1
$\begingroup$

My understanding is that, if an atom finds itself in an excited state, for instance following electron capture and gamma emission, the atom relaxes to the ground electronic state because of the perturbation caused by the room temperature. The atom may temporarily borrow energy in order to relax, if you like.

So in the absence of the kick provided by the temperature of the room, would excited electronic states be stable?

Radioactive decay is spontaneous because essentially, everything is at absolute zero for nuclear relaxations, because of the large energy changes involved in transitioning between nuclear states.

This is of course not true in stars.

Thanks in advance, I look forward to hearing the answer.

$\endgroup$
1
  • 1
    $\begingroup$ I don’t think a perturbation is necessary for spontaneous emission. For the transition to occur there must be wave function overlap between the initial and final states. en.m.wikipedia.org/wiki/Fermi%27s_golden_rule $\endgroup$
    – boyfarrell
    Mar 12, 2021 at 6:44

1 Answer 1

0
$\begingroup$

Usually, radioactive decay has nothing to do with temperature and does not at all depend on it. (*)

Radioactive transitions are probabilistic as they are best modeled just like the tunneling effect. There is some potential barrier, but the (quantum) state of the nucleus can tunnel through that and relax to another state. This has nothing to do with thermal energy at all. Thermal motion is kinetic motion of the entire nucleus, and thus does not provide any additional energy for the tunneling process. In the OP's picture, thermal motion kicks the entire nucleus, not just individual nucleons.

Also, typical kinetic energies at room temperature are k$_B$T $\sim 10^{-4}\mathrm{eV/K} \cdot 300\mathrm{K}\sim 10^{-2}$eV which is orders of magnitude below typical decay energies $\gg$keV. From this you also get the caveat to the initial statement: (*) ...Unless you go to crazy high temperatures and densities, or highly ionized states of atoms.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.