I reasoned as follow:

the probability of an $\alpha$ emission


is given by:


where the Gamow factor is given by:

$G\simeq \pi \sqrt{\frac{2\mu c^2}{E}}Z_1Z_2\frac{e^2}{\hbar c}$

so $G\propto \sqrt{\mu}$

where $\mu=\frac{m_{\alpha}m_{X^{'}}}{m_{X^{'}}+m_{\alpha}}\simeq m_{\alpha}$ because usually $m_{X^{'}}>>m_{\alpha}$

In the case of a spontaneous fission I have that


The only thing I can say is that the probability of tunneling is less than in the previous case because now the reduced mass is higher. Then I have some qualitative knolwledges like "the fission happen when the energy associated to proton repulsion become higher than the binding energy, so because of the repulsion energy is $\propto Z^2$ the fission is a mechanism that involves heavy nucleus"

What is the correct way to calculate the spontaneous fission probability and so compare it with the $\alpha$ emission probability?


1 Answer 1


The general process is called "cluster decay", see this article for details.

  • $\begingroup$ Can you elaborate a little bit? We prefer standalone answers to link-only answers here. $\endgroup$
    – rob
    Jun 10, 2014 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.