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What exactly does it mean for an isotope to be stable?

  1. Feb 24, 2015 #1
    The basic "nuclear physics for dummies" explanation of nuclear physics goes something like this:
    There are two dominant forces at play in atomic nuclei: the residual strong force (aka the nuclear force) which binds nucleons together and the electromagnetic force (or, more simply, the electrostatic force aka Coulomb repulsion) which causes positively charged protons to repel each other. The nuclear force is strong but only effective over short distances, while Coulomb repulsion is weaker but active over longer distances. Hence, for small nuclei the strong force is dominant and it is more energetically favourable to fuse together into larger nuclei. For larger nuclei, the repulsive force that wants to rip the nucleus apart dominates and its more energetically favourable to split apart into smaller nuclei. The "break even" point is around iron/nickel where isotopes with the highest nuclear binding energy per nucleon are found.

    There are simplifications here, of course, but if anything I've said is egregiously wrong please tell me.

    My question, then, is what do we mean when talk we about nuclides heavier than iron being "stable". They're clearly not purely energetically stable because they don't have the maximum possible binding energy per nucleon. Wikipedia defines a stable nuclide as one that "are not radioactive and so (unlike radionuclides) do not spontaneously undergo radioactive decay". Is it really case that so-called stable nuclides heavier than iron do not undergo spontaneous decay period, or do they merely have such long half-lives that they may be considered stable for all intents and purposes? Clearly, there is more to it than just the "for dummies" explanation in terms of binding energy, or else we would expect isotopes with close the same molecular weight to have almost the same half lives, which is obviously not the case.
     
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  3. Feb 24, 2015 #2

    phyzguy

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    I don't think we know for sure whether the stable isotopes are truly stable or whether they just have extremely long half lives. We don't even know for certain if the proton is truly stable. You can make the same argument for the proton, because energetically it can decay into a positron and a neutral pion, which is a lower energy state than the initial proton.
     
  4. Feb 24, 2015 #3

    SteamKing

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    Clearly, there are isotopes of elements beyond iron in the Periodic Table which have some truly long half-lives, for instance, U-238 with a half-life of about 4.5 billion years, or the age of the earth. There are a couple of isotopes of Zinc with estimated half-lives several orders of magnitude greater than the age of the universe, but there are three Zinc isotopes which are considered "Stable".

    http://en.wikipedia.org/wiki/Isotopes_of_zinc

    Gold and Silver have one and two stable isotopes, respectively, and a whole host of radioactive isotopes. Although Gold-197 is theoretically thought to decay by alpha emission, and both stable silver isotopes could spontaneously fission, none of this theoretical activity has been observed.
     
  5. Feb 25, 2015 #4
    So, again, my question is: when you say there are three zinc isotopes considered "stable" does that mean they are not believed to decay at all? If so, what, beyond simple energy considerations, determines whether an isotope can decay or not?
     
  6. Feb 25, 2015 #5

    SteamKing

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    It goes back to the fundamental question raised about the longevity of the proton. If the proton decays after umptey-ump gillion years, then no atom can be considered stable. Although certain isotopes are thought theoretically able to decay, the theory predicting such an occurrence cannot be confirmed unless evidence can be furnished that such an event has, in fact, been observed.

    The section entitled "Theoretical basis of decay phenomena" in this article:

    http://en.wikipedia.org/wiki/Radioactive_decay

    suggests that any upper limit on the half-life of a given isotope is limited only by the sensitivity of the instrument used to detect the radiation. It's also not clear that all the causes of nuclear instability have been identified and understood.
     
  7. Feb 25, 2015 #6
    Thanks, that's helpful. That said, even if nuclear stability is still not fully understood, I would assume we're not just stumbling around completely in the dark either, right? Is it at least broadly understood how it is that two isotopes with very close to the same binding energy per nucleon can have radically different half lives? I realize there's a lot we don't know about nuclear physics, but surely there's at least vague enough understanding to know why it's possible for some heavy nuclides to hold together for reasonably long periods even though it would be energetically favourable to split apart? I can see that it might be infeasible to work exactly when this is the case, but it would surprise me greatly if we don't even understand how it's possible in principle.
     
