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I have been searching for an explanation for Technetium's instability both here and across the net including the "Why is Technetium Unstable" and I haven't found an answer that is satisfying to me. However when researching this I did notice that Molybdenum and Ruthenium both have a high number of stable isotopes (6 and 7).

This is my current mental model of radioactive decay. If you consider all the arrangements of the nuclear particles as existing in a phase space. Then an atomic nucleus sits in a local minima of protentional energy. A stable nucleus is one where the walls of the the potential well are so high it is impossible to get out of.

The Molybdenum and Ruthenium isotopes have very deep potential wells which makes them stable. However because Technetium's close to them in phase space it doesn't take much for technetium to slip over the wall and decay into those other elements.

Is this model flawed? And what impact to the fact that Molybdenum and Ruthenium have on the stability of Technetium.

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  • $\begingroup$ That's a reasonable heuristic, IMHO. Take a look at the question John linked. There's probably not much more we can say here on why technetium fails to have a stable isotope. $\endgroup$
    – PM 2Ring
    Commented May 10 at 10:30
  • $\begingroup$ I had ready the linked answer. My question is about the relationship of the other elements to its stability which isn't covered in the linked question. $\endgroup$
    – Q the Platypus
    Commented May 10 at 11:05
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    $\begingroup$ The other elements don’t have anything to do with it, really. At most you can say that there’s somewhere downhill to go, but the strength of the barrier that you have to overcome really is about how tightly bound the nucleons are inTc atoms, not their neighbors. In other words, spontaneous radioactive decay probability really is a 1st order kinetic process that depends only on the identity of the nucleus decaying. $\endgroup$
    – Matt Hanson
    Commented May 10 at 11:13

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