Why is Palladium-108 "theoretically capable of spontaneous fission?" I was just reading Wikipedia's page on palladium isotopes and I noticed the note about Palladium-108. It says "theoretically capable of spontaneous fission", but I don't find that explained anywhere. the same note is on other isotopes of palladium, as well. So, I am just asking if someone could explain the note.
For example, how do we know it's capable of spontaneous fission? What makes it capable of spontaneous fission? And why is it still called stable instead of just giving it a half life absurdly long like 9.9x10^100 or something? Thanks for the help?
 A: Whether a reaction is theoretically possible or not is determined, firstly, by the $Q$ of the reaction. For nuclear reactions, $Q$ is the mass energy of the difference in reactant nuclear mass and product nuclear mass. If $Q$ is positive, the reaction is possible, as long as other conservation laws aren't violated.
With spontaneous fission, the reaction would be something like $$A\to B+C+Xn,$$
with $A$, $B$, $C$ being specific nuclides and $X$ being the number of neutrons. If you calculate the $Q$ for $^{108}$Pd splitting equally or near equally (1 proton high, 1 proton low), you will get a positive $Q$ close to $10$ MeV. So SF is theoretically possible, but for some reason it's inhibited, probably because the fission activation energy is too high. So, we don't observe it in reality.
This is similar to the case of alpha decay for $^{208}$Pb. It's theoretically possible, but the $Q$ is so small that the predicted halflife ( $2.6\times 10^{21}$ years) indicates we would probably never observe it, and the predictive model is not accurate enough to publish a value.
