
#1
Nov1112, 09:31 PM

P: 99

So I think, as a rule of thumb that for fission to be possible [itex]\frac{Z^2}{A}[/itex][itex]\geq[/itex]47
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[itex]\pi[/itex]R^{2}(1+[itex]\frac{2}{5}[/itex][itex]\epsilon^2[/itex] + ...) call this X And its Coulomb energy can be described as [itex]\frac{3Z^2}{20\pi\epsilon_0R}[/itex](1[itex]\frac{1}{5}[/itex][itex]\epsilon^2[/itex] + ...) call this Y And so ΔBE = XY 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 = r_{0}A^{1/3} thanks 



#2
Nov1112, 09:35 PM

P: 99

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




#3
Nov1212, 07:40 AM

Mentor
P: 10,853

You can use that formula for R and solve for ##\frac{\partial (XY)}{\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. 


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