  8. Feb 25, 2015 #7

    SteamKing

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    In the early part of the Twentieth Century, physicist George Gamow was able to use some of the aspects of quantum theory to derive a law relating to alpha decay which had previously been known empirically. This work was a major advance in providing a theoretical underpinning for radioactive decay:

    http://en.wikipedia.org/wiki/George_Gamow#Radioactive_decay

    There have been various predictive models put forward to assess the stability of various nuclei by the ratios of protons to neutrons, for instance, or those which hypothesize "islands of stability" for certain, as yet, super heavy elements which have not been synthesized. Only someone with a much more developed knowledge of quantum theory can answer your questions about nuclear stability.
     
  9. Feb 25, 2015 #8

    ChrisVer

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    For splitting a nucleus, I think it's good to think of the problem as a potential step... There is some possibility the particle will remain within the step on the left (small distances/within the nucleus), or there is a possibility the particle will undergo a quantum tunneling and escape from the right of the well (large distances/outside the nucleus).
    http://philschatz.com/physics-book/resources/Figure_32_07_02.jpg
    By that idea you can make some predictions about the time it takes for the particle of a given energy to escape the nucleus bound state.
    I am not sure how good this model is in giving answers.

    PS- In fact I just described somehow the Gamow's approach in alpha decay- I just didn't remember it was his model.
     
    Last edited: Feb 25, 2015
  10. Feb 25, 2015 #9
    Thank you both for the information about Gamow's model, that's very interesting. What about weak processes? Beta/positron emission are also viable decay mechanisms. Are there similar simplified models to explain why certain nuclides undergo isobaric decay processes more easily than others?

    I guess this is spirally a bit outward from my original question and I'm creeping into, "Please teach me about nuclear physics," territory. In which case, it might be more helpful at this point if someone could direct me to a good introductory resource I could work through on my own. I know enough quantum field theory to have a reasonably good understanding of the Standard Model, so something technical is fine. Any recommendations?
     
  11. Feb 25, 2015 #10

    mfb

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    stable nuclide

    There are isotopes that are called "observationally stable" - there is a possible decay mode but it has never been observed.
    Note that a smaller binding energy does not mean there has to be a possible decay channel. There is no way an isotope with "iron plus a few protons and neutrons" can split into 1.2 iron nuclei - every fission process would produce smaller products with a smaller binding energy, and can be forbidden.

    Spontaneous fission - if possible at all - is extremely unlikely for lighter elements, and would give those elements lifetimes so large we'll never see even a single decay.

    At least according to the wikipedia article, Zirconium-92 is the heaviest element with no possible decay mode (except proton decay, if it exists).
     
  12. Feb 25, 2015 #11
    @mfb, can you clarify what it means for there not to be a "possible decay mode"?
     
  13. Feb 26, 2015 #12

    ChrisVer

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    That it doesn't decay...
     
  14. Feb 26, 2015 #13
    Snarky answers don't help anyone. Obviously I know that, I'm asking what it means not what the significance of it is. The entire point of this thread is I'm trying to understand how it is that certain nuclides heavier than iron do not decay (or at least take such a ridiculously long time to do so that they might as well be considered not to decay), despite it be energetically favourable to do so. One explanation so far as been in terms of the potential barrier and the sensitivity of tunneling probability to its width. Now @mfb has said that the absence of a "possible decay mode" is a factor, and so I'm trying to understand what means.
     
    Last edited: Feb 26, 2015
  15. Feb 26, 2015 #14

    ChrisVer

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    I said what it means...
    A groundstate nuclide will decay under:
    a) alpha decay
    b) beta decay
    A stable nuclide will decay with neither (a) nor (b) and that's why it's called "stable".
    If you take a heavier nucleus than the Iron, you have to check whether it's energetically favorable for it to decay with either of these modes. If it is not, then that nucleus will be stable, even if the Iron has the highest binding energy per nucleon, there is no "path" the heavy nuclei will take in order to reach the Iron. And it's not only the "iron", which I am afraid you are misusing in this context (why do you think that it's energetically favorable for all heavier nuclides to decay??? If I tell you, give me 100$ and I'll give you 10$, you might get 10$ but you will lose a net amount of 90$). An example for that is to think that you are in a room that has neither a window (a) nor a door (b)- although you can be hungry, you cannot reach the kitchen in any way- you will remain to that room.
    Unstable nuclei, will start undergoing these decays until they'll reach a stable one. And there, they stop.
    I think "observational stable" means that although the decay mode is possible, it doesn't occur within an "observable" time for us to determine it.
     
