Rb83 is unstable and decays to Kr-83. Mode of decay is electron capture. Rb85 is stable.

The nuclei Rb83 and Rb85 have the same charge, but Rb85 is heavier than Rb83. While gravity acts more strongly on Rb85, this is probably not the factor producing the stability of Rb85. So, why does the orbital electron fall into the nucleus of Rb83, and Rb85 is stable?

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    $\begingroup$ Google for "Valley of Stability" $\endgroup$ – Georg Feb 2 '11 at 22:21

It is not a matter of "falling in": all s orbitals have non-trivial probability densities at the center.

It is about energy balance in the nucleus.

Kr-83 is a lower energy configuration than Rb-83 by enough to make up for the neutrino and the gamma(s). Evidently Kr-85 is not a sufficiently lower energy state than Rb-85.

  • $\begingroup$ can it be considered as elastic and inelastic collision of orbital electron and nucleus? $\endgroup$ – voix Feb 2 '11 at 19:30
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    $\begingroup$ @voix: I suspect that you are thinking in terms of classical trajectories (i.e. little billiard balls zooming around on well defined paths), and that picture is not very useful in this case. Quantum mechanics dominates here and the deal is that the electrons are---in some sense---fractionally in the nucleus all the time. The reaction $e + p \to n + \nu$ (and generally one or more subsequent gammas as the resulting nucleus rearranges itself) is allowed by that proximity. $\endgroup$ – dmckee --- ex-moderator kitten Feb 2 '11 at 19:47

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