Stable and Unstable Nuclei

Protons in a nucleus are separated by only a few femtometres ($10^{-15}$ m), so the electrostatic repulsion between them is enormous. Something has to overpower it, and that something is the $\text{strong nuclear force}$: a force that acts equally between any two nucleons (proton–proton, proton–neutron or neutron–neutron), is far stronger than the electrostatic force at nuclear separations, but has an extremely short range — beyond about $3$–$4$ fm it has essentially no effect.

At very short range the strong force is actually $\text{repulsive}$ (stopping nucleons collapsing into each other), becomes strongly attractive at a separation of around 1–2 fm, and then falls to zero beyond a few fm. This combination is why nuclei have a roughly constant density but a definite size, and why heavy nuclei — where not every proton is close enough to every other nucleon to feel the attraction, while every proton still repels every other proton over the whole nucleus — struggle to stay stable.

For light elements, stable nuclei have roughly equal numbers of protons and neutrons ($N\approx Z$). As nuclei get heavier, the number of neutrons needed for stability grows faster than the number of protons, because extra neutrons add short-range attraction without adding any long-range electrostatic repulsion — stable heavy nuclei need $N>Z$.

A nucleus becomes unstable if it has too many neutrons relative to protons (neutron-rich), too many protons relative to neutrons (proton-rich), or if it is simply too large overall for the strong force to bind everything together (every nucleus with more than 82 protons is unstable). An unstable nucleus decays by emitting radiation to move towards a more stable neutron-to-proton ratio.

In $\alpha$ decay, a nucleus too large or too proton-rich for the strong force to hold together ejects an alpha particle — two protons and two neutrons bound together, identical to a helium-4 nucleus. This reduces the nucleon number by 4 and the proton number by 2.

In ${\beta }^{-}$ decay, a neutron-rich nucleus converts a neutron into a proton, emitting a fast electron (the ${\beta }^{-}$ particle) and an antineutrino. The proton number increases by 1 while the nucleon number is unchanged. In ${\beta }^{+}$ decay — the mirror process, in a proton-rich nucleus — a proton converts into a neutron, emitting a positron and a neutrino; here the proton number decreases by 1. (You’ll meet the neutrino and antineutrino properly in the next lesson.)

Every decay equation must conserve both nucleon number and charge (proton number) between the left- and right-hand sides — that conservation is exactly what lets you balance an equation and identify the daughter nuclide.

Alpha and beta decay often leave the daughter nucleus in an $\text{excited state}$ — it has the right numbers of protons and neutrons but too much internal energy. The nucleus loses this excess energy by emitting a $\gamma$-ray photon, with no change to $A$ or $Z$ at all.

Because $\gamma$ emission changes neither the nucleon number nor the proton number, it is never the whole story of a decay — it always follows an $\alpha$ or $\beta$ decay (or occasionally a nuclear reaction) that has just left the nucleus in an excited state.