Instability, getting it done with
- Thread starter John M
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Why don't unstable atoms simply decay all at once and simultaneously attain a stable configuration without strung out delay.
Well their decay is a quantum effect, in order for it to happen there needs to be some quantum tunnelling from one state to another. For instance if it is alpha decay we are talking about then a helium nuclei needs to escape from the parent nuclei. There is a potential barrier between the bound and unbound states and the system needs to quantum tunnel to reach the lower energy state. This occurs only with a certain probability per second, thus there is a random chance that the thing will decay or not in a given time period.
No, as far as I know. Neutrons for instance decay if they are on their own but are stable as part of a nucleus, but I don't think a similar thing occurs for whole atoms.
I do believe some man made elements decay almost instantly.
Yes, certainly, this is because the barrier between the bound and unbound states is very small for these elements, so there is a very high probability of tunneling occurring in a short time.
Perhaps I should explain a little further:
I just pulled this from some random google image search so I'm not sure of it's original context, but it works well enough to explain the current situation. V in the diagram is the potential energy associated with the configuration of the system, i.e. where the various protons and neutrons are located relative to each other. For unstable nuclei, the bound configuration would be the shallow well and the unbound would be the deeper, lower energy well (except it wouldn't be a well, it'd be more like the graph just kept diving off on the RHS). There is a barrier between these configurations because you can't pull the nucleons apart without increasing their potential energy temporarily. The height and width of the barrier is proportional to the decay (tunneling) probability, which is proportional to the decay rate ($$ \Gamma $$) of a lump of material made of those atoms.
Why don't unstable atoms simply decay all at once and simultaneously attain a stable configuration without strung out delay.
It's not clear whether you're talking about decay from higher electron energy states or nuclear decay processes such as alpha, beta and gamma radiation, or nuclear fission, but for all of these the answer is similar.
Unstable atoms have a half life, which you can look at in two ways. If you have a whole heap of atoms, the half life is the time required for half of the atoms to decay. If you're looking at just one atom, the half life is the time after which there will be exactly a 50% chance that the atom will have decayed.
The decay of each individual atom is random, but subject to the half life. The reason there is a half life at all, I think has to do with quantum fluctuations of the nucleus (energies of nucleons). If you're talking nuclear decay, the strong nuclear force normally holds the nucleus together tightly, against the repulsive forces of the electromagnetism. However, fluctuations occasionally allow nucleons to gain enough energy to overcome the strong attraction momentarily, and then the nucleus decays.
If you're talking about nuclear processes, I think the answer is no. Remember that the nucleus is about 100,000 times smaller than the atomic diameter, and is usually "shielded" by electrons.
Fraggle Rocker
Staff member
I think this can be seen as a manifestation of the critically important loophole in the Second Law of Thermodynamics: Entropy only tends to increase. This is much different from increasing monotonically.
This provides every elementary particle with "a choice," as it were, about when to start moving to a state of increased entropy. The probability of any individual particle choosing to move to a higher state of entropy at any point in time is expressed by the specific half-life of that particle in its current state. But given that, the selection of which particles are going to move now and which are going to wait until later is governed strictly by chance. We know (approximately) how many are going to move in the first nanosecond and how many in each subsequent nanosecond, but we have no way of guessing in advance which ones they will be.
To pursue a point I have been making in other threads, the behavior of elementary particles digs us down into the realm of microcosmology. Cosmology, as I have said before without being corrected, is an uncomfortable mixture of physics, mathematics and philosophy. (Perhaps it's the advantage of being the Head Linguist around here: even if I'm wrong nobody else can say it better.
)In macrocosmology (the Big Bang, the radius of the Hubble volume, etc.) we find ourselves veering off into philosophy (what happened before the Big Bang? what's outside the Hubble volume?) But in microcosmology, it appears that we are veering off into mathematics, where we are dealing with pure abstractions like probability theory.
