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Looking at the band of stability, my first intuition is to conclude (erroneously) that there is a stable isotope of every element that lies close to the belt of stability. Why is this false? (For example, Uranium has no stable isotopes.) Or in particular, why are some number of protons inherently unstable in the nucleus?

Again, based purely on my poor intuition, I would assume that some number of neutrons could be arranged with that number of protons to create stability but that is not true in all cases. Why are these numbers of protons (e.g. 92 for Uranium) unstable?

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related: physics.stackexchange.com/questions/40960/… –  Ben Crowell Aug 28 '13 at 23:12

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up vote 5 down vote accepted

Because the Coulomb interaction is proportional to charge, one can say roughly that the energy released in alpha decay is proportional to $Z$, and the energy released in fission is proportional to $Z^2$. This guarantees that sufficiently heavy elements will be unstable with respect to alpha decay and fission.

Any even atomic number is guaranteed to have an even-even isotope that's stable against beta decay, because even-even isotopes are more stable than odd-odd ones, due to pairing.

In general we expect a a richer variety of stable isotopes for elements whose atomic numbers are near a magic proton number, or for elements such that for that element, the line of stability comes close to a magic neutron number.

Based on these considerations, if we were hoping to find a light element with no stable isotopes, it would be one that was an odd $Z$, one whose $Z$ was far from any magic proton number, and one for which the line of stability is not close to a magic neutron number. Technetium fulfills these requirements, and indeed has no stable isotopes.

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