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Where does nuclear energy come from?

Mass-energy equivalence E=mc2E = mc^2, mass defect and binding energy, the binding energy per nucleon curve, and energy release in nuclear fission and fusion.

A focused answer to the Edexcel 9PH0 nuclear energy content, covering mass-energy equivalence, mass defect and binding energy, the binding energy per nucleon curve, and the energy released in nuclear fission and fusion.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

Edexcel wants you to use mass-energy equivalence E=mc2E = mc^2, define and calculate mass defect and binding energy, interpret the binding energy per nucleon curve, and explain how fission and fusion release energy by moving towards the peak of that curve.

The answer

Mass-energy equivalence

Because c2c^2 is enormous, a tiny mass change releases a huge energy. This is why nuclear processes release millions of times more energy per kilogram than chemical reactions, which involve only electron rearrangement and negligible mass change.

Mass defect and binding energy

A nucleus has less mass than its parts because energy is released (and mass lost) when the strong nuclear force binds the nucleons together. To separate them again you must supply that binding energy back.

The binding energy per nucleon curve

This single curve explains where nuclear energy comes from. Any process that moves nucleons to a position higher on the curve (greater binding energy per nucleon) releases energy.

Fission and fusion

In fission, a heavy nucleus (such as uranium-235, struck by a neutron) splits into two medium-mass fragments plus a few neutrons. The fragments lie nearer the peak, so their binding energy per nucleon is higher; the released neutrons can trigger further fissions in a chain reaction. In fusion, light nuclei (such as hydrogen isotopes) combine to form a heavier nucleus nearer the peak. Fusion releases more energy per nucleon because the curve climbs most steeply on the light side, but it requires extreme temperatures and pressures to overcome the electrostatic repulsion between the positive nuclei, which is why it powers stars and is so hard to harness on Earth.

Examples in context

Nuclear power stations harness controlled fission of uranium-235 or plutonium-239, using control rods to absorb neutrons and keep the chain reaction steady. Stars shine by fusing hydrogen into helium in their cores, the source of sunlight. Experimental reactors such as tokamaks confine a hot plasma with magnetic fields to pursue controlled fusion. The PET scanner and radiotherapy both exploit mass-energy conversion when matter and antimatter annihilate or when nuclei decay.

Try this

Q1. Define the binding energy of a nucleus. [1 mark]

  • Cue. The energy needed to separate the nucleus completely into its individual nucleons, equal to the mass defect times c2c^2.

Q2. A reaction loses 1.5×10−291.5 \times 10^{-29} kg of mass. Find the energy released. [2 marks]

  • Cue. E=mc2=1.5×10−29×(3.0×108)2=1.4×10−12E = mc^2 = 1.5 \times 10^{-29} \times (3.0 \times 10^{8})^2 = 1.4 \times 10^{-12} J.

Q3. Explain why fusion releases more energy per nucleon than fission. [2 marks]

  • Cue. The binding energy per nucleon curve rises more steeply on the light-nucleus side, so moving towards the peak by fusion gives a larger increase per nucleon than fission does.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksCalculate the energy released when a mass defect of 0.200.20 u is converted to energy in a nuclear reaction. Take 11 u =1.66×10−27= 1.66 \times 10^{-27} kg and c=3.0×108c = 3.0 \times 10^{8} m per second.
Show worked answer →

Convert the mass defect to kilograms: Δm=0.20×1.66×10−27=3.32×10−28\Delta m = 0.20 \times 1.66 \times 10^{-27} = 3.32 \times 10^{-28} kg.

Energy released: E=Δmc2=3.32×10−28×(3.0×108)2=3.32×10−28×9.0×1016=3.0×10−11E = \Delta m c^2 = 3.32 \times 10^{-28} \times (3.0 \times 10^{8})^2 = 3.32 \times 10^{-28} \times 9.0 \times 10^{16} = 3.0 \times 10^{-11} J.

Markers reward conversion of u to kg, use of E=mc2E = mc^2, and the value about 3.0×10−113.0 \times 10^{-11} J (roughly 187187 MeV).

Edexcel 20225 marksExplain why both fission of heavy nuclei and fusion of light nuclei release energy, with reference to the binding energy per nucleon curve.
Show worked answer →

The binding energy per nucleon curve rises steeply for light nuclei, peaks near iron-56 (the most stable, highest binding energy per nucleon), then falls slowly for heavy nuclei.

Fission: a heavy nucleus splits into two medium-mass nuclei that lie closer to the peak, so they have a higher binding energy per nucleon; the increase in total binding energy is released as kinetic energy of the fragments.

Fusion: light nuclei combine to form a nucleus nearer the peak, again increasing the binding energy per nucleon; the difference is released. Fusion releases more energy per nucleon because the curve rises most steeply on the light side.

Markers reward the shape of the curve (peak at iron), moving towards the peak in both processes, and energy release from increased binding energy per nucleon.

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