If electron fails to get into the lowest energy orbit after neutron decay does it remain bonded to the proton?

Question

I have read this on Wikipedia:

A very small minority of neutron decays (about four per million) are so-called "two-body (neutron) decays", in which a proton, electron and antineutrino are produced as usual, but the electron fails to gain the 13.6 eV necessary energy to escape the proton (the ionization energy of hydrogen), and therefore simply remains bound to it, as a neutral hydrogen atom (one of the "two bodies"). In this type of free neutron decay, in essence all of the neutron decay energy is carried off by the antineutrino (the other "body").

Does anyone know what the source for this is since none was cited, also can anyone tell me if this is true? What are the properties of the two body atom can it form $\mathrm{H}_2$?

Answer 1

The Particle Data Group lists a branching ratio of $\Gamma_{n\to\rm H\bar\nu} < 2.7\times 10^{-3}$, with the non-discovery discussed in papers from 1990 and from 2018.

(It might be polite for someone to insert one or more of these references into the Wikipedia article. The PDG is the most solid, since that group updates their literature review every couple of years.)

All hydrogen atoms are described only by their quantum numbers. Suppose you have a "decay hydrogen," which has suddenly appeared in the vacuum volume of your neutron-decay experiment, and you also have an "old hydrogen" which you liberated from some ancient cometary ice molecule. If those two hydrogen atoms have the same quantum numbers, they are indistinguishable from each other.

Answer 2

What are the properties of the two body atom can it form H2?

It is just a hydrogen atom. It has all the properties of a hydrogen atom.

A hydrogen atom is one proton and one electron in a bound state.

If you hit a hydrogen atom hard enough (e.g., with an x-ray or gamma ray beam) you can break up the bound state and create a free electron and proton. It breaks up because you impart enough energy to the electron to escape the coulomb potential of the proton. Similarly, if an electron and proton are created (as in neutron decay) and they happen to have a low relative kinetic energy then the bound state (i.e., hydrogen) could form.

This all that they are saying.

Answer 3

You ask:

Does anyone know what the source for this is since none was cited, also can anyone tell me if this is true?

The source is knowledge of kinematics and quantum mechanics: Quantum mechanics is a probabilistic theory so in the calculation of the possible kinematics of the decay there will be, by construction, a probability that the proton and electron have such a four momentum vector that they can be trapped making a hydrogen atom. This probability can and has been calculated .

In the link given by rob:

A scenario similar to that of Fornal and Grinstein was envisioned by K. Green and D. Thompson [59] for the rare decay n → hydrogen + ¯ν. They used the different effects of that decay on beam and trap lifetimes to obtain a bound of < 3% for that branching ratio (to be compared with the $4·10^{−6}$ prediction [60–63]). Our general analysis employing τ trap n and g post2002 A in eq. (12) can be used to reduce that bound by an order of magnitude to < 0.27%.

So the prediction of the calculation for the probability of a two body decay for the neutron is too small for that decay to be seen in experiments. Only limits can be given.