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"By adding a neutron to $U^{235}_{92}$ an even-even nucleus is obtained. The binding energy increases in the process, so the energy gained in the process is greater."

I'm confused, if the binding energy increases, how can fission be favored? Furthermore, how can the energy gained in the process be greater if $B$ is a negative contribution to the nucleus energy?

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If the binding energy increases this makes the nucleus more stable as more energy will be needed to split the nucleus into its constituent parts.

When a neutron is added the U235 to form U236 there is a release of energy of about 6.8 MeV, ie it would take an extra 6.8 MeV of energy to split up a U236 nucleus into its constituent parts than a U235 nucleus or put another way it would take 6.8 MeV of energy to remove one neutron from a U236 nucleus.

This 6.8 MeV of energy is now available to the U236 nucleus to overcome the critical (activation) energy of about 6.5 Mev for it to undergo fission.

The critical energy for a fission is smaller if the nucleus is an even-even nucleus as can be seen from the table below which also shows that absorbed neutron has to have kinetic energy (1.5 MeV in the case of U238 absorbing a neutron to form U239) for the fission process to have a high probability of occurring.

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The answer is that $U_{92}^{236}$ is neither a fissile or fertile isotope, and is only fissionable by fast neutron capture. Oddness of both Z or N tend to lower the stability of the nucleus, this being explained by the nuclear shell model.

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  1. addition of one neutron is local profit
  2. system might get global profit from fission
  3. to get global profit system should overcome the barrier
  4. if the energy from reaction with neutron is more than needed for the barrier, fission is inevitable
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