OCR A Level Physics A: Mass-energy, binding energy, fission and fusion (H556)

This guide covers section 6.4.4 (Nuclear fission and fusion), outcomes (a) to (k), of the OCR A Level Physics A (H556) specification, Version 3.0 (March 2026). It is full A level content. The nuclear atom, nuclear size and fundamental particles (6.4.1 and 6.4.2) and radioactive decay, activity and half-life (6.4.3) are separate units and are not repeated here.

Use it with the revision notes and the practice questions. See also the course hub, the printable checklist and the free diagnostics.

What this unit covers

Outcome	What you must be able to do
6.4.4(a)	Use Einstein’s mass-energy equation ΔE = Δmc²
6.4.4(b)	Find the energy released or absorbed in simple nuclear reactions
6.4.4(c)	Describe creation and annihilation of particle-antiparticle pairs
6.4.4(d)	Define mass defect, binding energy and binding energy per nucleon
6.4.4(e)	Use the curve of binding energy per nucleon against nucleon number to explain energy changes
6.4.4(f)	Calculate binding energy from the masses of nuclei
6.4.4(g)	Describe induced fission and a chain reaction
6.4.4(h)	Describe a fission reactor: fuel rods, control rods, moderator
6.4.4(i)	Discuss the environmental impact of nuclear waste
6.4.4(j)	Describe fusion and why it needs very high temperatures
6.4.4(k)	Balance nuclear transformation equations

6.4.4(a) Mass-energy: ΔE = Δmc²

Mass and energy are linked. A change in the energy of a system is accompanied by a change in its mass:

ΔE = Δmc²

ΔE is in joules, Δm in kilograms and c = 3.00 × 10⁸ m s⁻¹. Because c² is huge, a tiny mass change is a large energy change.

Worked example. Find the energy equivalent of 1 u, then in MeV. Data supplied: 1 u = 1.661 × 10⁻²⁷ kg, 1 eV = 1.60 × 10⁻¹⁹ J.

E = 1.661e-27 × (3.00e8)² = 1.49 × 10⁻¹⁰ J
E in MeV = 1.49e-10 / 1.60e-13 = 934 MeV

With more precise constants the figure is 931.5 MeV per u. Use the constants you are given and do not mix routes.

6.4.4(b) Energy released or absorbed in reactions

Compare total mass before and after. If the products have less mass, the difference is released as energy. If they have more, energy must be supplied.

energy released = (total mass before − total mass after) × c²

Worked example (energy absorbed). Nitrogen-14 is hit by an alpha particle: ¹⁴₇N + ⁴₂He → ¹⁷₈O + ¹₁H. Atomic masses supplied: N-14 14.003074 u, He-4 4.002603 u, O-17 16.999132 u, H-1 1.007825 u.

before = 14.003074 + 4.002603 = 18.005677 u
after  = 16.999132 + 1.007825 = 18.006957 u
Δm = 0.001280 u more after the reaction

The products are heavier, so energy is absorbed: 0.001280 u × 931.5 MeV per u ≈ 1.2 MeV must come from the kinetic energy of the alpha particle. Atomic masses work here because the electron count is the same on both sides.

6.4.4(c) Pair creation and annihilation

Every particle has an antiparticle of equal mass. When a photon of enough energy passes close to a nucleus, its energy can become a particle and its antiparticle: pair creation. The minimum photon energy is the combined rest energy, 2mc².

When a particle meets its antiparticle both disappear and their mass becomes photons: annihilation. Two photons are usually produced, travelling in opposite directions, so that total momentum stays zero.

Worked example. (i) Minimum photon energy to make an electron-positron pair, and its longest wavelength. (ii) Photon energy when a proton and antiproton annihilate at rest. Data: m(electron) = 9.11 × 10⁻³¹ kg, m(proton) = 1.673 × 10⁻²⁷ kg, h = 6.63 × 10⁻³⁴ J s.

