# Nuclear fission mechanism: neutron capture

Consider this nuclear fission reaction:

$\mathrm{^{235}U+{}^1n\to{}^{236}U \text{ (excited)}\to{} ^{92}Kr+{}^{141}Ba+3{}^1n}$

I have not understand why a thermal neutron ($^1$n with $E\simeq 0.025$ eV) has enough energy to start the reaction. I mean, to start the reaction the $^1$n should be captured by the nuclei of $^{235}$U.

What are the conditions under which this neutron is captured?

My reasoning is the following: the energy that the neutron must have to be captured by the $^{235}$U should be equal to the $S_\mathrm n$ (separation energy for a neutron) of $^{236}$U which I found is 6.34 MeV, higher than the energy of the thermal neutron. So my reasoning must be wrong but I cannot understand where I make mistakes...

I looked on the Povh Rith Particles and Nuclei but it does not explain the reaction well. Could anyone give me a reference or explain here the fission reaction ?

You're misinterpreting the separation energy $S_\mathrm n$. If you wanted to start with $^{236}$U and end up with $^{235}$U and a neutron at rest, you'd have to add 6 MeV. When the neutron is captured, it's into some nucleon orbital 6 MeV above the $^{236}$U ground state; as the excited nucleus cools that energy gets distributed among all the nucleons. Lighter nuclei throw off the energy of neutron capture by emitting a cascade of photons; in uranium, fission is another available pathway for the nucleons to reduce their interaction energy.
A nucleus with a negative neutron separation energy $S_\mathrm n < 0$ is beyond the neutron drip line.