Nuclear Reactions and Binding Energy

One of the most profound ideas in physics is that mass and energy are interchangeable. Einstein’s equation E = mc² tells us that a small amount of mass corresponds to an enormous amount of energy. Nowhere is this more apparent than in nuclear physics, where the mass of a nucleus is measurably less than the mass of its individual nucleons — and this "missing mass" is the key to understanding nuclear energy.

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Mass Defect

If you could weigh a nucleus and then separately weigh all its individual protons and neutrons, you would find that the nucleus weighs less. This difference is called the mass defect (Δm):

Δm = (Z × mₚ + N × mₙ) − M_nucleus

where:

• Z = number of protons, mₚ = mass of a proton
• N = number of neutrons, mₙ = mass of a neutron
• M_nucleus = measured mass of the nucleus

The mass defect is always positive for bound nuclei. The "missing" mass has been converted into energy — the energy that was released when the nucleons were brought together to form the nucleus.

Example: Helium-4

• 2 protons: 2 × 1.00728 u = 2.01456 u
• 2 neutrons: 2 × 1.00866 u = 2.01732 u
• Total separate mass: 4.03188 u
• Mass of He-4 nucleus: 4.00151 u
• Mass defect: 4.03188 − 4.00151 = 0.03037 u

This 0.03037 u of "missing" mass has been converted to binding energy.

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Binding Energy

The binding energy of a nucleus is the energy required to completely separate the nucleus into individual protons and neutrons. Equivalently, it is the energy released when individual nucleons come together to form the nucleus.

Using E = mc²:

Binding energy = Δm × c²

The conversion factor is: 1 u = 931.5 MeV/c²

So for helium-4: Binding energy = 0.03037 × 931.5 = 28.3 MeV

This means you would need to supply 28.3 MeV of energy to completely disassemble a helium-4 nucleus into two separate protons and two separate neutrons.

Worked Example: Binding Energy of Iron-56

Iron-56 contains 26 protons and 30 neutrons:

• 26 protons: 26 × 1.00728 = 26.18928 u
• 30 neutrons: 30 × 1.00866 = 30.25980 u
• Total separate mass: 56.44908 u
• Mass of Fe-56 nucleus: 55.92068 u
• Mass defect: 56.44908 − 55.92068 = 0.52840 u
• Total binding energy: 0.52840 × 931.5 = 492.3 MeV
• Binding energy per nucleon: 492.3 / 56 = 8.79 MeV per nucleon

This is near the peak of the binding energy curve, confirming iron-56 as one of the most stable nuclei.

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Binding Energy Per Nucleon

The total binding energy tells us how tightly bound the whole nucleus is, but to compare the stability of different nuclei, we use binding energy per nucleon:

Binding energy per nucleon = Total binding energy / A

For helium-4: 28.3 / 4 = 7.08 MeV per nucleon

A higher binding energy per nucleon means the nucleons are more tightly bound, and the nucleus is more stable. Nuclei with the highest binding energy per nucleon are the most stable and require the most energy per nucleon to disassemble.

Key Binding Energy Per Nucleon Values

Nucleus	A	BE per nucleon (MeV)	Notes
²H (deuterium)	2	1.11	Very weakly bound
⁴He	4	7.08	Exceptionally stable for its size
¹²C	12	7.68	Peak in light nuclei
¹⁶O	16	7.98	Another peak
⁵⁶Fe	56	8.79	Near the overall peak
⁶²Ni	62	8.79	Highest BE per nucleon of all
²³⁵U	235	7.59	Fission fuel
²³⁸U	238	7.57	Most common uranium isotope

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The Binding Energy Per Nucleon Curve

When binding energy per nucleon is plotted against nucleon number A, one of the most important graphs in physics is produced:

Key Features of the Curve

Rising region (A = 1 to ~56):

• Binding energy per nucleon increases steeply for light nuclei
• He-4, C-12, and O-16 are particularly stable (peaks in the curve)
• The increase occurs because adding nucleons means each nucleon has more neighbours to interact with via the strong force

Peak region (A ≈ 56):

• Iron-56 (and nearby nickel-62) has the highest binding energy per nucleon at approximately 8.8 MeV
• This is the most stable nucleus
• It sits at the "valley of stability"

Declining region (A > 56):

• Binding energy per nucleon gradually decreases for heavier nuclei
• The decline occurs because the electromagnetic repulsion between protons (which has infinite range) increasingly competes with the strong nuclear force (which has a short range of ~3 fm)
• Uranium-238 has about 7.6 MeV per nucleon

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Why Nuclear Reactions Release Energy

The binding energy per nucleon curve explains why both fission and fusion can release energy:

Fusion of Light Nuclei

When light nuclei (left side of the curve) fuse to form a heavier nucleus, the product has a higher binding energy per nucleon. The nucleons become more tightly bound, meaning mass is lost and energy is released.