(i)  E = 2mc² = 2 × 9.11e-31 × (3.00e8)² = 1.64 × 10⁻¹³ J (about 1.02 MeV)
     λ = hc/E = 6.63e-34 × 3.00e8 / 1.64e-13 = 1.21 × 10⁻¹² m
(ii) total energy = 2 × 1.673e-27 × (3.00e8)² = 3.01 × 10⁻¹⁰ J (about 1880 MeV)
     two photons, so each carries 1.51 × 10⁻¹⁰ J

6.4.4(d) Mass defect and binding energy

A nucleus has less mass than its separated nucleons. The mass defect is the total mass of the free protons and neutrons less the mass of the nucleus. The energy equivalent of the mass defect is the binding energy: the energy required to pull the nucleus apart into free nucleons (equivalently, the energy released when they come together).

Binding energy per nucleon is the binding energy divided by the nucleon number A. A higher value means a more stable nucleus.

6.4.4(f) Binding energy from nuclear masses

Method: (1) add the masses of Z protons and (A − Z) neutrons; (2) subtract the nuclear mass; (3) convert with ΔE = Δmc²; (4) divide by A.

Worked example. Oxygen-16 has Z = 8. Data supplied: nuclear mass of O-16 = 15.990523 u, proton 1.007276 u, neutron 1.008665 u, 1 u = 1.661 × 10⁻²⁷ kg.

nucleons = 8 × 1.007276 + 8 × 1.008665 = 16.127528 u
mass defect = 16.127528 − 15.990523 = 0.137005 u = 2.276 × 10⁻²⁸ kg
binding energy = 2.276e-28 × (3.00e8)² = 2.05 × 10⁻¹¹ J = 128 MeV
per nucleon = 128 / 16 = 8.00 MeV

With atomic masses, use hydrogen atom masses in place of protons: 8 × 1.007825 + 8 × 1.008665 − 15.994915 also gives 0.137005 u, because the electron masses cancel.

6.4.4(e) The binding energy per nucleon curve

The curve plots binding energy per nucleon (vertical axis) against nucleon number (horizontal axis).

Region	Shape	Meaning
Very light nuclei	Rises steeply, with a local peak at helium-4	Joining light nuclei gains a lot
Around A = 56	Maximum, about 8.8 MeV per nucleon (iron region)	Most tightly bound
Heavy nuclei	Falls slowly, to about 7.6 MeV at uranium	Splitting a heavy nucleus gains

Energy is released when the products have a higher binding energy per nucleon than the starting nuclei: fusion of light nuclei (left of the peak) and fission of heavy nuclei (right of the peak) both qualify. Energy released = total binding energy of products − total binding energy of reactants. Free neutrons have no binding energy.

Worked example (fission). U-235 absorbs a neutron and splits into Xe-140, Sr-94 and 2 neutrons. Binding energy per nucleon supplied: U-235 7.59 MeV, Xe-140 8.29 MeV, Sr-94 8.59 MeV.

products: 140 × 8.29 + 94 × 8.59 = 1160.6 + 807.5 = 1968.1 MeV
reactant:  235 × 7.59 = 1783.7 MeV
energy released = 184 MeV

Worked example (fusion). ²₁H + ³₁H → ⁴₂He + ¹₀n. Total binding energies: H-2 2.22 MeV, H-3 8.48 MeV, He-4 28.3 MeV. Released = 28.3 − (2.22 + 8.48) = 17.6 MeV.

6.4.4(g) Induced fission and the chain reaction

In a reactor fission is induced. A slow (thermal) neutron is absorbed by a uranium-235 nucleus. The nucleus becomes unstable, splits into two smaller nuclei (fission fragments) and releases typically two or three neutrons. Most of the energy is the kinetic energy of the fragments, which becomes thermal energy in collisions.

In a chain reaction, neutrons released by one fission cause further fissions. If, on average, one neutron per fission goes on to cause another, the reaction is steady and controlled. If it is more than one, the rate grows rapidly; if less, the reaction dies away.

6.4.4(h) The fission reactor

Component	Job	Notes
Fuel rods	Hold the fissile material, usually uranium enriched in U-235	Heat is produced here
Moderator	Slows fast neutrons to thermal speeds by collisions	Water or graphite: light nuclei that absorb few neutrons
Control rods	Absorb neutrons to hold the chain reaction steady	Boron or cadmium; lowered in to slow the reaction, raised to speed it up, fully inserted to shut down

A coolant carries thermal energy from the core to a heat exchanger, which produces steam for the turbines, and shielding surrounds the core. Do not mix up the roles: the moderator slows neutrons, the control rods absorb them.

