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How does Einstein's mass-energy equivalence explain energy released in nuclear reactions?

Apply mass-energy equivalence to calculate energy changes in nuclear reactions

A focused answer to the WACE Year 12 Physics Unit 4 content point on mass-energy equivalence. The meaning of E=mc^2, the unified atomic mass unit, converting a mass defect into energy, and the link to relativistic energy at high speed.

Generated by Claude Opus 4.77 min answer

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What this dot point is asking

WACE wants you to apply E=mc2E=mc^2 to nuclear reactions, calculate the energy from a mass change, and use the unified atomic mass unit. This is the quantitative heart of nuclear energy.

The equivalence

Einstein's relation states that mass and energy are two forms of the same thing:

E=mc2.E=mc^2.

Because c29×1016 m2s2c^2\approx9\times10^{16}\ \text{m}^2\text{s}^{-2} is huge, even a tiny mass corresponds to a vast energy. One gram of matter, fully converted, would release about 9×1013 J9\times10^{13}\ \text{J}. This is why nuclear processes, which convert a small fraction of mass, release millions of times more energy per reaction than chemical processes.

Mass defect and energy release

In a nuclear reaction the total mass of the products is slightly less than the total mass of the reactants. This missing mass, the mass defect Δm\Delta m, has been converted to energy:

ΔE=Δmc2.\Delta E=\Delta m\,c^2.

Whether the process is fission, fusion or radioactive decay, the released energy comes from this mass difference. The released energy appears as kinetic energy of the products and as photons.

Atomic mass units

Nuclear masses are conveniently measured in unified atomic mass units, where 1 u=1.661×1027 kg1\ \text{u}=1.661\times10^{-27}\ \text{kg}. Converting a mass of 1 u1\ \text{u} entirely to energy gives

E=(1.661×1027)(3.0×108)21.49×1010 J931.5 MeV.E=(1.661\times10^{-27})(3.0\times10^8)^2\approx1.49\times10^{-10}\ \text{J}\approx931.5\ \text{MeV}.

So a quick route is to find the mass defect in atomic mass units and multiply by 931.5 MeV per u931.5\ \text{MeV per u}.

Relativistic energy

At high speed the total energy of a particle is E=γmc2E=\gamma mc^2, which separates into the rest energy mc2mc^2 (present even at rest) and the kinetic energy. At low speeds this reduces to the familiar 12mv2\tfrac{1}{2}mv^2. The rest energy term is what becomes available when mass is destroyed in a reaction.

Choosing units

For nuclear questions, working in atomic mass units and multiplying by 931.5 MeV per u931.5\ \text{MeV per u} is fastest. If asked for joules, convert at the end. Always take the mass change as products minus reactants, and use its magnitude for the energy released.