For example, when two deuterium nuclei fuse:

• Reactants: lower binding energy per nucleon
• Product: higher binding energy per nucleon
• The increase in binding energy per nucleon × total nucleons = energy released

Fission of Heavy Nuclei

When a heavy nucleus (right side of the curve) splits into two medium-mass fragments, the products have a higher binding energy per nucleon. Again, mass is lost and energy is released.

For example, when uranium-235 undergoes fission:

• Reactant: ~7.6 MeV per nucleon
• Products (e.g., barium-141, krypton-92): ~8.5 MeV per nucleon
• Energy released ≈ (8.5 − 7.6) × 235 ≈ 200 MeV per fission

The Central Principle

Energy is released in any nuclear reaction where the products have a higher binding energy per nucleon than the reactants. The "missing" mass (the increase in total binding energy) appears as kinetic energy of the products.

This is why fusion works for light nuclei (moving right up the curve) and fission works for heavy nuclei (moving left up the curve). Both move towards iron-56, the peak of stability. You cannot gain energy by fusing nuclei heavier than iron, and you cannot gain energy by splitting nuclei lighter than iron.

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Energy Calculations

To calculate the energy released in a nuclear reaction:

1. Find the total mass of the reactants
2. Find the total mass of the products
3. Calculate the mass difference: Δm = mass of reactants − mass of products
4. Convert to energy: E = Δm × 931.5 MeV (if Δm is in atomic mass units)

If Δm is positive, energy is released (exothermic). If Δm is negative, energy must be supplied (endothermic).

Worked Example: Energy from Tritium Production

Calculate the energy released when lithium-6 absorbs a neutron to produce tritium and an alpha particle:

⁶₃Li + ¹₀n → ³₁H + ⁴₂He

Masses: Li-6 = 6.01512 u, n = 1.00866 u, H-3 = 3.01605 u, He-4 = 4.00260 u

Mass of reactants = 6.01512 + 1.00866 = 7.02378 u Mass of products = 3.01605 + 4.00260 = 7.01865 u Δm = 7.02378 − 7.01865 = 0.00513 u

Energy released = 0.00513 × 931.5 = 4.78 MeV

This reaction is important because it is used to breed tritium fuel for D-T fusion reactors.

Worked Example: Energy Released Per Nucleon in Fission vs Fusion

Fission: U-235 → products with Δ(BE per nucleon) ≈ 0.9 MeV Energy per nucleon ≈ 0.9 MeV

Fusion: D + T → He-4 + n, releasing 17.6 MeV for 5 nucleons Energy per nucleon ≈ 17.6/5 = 3.5 MeV

Fusion releases approximately 4 times more energy per nucleon than fission. This is why fusion fuel has a much higher energy density than fission fuel.

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Common Exam Mistakes

• Confusing total binding energy with binding energy per nucleon. The question may ask for either — read carefully.
• Using the wrong sign convention. Mass defect is positive (separate nucleons have more mass). Energy is released when products are more tightly bound.
• Forgetting to account for all particles. In energy calculations, include every particle on both sides of the equation (don’t forget the neutrons).
• Thinking binding energy is energy stored in the nucleus. Binding energy is the energy you must supply to break the nucleus apart — it is not available to be released. Energy is released when nuclei form from separate nucleons (or when they rearrange into more tightly bound configurations).

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Summary

The mass defect is the difference between the total mass of individual nucleons and the mass of the assembled nucleus. This missing mass, converted via E = mc², gives the binding energy. Binding energy per nucleon peaks at iron-56 (~8.8 MeV per nucleon). Fusion of light nuclei and fission of heavy nuclei both release energy because the products have higher binding energy per nucleon than the reactants — both processes move nuclei towards the peak of the curve. Fusion releases roughly four times more energy per nucleon than fission.

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A-Level Deep Dive: Nuclear Reactions and Binding Energy

Spec mapping

Edexcel 9PH0 specification Topic 8 — Nuclear and Particle Physics covers nuclear binding energy, mass defect, the unified atomic mass unit, the binding-energy-per-nucleon curve and its implications for fission and fusion (refer to the official specification document for exact wording). The mass-energy relationship $E = \Delta m c^2$E=Δmc2 underpins every quantitative nuclear question on Paper 2. This material is examined in Paper 2 (Section 8 — Nuclear and Particle Physics) but is also synoptic with Topic 12 (Space) through stellar nucleosynthesis and with Topic 11 (Quantum Phenomena) through the energy quantisation that governs decay products. The Edexcel Physics formula booklet provides $E = c^2 \Delta m$E=c2Δm and $1\,\text{u} = 1.66 \times 10^{-27}\,\text{kg}$1u=1.66×10−27kg, but the conversion $1\,\text{u} \approx 931.5\,\text{MeV}/c^2$1u≈931.5MeV/c2 must be derived or recalled.

