# Alpha decay condition using semi-empirical mass formula

I have to find a equation of Z as a function of A, where Z denotes the atomic number and A denotes the mass number of an atom, for odd-even nucleii which can barely undergo alpha decay (by themselves) .

$^A_ZX_N \rightarrow _{Z-2}^{A-4}Y_{N-2}+_2^4He_2$

I use the semi-empirical mass formula $W_b(A,Z)=-A\space w_0+A^{2/3} w_1+w_2\frac{ Z^2}{\sqrt[3]{A}}+w_3\frac{ (A-2 Z)^2}{A}$

and the Energy-balance (The Energy balance should be $<0$ for the decay to take place) : $T<-W^{He}_b-W_b^{Y}+W_b^X=-W^{He}_b-W_b(A-4,Z-2)+W_b(A,Z)$ , where $W_b$ are binding energies and T is the kinetic energy of the alpha particle. $W_b^{He}$ is given and equals $W_b^{He}=-28.3 MeV$.

My question: Can I just set $T=0$ so that the kinetic energy of the emitted alpha particle is 0, which basically means that the energy difference between the starting nucleus and the ending nucleus is minimal and those nucleii barely undergo alpha decay by themselves?

In one of the notes I found they differentiated $-W^{He}_b-W_b(A-4,Z-2)+W_b(A,Z)$ with respect to Z and thus found a minimum of T in regard to Z... But why didn't they simply set the kinetic energy to 0 ?

• I think the question is a bit confusing. 1/ You use a semi-empirical formula for masses. 2/ You can deduce if the 'alpha decayed final state' is favoured/possible by energy. Why you should put T to something? You may search for best candidates for alpha decay by minimizing T... – jaromrax Apr 24 '17 at 18:12