Nuclear beta decay to hydrogen

This obviously cannot happen[1] in nuclear beta decay, since the proton stays bound to the nucleus while the electron and antineutrino $\bar \nu$ are emitted with a high kinetic energy. So the proton cannot combine with the ejected electron to form a hydrogen atom.

But this can and does happen rarely for free neutrons and not neutrons which are bound to a nucleus.

This article here talks about this process called free neutron decay. As per the link, for every one million neutron decay events, on average only four will result in the formation of a hydrogen atom. In such cases, the electron resulting from the decay has energy smaller than $13.6 \ eV$ (binding energy of electron in hydrogen atom) and so can bind itself to the proton.

But in a significant majority of free neutron decay events, the energy of the resulting electron has energy $\approx 0.80 \ MeV$ which is significantly higher than that for the binding energy for the proton + electron state mentioned above.

[1]Having said all this, I noted in a comment above by rob, that there is a hypothetical and non-verified mechanism for bound nuclear beta decay and the formation of hydrogen. The abstract reads

For many years neutron decay has been investigated as a possible pathway to the exploration of new physics. One such example is the bound beta-decay (BoB) of the neutron into a hydrogen atom and an anti-neutrino. This two-body decay mode offers a very elegant method to study neutrino helicities, just as the Goldhaber experiment has done. However, this rare decay has not yet been observed so far owing to the challenges of measuring a decay involving only electrically neutral particles with an estimated branching ratio of only 10-6 of the three-body decay mode. Specifically an intense source of thermal neutrons would be required for such an experiment, such as the FRMII in Garching, the ILL in Grenoble or the ESS in Lund. This paper provides a summary of the novel experimental scheme that we propose to observe the BoB neutron decay, addressing all necessary problems in a very coherent way.

And in the paper

In 1947 Daudel, Jean and Lecoin predicted the existence of a two-body beta-decay mode in which the daughter nucleus and the electron remain bound (Daudel, Jean and Lecoin (1947)). For the beta-decay of the free neutron, this is referred to as “bound beta-decay” or “BoB”.

I have never noticed this area of research and it is very interesting.

It appears you are asking about the decay of a free neutron, not the beta decay of a radionuclide. Neutron decay results in the release of a proton, electron and an antineutrino each with kinetic energy, since this is an exothermic process (rest mass of neutron greater than rest masses of proton plus electron, antineutrino has zero rest mass). 0.78 MeV is the total kinetic energy of the proton, electron, and antineutrino. Since the electron has kinetic energy it "escapes" its point of origin and has a very low probability of combining with the proton to form a hydrogen atom. If the electron does not escape the surrounding medium, it will eventually be captured and form an ion within the medium (same for the proton).

When a particle at rest decays, the momentum of the fragments has to add up to zero, because momentum is a constant when there isn’t any external force. In a two-body decay this means the two fragments have equal and opposite momenta. In a three-body decay, the magnitudes of the different momenta take on different values depending on the angles between them. Computing the details of the spectrum is hard, but the hand-waving approximation is that each fragment carries about the same amount of momentum.

This means that nearly all of the energy in the decay is carried away by the low-mass electron and the ultra-relativistic neutrino: the poor nucleus only gets to carry kinetic energy $\sim p^2/2M$, while the electron gets to carry $\sim p^2/2m_e$.

The reason that we can separate nuclear physics from atomic physics is that the energy scales involved in the interactions are very different. In order to separate an electron from a hydrogen atom, you have to supply it with a minimum of 13 electron-volts (eV) of energy. But the typical energy in a nuclear decay is $10^6$ eV. So in the vast majority of decays, the electron and the nucleus go in different directions, with too much energy for the electromagnetic force to bind them.

However, there is a very small corner of the parameter space where nearly all of the energy is carried away by the neutrino, leaving the daughter nucleus and the decay electron nearly at rest. This is called a “two-body beta decay” or a “bound beta decay.” For the free neutron, whose beta-decay energy is around 0.8 MeV, the bound decay $$\require{mhchem} \ce{n \to H + \nu}$$ is predicted to occur a few times out of every million decays. This 2014 paper outlines a proposed attempt to measure it, but the experiment is tricky and I wouldn’t be surprised if there were no result yet —— they hadn’t even picked a site for the experiment. The goal would be not just to detect the rare decay mode, but to measure the total spins of the produced hydrogen atoms, which tell you in a direct way about the spins of the invisible neutrinos.

You could in principle apply the same logic to heavier beta emitters. One candidate might be bound tritium decay, $$\ce{^3H \to {}^3He + \nu},$$ where the beta decay energy is much smaller (around 15 keV) and the ionization energy well is deeper: you can imagine the odds of the neutrino carrying away “all” of the energy might be many per million decays, instead of a few per million decays. But [experimentalist rabbit hole deleted] it’s not clear to me that a higher branching ratio would immediately make for a better experiment.

You would never expect to find a decay like

$$\ce{ ^{14}C \not\to {}^13C + {}^1H + \nu }$$

because it takes at least 10 MeV to knock a proton or neutron out of a stable nucleus, and beta decays are typically not that energetic.

tl;dr summary: such decays are predicted, rare, not yet observed, but not really in doubt.