What happens when a neutron hits a protium nucleus?

Deuterium appears to have a positive binding energy versus a proton and a neutron. So what happens when a free neutron hits a proton? Do they fuse? If they don’t, why not, since there is a positive binding energy? Or is the neutron cross section just so small that it rarely captures a neutron?

If the neutron has low enough kinetic energy, they fuse, releasing a gamma ray:

n + p --> d + [symbol]g[/symbol]

(In particle detectors, the characteristic energy of this gamma ray is one way to tell that a neutron was around.) Other reactions compete with “neutron capture”, especially at higher energies:

n + p --> n + p (elastic scattering, like billiard balls)
n + p --> n + [symbol]D[/symbol][sup]+[/sup] (excitation of the proton to a Delta resonance (or other resonance – there are many) )
n + p --> [symbol]D[/symbol][sup]0[/sup] + p (excitation of the neutron to a resonance)
n + p --> n + p + [symbol]p[/symbol][sup]0[/sup] (non-resonant pion production, a.k.a. inelastic scattering)
and more…

I did read that the gamma production is the reason it is harder to harness the energy for commercial purposes, but it still doesnt make sense how little deuterium there is in the universe. Or do the neutrons produced by fusion just never get that far beyond a star’s core?

Lots of deuterium gets made in stars and got made in the Big Bang, but it is/was quickly used up as fuel to make helium:

d + p --> [sup]3[/sup]He, or
d + d --> [sup]4[/sup]He, or
d + n --> t, t + p --> [sup]4[/sup]He

So by the time you run out of neutrons, you’ve just got helium and whatever leftover protons you still have, but very little deuterium.

I should add:
You have fewer neutrons than protons to work with after the Big Bang because the neutron can decay and because the interconversion of p and n slightly favors n–>p over p–>n since the proton is lighter. The ratio at the time of nucleosynthesis was about n : p = 1 : 7.

Thanks: the latter fact does explain a lot of the lack of early neutrons (although I had thought of the first fact but it doesnt seem they would be alone long enough to decay.)

Does this work with H as well, or does it have to be a free proton?

I should mention that as a chemist your notation of p for H[sup]+[/sup] and d for [sup]2[/sup]H[sup]+[/sup] drives me nuts. Is this standard nuclear chemist notation?

The piddling little electron on H doesn’t cause any trouble.

I should mention that as a particle physicist, your notation of H[sup]+[/sup] for p and [sup]2[/sup]H[sup]+[/sup] for d seems unnecessarily bulky. :smiley:

This is related to the question of the electron’s influence. Specifying H[sup]+[/sup] is overly specific, since the proton may not be isolated and, in earthly applications of neutron capture, is often bound up in a complex molecule. All that matters is that there’s a proton, hence “p”. A nuclear physicist would perhaps have chosen [sup]1[/sup]H and [sup]2[/sup]H (no charge label), as they typically deal with heavier isotopes as well. But a particle physicist would go with simply “p” and “d”. Consider the two primary decay modes of the above-mentioned [symbol]D[/symbol][sup]+[/sup]:

[symbol]D[/symbol][sup]+[/sup] --> n + [symbol]p[/symbol][sup]+[/sup]
[symbol]D[/symbol][sup]+[/sup] --> p + [symbol]p[/symbol][sup]0[/sup]

It would be odd to invoke “elemental” notation here:

[symbol]D[/symbol][sup]+[/sup] --> n + [symbol]p[/symbol][sup]+[/sup]
[symbol]D[/symbol][sup]+[/sup] --> H[sup]+[/sup] + [symbol]p[/symbol][sup]0[/sup]

So it doesn’t matter what compound the nucleous is in as long as the nucleous is accesible. Or do the electrons get in the way for heavier elements.

Nope. Since the neutron is neutral, the electrons are essentially meaningless to it.

(I say “essentially” because there are a couple of very small corrections that are usually negligible when dealing with the nuclear phenomena, namely (a) the electrons add a small amount of mass to the target system and (b) the neutron has a magnetic dipole moment, so it has a small chance of scattering off the electrons.)