Why is it that some elements do not have any stable isotopes, while others of very similar mass have multiple stable isotopes? What determines this phenomenon and what determines the strict periodicity of an unstable isotope's half-life?
Isotope Stability – What Determines the Half-Life or Stability of an Isotope?
binding-energyhalf-lifeisotopesnuclear-physicsradioactivity
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This is really a comment, since I don't think there is an answer to your question, but it got a bit long to put in as a comment.
If you Google for "Why is technetium unstable" you'll find the question has been asked many times in different forums, but I've never seen a satisfactory answer. The problem is that nuclear structure is much more complex than electronic structure and there are few simple rules.
Actually the question isn't really "why is technetium unstable", but rather "why is technetium less stable than molybdenum and ruthenium", those being the major decay products. Presumably given enough computer time you could calculate the energies of these three nuclei, though whether that would really answer the "why" question is debatable.
Response to comment:
The two common (relatively) simple models of the nucleus are the liquid drop and the shell models. There is a reasonably basic description of the shell model here, and of the liquid drop model here (there's no special significance to this site other than after much Googling it seemed to give the best descriptions).
However if you look at the sction of this web site on beta decay, at the end of paragraph 14.19.2 you'll find the statement:
Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a qualitative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements.
So these models fail to explain why no isotopes of Tc are stable, even though they generally work pretty well. This just shows how hard the problem is.
If protons decay, then what you say is true: all atomic nuclei are indeed unstable, and a so-called "stable" nucleus simply has too long a half-life for its decay to be observed.
The most tightly bound nucleus is $^{62}$Ni, with a binding energy per nucleon of 8.79 MeV [source], which is less than 1% of the mass of a nucleon. On the other hand, the decay of a proton through a process such as
$$p \to e^+ + \pi^0$$
results in the loss of most of the mass of the proton. So if the proton can decay then it's pretty clear that an atomic nucleus always has more much more mass than a hypothetical final state in which some or all of the protons have decayed. In other words, while neutrons do not decay inside "stable" atomic nuclei because of the binding energy of the nucleus, protons cannot be so protected because their decay would be much more energetically favourable (than that of a neutron to a proton).
The question of whether protons do decay is still unresolved, as far as I know.
If protons do not decay, then the $^1$H nucleus, by definition, is stable, so there is at least one stable nucleus.
Now, you might be wondering how we can establish that a nucleus is stable (assuming no proton decay). We make the assumption that energy is conserved, and it's impossible for a nucleus to be created if there isn't enough energy in the system to make up its rest mass. Given that assumption, say we have a nucleus. If we know the masses of the ground states of all nuclei with an equal or smaller number of nucleons, then we can rule out the possibility of there being a state that the given nucleus can transform into with less total mass. That in turn guarantees that the given nucleus is stable, since it can't decay into a final state with greater mass without violating conservation of energy. For a simple example, consider a deuteron, $^2$H. Its minimal possible decay products would be:
- a proton plus a neutron;
- two protons (plus an electron and an electron antineutrino)
- two neutrons (plus a positron and an electron neutrino)
- a diproton (plus an electron and an electron antineutrino)
- a dineutron (plus a positron and an electron neutrino)
But all of those states have higher mass than the deuteron, so the deuteron is stable; it has no decay channel.
Of course, you might wonder whether there are possible daughter nuclei whose masses we don't know because we've never observed them. Could, say, the "stable" $^{32}$S decay into $^{16}$P (with 15 protons and 1 neutron) and $^{16}$H (with 1 proton and 15 neutrons)? After all, we don't know the masses of these hypothetical nuclei. But if nuclei so far away from the drip line actually have masses low enough for that to happen, then there would have to be some radically new, unknown nuclear physics that would allow this to happen. Within anything remotely similar to existing models, this simply isn't possible.
Best Answer
Here is a somewhat simplified explanation.
The nucleus of an atom is populated with protons and neutrons, which behave as if they were tiny little spheres with well-defined radii. The strong force holds them all together, and in this state they are all essentially touching their nearest neighbors inside the nucleus.
Like the electron shells surrounding them, nuclei too have quantized energy levels inside them and the protons and neutrons occupy those levels. Since those nucleons resist being deformed, and since it is not possible in all cases to stack them all together into perfectly spherical shells, the resulting nuclei are not perfectly spherical either- and the amount of energy stored in those imperfect packing schemes varies from one type of nucleus to another.
Finally, consider the fact that when bound closely together by the strong force, two adjacent protons are actually repelling each other very strongly- so strongly, in fact, that neutrons must be included in the nuclear mix to help glue those protons together. But since the nucleon packing is imperfect, so also is the distribution of attractions and repulsions between nearest neighbors. This in turn means that some nuclei are naturally less stable to perturbations and upsets than others- and some of them would for example very much like to be rid of a proton or two. But getting rid of a couple of protons is easiest if you also get rid of a couple of neutrons at the same time, since two protons and two neutrons in one glob (an alpha particle) is an exceptionally stable configuration.
To spit out an alpha particle in the interests of reducing the energy of the nucleus and make it more stable requires a certain amount of activation energy, which over time will be furnished by random chance. The lower the activation energy, the more likely in any given slice of time it will be that the alpha gets ejected, which yields a short half-life. The higher the activation energy, the less likely, and you have a longer half-life.