Neutron Decay

A free neutron will decay with a half-life of about 14.8 minutes, but it is stable if combined into a nucleus. Why would the neutron be stable in the nucleus?

Answer

The failure of the neutron in a nucleus to decay is a quantum mechanical effect. According to quantum mechanics, the rate of decay is dictated by Fermi’s Golden Rule, which states that the rate is proportional to the probability of decay (i.e., the absolute value of the square of the matrix element connecting the initial and final states) times the density of final states. Because the free neutron decays to a proton plus electron plus electron antineutrino, we know that the probability for this beta decay process is not zero and that there are available final states for the three product particles. Energy conservation dictates that the total final state energy equals the total initial energy of the free neutron.

Inside a nucleus, the decay of a neutron is a transition from an initial energy state, the particular bound neutron state that the neutron occupies, to a final state consisting of a proton in some final proton energy state plus a free electron and a free electron antineutrino, the latter two particles contributing to the energy of the final state. Therefore, energy conservation dictates that the proton will be in a proton energy state that is lower in energy than the initial energy of the neutron. In many nuclei all available proton states that is, those that are not occupied by protons have higher energies than the energy of the initial neutron state, so the decay cannot occur.

The equivalent energy levels of the protons in nuclei are higher than for the neutrons because their energies include the Coulomb repulsion between two protons and other properties of the nuclear force, especially the spin dependence. Obviously the stable nuclei will include those for which neutron and proton decays do not occur!

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