- #1

unscientific

- 1,734

- 13

**Z**for a given

**A**by 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.