Why is gallium 70 suddenly unstable, whereas it becomes stable again when you add another neutron? Shouldn't isotopes become unstable when there are either too few or too many neutrons? Why is gallium then unstable again in the middle?

  • 5
    $\begingroup$ You may want to read en.wikipedia.org/wiki/Even_and_odd_atomic_nuclei. $\endgroup$ Dec 29, 2021 at 14:33
  • 4
    $\begingroup$ Related: Why is tin-112 stable, but indium-112 radioactive? $\endgroup$
    – Loong
    Dec 29, 2021 at 14:42
  • 1
    $\begingroup$ @Poutnik - well, that "rule" is a result of nuclear physics as you note. It isn't like we imposed the rule, and 70Ga was a bit late to the party so was the odd isotope out once 69Ga and 71Ga took the spots... $\endgroup$
    – Jon Custer
    Dec 29, 2021 at 16:21
  • $\begingroup$ @JonCuster Well, even if it were not odd-proton number element, 70Zn and 70Ge took the place already. // For most of chemists, the neighbors rule and the max 2 for odd ones rule are taken as they are, regardless of if they are results of nuclear physics reasoning or not. Like they were empirical rules that may or may not have theoretical background. $\endgroup$
    – Poutnik
    Dec 29, 2021 at 16:29
  • 1
    $\begingroup$ @Poutnik - indeed, as can be seen from many questions here, chemistry "rules" range from pretty fundamental to suggestions to vague hints. Early chemistry courses should note that isotopes exist, and move on quickly. $\endgroup$
    – Jon Custer
    Dec 29, 2021 at 17:55

1 Answer 1


There is the general neutron/proton ratio rule, regarding stability of nuclides, described by the Valley of stability.

Additionally, in beta decay context, there are 3 easily remembered rules, based on reasoning of nuclear physics:

  1. Elements with the odd proton number have maximally 2 stable nuclides. (Some have just one, technetium and promethium have none.)
  2. If there are two isobars (nuclides with the same nucleon number) of elements that differ by 1 proton (and 1 neutron) then maximally one of them is stable.
  3. The least stable are nuclides with both odd numbers of protons and neutrons. More stable are odd-even nuclides and the most stable are even-even nuclides. The reason is the ​nucleon pairing stabilization, increasing bonding energy per nucleon.

The only stable odd-odd nuclides are $\ce{^2_1H}$, $\ce{^6_3Li}$, $\ce{^{10}_5B}$, $\ce{^{14}_7N}$,$\ce{ ^{180m}_{73}Ta}$.

  • For very light elements, the n/p ratio based stability overrules parity based stability. As the valley of stability is initially very narrow. The deviation from the ideal ratio causes a higher nucleus energy raise than would be the energy release by reaching the parity.
  • For tantalum, there is the interesting fact the stable one is the naturally occurring meta-stable nuclear isomer, while the lower energy isomer is unstable.

Back to gallium-70:

  1. Gallium is the odd-proton element and there are already stable gallium-69 and gallium-71.
  2. Gallium-70 has an odd-odd nucleus, generally predicted to be unstable.
  3. Both even-even neighbor isobars are stable - zinc-70 and germanium-70.

Gallium-70 has therefore 3 reasons why it is not stable, beta decaying to both zinc (electron capture) and germanium (electron emission).

  • $\begingroup$ Care to add a comment about the pathological stability of $N_{14}$? $\endgroup$ Dec 30, 2021 at 11:51
  • $\begingroup$ @PrimeMover Why pathological? By chance, just seconds ago I added the list of 5 known stable odd-odd nuclides. For very light elements, n/p ratio based stability overrules parity based stability. $\endgroup$
    – Poutnik
    Dec 30, 2021 at 11:56
  • $\begingroup$ P.S.: You must mean $\ce{^{14}N}$. $\endgroup$
    – Poutnik
    Dec 30, 2021 at 12:35
  • $\begingroup$ Wasn't there when I posted it. Whatever the notation is, nobody really cares. It's pathological because it's one of those exceptions. $\endgroup$ Dec 31, 2021 at 0:20
  • $\begingroup$ @PrimeMover That does not make it pathological. The least stable does not mean unstable, especially if parity is not the only contribution to stability. // Chemists care, as $\ce{N_{14}}$ means a nonexistent nitrogen molecule with 14 atoms. $\endgroup$
    – Poutnik
    Dec 31, 2021 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.