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

What this dot point is asking

WACE wants you to apply $E=mc^2$ 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:

Because $c^2\approx9\times10^{16}\ \text{m}^2\text{s}^{-2}$ is huge, even a tiny mass corresponds to a vast energy. One gram of matter, fully converted, would release about $9\times10^{13}\ \text{J}$. 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 $\Delta m$, has been converted to energy:

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 $1\ \text{u}=1.661\times10^{-27}\ \text{kg}$. Converting a mass of $1\ \text{u}$ entirely to energy gives

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

Relativistic energy

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

Rest energy and where the mass goes

A point worth stating clearly in extended answers is that mass is not simply "destroyed" in a vague sense; the rest energy locked up in the missing mass is transferred to the kinetic energy of the products and to emitted photons. The total energy and total mass-energy of the system are conserved. In fission, the fragments fly apart with large kinetic energy and the system as a whole is lighter than before by the mass defect. In fusion, the combined nucleus plus any ejected particles are lighter than the original separate nuclei. Framing the energy release as a transfer of rest energy, rather than a violation of conservation, is what distinguishes a high-band answer.

The atomic-mass-unit shortcut in practice

Because nuclear masses are tabulated in unified atomic mass units, the fastest route to energy is almost always to work in those units and use the conversion $1\ \text{u}\to931.5\ \text{MeV}$. Find the total mass of the reactants and the total mass of the products in atomic mass units, take the difference (the mass defect), and multiply by $931.5$ to get the energy released in MeV directly, with no powers of ten to juggle. Only convert to joules at the very end if the question demands SI units, using $1\ \text{MeV}=1.6\times10^{-13}\ \text{J}$. Keeping enough significant figures in the masses is essential, because the mass defect is a small difference between larger numbers, and rounding the input masses too early can change the final energy substantially.

Choosing units

For nuclear questions, working in atomic mass units and multiplying by $931.5\ \text{MeV per u}$ 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.