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If at high temperatures atoms are more intensely interacting with each other or emitted photons that also could make the core vibrate. Is in these circumstances the radioactive material more likely to fission faster? Can this be used to get rid of radioactive garbage?

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In the years following the discovery of radioactivity, physicists and chemists (recall that Rutherford was given the Nobel prize for Chemistry!) investigated the effect of heating radioactive substances. They could detect no effect on the activity, and therefore none on the half life. This was interpreted (as soon the atom had been established as a nucleus surrounded by electrons) as evidence that the radiation came from the nucleus.

The argument was – and still is – that even at furnace temperatures (say up to 3000 K) there will be disturbance to the electron configurations but it will be rare for atoms to be totally stripped of electrons, and violent internuclear collisions will be very rare. Only such collisions would be likely to influence the emission of a particle from an unstable nucleus.

At much higher temperatures and densities (e.g. in a tokamak or in a star) violent internuclear collisions will be common, and I'd guess that the half lives of unstable nuclei would be reduced, but this is not, as far as I know, detectable at 'ordinary' terrestrial temperatures.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Aug 17 at 21:31
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    $\begingroup$ To add some numbers kT at room temperature (~300K) is about 25meV (0.025eV). The binding energy of a uranium nucleus is around 7MeV (7,000,000eV). For ambient temperature to approach the energy density required to dissociate a nucleus, the temperature would need to be not 300K or 3000K, but about five orders of magnitude higher, closer to 30,000,000K. 30 million degrees is pretty hot - and I think it gives a good sense of the scales we're talking about. $\endgroup$ – J... Aug 18 at 15:55
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There are already two good and correct answers. Especially considering that the OP mainly asks about fission processes, these answers capture the main physics. I would just like to point out that there exist decay processes in the nucleus that do get affected by temperature, even at room temperature scale.

A prominent example are the famous Mössbauer nuclei, which feature recoilless gamma-decay. Let us look at a typical example isotope and it’s decay chain. 57Co decays radioactively (indeed by electron capture, which was given as another example in another answer) to 57Fe. What’s cool is that it ends up in an excited nuclear state of 57Fe, which subsequently decays by releasing a gamma-photon.

These transitions are used in Mössbauer spectroscopy and have many applications. One is to study phonon spectra and lattice vibrations, which are strongly affected by temperature.

For example the so called Lamb-Mössbauer factor is often directly temperature dependent, and is in turn directly related to the broadening of the natural line width and hence to the half-life/decay time.

Note that this effect does not come from a direct influence on the nucleus, but from an influence on the decay channels and the resulting nuclear recoil. This explains why the energy scales of the temperature variation do not have to be the nuclear ones.

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  • $\begingroup$ I like this answer and explanation very much! I suppose if you heated the material sufficiently that its thermal blackbody radiation extended to X-rays, you could get <sup>178m2<\sup>Hf to decay faster, mentioned in this answer to How did Northrop Grumman propose to make the Global Hawk nuclear powered? Also see Hafnium controversy and Induced gamma emission Note that one has to carefully separate false claims from the science $\endgroup$ – uhoh Aug 17 at 0:34
  • $\begingroup$ @uhoh Thanks for the kind words! I’m currently reconsidering if my answer actually addresses the question or is rather misleading for readers. After all, the half-life is not actually changed in this process, it is only the decay products where you see a change. I will leave it up for now but might delete. More feedback welcome. $\endgroup$ – Wolpertinger Aug 17 at 9:03
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    $\begingroup$ Well the answer is well-received and I think your disclaimer in the beginning is quite clear, so since the community values your answer and you've explained its limitations, there is no Stack Exchange reason to delete it. Also I think you could cook up some effect where you have a metastable isomer and some off-tuned X-rays, and if you heat it up the doppler broadening will bring some of the X-rays to the right frequency to stimulate decay, thereby increasing the decay rate and shortening the half-life of the metastable state while X-rays and heat were applied. $\endgroup$ – uhoh Aug 17 at 10:52
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    $\begingroup$ @uhoh you’re right, I’ll keep it :) Also about that x-ray plan of yours: using laser fields for nuclear transmutation and the like is quite an active field of research. It may be possible with upcoming laser facilities but to my knowledge it is not at the moment. See e.g. arxiv.org/abs/2004.07953 $\endgroup$ – Wolpertinger Aug 17 at 11:23
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    $\begingroup$ Also note my answer here: physics.stackexchange.com/a/574060/101770 which provides another example of a nuclear transition that would be relevant for this question, since it is affected by temperature. $\endgroup$ – Wolpertinger Aug 18 at 9:59
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You seem to be confusing two separate concepts. The half-life of a radioactive isotope gives the rate at which individual atoms will spontaneously decay. The likelihood that a fissile material will undergo a chain reaction is quite different from its half-life.