In both cases we see physics receding in our rear-view mirrors, reminding us that cosmology is not exactly a science. As I've suggested before, it's theories may eventually be found robust enough to be listed in the Canon of Science, but with tiny asterisks next to them.
Mm, interesting. I'm not sure the answers really explain my question, but they put the problem in a different perspective I hadn't thought of before. kurros, thanks for the insight on quantum tunneling and quantum fluctuations in the -I presume you mean binding energy- nucleus.
Hmm, I'm not actually sure of the relationship between the binding energy of the nucleus and the sketch I included. It is a bit complex because binding energy by itself doesn't tell you anything about the stability of a nucleus, there is a bunch of junk about how nucleons arrange themselves in "shells", in an analogous fashion to electron shells. For instance Helium has a low binding energy, Iron has the maximum binding energy (which explains why it is very stable), but Uranium also has a low binding energy, but higher than Helium. So you see the stability is all over the place. The barrier in the diagram is called the activation energy, and works pretty much the same way as in chemistry, except that the energy to overcome the barrier comes from quantum fluctations rather than heat or a catalysing material.
Actually I guess you could say the barrier was more or less the binding energy between (say for alpha decay) the alpha particle and the sibling nuclei, but this is not quite true because there are entropy factors at work, i.e. the alpha particle could be made of any of the nucleons, not some chosen 4. But it is something like that.
A more appropriate picture for an alpha decay potential is this:
Yes, but how does that relate to the half-life: i.e. what makes it decay at varying rates? Different alpha-emitters have vastly varying half-lifes, even though their potential wells (as depicted in your diagram above) are all about the same (high-Z nuclei with high Coulomb barrier).
It is far more subtle than you imply by your diagram, Prom. Currently, we have no model that accurately predicts half-lifes -- rather we have vague models that have broad expected ranges.
It is far more subtle than you imply by your diagram, Prom. Currently, we have no model that accurately predicts half-lifes -- rather we have vague models that have broad expected ranges.
I haven't said or implied anything about half life. All I was doing was providing a better picture than the one above.
Didn't I explain that already? Even though the potentials have a similar form for different nuclei they are different in the details, i.e. different heights and widths, which affect the tunneling probability and decay rate. Sure it was imprecise but I didn't think anyone was looking for rigor.
Cheers for the better picture, the other one was pretty vague. Also looks like I was wrong that the unbound state was more stable, the short range potential looks like it goes right down to negative infinity. I think for nuclei with more angular momentum there may not be a pole at zero distance, can't remember though. Anyway I guess this is where entropy comes in, even though the bound state is more stable, once a decay occurs it is extremely unlikely that it will ever reverse due to the vastly greater number of possible unbound states (i.e. the alpha particle can shoot off to anywhere).
Ahh, actually, even though the potential has a pole, the energy of the nucleons remains high due to the uncertainty principle (they are confined in a small space, so they have high uncertainty in their momentum, i.e. they rattle around with a lot of energy in there), so they sit up quite high in the well. This is why E in Prom's diagram is labelled so high.
John M: others have sufficiently answered this but if you're looking for a real-world analogy, consider the decaying particles as popcorn kernels. Let's say each individual kernel has "some percentage" chance of popping while in the microwave after 10 seconds. Put a bunch of them together and you will see what appears to be heavy popping at the beginning of the cooking with a gentle tapering of popped kernels near the end; you may in fact get an occasional pop hours, days, or even years after you've started (not very likely in this analogy, but just roll with it). This is all a function of the kernels exhibiting that "percentage chance" of popping after 10 seconds (see: Poisson Distribution). In this analogy, as with decaying matter, the half-life behavior is an emergent property that comes from combining the law of large numbers with the Poisson distribution of the "firing event". It has nothing to do with the fact that the elements are clumped together.*
*This may not be true with popcorn because some the kernels may have a heating effect on neighboring kernels, which may increase their chances of popping.