In induced fission of U-235, neutrons are bombarded at the U-235, producing U-236. This U-236 then undergoes fission:

U-235 + n --> U-236 --> Ba-141 + Kr-92 + 3n

As far as I understand, the energy released in fission is gained as kinetic energy of the products, and also released as gamma photons/beta particles and neutrinos when the products decay. My confusion lies in calculating the energy released, as I do not think the method in the textbook is correct.

The absorbed neutron loses nuclear potential energy. This causes the binding energy of U-236>U-235, meaning the rest mass of U-236 is less than the rest mass of U-235 + n. This increase in binding energy is then used to deform the nucleus into a double-lobed drop allowing the two fragments to separate due to electrostatic repulsion.

The two fragments formed have a greater binding energy per nucleon than U-236, and hence the binding energy of the fragments is greater than U-236. (In turn causing the mass of the products to decrease). However this increase in binding energy is gained as kinetic energy of the products/released as gamma photons etc.

Mo1 = Mass(U235+n)

Mo2= MassU236

Mo3 = Massfissionproducts

B1=binding energy of U235

B2=binding energy of U236

B3 = binding energy of fission products

The energy released in fission is due to the increase in binding energy B3-B2. The increase in binding energy B2-B1 is used to deform the nucleus - it is not 'released'. Therefore the energy released in the fission process = (Mo2-Mo3)c^2.

However my textbook states that the energy released in the fission process =(Mo1-Mo3)c^2. I don't understand this as they are including the energy to deform the nucleus, (Mo1-Mo2)c^2, when this is not actually released.

Any help is greatly appreciated!


1 Answer 1


In principle the energy released is the total energy in (mass of $^{235}U$ + energy of thermal neutron) minus the total rest mass of the fission products. In practice, the neutron coming in is thermal (moderated), and has an energy of a few eV, which we can neglect. This means we can compute the energy released from the difference in the rest masses of the "in" and "out" particles. The rest mass of $^{236}U$ is not relevant, because what is formed is an excited state.

  • $\begingroup$ Thanks for the answer. But the mass has decreased partly because the neutron has lost NPE, which has released energy to deform the nucleus. So why are we saying that the total decrease in mass is the energy released? $\endgroup$
    – John
    Feb 19, 2017 at 14:59
  • $\begingroup$ Your initial state is uranium and a well separated neutron. There is no potential energy. $\endgroup$
    – Thomas
    Feb 19, 2017 at 15:28
  • $\begingroup$ But doesn't the neutron binding to the nucleus release energy? I thought that when the neutron gets in range to be acted on by the SNF, then it loses potential energy. $\endgroup$
    – John
    Feb 19, 2017 at 15:29
  • $\begingroup$ "The absorbed neutron loses nuclear potential energy. " That is incorrect for those nuclei which fission. The initially formed $^{236}$U is not in the ground state, but has about 6.6 MeV of extra energy known as the activation energy. That extra energy is what produces the fission, and is only a part of the energy released. But you can't double-count it. Occasionally, the system throws off this energy in the form of gamma rays in a process known as radiative absorption, and no fission takes place for that specific nucleus. $\endgroup$
    – Bill N
    Feb 19, 2017 at 15:38
  • $\begingroup$ I thought that if a nucleon was acted on by the SNF then it loses nuclear potential energy. Is the U236 nucleus formed in an excited state just due the kinetic energy of the neutron? $\endgroup$
    – John
    Feb 19, 2017 at 15:43

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