Is there radioactivity at absolute zero?

Question

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)?

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? — Mithoron, Oct 19 '18 at 19:16
@Mithoron I would think radioactivity and nuclear phenomena are borderline, but would work here in a lot of cases. — Tyberius, Oct 19 '18 at 21:45
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. — matt_black, Oct 20 '18 at 19:32

score 43 · Answer 1 · answered Oct 19 '18 at 11:49

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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.

answered Oct 19 '18 at 11:49

Ivan Neretin

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I meant nuclear vibration – Harsh jain Oct 19 '18 at 13:20
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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. – Ivan Neretin Oct 19 '18 at 13:21
Do you mean nuclear internal degrees of freedom when you say that nucleus sits in the ground state? Or do you assert that there're no phonons at room temperature? – Ruslan Oct 19 '18 at 14:25
I mean internal processes, of course. Phonons do exist (and then again, some things are liquid or even gaseous at room temperature, so their atoms move even more), but that's irrelevant to radioactivity. – Ivan Neretin Oct 19 '18 at 14:26
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This has been covered on Physics SE a few times. – Jon Custer Oct 19 '18 at 15:34
Would have been surprising if it hadn't. – Ivan Neretin Oct 19 '18 at 15:36
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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? – Klom Dark Oct 19 '18 at 16:02
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@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. – Oct 19 '18 at 17:11
Dividing the temperature of the core of the Sun by room temperature yields a pretty big number, but how does that compare to dividing room temperature by zero? – Solomon Slow Oct 19 '18 at 22:03
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@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. – Ivan Neretin Oct 19 '18 at 22:05
This is a bit errant. The nucleus is not internally excited, but it is moving as a complete unit as its attendant atom vibrates. Indeed, if it could not do so, I could not throw a block of matter with nuclei in it, as they could not move through space as part of the block. Motion of the atom in thermal vibration is still motion. – The_Sympathizer Oct 21 '18 at 04:54
The reason it can move as a whole without being internally excited is in fact the same reason that cooling something down will not affect the radioactivity. – The_Sympathizer Oct 21 '18 at 04:54
I know little about nuclear physics, but a little intuition suggests your answer is wrong. A system at 0K is incapable of work, because it has zero entropy. Since nuclear decay can be used to perform work, this means a radioactive material is not at 0 K. qed. – Buck Thorn Nov 14 '18 at 19:20
Ok, but the implications of what I laid out are interesting: it implies (obvious perhaps in hindsight) that a radioactive material has residual entropy, due to the structure of the nucleus. Care to comment? Congratulations on being right, by the way! – Buck Thorn Nov 15 '18 at 15:25
"Then, and not before, your nuclear processes will show first feeble indication of thermal dependence of any sort." Why should the energy of the nucleus show any dependence on the electronic structure? – Buck Thorn Nov 15 '18 at 15:29
Many things have some residual entropy, due to nuclear or other reasons. I don't see how is that relevant. 2. The energy of the nucleus does not depend on the electronic structure.

Ivan Neretin

Nov 15 '18 at 15:31

There are a few examples of substances exhibiting interesting properties near 0 K due to residual entropy, or despite zero entropy (liquid $He^{II}$). It's been a potential source of confusion for chemists for instance during the application of the 3rd law. It's not relevant to radioactivity, but it is a thought exercise. For instance, another answer suggests applying pressure to contain nuclear decay. But at 0 K, P goes to zero.

– Buck Thorn Nov 15 '18 at 15:40

The_Sympathizer · Answer 2 · 2018-10-21T05:06:08.127

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.)

score 3 · Answer 3 · answered Mar 14 '19 at 15:56

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.

KDodd · Answer 4 · 2019-04-09T03:19:33.267

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.
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.

score 0 · Answer 5 · answered May 10 '23 at 05:08

I think the premise you have that absolute zero is somehow decreasing energy within a system is true to a point. We are removing energy from the system at the macro level but the conditions of an atom for example blend to a single state as we move near absolute zero.

Strings, or whatever current theory says is at the heart of matter spread out to a resonating mass with but a single state.

Its at this point that many of us think is the key to readjusting proton counts in the effort to neutralize radiative material. Making matter stable or configure din such a way as to be more controllable, creating elements that are stable at higher counts of the periodic table.

Is there radioactivity at absolute zero?

5 Answers5