44
$\begingroup$

I've read that tellurium-128 has an half-life of $2.2 \times 10^{24}$ years, much bigger than the age of the universe.

So I've thought that maybe every single isotope of every single atom are radioactive, and isotopes which we call "stable" are actually unstable but their half-life are immensely big (but not infinite), like $10^{100}$ years.

Is this a possible theory or are we truly $100$% sure that stable isotopes are really eternal?

$\endgroup$
  • 8
    $\begingroup$ The answer may well depend on the question of the proton's stability (posited to be unstable in may next-generation models). $\endgroup$ – dmckee Mar 30 '15 at 21:25
  • 2
    $\begingroup$ I guess you could argue on quantum grounds than any nucleus has a small probability of spontaneously flying apart. $\endgroup$ – Jiminion Mar 31 '15 at 21:11
46
$\begingroup$

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:

  1. a proton plus a neutron;
  2. two protons (plus an electron and an electron antineutrino)
  3. two neutrons (plus a positron and an electron neutrino)
  4. a diproton (plus an electron and an electron antineutrino)
  5. 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.

$\endgroup$
  • $\begingroup$ The p->n decay indeed is the only one that really matters, because the result is not an isotope. All other decays are from one isotope to another with lower energy. That means there's a partial order with at least one minimum. Makes sense: you can't have a cycle of unstable isotopes A->B->C->A, that would be a perpetuum mobile. $\endgroup$ – MSalters Mar 31 '15 at 14:46
  • 3
    $\begingroup$ Protons can't "decay" into neutrons because that is going uphill energetically. $\endgroup$ – Jiminion Mar 31 '15 at 21:12
  • $\begingroup$ @Jiminion Counterintuitive as it is, I think it does happen in some radionuclei. See: en.wikipedia.org/wiki/Positron_emission $\endgroup$ – Darth Wedgius Apr 1 '15 at 19:55
25
$\begingroup$

We are never 100% certain of anything. The scientific method falsifies wrong theories, but it does not verify those we colloquially call "correct" or "true"

If we tomorrow detect a normal oxygen atom decaying, we'll have to devise new theories to explain it.

But we don't expect the things we call stable to ever decay (that's why they're called stable). We have never seen them decay, and we - within the theories we currently accept as true - see no way how they could decay. Since those theories have done well by us in other cases, there is no reason to not trust them in this case (until evidence comes in that they are indeed false).

As a side note, Tellurium-128 is simply the nuclide with the longest half-life we have ever observed decaying. There are others, thought to be unstable with much longer half-lives, which are "observationally stable" in the sense that we've never observed them in sufficient quantity and duration to see them decay.

$\endgroup$
  • 9
    $\begingroup$ Might be worth making it more clear why the standard model implies certain isotopes should be stable, and hence that any decay implies new physics. $\endgroup$ – Keith Mar 31 '15 at 0:24
  • $\begingroup$ Note that a lack of certainty in details doesn't necessarily matter at a higher level. Say we have two isotopes A and B which might decay as A->B+stuff or B->A+stuff. Now it's possible that neither decay is possible, or that one of the two is possible, but in any of the 3 scenario's there is a stable isotope. When we see an actual decay, we know which of the two would be stable. (This is what math knows as non-constructive proofs: we can prove that some X exists even if we don't know all properties of X) $\endgroup$ – MSalters Mar 31 '15 at 14:52
10
$\begingroup$

In regards to a single element, this actually happened.
Natural Bismut is about 100% 209Bi, which shows no obvious indication of being radioactive.

But it turned out that all existing bismuth is radioactive.

The isotope 209 had been suspected to be unstable before, but that was experimentally verified as recently as 2003, finding a half life of 1.9×1019 y. All other isotopes are unstable, too.

As the half life is very long - meaning that relatively very few atoms decay per time, bismuth is still stable for all practical purposes.

Actually, I'm not afraid of the 200 g bismuth in my left hand.

$\endgroup$

protected by Qmechanic Mar 31 '15 at 23:43

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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