# nuclear fission and half life

Why is the alpha, beta or gamma decay of an unstable nucleus unaffected by the chemical situation of an atom, such as the nature of the molecule or solid in which it is bound? The chemical situation of the atom can, however, have an effect on the half life in electron capture. Why is this so?

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The cross-section for electron capture depends (among other things) on the likelihood that there is an electron in proximity to a proton.

Orbital states with angular momentum other than 0 (that is all states except s-states) have a node at $r=0$ and because the nucleus is several order of magnitude smaller than a typical atomic orbital this means that electrons from those states have essentially no chance to be in proximity.

However, all s-states have a local maximum at $r=0$, and therefore have non-trivial probability to be available for capture. Now if an outer electron that in an atomic context was in a s-state is instead involved in a molecular bond, then it probably does not have a high probability of being in the nucleus and the total rate for the process can be expected to go down.

The other decay processes you list don't depend on the presence of an electron.

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A qualitative argument:

The $\alpha$, $\beta$ or $\gamma$ decay are phenomena that are associated with the weak nuclear, and strong nuclear forces which rule in the nucleus. The electromagnetic force, although it plays an important part in the stability of the nucleus, it does not completely determine what and how will happen inside the nucleus. The $\beta$ decay is determined by the weak nuclear force which goes something like this:

$n\rightarrow p+e^-+\bar\nu_e$

$p\rightarrow n+e^++\nu_e$

The $\gamma$ decay results from a rearrangement of the nucleons form higher to lower energy shell configuration, with the emission of the excess energy as $\gamma$ light. The energies involved here are from several eV (Th-229) to several MeV (Co-60).

The events that take place in the outside the nucleus part of the atom/molecule is simply rearrangements of the electrons in the various energy levels. Very low energy involved. If you can understand that the electromagnetic interaction of the protons inside the nucleus cannot have an effect in a $\beta$ decay, then you can understand why the electrons outside the nucleus do not have any effect on what is going on inside the nucleus.

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The process called "electron capture" is not one of the ones you list. It means $p + e^- \to n + \nu_e$. –  dmckee Mar 14 '13 at 14:26
@dmckee Thank you for reminding me of this very interesting process. You have covered this possibility very well in you answer. The only comment I would like to make, is to mention the exceptional case of Be-7 which decays purely by electron capture (energetically more favorable.) But this is interesting from astrophysics point of view where Be-7 is under immense pressure. You were most likely talikng about nuclei under "normal conditons." Correct me if I misred your answer. –  JKL Mar 14 '13 at 21:36