53
$\begingroup$

So, based on this question, a molecule containing a radioactive atom will break when the atom decays. But suppose you need a lot energy to break the compound apart --- as in, more energy than the decay of the atom will release (obviously, a molecule this stable isn't actually possible... right?). Will the atom just be forced to stay static, or would something else happen?

I can't think of a way for the compound to break, since that would probably need free energy. But maybe the compound can "soak up" energy, so a sharp jolt or high heat can cause the atom to decay and the bonds to break?

$\endgroup$
  • 1
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Aug 17 at 20:46
49
$\begingroup$

In principle, yes. If the would-be decay products have a higher energy than the original molecule, the decay cannot occur.

In practice, chemical binding energies (typically in the $\rm eV$ range) are much, much smaller than nuclear decay energies (typically in the $\rm MeV$ range), and so this does not occur in any cases that I am aware of. This is not a coincidence, but just a natural consequence of the relative strength of nuclear and electromagnetic interactions.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Chris Aug 19 at 5:02
60
$\begingroup$

While the arguments put forward in the other answers are in principle correct, it is important to note that nuclear decay processes span a huge parameter space, both in energy and half-life.

So to provide a counter example, let us look at the most special nuclear transition in this regard: Thorium 229, which has an isomeric state Thorium 229m, which can be obtained as a decay product of Uranium 233. The transition energy of this state is 8.28 +- 0.17 eV (source). Yes, eV! This transition is in the optical regime.

As a result, it gets affected by all kinds of electronic processes , for example, internal conversion . Also the chemical environment or rather the crystal structure is relevant (as stated here). Note that there is a huge body of literature on this subject and I only give examples here which are by no means representative of the whole work. For further reading see this and references therein.

Note that this is an extremely exotic transition, but also a very important one. Much effort is being invested into building an extremely precise nuclear clock using these nuclei.

So at least a weaker version of the question in the OP can be answered in the affirmative: there are radioactive decay processes that get strongly affected by the electronic environment.


Edit for clarity

I have been asked in the comments to clarify my answer with regards to how it addresses the question and what kind of nuclear transition we are talking about.

  • (in response to @Helen's request) My answer points out a particular nuclear transition that is affected by the electronic environment. The transition may be considered exotic, most other nuclear decay processes (especially alpha and beta decay) will not be affected in such a way, as pointed out in the currently accepted answer. Whether this constitutes a "yes"-answer to the question can be debated.
  • (in response to @Emilio Pisanty's request) The Thorium transition is a very special gamma transition of an isomeric state that has an untypically low transition energy (see Figure 3 in this open access paper for a nice visualization). Indeed, it is the lowest lying known transition. The excited state may decay also via other decay channels, such as internal conversion, where a shell-electron is ejected instead of the emission of a gamma photon. The mass and charge of the nucleus is not changed in either of these, as usual for a gamma transition.

Also check out @BCS's answer for another nice example that works via electron capture.

| cite | improve this answer | |
$\endgroup$
  • 17
    $\begingroup$ +1 because this is way more fascinating than anything I should be doing today! $\endgroup$ – uhoh Aug 18 at 10:05
  • 3
    $\begingroup$ Note that this provisional "yes" applies only to a few specific nuclear decays. There is no chemical bond that will have significant effect on the decay of Uranium or most other radioactive isotopes. $\endgroup$ – user3294068 Aug 19 at 15:36
  • 2
    $\begingroup$ @user3294068 comment upvoted, I meant to say that but could have probably been clearer. This is also what the other answers talk about, I just wanted to provide some detail on the richness of nuclear decay processes with this exotic transition. $\endgroup$ – Wolpertinger Aug 19 at 16:18
  • 1
    $\begingroup$ On one hand, I can't help admiring this answer. On the other, I can't help noticing it doesn't answer the question. (I know that the user acknowledges this, I'm just wondering why it is selected as the answer.) $\endgroup$ – Helen Aug 26 at 7:17
  • 1
    $\begingroup$ I find this answer misleading, due to the lack of emphasis on, and explanation of, the nature of the decay involved. The decay in question is not one decay channel but two: one radiative gamma-decay channel, and one internal conversion decay (unusual, and generally unknown to this thread's audience) in which the energy is used to kick out an electron from the ion. These behave very differently with respect to the OP's question: $\endgroup$ – Emilio Pisanty Aug 28 at 12:57
14
$\begingroup$

The usual answer is that chemical reactions cannot affect processes taking place inside the nucleus, because chemical processes only involve the outermost electron orbitals in the atom or molecule involved, and the nucleus is smaller than that by a factor of order ~10^-5 which means it is completely out of the picture as far as chemical reactions are concerned.

The only possible exceptions are for those nuclear processes involving electron capture, as pointed out by others here in the comments section.

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

7Be

The 2s electrons of beryllium may contribute to chemical bonding. Therefore, when 7Be decays by L-electron capture, it does so by taking electrons from its atomic orbitals that may be participating in bonding. This makes its decay rate dependent to a measurable degree upon its chemical surroundings – a rare occurrence in nuclear decay.

link

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

The answer is definitely yes in principle, because we have a perfect example from a related field -- nuclear physics. A bare neutron is unstable; it will radioactively decay (beta decay) to a proton, an electron and a neutrino and release a tiny bit of energy, with a half life of about 1000 seconds. The universe is a lot more than 1000 seconds old, so why are there any neutrons left? Because of the strong bonds between neutrons and protons in nuclei (using the strong force, not the electromagnetic force as in chemistry bonds). In the vast majority of everyday nuclei, it is energetically unfavorable for the decay to happen and make a less-stable nucleus, with too many protons and not enough neutrons. The few nuclei where this is not true are the radioactive nuclei which do undergo beta decay.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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