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I'm wondering if the temperature could affect the half-life in an element. For example, Carbon 14 has a half-life of 5,730 years. Is this always true or only true for standard conditions of temperature and pressure? Also, if that does affect, how does it affect?

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    $\begingroup$ Temperature is not very meaningful for atomic nuclei. This is because a nucleon can be in an excited state (just like an electron orbiting an atom can be excited to a higher orbital), but the energy of this excitation corresponds to at least 10^7K, normally even much higher (google nucleon resonances for more info). Therefore, for most practical purposes a high temperature will not affect the nucleon too much. But if the temperature is >10^7K, all sorts of nuclear reactions begin to happen, as they do inside stars. $\endgroup$
    – LLlAMnYP
    Commented Apr 16, 2018 at 7:15
  • $\begingroup$ @LLlAMnYP And what about very low temperatures, close to zero kelvin? $\endgroup$
    – Guilherme
    Commented Apr 16, 2018 at 7:21
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    $\begingroup$ As I said, since the minimum excitation energy for a nucleon is in the ballpark of millions of kelvin, there is virtually no observable difference for nucleons at 1, 10, 100, 1000 or even 10000 K. Only an exponentially small fraction of them (e.g $e^{-1000}$) will be in an excited state. $e^{1000}$ is a huge number, orders of magnitude more than the number of atoms in the universe. So far all intents and purposes - zero temperature. (cont'd) $\endgroup$
    – LLlAMnYP
    Commented Apr 16, 2018 at 7:26
  • $\begingroup$ I can imagine that maybe, excited electronic orbitals with their altered EM field interact ever so slightly with the nucleus, but the general behavior of a nucleus is governed by strong and weak nuclear forces, whereas this interaction is electromagnetic, so also not much of an avenue to change things. $\endgroup$
    – LLlAMnYP
    Commented Apr 16, 2018 at 7:29

2 Answers 2

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Temperature affects the half-life via time dilation. If the half-life of a nucleus as observed in its rest frame is $\tau(0)$, then in some other frame in which it moves at a speed of $v$, it will be:

$$\tau(v) =\frac{\tau(0)}{\sqrt{1-\dfrac{v^2}{c^2}}}$$

Because the typical thermal speeds of atoms at room temperature are much smaller than the speed of light, we may expand the above expression to leading order. This yields:

$$\tau(v) =\tau(0)\left[1 + \frac{v^2}{2c^2}+\cdots \right]$$

The average of the square of the speed over the velocity distribution of atoms at temperature $T$ follows from the equipartition theorem. The average energy in each degree of freedom of a gas is $\frac{1}{2} k T$, this means that the average of the kinetic energy is $\frac{3}{2} k T$, it then follows that:

$$\left\langle v^2\right\rangle = \frac{3 k T}{m}$$

Therefore, the leading temperature dependence of the half-life is given by:

$$\tau =\tau(0)\left[1 + \frac{3 k T}{2mc^2}+\cdots \right]$$

For carbon-14 at $25 ^{\circ}$ C the term $ \dfrac{3 k T}{2mc^2}$ is $2.95\times 10^{-12}$. Since the experimental error in the half life is about 0.7% this effect is too small to be observed.

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  • $\begingroup$ Very interesting. So, if I understand correctly, any stationary object can suffer time dilation via (extremely high) temperatures? $\endgroup$
    – Guilherme
    Commented Apr 16, 2018 at 8:15
  • $\begingroup$ Nevermind: physics.stackexchange.com/questions/258278/… $\endgroup$
    – Guilherme
    Commented Apr 16, 2018 at 8:18
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Pierre Curie checked this out in 1913 by cooling radium in liquid hydrogen and found that the half life was the same as room temperature...radioactive decay is independent of temperature. I am seeing other answers that provide more detail, but this I would say is the simple answer.

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