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Worked example with full mark scheme

Question (8 marks):

The mass of a helium-4 nucleus is $4.001506\,\text{u}$4.001506u. The mass of a proton is $1.007276\,\text{u}$1.007276u and the mass of a neutron is $1.008665\,\text{u}$1.008665u.

(a) Calculate the mass defect $\Delta m$Δm for helium-4 in kilograms. (3)

(b) Calculate the binding energy of the helium-4 nucleus in joules and in MeV. (3)

(c) Calculate the binding energy per nucleon and explain what this value tells you about the position of helium-4 on the binding-energy curve relative to iron-56. (2)

Solution with mark scheme:

(a) Step 1 — sum the constituent masses.

Helium-4 contains 2 protons and 2 neutrons:

$\Sigma m = 2(1.007276) + 2(1.008665) = 2.014552 + 2.017330 = 4.031882\,\text{u}$Σm=2(1.007276)+2(1.008665)=2.014552+2.017330=4.031882u

M1 — correct identification of 2 protons + 2 neutrons and addition of constituent masses.

Step 2 — compute the mass defect.

$\Delta m = \Sigma m - m_{\text{nucleus}} = 4.031882 - 4.001506 = 0.030376\,\text{u}$Δm=Σm−mnucleus​=4.031882−4.001506=0.030376u

M1 — subtraction in correct sense (constituents minus nucleus); a negative value here indicates a sign error.

Step 3 — convert to kilograms.

$\Delta m = 0.030376 \times 1.66 \times 10^{-27} = 5.04 \times 10^{-29}\,\text{kg}$Δm=0.030376×1.66×10−27=5.04×10−29kg

A1 — correct value to 3 significant figures with units.

(b) Step 1 — apply $E = \Delta m c^2$E=Δmc2.

$E = (5.04 \times 10^{-29})(3.00 \times 10^{8})^2 = (5.04 \times 10^{-29})(9.00 \times 10^{16})$E=(5.04×10−29)(3.00×108)2=(5.04×10−29)(9.00×1016) $E = 4.54 \times 10^{-12}\,\text{J}$E=4.54×10−12J

M1 — correct substitution into $E = \Delta m c^2$E=Δmc2.

A1 — $E \approx 4.5 \times 10^{-12}\,\text{J}$E≈4.5×10−12J to 2 sf.

Step 2 — convert to MeV.

$E = \frac{4.54 \times 10^{-12}}{1.60 \times 10^{-13}} \approx 28.4\,\text{MeV}$E=1.60×10−134.54×10−12​≈28.4MeV

A1 — $\approx 28\,\text{MeV}$≈28MeV accepted (also obtainable directly via $0.030376 \times 931.5 \approx 28.3\,\text{MeV}$0.030376×931.5≈28.3MeV).

(c) Binding energy per nucleon:

$\frac{E}{A} = \frac{28.4}{4} \approx 7.1\,\text{MeV per nucleon}$AE​=428.4​≈7.1MeV per nucleon

B1 — correct numerical value.

B1 — comment that helium-4 sits well below the peak of the binding-energy curve at iron-56 ($\approx 8.8\,\text{MeV/nucleon}$≈8.8MeV/nucleon), but is locally elevated relative to its lithium and beryllium neighbours, which is why $\alpha$α-particles are unusually stable products of decay and fusion.

Total: 8 marks (M3 A2 B2, plus structured M1 in (a) for setup).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A deuterium nucleus ($^2_1\text{H}$12​H, mass $2.013553\,\text{u}$2.013553u) and a tritium nucleus ($^3_1\text{H}$13​H, mass $3.015500\,\text{u}$3.015500u) fuse to produce a helium-4 nucleus (mass $4.001506\,\text{u}$4.001506u) and a free neutron (mass $1.008665\,\text{u}$1.008665u).

(a) Write the balanced nuclear equation for this reaction. (2)

(b) Calculate the energy released in MeV. (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1) — recognise that nucleon number and proton number must both balance: $2 + 3 = 4 + 1$2+3=4+1 and $1 + 1 = 2 + 0$1+1=2+0.
• A1 (AO1.2) — equation written as $^2_1\text{H} + ^3_1\text{H} \rightarrow ^4_2\text{He} + ^1_0\text{n}$12​H+13​H→24​He+01​n.