Worked example. A reactor has thermal power 3.0 × 10⁹ W. Each fission releases 200 MeV. Find the fissions per second and the mass of U-235 fissioned in one day. Data: 1 MeV = 1.60 × 10⁻¹³ J; mass of one U-235 nucleus = 235 u; 1 u = 1.661 × 10⁻²⁷ kg.

energy per fission = 200 × 1.60e-13 = 3.2 × 10⁻¹¹ J
fissions per second = 3.0e9 / 3.2e-11 = 9.4 × 10¹⁹
mass per nucleus = 235 × 1.661e-27 = 3.90 × 10⁻²⁵ kg
mass per day = 9.375e19 × 3.90e-25 × 86 400 = 3.2 kg

6.4.4(i) Environmental impact of nuclear waste

Spent fuel contains radioactive, hot fission fragments and heavier nuclei formed by neutron capture.

• Problems: some nuclides have half-lives of thousands of years (plutonium-239 is about 24 000 years), so waste must be kept safe for very long periods; early decay products are strongly ionising and generate heat; transport and storage carry leak and security risks.
• Management: spent fuel is first stored in cooling ponds; high-level waste can be sealed in glass, stored in shielded containers and placed deep underground in stable geology. Lower-level waste is sorted by activity and stored accordingly.
• Balance: fission produces no carbon dioxide in operation and the waste is small in volume. Decisions on new stations weigh this against waste, accident risk, cost and public opinion.

Strong answers give a reason with each statement and set an advantage against a disadvantage.

6.4.4(j) Fusion and temperature

In fusion, light nuclei join to form a heavier nucleus and release energy because the product has a higher binding energy per nucleon. Both nuclei are positive, so they repel. The nuclei must approach to within about 10⁻¹⁵ m, where the strong nuclear force takes over. That needs very high kinetic energy, so very high temperature, and the fuel is a plasma. The mean kinetic energy of a particle is (3/2)kT, with k = 1.38 × 10⁻²³ J K⁻¹ (from 5.1.4).

Worked example. Two protons approach head-on to a separation of 2.0 × 10⁻¹⁵ m. Estimate the temperature at which they have enough kinetic energy. Data: e = 1.60 × 10⁻¹⁹ C, 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻².

electric potential energy = 8.99e9 × (1.60e-19)² / 2.0e-15 = 1.15 × 10⁻¹³ J
each proton supplies half: 5.75 × 10⁻¹⁴ J = (3/2)kT
T = 2 × 5.75e-14 / (3 × 1.38e-23) = 2.8 × 10⁹ K

This classical estimate is far above the temperature at the centre of the Sun, because not every particle needs the average energy and nuclei can sometimes get through the barrier. Fusion still needs millions of kelvin, and holding a plasma that hot is the main engineering difficulty.

6.4.4(k) Balancing nuclear equations

Both totals must match on each side: nucleon number A (top) and proton number Z (bottom, which is charge).

Worked example. ²³⁵₉₂U + ¹₀n → ¹⁴⁰₅₅Cs + ⁹⁴₃₇Rb + x ¹₀n.

A: 235 + 1 = 236 = 140 + 94 + x, so x = 2
Z: 92 + 0 = 92 = 55 + 37 + 0, which balances

For decays use the same rule: in beta-minus decay a neutron becomes a proton, so Z rises by 1 and A is unchanged, with an electron and an antineutrino emitted.

Common errors

• Using grams or u in ΔE = Δmc² without converting to kilograms.
• Dividing by Z, not A, to get binding energy per nucleon.
• Reversing the sum: binding energy released is products minus reactants, but mass lost is reactants minus products.
• Saying the control rods slow neutrons down.

Official syllabus

OCR Level 3 Advanced GCE in Physics A (H556), Version 3.0 (March 2026), first assessment 2017, published by OCR. Section 6.4.4, outcomes (a) to (k).