Both ##^{87}_{37}Rb## and ##^{87}_{38}Sr## are odd-even nuclei, so we can ignore the pairing term ##\delta##. I tried to calculate the most stable(adsbygoogle = window.adsbygoogle || []).push({}); Zfor a givenAby finding ##\frac{\partial B}{\partial Z} = 0##. That gives the most Z-stable value of ##Z_0 = \frac{2\gamma A}{4\gamma + \epsilon A^{\frac{2}{3}}}## which is ##38## for ##A=87##.

If that's the case, then why wouldn't Rb beta decay to Strontium as these are naturally occurring isobars.

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# Why doesn't Rubidium decay to Strontium?

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