What minimum half-life an isotope should have to be considered stable?

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    $\begingroup$ I don't think there is any minimum. Any detectable decay means the element is unstable. For example Bismuth 209 was thought to be unstable but actual decays weren't detected until 2003. $\endgroup$ – Brandon Enright May 5 '14 at 22:47
  • $\begingroup$ @Brandon Enright at what minimum half-life a decay considered detectable? $\endgroup$ – Anixx May 5 '14 at 22:51
  • $\begingroup$ No idea but I bet if you put a chunk of Bismuth in a isolated environment like a Neutrino detector it would be quite easy to detect. You only need a statistically significant detection against background so getting more material or waiting more time (or both!) should allow you to detect pretty much any half-life with statistical significance. $\endgroup$ – Brandon Enright May 5 '14 at 23:03
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    $\begingroup$ You don't "consider" things detectable. You look at a detector and say "the limit for this apparatus is about ...". $\endgroup$ – dmckee May 6 '14 at 2:29
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    $\begingroup$ Keep in mind that $10^{16}$ seconds is "only" 300 million years. $\endgroup$ – rob May 6 '14 at 15:19

It depends on the context.

For a given number of nucleons, there are typically one or two charge states that have much lower energy than any of their neighbors. For example, in the $A=12$ isobar, the weak decays $^{12}\text{B}\to{}^{12}\text{C}$ and $^{12}\text N\to{}^{12}\text C$ both release around 15 MeV excess binding energy. Carbon-12 is at the bottom of that valley. Carbon-12 also is more tightly bound than any $A=11$ system with a free proton or neutron at rest. So carbon-12 is really and truly stable: unless there is proton decay (for which there's no evidence in the free proton, and which we'd expect to be even slower for a bound proton) there's no known final state to which it could release energy by decaying spontaneously. This is the case for most of the isotopes with black squares in the table of nuclides.

Isotopes with lifetimes comparable to or longer than the age of the earth occur naturally and can be considered stable for most practical purposes: for instance, uranium-containing minerals can be smelted into uranium metal and machined.

In fast reactions, an long-lived unstable nuclide may effectively be considered stable. This is, for instance, why cosmic rays at sea level consist of muons and not pions. It's also why the entire ensemble of neutron-rich isotopes is necessary to understand the rapid fusion process in supernovae.

There do exist nuclei for which decays are allowed energetically but have not been observed. Brandon Enright mentions bismuth 209. Another example is xenon 124, which is too tightly bound to decay to iodine 124 and a positron-neutrino pair at rest, but can energetically decay to tellurium 124 and two positrons at once. This is a system that is used to search for evidence that the neutrino is its own antiparticle.

  • $\begingroup$ U238 is black in your table. Do you seriously claim there is no known final state to which U238 can decay? And even the description says they colored black all nuclides who has half-life greater than 10^15 seconds. $\endgroup$ – Anixx May 6 '14 at 9:14
  • $\begingroup$ Sorry, it's been a few years since I actually read the description on the table. I considered writing something about how nucleon or alpha emission is possible for A >~ 60, but I didn't. Should I edit? $\endgroup$ – rob May 6 '14 at 14:12
  • $\begingroup$ @rob: In the table of nuclides link, what physical phenomenon explains the pink/yellow "gouge" (very short half-lives) for those elements just to the "north east" of Pb-206? It's almost creepy how obvious that is even when zoomed out all the way. $\endgroup$ – pr1268 Aug 30 '16 at 10:52
  • $\begingroup$ @pr1268 Sounds like a follow-up question. $\endgroup$ – rob Aug 30 '16 at 14:44
  • $\begingroup$ oh, look! hi, xenon-124! $\endgroup$ – Emilio Pisanty May 12 at 18:16

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