Stable and unstable nucleiAQA A-Level Physics: Revision notes

The protons in a nucleus repel each other with the electrostatic force, which has a long range. Gravity between nucleons is negligible. Nuclei stay together because of the strong nuclear force, which acts between nucleons (protons and neutrons).

• It is attractive at separations up to about 3 fm, and negligible beyond that.
• It is repulsive at separations closer than about 0.5 fm, which stops the nucleons collapsing together.
• It is much stronger than the electrostatic force at nuclear distances.

Because the strong force has such a short range, each nucleon is attracted only by its neighbours, while the electrostatic repulsion acts between all the protons. Large nuclei therefore need extra neutrons, which add attraction without adding repulsion.

A nucleus is unstable when the balance of forces does not favour staying together, for example when it is very large (too many protons repelling each other) or has too many neutrons for its number of protons. An unstable nucleus decays spontaneously, emitting radiation to become more stable.

The two decays you need are alpha decay, which happens in large nuclei, and beta-minus decay, which happens in nuclei with too many neutrons.

An alpha particle is a helium nucleus, ⁴₂He: 2 protons and 2 neutrons. The parent nucleus loses 4 nucleons, so A falls by 4 and Z falls by 2.

Worked example. Radium-226 decays by alpha emission to radon:

$^{226}_{88}\mathrm{Ra}$ → $^{222}_{86}\mathrm{Rn}$ + $^{4}_{2}\mathrm{He}$

Check both: nucleon number 226 = 222 + 4, and proton number 88 = 86 + 2. Nucleon number and charge are always conserved.

In beta-minus decay a neutron in the nucleus changes into a proton, and an electron and an antineutrino are emitted: n → p + e⁻ + ν̄ₑ. The nucleon number is unchanged and the proton number rises by 1. The electron is created in the decay; it is not an orbital electron.

Worked example. Strontium-90:

$^{90}_{38}\mathrm{Sr}$ → $^{90}_{39}\mathrm{Y}$ + $^{0}_{-1}\mathrm{e}$ + $\bar{\nu}_\mathrm{e}$

Check: nucleon number 90 = 90 + 0, and charge 38 = 39 − 1 + 0.

When beta decay was first studied, the emitted electrons were found to have a continuous range of kinetic energies up to a maximum. If only the electron and the daughter nucleus were emitted, each decay would release a fixed energy and all electrons would have the same energy. Energy appeared not to be conserved.

The neutrino was hypothesised as a third particle, sharing the energy with the electron so that energy is conserved. It has no charge and almost no mass, so was very hard to detect. The electron and neutrino share the energy in different proportions in different decays, which explains the range.