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i heard that fission activation energy of (235)U is less than of neutron separation energy of (236)U so this must the reason that (235)U is fission able

$$E_s+(236)U\to (235)U+n$$ in this interaction $E_s$ is neutron separation energy ( energy required to separate neutron )

activation-energy

  1. how to measure activation energy of fission?

  2. is there any relation or equation?

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2 Answers 2

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Here is a tutorial on fission .

The concept of "activation energy" is not utilized in the study of fission. There are energy levels that can be modeled with nuclear potential well models.Mass is turned into energy in fission.

What happens is that there is a high cross section for U235 to capture a thermal neutron and become a U236 in an excited state. There are more than one energy levels where this could happen.

The configuration is unstable enough that instead of falling down with gamma rays to a ground state of U236, which is fairly stable, the system (U235+neutron) breaks up into Kr92 Ba141 and 3 neutrons which can with suitable engineering be thermalized to create a chain reaction in a fission reactor.

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Anna gives an excellent description of fission by neutron capture, and I'd guess that is what you were asking about. However I thought it worth adding that nuclei can also undergo spontaneous fission, though only very heavy nuclei do this. In principle uranium (235 and 238) can undergo spontaneous fission but it's very slow compared to alpha or beta decay.

In spontaneous fission the idea of a potential barrier is important for obvious reasons. The energy of a U238 nucleus is greater than the energy of it's fission products, so it's energetically favourable for the nuclei to split. If there were no potential barrier U238 nuclei would all fission instantly and there wouldn't be any left. However in order to split the U238 nucleus has to go through an intermediate state with a higher energy, and this creates the potential barrier, just like the diagram you draw.

Well, your diagram isn't really correct for fission because fission is a quantum process and the nucleus splits by tunneling through the barrier. It's not like a chemical reaction where you have a Boltzmann distribution of energies in the reagents so a proportion of the reacting molecules do have enough energy to get over the barrier. Still, the end result is much the same.

According to http://www.eng.fsu.edu/~dommelen/quantum/style_a/ntcsf.html the activation energy for spontaneous fission of a uranium nulceus is 6.5MeV.

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Interesting mention of spontaneous fission - I believe that all neutron induced fissions can be represented as SF of the target U nucleus with one more neutron. The trouble with a simple chemical analogy is that neutrons induce fission at thermal energies (for the most attractive nuclei). A neutron with no momentum would gladly tunnel right into U-235. The intermediate species could be a range of quantum energy states, as many of these correspond to the resonances peaks which are numerous. –  Alan Rominger Jun 5 '12 at 19:59
    
@John Rennie "Anna gives an excellent description of fission by neutron capture," ok –  user8784 Jun 6 '12 at 9:19

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