# Is there a direct relationship between an isotope's neutron count and radioactivity?

In my textbook, it lists isotopes of Carbon: C-12, C-13, and C-14. It noted that C-14 is radioactive (C-12 and C-13 are not).

Is there a direct relationship between the number of neutrons and an element's radioactivity?

In other words, since we know C-14 is radioactive, does that mean we also know that C-15 (if such a thing exists) would also be radioactive?

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Useful link: interactive table of the isotopes with a periodic table of the elements interface. In it you will find that C-15 has a 2.4 second half-life and decays by $\beta^-$ among other things. – dmckee May 22 '12 at 15:38
If you look in the "Related" sidebar, you will find a number of other questions which touch on this point, though I don't believe that any of them address it exactly. The full answer is complicated. – dmckee May 22 '12 at 15:41
Related - Having the same number of neutrons – voix May 22 '12 at 17:14
Out of curiosity, did you mean the neutron count? Because $^{12}$C, $^{13}$C and $^{14}$C have different numbers of neutrons. – Warrick May 23 '12 at 8:19
Thanks Warrick - you're correct! Brain-hiccup on my part - will edit and correct the OP. – mikemanne May 23 '12 at 12:58

As @dmckee says the problem is complicated. It is complicated because it is not a solution of a potential describing one force, but a balance between electromagnetic forces and the strong force that is keeping the quarks within the nucleons. (In the nucleus the strong force is like a type of Van der waals potential, a higher order interaction, overflowing from the QCD dynamics of the nucleons). In addition there is the fermi exclusion principle because both protons and neutrons have spin 1/2.

All these have been approximated with the shell model of the nucleus, and you could maybe spend some time reading the link.

The shell model is partly analogous to the atomic shell model which describes the arrangement of electrons in an atom, in that a filled shell results in greater stability. When adding nucleons (protons or neutrons) to a nucleus, there are certain points where the binding energy of the next nucleon is significantly less than the last one. This observation, that there are certain magic numbers of nucleons: 2, 8, 20, 28, 50, 82, 126 which are more tightly bound than the next higher number, is the origin of the shell model.

Note that the shells exist for both protons and neutrons individually, so that we can speak of "magic nuclei" where one nucleon type is at a magic number, and "doubly magic nuclei", where both are. Due to some variations in orbital filling, the upper magic numbers are 126 and, speculatively, 184 for neutrons but only 114 for protons, playing a role in the search of the so-called island of stability. There have been found some semimagic numbers, notably Z=40.2 16 may also be a magic number.3

So there are stable nuclei and the various models do a good job of predicting them. There exists a band of instability for the various isotopes and the island of stability for high Z.

So the answer is no, there is no general rule, except solutions of the shell model, though adding or subtracting neutrons to a stable isotope one expects a high probability that it will become unstable,as an examination of the table of nuclides shows..

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Thanks very much for your quick & thorough response! As a relatively new (though not young) student, some of it went over my head. I'm marking it as the answer, and I'm sure I'll "grow into" more of the details as I progress in my classes. – mikemanne May 22 '12 at 16:27
Not just E&M and the strong force, but also the weak force, which is minimized for equal numbers of protons and neutrons. – Jerry Schirmer May 23 '12 at 13:22
@JerrySchirmer do you have a link for this? The weak interaction is so much weaker than the electromagnetic and the strong that I am surprised by this statement. en.wikipedia.org/wiki/Fundamental_forces – anna v May 23 '12 at 14:27
I can try and look this up when I have more time, but I'm certain that the weak force is the driving force behind beta decay, which is driven by $Z-1$ or $Z+1$ molecules having lower mass than the original $Z$ nucleus. – Jerry Schirmer May 23 '12 at 23:51
@JerrySchirmer No quarrel with the decay once the binding energy is off optimum, certainly it is due to the weak force, but it is the second stage. I think that the reason the binding energy is less is on how the repulsions of the positive masses balance around the attraction of the strong force. A free neutron decays, two neutrons do not have enough strong binding energy to stop the decay.equally two protons do not bind. The strong energetics ( used to be called pion exchange) allow for a lower binding energy in a proton neutron combination ( deterium), is the way I see it. – anna v May 24 '12 at 3:44

Thanks to @dmckee, and the link he suggested: interactive table of the isotopes. Looking at that table, it seems to me that there is not a reliably direct relationship between number of neutrons and radioactivity. Using Calcium (Ca) as an example (assuming I'm reading the chart correctly):

• Ca-40: stable
• Ca-41: radioactive (with a relatively long half-life)
• Ca-42: stable
• Ca-43: stable
• Ca-44: stable
• Ca-45: radioactive (with a relatively short half-life)
• Ca-46: stable
• Ca-47: radioactive (with a relatively short half-life)
• Ca-48: radioactive (with a relatively long half-life)
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Good find. I knew there were several elements that had this kind of "mixed-up" situation, but was waiting to recall which ones there were before I went hunting. – dmckee May 22 '12 at 16:28
Note the relative stability of odd-even pairs... – Warrick May 23 '12 at 8:17