When a nucleus decays it is finding a more stable configuration, as all nature is. But is there a way of finding whether a specific configuration will decay?

I know that there is a specific proton-to-neutron ratio range, of which outside nucleus decay. This can also be seen on the nuclides chart. I have also looked into concepts like nuclear binding energy and gluons. However, I never found a definitive answer to my question.

My question: Is there a way, given the number of protons and neutrons (and electrons if need be), to find if the nucleus in this configuration will decay?

(Please don't hesitate to use math to explain the concepts. In fact, I'd prefer it if I ended up with a formula.)

  • $\begingroup$ The field of nuclear physics is both broad and deep. An textbook will have various ways of looking at the stability of a nucleus. And, yes, in some cases the electrons do impact nuclear decay. $\endgroup$
    – Jon Custer
    Jun 30, 2023 at 12:46
  • $\begingroup$ In principle, you can look at all possible configurations with the same baryon number and the same lepton number and find the total mass for each one. If you don't find such a configurations with a lower mass, then your nuclide is stable. But that may not be a terribly satisfying answer; and even then, there are some nuclides that are "observationally stable": they "should" be able to decay but have never been observed to do so, presumably due to extremely long half-lives. $\endgroup$ Jun 30, 2023 at 12:51
  • $\begingroup$ @MichaelSeifert I understand. Is there a mathematical formula/algorithm to predict such a decay? $\endgroup$
    – Saksham
    Jun 30, 2023 at 19:55
  • $\begingroup$ "When a nucleus decays it is finding a more stable configuration" Not quite. Eg, radium-226 has a half-life of 1602 years, but it decays to radon-222, which has a half-life of 3.82 days. See en.wikipedia.org/wiki/Decay_chain for numerous other examples. $\endgroup$
    – PM 2Ring
    Jun 30, 2023 at 20:11

1 Answer 1


To determine whether a nucleus is allowed to decay into another system with the same baryon number, you weigh it. If its mass is more than the mass of the products at the end of the decay, then the decay is allowed. Symmetries which "forbid" the decay really just make it slower — for example, a "third-forbidden" beta decay is faster than a "fourth-forbidden" beta decay.

For example, a neutral carbon-14 atom has a mass of $14.003\ 242$ dalton, while a neutral nitrogen-14 atom has a mass of $14.003\ 074$ dalton. We can look at just the "mass excess," the part of the mass that's different from the mass number $A=14$, and see that there is 0.168 millidalton of mass, or about 150 keV of energy, to be liberated in that decay.

Some energetically-allowed decays may have extremely long lifetimes. For example, potassium-40 may decay to either calcium or argon, via positive or negative beta decay, but the calcium decay is less energetic. This means it is theoretically possible for calcium-40 to "double-beta decay" to argon-40. However, that decay is so slow that it has never been observed.

There is currently no evidence for any decays which change the baryon number (total protons plus neutrons) of a system. The best-studied example would be proton decay where the half-life is no shorter than $10^{34}$ years. This limit was set, roughly, by putting $10^{33}$ protons, as the hydrogen nuclei in water, in a mine full of radiation detectors, and then observing zero proton decays over a decade.


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