    Last edited: Feb 26, 2015
  16. Feb 26, 2015 #15
    This is a bit frustrating. I feel like I've been pretty clear about my questions—and everyone else seems to have understood me just fine—but you're pointedly not answering exactly the question I'm asking. How can it not be energetically favourable for a heavy element to decay via alpha decay? If binding energy per nucleon decreases with number of nucleons past iron, it seems like such a decay is always energetically favourable.
    I obviously understand that in order to not decay, it must not be possible to decay via each decay mechanism. That's just basic logic, and I think you're being extremely uncharitable in your interpretation of my questions by not assuming I understand such obvious facts. It comes off as very patronizing, and I don't appreciate it.
     
  17. Feb 26, 2015 #16

    ChrisVer

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    It's not patronizing. And I added something in my last answer.
    So even if you are to win something, you are losing more. I cannot make it more simple than that, you have to take yourself, a stable heavier than iron nucleus [itex]X[/itex], and check:
    [itex]X \rightarrow Y + \alpha [/itex]
    or
    [itex] X \rightarrow Z + \beta + \nu [/itex]
    to see whether they are possible or not. And they don't decay immediately to Iron, they decay to iron via steps.
     
  18. Feb 26, 2015 #17
    The plot of binding energy per nucleon indicates you are not losing more. I'll go through the patronizing exercise you're insisting I do to show exactly what I mean:

    The binding energy per nucleon of Pb-206 is 7.875MeV. The binding energy per nucleon of Hg-202 is 7.897MeV and of He-4 is 7.074MeV. Therefore, the total binding energy of Pb-206 is 1622.324MeV and of Hg-202 and an alpha particle is 1623.46MeV (computed with Wolfram Alpha). Thus, Hg-202 and an alpha particle together has higher binding energy per nucleon, and therefore less total potential energy, than Pb-206.

    Thus, Pb-206 > Hg-202 + alpha is energetically favourable and possible in theory. This is confirmed by the Wikipedia article on lead which says, "Lead occurs naturally on Earth exclusively in the form of four observationally stable isotopes: lead-204, -206, -207, and -208. All four could theoretically undergo alpha decay with release of energy, but this has not been observed for any of them." Pb-206 is considered stable and the end point of U-238 chain, despite it being energetically possible to decay into mercury. As I have said, now quite a few times, I am trying to understand why such a decay does not happen. I have done these calculation and they are the reason for my question.

    Honestly, I'd prefer it if someone else answered and that you just left this thread alone. You are being unhelpful and, frankly, quite rude.
     
    Last edited: Feb 26, 2015
  19. Feb 26, 2015 #18

    ChrisVer

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    The lead that you are talking about is categorized as "observationally stable".
    http://en.wikipedia.org/wiki/Stable_nuclide#Still-unobserved_decay
    In other words are considered stable because they haven't been observed to decay yet. "why that is the case" is a different question to what you asked and I answered so far.
    I have been trying to answer your ill-defined questions (from what does it mean for a decay mode not-to-exist up to why heavier than Iron nuclei can be stable) and your misinterpretation of the binding energy/nucleon graph . If that graph is enough to tell you which decays are possible, I'd say that you are wrong, it only gives you what you can gain, and not what you spend for it. In a reversed example, we wouldn't have lighter than iron nuclei because they would prefer "fusing". If you don't want to listen, but consider this as "patronizing" or "rude", it's not my problem. You were the one who was rude, and because of that this is my last answer to any of your questions or threads. "finito"
     
  20. Feb 26, 2015 #19
    It is explicitly my original question, as everyone but you has managed to understand.

    Are you serious? I explicitly do the exercise you insist on to show an energetically favourable but unobserved decay based on the binding energy graph, and you're still insisting on your irrelevant "what you spend for it" explanation?

    Good riddance.
     
  21. Feb 26, 2015 #20

    mfb

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    It means there is no decay mode allowed by energy conservation.

    Pb-206 will decay if you wait long enough. But its lifetime is orders of magnitude above the age of the universe.
    And there are isotopes where the lower binding energy of the alpha particle makes this decay impossible.

    It is not just alpha and beta, you also have to check double beta, cluster decay and fission.
    That is wrong. The graph is sufficient (if you consider every isotope separately), it just does not tell you the lifetime of the isotopes.
     
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