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. $\endgroup$ Oct 22, 2015 at 8:22

2 Answers 2


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)."


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.

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


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