Another answer to your question requires some working knowledge of the shell model and is more conceptual in nature. Anyway, the shell model for the nucleus is based on similar concepts as that of the electrons. There are multiple levels corresponding to different energies inside of a potential well. Wikipedia does a better job at explaining it: http://en.wikipedia.org/wiki/Shell_model
There are similar, yet separate shells for the protons and for the neutrons. For example, it is possible to have a proton in the s1/2
shell with an identical spin as a neutron in the s1/2
shell without violating any laws. However, due to the Coulomb repulsion of the protons, there is a slight increase in the spacing of the proton shell levels compared to those of neutrons.
With a low number of nucleons, this spacing is not very important and you end up with equal numbers of protons and neutrons. But as you increase the number of nucleons, then you have more neutron shells within the same energy region, leading to stable nuclei with more neutrons than protons.
For example: The 2d5/2
for protons will be higher in energy than the corresponding 2d5/2
level for neutrons. As such, if there is one proton in the 2d5/2
, and a free hole in the neutron 2d5/2
, then the proton will decay with a Q value proportional to the difference in the level energies.
How does this relate to the question in the original post? Well, the same is true for neutron rich nuclei. If you have a neutron that is in a level that has a higher energy than an unfilled proton level, then the nucleus will decay via beta decay.
Probably not the best explanation, but interesting none the less