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Theoretically, will a radioactive material still be radioactive at absolute zero? What would happen at the lowest realistic temperatures we have ever achieved?

Will radioactivity stop at absolute zero, since it is a nuclear phenomenon and nuclear motion slows down as we approach absolute zero (and theoretically stopping entirely at absolute zero)?

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  • $\begingroup$ Your premise is that the nuclear decay is supposed to be caused by their kinetic energy? Nuclei clashing with each other or what, because this seems rather unclear? $\endgroup$ – Mithoron Oct 19 '18 at 19:16
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    $\begingroup$ @Mithoron I would think radioactivity and nuclear phenomena are borderline, but would work here in a lot of cases. $\endgroup$ – Tyberius Oct 19 '18 at 21:45
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    $\begingroup$ @Tyberius Yeah, maybe "unclear" is better reason here. $\endgroup$ – Mithoron Oct 19 '18 at 22:11
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    $\begingroup$ Thermal vibration in molecules (which could be described as "nuclear vibration") is irrelevant to radioactivity which depends on internal quantum processes in the nucleus that have no meaningful dependence on temperature. Or any notable dependence on any other chemical process. $\endgroup$ – matt_black Oct 20 '18 at 19:32
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Theoretically, a radioactive material will still be radioactive at absolute zero, and its rate of decay will be $100.00\%$ of that at room temperature. Practically, at the lowest achievable temperatures we observe the same thing: radioactivity is still there, not affected the slightest bit.

Nuclear motion does not slow down as we approach absolute zero, because there is no such thing as nuclear motion in the first place. In a way, all nuclear motion has stopped already at room temperature. Each nucleus just sits there in the ground state and does not know what happens in the chemical world above. From its point of view, the room temperature is the same as absolute zero. To reach its first excited state, it would need energies a great deal greater than that.

Say, you heat your radioactive sample until it melts. Then you heat it a few more thousand degrees, until all materials, including tungsten, melt and then evaporate. Then you heat it some more, until even the strongest chemical bonds are broken and there are no more molecules, just atoms. Then you heat it about ten times more, until atoms lose much of their valent electrons and you have a highly ionized plasma. Then you heat it about a hundred times more, until all atoms lose all their electrons and you have something like a stellar plasma. Then you heat it some more, just in case. Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort.

Short of that, you could just as well have asked if the radioactivity in a sample stops when you paint it blue.

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    $\begingroup$ Yes, I got that. What I said still applies. Then again, you might point out that atoms do move, and so do nuclei, being part of them. That's right, but that's irrelevant. $\endgroup$ – Ivan Neretin Oct 19 '18 at 13:21
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    $\begingroup$ This has been covered on Physics SE a few times. $\endgroup$ – Jon Custer Oct 19 '18 at 15:34
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    $\begingroup$ I'd just like to point out that hyper-heating a sample really seems to get as far away from proving the point as possible. Would be better to say something about why it would retain its radioactive properties at absolute zero. Sure, chemical processes stop there, but then why don't the quantum properties also stop? What are the rules in play that make radioactivity not stop at absolute zero? $\endgroup$ – Klom Dark Oct 19 '18 at 16:02
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    $\begingroup$ @KlomDark It sort of does prove the point. What Ivan is saying is that the temperatures required to start affecting nuclear processes is very high. He is saying that absolute zero is effectively the same as room temperature on the scale of nuclear processes. Absolute zero does not mean zero energy in the ground state. Ivan is saying that temperatures required to bring nuclear states out of ground state are very very high. So absolute zero is effectively the same as room temperature. $\endgroup$ – aidan.plenert.macdonald Oct 19 '18 at 17:11
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    $\begingroup$ @SolomonSlow Who said you should divide? Who said it should make any sense? In my book, you should compare things like $e^{-{E\over RT}}$, and those behave quite differently. $\endgroup$ – Ivan Neretin Oct 19 '18 at 22:05
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Yes, it will continue to be radioactive. This is because the instability of the nucleus is due to the balance of forces internal to it, not to the motion of it as a whole object. It's just like how that throwing a red ball doesn't stop it from being colored red anymore while it is in motion. Likewise, no chemical process can stabilize a radioactive nucleus because chemistry deals with how the atoms containing the nuclei fit together due to interactions of their electron clouds, and this is another factor external to the nucleus itself and so does not affect that instability originating from within.

That said, in theory it is possible to stabilize atomic nuclei with the application of pressure, as it is a question of balance of forces and thus with enough force you can squeeze on the nucleus and keep it together (roughly speaking - from a quantum mechanics POV it's more like "crowd out all the other energy states so that it cannot decay without entering one that's already occupied by somebody else".). But the pressure required here is profound: it is comparable to those in a neutron star, $10^{30}\ \mathrm{Pa}$, around which any more would cause it and the other nuclei in the heavily compressed matter to "melt" into a stew of neutrons and (some) protons. Within a neutron star, there is a layer near the surface - shortly before you get to the neutron stew - of nuclei too heavy to exist at ordinary pressure without being extremely radioactive, but here are stable and persistent.

(There is also one other way you could "stabilize" a radioactive material and that is if you send it off moving through space close to the speed of light. There, time dilation will expand the apparent half-life of the constituent nuclei relative to you on the ground. But you will, of course, not be able to use it for anything while it's flying away from you at extreme speed. No internal properties of the material have been changed, so in its own frame it is still as unstable as ever, but the geometry of spacetime allows you to experience its evolution as being slowed down. This can be observed realistically with unstable single subatomic particles like muons, and these observations provide a good test of Einstein's special theory of relativity.)

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  • $\begingroup$ "a stew of neutrons and (some) protons" -- don't forget embedded electrons, which form with the protons via the (reversible, inside the neutron star) neutron decay and maintain conservation of charge. $\endgroup$ – Oscar Lanzi Mar 14 at 15:30
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Another way of lookung at this problem is to identify nuclei, at least heavier ones, as having their own internal temperature. Dynamic processes take place within the nuclei, and the energy exchanged in these processes corresponds to roughly $10^{11}$ Kelvins when translated to a degree scale -- a temperature locked within the nuclei and not generally seen by us outside observers. Compared with it, the temperatures we do access, driven by long-range but inherently weaker interactions, are just a tiny drop in a huge bucket, scarcely impacting the equilibrium within the nucleus.

We can alter the equilibrium by altering the externally applied temperature on this scale. Say we use adiabatic compression. Inside the Earth gravity produces pressures in the megabar range, between $10^{11}$ and $10^{12}$ pascals. But the corresponding temperature there is only a few thousand Kelvins, not enough to materially alter what goes on within the internally much hotter nuclei. To get the right order of magnitude temperature effect requires those neutron star pressures described by The Sympathasizer.

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  1. Radioactive materials by definition cannot reach absolute zero since internal energy in/out is non-equilibrium. (First Law of Thermodynamics) The stellar events that create these internal nuclear energies and their inherent quantum properties can't be overcome by simply reducing electron energy to ground state.

  2. The only way to stop the decay of radioactive materials is conversion to pure energy (#), or implementation into a system of extreme gravity that nuclei escape velocities cannot be exceeded by decay or emission velocities (*).

(#) Reaction by conventional nuclear fission/fusion/fission devices results in less than 1 gram per 2200 grams of Uranium or Plutonium isotopes converted into pure energy.

(*) It has been observed that in these types of extreme systems, matter and energy is converted to gamma ray emissions that violently exceed Black Hole, Quasar, Pulsar and Magnetar gravity.

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