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It's often explained how $\ce{^14C}$ is formed in the upper atmosphere, a neutron hits a nitrogen atom and ejects a proton. Source. Since this is likely to happen to an $\ce{N2}$, I'm curious what the chemistry that follows is.

Does the energy of impact split the molecule, leaving a lone $\ce{^14C}$ and a lone $\ce{^14N}$ (and a lone hydrogen), (each highly eager to bond with something) or does it leave a $\ce{C-N}$ which might also readily bond with $\ce{O2}$ or maybe something else. I'm mostly curious what the immediate chemical reactions are after the $\ce{^14C}$ forms cause it seems to me that it should be initially pretty reactive.

Edit & maybe a partial answer:

It occurs to me after giving it some thought that if an $\ce{N2}$ is split into a free $\ce{C}$ and a free $\ce{N}$, each would be most likely to bond to an $\ce{O2}$, or a single $\ce{O}$ from $\ce{O3}$ if the split happens near the ozone layer (which is possible given the listed height where most $\ce{^14C}$ forms, "altitudes of 9 to 15 km" - same link as above). Initially I was thinking it might be more exotic than that, but having given this some thought I think that's what happens. The $\ce{C}$ and the $\ce{N}$ split and mostly each binds with $\ce{O2}$, 2nd most common they bind with $\ce{O}$. $\ce{CN}$ is probably pretty uncommon, now that I think about it.

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    $\begingroup$ My initial guess would be that the nitrogen molecule would form cyanogen radical. $\endgroup$ – user1945827 Oct 22 '15 at 8:22
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wikipedia article cites: "After production in the upper atmosphere, the carbon-14 atoms react rapidly to form mostly (about 93%) 14CO (carbon monoxide), which subsequently oxidizes at a slower rate to form 14CO2, radioactive carbon dioxide. The gas mixes rapidly and becomes evenly distributed throughout the atmosphere (the mixing timescale in the order of weeks)."

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The main reaction forming C-14 in the air is the reaction of N-14 with neutrons to form C-14.

I know that if fast neutrons from fission events are allowed to strike a nitrogen nucleous that a very high energy gamma photon (higher than 1.6 MeV) are emitted. Now if we consider for a moment the emission of a gamma photon. We must conserve momentum.

The momentum of a gamma photon is given by planks's constant divided by the wavelength. According to J. M. Hendrie (The Journal of Chemical Physics 22, 1503 (1954); https://doi.org/10.1063/1.1740449) to break a N2 molecule into two nitrogen atoms requires about 8.8 eV.

I have calculated the energy of the recoiling atom of C14 which would be for a moment inside a CN radical when the reaction occurs. Thus we will break the molecule up to form carbon atoms.

enter image description here

I imagine that a carbon atom will be able to react with oxygen with ease to form a carbon oxide.

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