Is there any way to predict the half-lives of radioactive isotopes from theory (that is, using only theoretical considerations, without using data about the decay)? For example, could we predict that the half life of Carbon-14 is roughly 5700 years?
[Physics] Can we predict the half-lives of radioactive isotopes from theory
nuclear-physicsradioactivity
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The short answer is no: halflives are constant.
However, let's discuss a situation in which that comment might have some kind of truth behind it. If you have a parent nucleus that decays to a radioactive daughter so that there will be two (or more) decays before stability. In general there are two possibilities for this:
- The daughter has a shorter halflife than the parent. In this case the concentration of the daughter is always $\displaystyle\frac{\tau_\text{daughter}}{\tau_\text{parent}}$ of the parent concentration. This means that the concentration of the daughter actually decays on the parent's halflife (because the daughter is constantly refreshed from the parent).
- The daughter has a longer half life than the parent. In this case the daughter will accumulate steadily as the parent decays away.
The latter case is interesting to us here because at the start the sample will register an activity that decays with the parent's (short) halflife, but after a number of those halflifes have passed the activity of the sample will be dominated by the daughter and exhibit a longer halflife.
That is something that your instructor could have meant which would not be wrong. However, the halflife of each isotope remains the same: it is only the halflife of the sample (which contains more than one isotope) that varies.
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
Answering this question is one of the major successes of 20th-century physics.
For strong decays, Gamow's alpha-tunneling model is quite successful. It relates the lifetime of an alpha emitter to the energy released in the decay using the approximately-valid assumption that nuclear density is constant and that the nucleus has a relatively sharp edge.
For beta decays there is quantity "$ft$" which convolves the half-life of the decay with the electrical interaction between the emitted electron and the positively-charged daughter nucleus. The $ft$-values are related in a relatively simple way to the matrix element for the decay, and for a given class of decay ("allowed", "superallowed", "first forbidden", etc., which are determined by the quantum numbers of the parent and daughter nucleus) the $ft$ values for most nuclei fall into a pretty narrow range.
Any nuclear physics textbook should have a chapter on characterizing decays. (I happen to be looking at Wong.)