In reading about nuclear beta decay:

$$n \longrightarrow p + e^{-} + \bar \nu$$

It occurred to me that two of the particles resulting from this decay are the constituents of the hydrogen atom. So why do we never see

$$n \longrightarrow H + \bar \nu$$

where $H$ is a hydrogen atom? Can a neutron turn into a hydrogen atom?

  • $\begingroup$ This closely related question contains an answer with a link to a 2014 literature review; I don’t know whether any search for bound neutron beta decay has succeeded since then. $\endgroup$
    – rob
    Dec 12, 2020 at 3:50
  • $\begingroup$ thanks.but this has never been measured? $\endgroup$
    – user280085
    Dec 12, 2020 at 4:06

4 Answers 4


During nuclear beta decay, the proton stays bound to the nucleus while the electron and antineutrino $\bar \nu$ are emitted with a high kinetic energy. So, in such processes, the proton will not combine with the ejected electron to form a hydrogen atom.

But it is thought to happen rarely for free neutrons, in a process called bound beta decay.

The article here talks about bound beta decay, where it is proposed that for every one million free 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 (hence the term "bound beta decay"). 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.

In the following paper bound nuclear beta decay and the formation of hydrogen, details an experimental method to detect and observe bound beta decay, or BoB. 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.

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

  • 1
    $\begingroup$ This answer refers to “two-body decay” and “bound beta decay” as if they are different things, which I think is incorrect: the Wikipedia article simply reports the predicted branching ratio without mentioning whether the decay mode has been detected, while the McAndrew article describes a measurement effort. Furthermore, the bound-decay kinematics are fundamentally the same in heavier nuclei. For example, $\rm^3H\to{}^3He\nu$ might have a larger branching ratio than $\rm n\to H\nu$, since the decay energy is 50x smaller and the ionization energy well is twice as deep. Downvoted for now. $\endgroup$
    – rob
    Dec 12, 2020 at 5:28
  • $\begingroup$ That ... didn’t really address either of my complaints, so I wrote a complementary answer. $\endgroup$
    – rob
    Dec 12, 2020 at 6:56
  • $\begingroup$ I’m not convinced that you are aware what the op is really asking. $\endgroup$
    – joseph h
    Dec 12, 2020 at 7:14
  • $\begingroup$ @rob I can see how you may see it that way. To be clear "two-body decay" and "bound beta decay", refers to a process where we have "two bodies" as the resulting particles. So, for nuclear beta decay, nucleus(neutron)$\rightarrow p^+ + e^- +\bar\nu$ the proton resides in the nucleus, so we have a free electron and antineutrino. I made an edit. One thing regarding terminology still concerns me: Is the author referring to free neutron decay as two body beta decay? Is this what you are suggesting too? $\endgroup$
    – joseph h
    Oct 13, 2023 at 23:55
  • 1
    $\begingroup$ No. The "bound" in "bound beta decay" refers to the electron in the final state remaining bound to the nucleus, so that there are two objects (atom plus neutrino) in the final state, rather than the usual three (ion plus electron plus neutrino). There is never a question of the baryon in a beta decay being ejected from a nucleus, because beta decay energies are much smaller than nucleon separation energies. $\endgroup$
    – rob
    Oct 14, 2023 at 5:09

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.

  • $\begingroup$ I want to read about the experimentalist rabbit hole that was deleted. $\endgroup$
    – alessandro
    Dec 15, 2020 at 18:09
  • $\begingroup$ Too ill-formed to write down. Find your local precision-measurement nuclear physics nerd and buy her a beer. $\endgroup$
    – rob
    Dec 15, 2020 at 18:51
  • $\begingroup$ I don't know, the bars are all closed around here. I was thinking you could do this with Cyclotron Radiation Emission Spectroscopy. Look at the tritium beta decay electron energy spectrum for a sharp cutoff at low energies. Then see if that cutoff agrees with the ionization energy for 3He. If they don't agree you probably forgot to account for something. $\endgroup$
    – alessandro
    Dec 15, 2020 at 19:27
  • $\begingroup$ For some idea of the challenges, read about the KATRIN tritium-decay spectrometer, which is searching for a massive-$\nu_e$ correction to the high-energy end of the spectrum. Everything at low energy is harder. A positive detection of the neutral daughter atom with the right kinetic energy would be much more convincing. $\endgroup$
    – rob
    Dec 15, 2020 at 19:57
  • $\begingroup$ I think it's easier to spot the low energy cutoff than it is to measure the high energy cutoff (which is continuous) with the cyclotron technique. With a 1 T field there's about 1.3 MHz difference in cyclotron frequency between a ~0 eV electron and a 24 eV electron. And no matter the electron's velocity you're going to get a ~30 GHz signal. I agree, trying to do this with KATRIN would be very difficult indeed. $\endgroup$
    – alessandro
    Dec 15, 2020 at 20:25

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).

  • $\begingroup$ It's not certain that the antineutrino has zero rest mass in reality. AFAIK the mass is currently assumed to be between 0 and 1,1 eV. Of course the possible small mass doesn't change very much in regard to the question. $\endgroup$
    – cg909
    Dec 13, 2020 at 0:37
  • $\begingroup$ @cg909 I did not know that; thanks for the information. $\endgroup$
    – John Darby
    Dec 13, 2020 at 3:16

Neutron decay gives 0.782 MeV, in a form of kinetic energy of its parts.

The hydrogen ionization energy is 13.6 eV.

So, decayed parts have about a 50k times more energy than a hydrogen can tolerate before ionizing. And momentum conservation will make these particles fly apart, further from each other. Since they do not stay together, they are not called hydrogen.


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