeV at minimal cost.
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
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.
| Region of curve | Nucleon number (A) | BE per nucleon (MeV) | Behaviour |
|---|---|---|---|
| Very light nuclei | 1–4 | 0 to ~7 | Rises steeply |
| Light to medium nuclei | 4–56 | ~7 to ~8.8 | Rises gradually |
| Iron-56 (peak) | 56 | ~8.79 | Maximum — most stable |
| Medium to heavy nuclei | 56–238 | ~8.8 to ~7.6 | Decreases gradually |
Iron-56 (and nickel-62, which is very close) has the highest binding energy per nucleon, making it the most stable nucleus. This peak is the reason why:
Key Point: Energy is released in any nuclear reaction that moves the products closer to the peak of the binding energy per nucleon curve (i.e., towards iron-56). The products are more tightly bound than the reactants.
The energy released in a nuclear reaction equals the increase in total binding energy:
Energy released = Total BE of products − Total BE of reactants
Alternatively, using masses:
Energy released = (Total mass of reactants − Total mass of products) × c²
Both methods give the same answer and both are acceptable in exams.
Consider the fission of uranium-235:
²³⁵₉₂U + ¹₀n → ⁹²₃₆Kr + ¹⁴¹₅₆Ba + 3¹₀n
Given: BE per nucleon of U-235 = 7.59 MeV, Kr-92 = 8.71 MeV, Ba-141 = 8.29 MeV.
Total BE of reactants = 235 × 7.59 = 1783.7 MeV (neutrons have zero binding energy)
Total BE of products = 92 × 8.71 + 141 × 8.29 = 801.3 + 1168.9 = 1970.2 MeV
Energy released = 1970.2 − 1783.7 = 186.5 MeV
(The actual energy released per fission event is approximately 200 MeV when the kinetic energy of all products is included.)
Consider the fusion of deuterium and tritium:
²₁H + ³₁H → ⁴₂He + ¹₀n
Given: BE per nucleon of H-2 = 1.11 MeV, H-3 = 2.83 MeV, He-4 = 7.07 MeV.
Total BE of reactants = 2 × 1.11 + 3 × 2.83 = 2.22 + 8.49 = 10.71 MeV
Total BE of products = 4 × 7.07 = 28.28 MeV
Energy released = 28.28 − 10.71 = 17.57 MeV
Although this is less total energy per reaction than fission, fusion releases much more energy per unit mass of fuel because the fuel nuclei are so light.
Exam Tip: When calculating energy released, always find the total binding energy (not per nucleon) for each species by multiplying BE per nucleon by A. Students frequently lose marks by comparing BE per nucleon directly without multiplying by A. Remember: free neutrons have zero binding energy and are not included in BE totals.
Iron-56 sits at the peak of the binding energy per nucleon curve. This means:
Common Misconception: Students sometimes say "iron has the most binding energy." This is wrong — uranium-235 has a much larger total binding energy (1784 MeV vs 492 MeV). The key quantity is binding energy per nucleon. Iron-56 has the highest binding energy per nucleon, making each nucleon in iron the most tightly bound.
Specimen question modelled on the AQA 7408 paper format (9 marks):
The induced fission of uranium-235 can proceed through many channels. One representative reaction is
²³⁵₉₂U + ¹₀n → ⁹²₃₆Kr + ¹⁴¹₅₆Ba + 3 ¹₀n
Nuclear masses (in atomic mass units): m(U-235) = 234.99346 u; m(Kr-92) = 91.92616 u; m(Ba-141) = 140.91440 u; m(neutron) = 1.00866 u; 1 u = 931.5 MeV/c².
(a) Define the binding energy per nucleon of a nucleus and state, with reasoning, which of the four nuclei in the reaction has the highest binding energy per nucleon. (3 marks)
(b) Calculate the energy released per fission event in MeV, using the mass-defect method. (4 marks)
(c) Explain why the peak of the binding-energy-per-nucleon curve at Fe-56 places a fundamental limit on the energy that can be obtained from any nuclear reaction. (2 marks)
| Part | Marks | AO1 | AO2 | AO3 |
|---|---|---|---|---|
| (a) | 3 | 2 (define BE per nucleon; recall iron peak) | — | 1 (justify which of the four has highest BE/A) |
| (b) | 4 | 1 (recall E = Δmc²) | 2 (compute Δm with correct counts) | 1 (convert correctly to MeV) |
| (c) | 2 | 1 (recall peak at Fe-56) | — | 1 (link to reactions producing iron) |
For 9-mark mass-defect calculations, examiners reward candidates who show every step: total mass of reactants, total mass of products, difference, multiplied by 931.5 MeV/c² per u. Sign errors and forgetting that there are 3 final neutrons (not 1) are the most common mark losses.
(a) The binding energy per nucleon is the total binding energy of the nucleus divided by the nucleon number A. Higher BE/A means the nucleus is more stable. The peak is at Fe-56. Of the four nuclei in the reaction, Kr-92 has the highest BE/A because it is closest to iron.
(b) Mass of reactants: 234.99346 + 1.00866 = 236.00212 u. Mass of products: 91.92616 + 140.91440 + 3 × 1.00866 = 235.86654 u. Δm = 236.00212 − 235.86654 = 0.13558 u. E = 0.13558 × 931.5 = 126.3 MeV.
(c) Fe-56 is at the peak, so no reaction starting from Fe-56 can release energy because the products would have lower BE/A than the reactants.
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