For most modes of radioactive decay the half-life of a radioactive isotope is independent of environmental factors such as temperature, pressure, chemical bonds, electric or magnetic fields. This has been confirmed by very accurate experiments.

The only known exception is that some modes of radioactive decay that involve the electrons in the atom (such as electron capture) are slightly affected by chemical bonds which may change the shape of the electron shells around an atom. For more details see this Wikipedia article.

What is dependent on temperature (and on many other environmental factors) is the neutron cross section of a fissile material - the probability that a neutron emitted in the decay of one nucleus will interact with another nucleus. This in turn determines whether or not a chain reaction will take place.

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The other answers have come up with a few exotic cases where external factors such as temperature can affect some aspects of nuclear processes (neutron capture cross-section). However, the overall answer is no, temperature does not affect half-life of an isotope.

To expand on why there is no effect, consider that (as you mention in your question) what we perceive as temperature is actually vibration of atoms. You can calculate the vibrational energy of atoms at various temperature and you will find that for typical temperatures achieved in chemical reactions, the energies are of the order of several electron-volts (eV). Nuclear reactions, on the other hand, occur at energies of a few mega-electron-volts (MeV).

So nuclear reactions are around six orders of magnitude more energetic than chemical reactions.

However, there is a way to accelerate nuclear decay by adding energy. You just have to add energy at the scale of MeVs. You can do this using an intense particle beam. The idea is theoretically sound, but it has not yet been experimentally developed.

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    $\begingroup$ The energy of the nuclear reaction is not the correct scale for the question. The half-life is related to the width of the transition in energy space, which can be very narrow compared to the transition energy. Since nuclear decay processes span many orders of magnitudes in lifetimes, this argument seems to be coming from the wrong angle. E.g. say you have an MeV decay process and you shift/broaden it by some eV, but the width is only neV, this would be a huge change. $\endgroup$ – Wolpertinger Aug 17 at 17:52
  • $\begingroup$ @Wolpertinger The OP is wondering if you can trigger nuclear reactions by heating up isotopes. He doesn't actually mention the scale, but we assume he's talking about furnace-like conditions. As I think we agree, this won't work and the reason it won't is that there isn't anywhere near enough thermal energy to overcome the potential barrier for radioactive decay. So I'd dispute saying that the decay energy isn't the correct scale to use. $\endgroup$ – Oscar Bravo Aug 18 at 12:23
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There is a relativistic effect.

According to special relativity, a (relatively) moving clock ticks slower. What this means is that at high speeds, a particle will survive a little bit longer on average before decaying.

At a higher temperature, the particles in a gas will be traveling faster, so they will decay a little bit slower. The effect will be really small until their speeds approach an appreciable fraction of the speed of light.

I have only heard of this effect being observed in particle accelerators and cosmic rays. The theory should hold if you could heat up a gas enough that the relativistic effects became observable (which is difficult, to say the least), but at that temperature you're going to have all kinds of other nuclear effects.

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    $\begingroup$ According to en.wikipedia.org/wiki/Relativistic_plasma you need 5.6 trillion kelvins to make hydrogen plasma that's hot enough for around 10% of the protons to have $\gamma=2$, i.e., kinetic energy equal to their rest mass-energy. If your plasma is made from heavy elements, the temperature needs to be even higher. At those temperatures, a lot of nuclear reactions are going to happen, eg photons causing pair production & photodisintegration, electrons converting protons to neutrons, nuclei fusing, etc. $\endgroup$ – PM 2Ring Aug 17 at 17:33
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    $\begingroup$ Yeah it's practically impossible, but the effect is still there even if it's too small for us to notice. $\endgroup$ – AlexQueue Aug 17 at 19:54

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