If we consider the reaction:

$$\require{mhchem} \ce{^1H + ^1n -> ^2H}$$

There is a net decrease in mass, so the process releases energy. I did a few calculations and got the energy of the process to be $6.405109\text{ MeV}$. Is this reaction feasible?

And as a follow-up, can this reaction be used practically for fusion? This reaction removes the difficulty in fusion of bringing like charged nuclei together, so I think the chance of proton colliding with a neutron is high.

  • $\begingroup$ Well yes this is the first step in the fusion cycle process used by the Sun. $\endgroup$ – John Rennie May 15 '17 at 9:35
  • $\begingroup$ @JohnRennie - I think you mean $p+p$. This one rather takes place after cooling the proton-neutron soup after Big Bang. $\endgroup$ – jaromrax May 15 '17 at 10:53
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    $\begingroup$ @jaromrax: No I mean $p+n$. In the Sun protons convert to neutrons via a (very slow) weak interaction then there is a very fast $p+n$ reaction to form a deuteron then another to form $^3He$. $\endgroup$ – John Rennie May 15 '17 at 10:56
  • $\begingroup$ @JohnRennie for some reason the \ce notation isnt working here. $\endgroup$ – Pritt Balagopal May 15 '17 at 11:05
  • $\begingroup$ @JohnRennie - ok, so you mean what is usually written as $^1H+^1H \rightarrow ^2H + e^+ + \nu_e$. One point of view is that diproton scattering state decays to $^2H$. But to convert a proton, one would need an interaction with a neutrino (that usually escape Sun immediately). I am not starting a fight, I am curious about this.... $\endgroup$ – jaromrax May 15 '17 at 11:06

Yes, hydrogen can absorb neutrons and emit the released energy as a gamma-ray photon. Your value is a bit too large, however. The Table of Isotopes (Firestone 1996) lists the neutron separation energy of deuterium (which is the same process in the other direction) as 2224.57 keV.

As to the practical feasibility for fusion, where would you get the neutrons from? Assuming you somehow got hold of some, you could let them react with hydrogen, e.g. in a water basin. This happens in fission reactors, but only as a side effect. The energy released by neutron absorption is only a small fraction of the total. One 235U fission releases about 200 MeV, the few extra neutrons get you a few MeV only.

Free neutrons are produced in D+T fusion, but if you can do that, your argument is kind of unnecessary, since you have fusion anyway. So the question is, can you get free neutrons for less energy input than you get back out from their absorption minus conversion losses? I don’t think so.

  • $\begingroup$ Well, if in uranium fission, they use neutrons to split the nucleus, so I guess the same source can be used to hit protons $\endgroup$ – Pritt Balagopal May 15 '17 at 13:16
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    $\begingroup$ Yes, but they contribute little energy, only a few percent. If neutron absorption did not yield any usable energy, fission reactors would still work. $\endgroup$ – Karsten Kretschmer May 15 '17 at 14:54

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