Based on my sources gallium 68 undergoes a beta + decay. I've read that if an isotope has "too many neutrons" it undergoes beta - decay; if it has "too many protons" it undergoes beta + decay. Initially, I thought: "gallium 68 has 31 protons and 37 neutrons, so it has too many neutrons" (comparing it to a 1:1 ratio). But, if that was the correct reasoning it should undergo a beta - decay.
Therefore, I'm now thinking that to establish if a nucleus has "too many neutrons" or "too many protons" you have to compare it to the most stable isotope (could that be it?!). If we take the most stable isotope (gallium 69) as a reference:
Gallium 69: 31 protons and 38 neutrons;
Gallium 68: 31 protons and 37 neutrons...
...then we can say it has too many protons (that is, compared to the ratio found in the most stable isotope). Is that right or what am I missing?
Thank you in advance!
Yes, but not compared with 1:1 ratio. With many protons the most stable nucleus has extra neutrons. The reason is the unfavourable electrical repulsion of many protons - that does not happen for many neutrons. Then K40 can decay both ways, to Ar40 and to Ca40 because the nucleus prefers even number of n and even numbers of p
Thank you for your answer Professor Wade William Magill Allison! I gather that the rule "too many neutrons" or "too many protons" to justify beta - and + decays, respectively, is a bit of a simplification. Thank you for clarifying that this is not in comparison with the 1:1 ratio.
You should add to the rule "too many n" and "too many p" the words "compared to the stable isotopes of the same element", i.e. 69Ga and 71Ga.
But this remains a gross rule. Some nuclei with "too many protons" undergo electron capture , not beta+ decay.
The exact rule allowing beta+ is that the mass of the parent 68Ga is higher than the sum of the masses of the progeny 68Zn + 1 positron.
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