Nuclear instabilityAQA A-Level Physics: Revision notes

On a graph of neutron number N against proton number Z, stable nuclei lie in a narrow band. For light nuclei, N ≈ Z; as Z increases, more neutrons are needed to offset the repulsion between protons, so the band curves above the line N = Z.

Nuclei outside the band are unstable and decay to move towards it:

• above the band: too many neutrons (neutron-rich)
• below the band: too many protons (proton-rich)
• beyond the top end: too large, with too many nucleons in total

• Alpha (α): large nuclei. N falls by 2 and Z falls by 2 (nucleon number falls by 4).
• Beta-minus (β⁻): neutron-rich nuclei. A neutron becomes a proton, so N falls by 1 and Z rises by 1. An electron and an electron antineutrino are emitted.
• Beta-plus (β⁺): proton-rich nuclei. A proton becomes a neutron, so N rises by 1 and Z falls by 1. A positron and an electron neutrino are emitted.
• Electron capture: proton-rich nuclei. A proton captures an inner-shell electron and becomes a neutron, so N rises by 1 and Z falls by 1. An electron neutrino is emitted.

In any decay, nucleon number and charge are conserved, and so is lepton number (the neutrino or antineutrino balances the electron or positron):

$^{14}_{6}\mathrm{C} \to {}^{14}_{7}\mathrm{N} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e$

$^{11}_{6}\mathrm{C} \to {}^{11}_{5}\mathrm{B} + {}^{0}_{+1}\mathrm{e} + \nu_e$

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

Check the top numbers add to the same total on each side, and so do the bottom numbers.

After alpha or beta decay the daughter nucleus is often left in an excited state. It returns to a lower energy state by emitting a gamma photon, whose energy equals the difference between the nuclear energy levels: $E = hf = \frac{hc}{\lambda}$. N and Z do not change in gamma emission.

A nuclear energy level diagram shows the parent nucleus, the energy levels of the daughter, and gamma transitions between them. A nucleus may emit one photon, or several photons in turn, and the total energy released is the sum of the photon energies.

Technetium-99m is a metastable excited state of technetium-99 that emits a 140 keV gamma photon, with a half-life of 6 hours. It is widely used as a gamma source for diagnosis, injected as a tracer and detected with a gamma camera. It is suitable because:

• it emits only gamma rays, which leave the body, so the patient's dose is low
• the half-life is long enough to carry out the scan but short enough that the radioactivity quickly disappears
• it is made on site from molybdenum-99 as it is needed

Identify the decay of iodine-131 (Z = 53, N = 78), which has too many neutrons.

It decays by β⁻ emission, so Z rises to 54 and N falls to 77:

$^{131}_{53}\mathrm{I} \to {}^{131}_{54}\mathrm{Xe} + {}^{0}_{-1}\mathrm{e} + \bar{\nu}_e$

Nucleon number: 131 = 131 + 0. Charge: 53 = 54 − 1.