Mass–Energy and Binding Energy

Einstein’s relation $\Delta E=c^2\Delta m$ says that any change in energy is accompanied by a change in mass. Because $c^2=9.0\times 10^{16}{\mathrm{m}}^2{\mathrm{s}}^{-2}$, a tiny mass corresponds to a vast energy.

Nuclear masses are measured in $\text{atomic mass units}$: $1\mathrm{u}=1.661\times 10^{-27}$ kg, one-twelfth the mass of a carbon-12 atom.

It is convenient to know that $1\mathrm{u}$ is equivalent to $931.5$ MeV. So a mass difference in u, multiplied by 931.5, gives energy in MeV directly.

The $\text{mass defect}$ of a nucleus is the difference between the total mass of its separate protons and neutrons and the mass of the nucleus itself.

The $\text{binding energy}$ is the energy needed to separate a nucleus into its individual nucleons. Equivalently, it is the energy released when the nucleus forms from them — which is why the nucleus has less mass.

Binding energy $=\Delta m\times c^2$, where $\Delta m$ is the mass defect.

Total binding energy grows with nucleus size, so to compare stability we divide by nucleon number: $\text{binding energy per nucleon}$. The higher it is, the more tightly bound and stable the nucleus.

The graph of binding energy per nucleon against $A$ rises steeply for light nuclei, peaks at about 8.8 MeV near iron-56, then falls slowly for heavier nuclei.

Helium-4 sits well above its neighbours — an unusually stable arrangement, which is why alpha particles are emitted as a unit.

Any nuclear reaction that moves products $\text{up}$ the binding energy per nucleon curve releases energy, because the products are more tightly bound than the reactants.

$\text{Fusion}$ joins light nuclei, left of the peak, into heavier ones with higher binding energy per nucleon. $\text{Fission}$ splits heavy nuclei, right of the peak, into medium ones with higher binding energy per nucleon.

The energy released equals the increase in total binding energy, which equals the decrease in total mass multiplied by $c^2$.