22
$\begingroup$

When a free neutron decays, it is transformed into a proton, an electron, and a neutrino. Does this electron begin to "orbit" the proton, forming a hydrogen atom? Or does the electron run off and do its own thing like the neutrino?

$\endgroup$
2

3 Answers 3

19
$\begingroup$

Since I have already gone to the trouble of computing this value to great detail, I might as well present it.

Matt Hanson cited a relevant Wikipedia article, but it links to another that is even more relevant, and it even states that the following only happens about 4 times in a million, i.e. extremely rarely (as should be expected).

I did not expect this case to be possible either, since there is actually a lot of energy available to ionise any such Hydrogen atom (and as mentioned, it usually does so, i.e. proton, electron and anti-neutrino flying apart from each other), so it is worthwhile checking that this decay mode is actually possible, rare that it may be. The SR 4-momentum computation relevant to this decay is, with $c=1$ $$ \begin{align} \begin{pmatrix}m_H+\Delta\\0 \end {pmatrix}&= \begin{pmatrix}\sqrt{m_H^2+p^2}\\-p \end {pmatrix}+ \begin{pmatrix}p\\p \end {pmatrix} \end {align} $$ where the mass of the ground state H atom is $m_H=$ 938.7830735MeV, and this is the mass of proton plus mass of electron minus the binding energy 13.6eV between them. $m_H+\Delta$ is the mass of the neutron, which thus defines that $\Delta=$ 0.7823470157MeV.

These will thus immediately fix the momentum = energy = $p$ of the anti-neutrino (I assumed it is massless, which is by far a reasonable approximation here because of how inconsequential it will be), and this gives us $p=$ 0.7820212976MeV. This is the energy that the anti-neutrino carries away, leaving just 325.7180673eV for the ground state H atom to recoil away with. The ground state H atom is thus going to recoil away non-relativistically.

Very little will change in this computation if you assumed that the H atom is excited rather than ground state. It will just be slight differences in the $m_H$ value being used, thereby inducing small changes in all the other values, since the mass of the neutron is fixed.

Anyway, it is surprising that this decay mode is even possible, so thanks for the interesting question.

$\endgroup$
6
  • 2
    $\begingroup$ We still haven't observed it, given the difficulty of detecting low energy neutral hydrogen atoms. But here is a proposal to do so: arxiv.org/abs/2210.02314 It would be trivial to hydrogen spectral lines since the atom is moving at only 326 eV (it will have plenty of time to de-excite before it flies away). I guess decay to an excited H atom is extremely rare (much rarer still than the 4ppm hydrogen decay)? $\endgroup$ Oct 14, 2023 at 2:08
  • $\begingroup$ Thanks, @KevinKostlan, but a nitpick: the difficulty is in detecting extremely low concentrations, few atoms worth of these decay photons, that can decay away in all directions and thus isn't easy to focus into an image. Experiments are going to be extremely difficult to do in such a situation. I'd not want to try to do it. hehe $\endgroup$ Oct 14, 2023 at 2:15
  • 1
    $\begingroup$ Kudos to improving my answer with actual relative probabilities! $\endgroup$ Oct 14, 2023 at 2:25
  • 4
    $\begingroup$ Ha, as a chemist I was actually surprised the result is less common! I suppose we are not particularly used to the insane energy scales at play in decay reactions. $\endgroup$ Oct 14, 2023 at 2:37
  • 4
    $\begingroup$ @MattHanson ha, that just means ya chemists dont have enough explosions in da labs $\endgroup$ Oct 14, 2023 at 8:44
4
$\begingroup$

The simple answer is that it is quite normal for free neutron decay to produce hydrogen. As noted in the relevant Wikipedia article, if the electron does not have sufficient energy to escape the electromagnetic attraction of the nucleus, it will remain bound and produce a stable hydrogen atom. In this case, effectively all the rest of the energy is carried off by the neutrino. The resultant hydrogen atom is not even moving at an abnormally fast speed, so this process is not remarkable.

A comment on the “orbit” phrasing is in order, though. The Bohr model is fundamentally and spectacularly wrong. The electron is not a little ball orbiting the nucleus like a planet. If it were, it would be constantly losing energy due to loss of electromagnetic radiation and would immediately spiral into the nucleus forever. The behavior of an electron is at least non-relativistically modeled via the Schrödinger equation, which describes the location of the electron as a wave that is smoothly smeared out over space. We call the spatial distributions of the electrons in the hydrogen atom “orbitals” as a nod to Bohr’s model.

$\endgroup$
2
  • 1
    $\begingroup$ "if the electron does not have sufficient energy to escape the electromagnetic attraction of the nucleus, it will remain bound and produce a stable hydrogen atom".Which always has? $\endgroup$
    – Cerise
    Oct 13, 2023 at 11:33
  • 1
    $\begingroup$ Minor side note we alway forget: the proton is also in orbital just like the electron....it's about $30\times$ bigger than the proton. $\endgroup$
    – JEB
    Oct 13, 2023 at 16:19
-3
$\begingroup$

No. The energy of the electron is just too big for a bound state (hydrogen atom) to be created. It flies away from the proton, just like the antineutrino.

$\endgroup$
1
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Oct 14, 2023 at 4:16

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.