What is the relationship between nucleus deformation and fission requirements?

[mitch_1211]

So I think, as a rule of thumb that for fission to be possible \frac{Z^2}{A}\geq47

I want to be able to derive this relationship though..

If a nucleus deforms into an ellipsoid, its surface area can be described by

4\piR²(1+\frac{2}{5}\epsilon^2 + ...) call this X

And its Coulomb energy can be described as

\frac{3Z^2}{20\pi\epsilon_0R}(1-\frac{1}{5}\epsilon^2 + ...) call this Y

And so ΔBE = X-Y

For fission to occur set ΔBE ≤ 0

Once I've done all this I'm not sure how to get ΔBE in terms of A and Z only. I know R = r₀A^1/3

thanks

 

[mitch_1211]

I should mention that \epsilon is an arbitrary deformation parameter associated with the ellipsoid and can be factored out

 

[mfb]

You can use that formula for R and solve for $\frac{\partial (X-Y)}{\partial \epsilon}=0$.
I would expect that this overestimates the required Z^2/A - if that derivative is negative, the nucleus should not form at all or decay within less than a femtosecond.