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Binding Energy in U-235 and daughter atoms

  1. Sep 22, 2007 #1
    I am a little bit confused.
    If U235 decays 2 neutrons, and the two daughter atoms posses a larger amount of binding energy per nucleon, then why is there excess energy? Why doesn't this process require the input of additional energy? I know I am missing something fairly simple.
  2. jcsd
  3. Sep 22, 2007 #2


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    Because binding energy is conted as "negative".
  4. Sep 22, 2007 #3
    So, wouldn't that imply that this would require energy from the surroundings in order to make the reaction to take place? I know that we can safely assume that the kinetic energy of the incoming neutron to be negligible. So, how can this happen without the addition of external energy?
  5. Sep 22, 2007 #4


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    Well now there is many different ways for the U-235 to decay and being fissioned. Just tell the specific reaction that you have questions about and I will try to explain it.

    If the daugher nucleis has greater binding energy per nucleon (i.e. more negative energy), then energy is released (in general, depends on the neutrons also etc but basically this is what happens)

    Mother nucleus A' has 20 nucleons, 5MeV / nucleon; total energy: -100MeV

    Decays to daughter nuclei B' and C', they have binding energy 6MeV / nucleon; total energy: -120MeV so here you have less energy, so in order to obey energy conservation; 20 MeV is then kinetic energy of daugher nucleis (NB! this is only a crude model for explanation)
  6. Sep 22, 2007 #5
    Okay, from lecture, the professor uses the following example. U-235 decays to two daughter nuclei X and Y each with A=117 with the release of two free neutrons. He says the change in binding energy = (7.6 Mev * 235 - 8.3 Mev * 234) = (approximately equals) (0.7)(235) = (approximately) 200 MeV.

    The above equation will yield a negative energy. Where does the energy come from that allows more overall binding energy between the two daughter nuclei?
  7. Sep 23, 2007 #6


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    Spontaneous fission is that reaction called.

    Binding energy is counted as negative, so You have to be aware of this and count backwards.

    U-235; 7.6MeV / nucleon
    X; Y : 8.3MeV / nucleon respectevly


    235*(-7.6) = -1786MeV initally
    2*117*(-8.3) = -1942MeV final

    You have more energy in the begining then in the final state, so you must add energy to obey energy conservation.

    So there is relased 256MeV, do the comparison with a hydrogen atom, that is excited and the electron has 1.51eV binding energy; deexcites to ground state that has binding energy 13.6eV; there is realeased 12.89eV as a photon (and recoil energy of atom). So then things is rather clear, cos' binding energy is counted negative, but in nuclear physics, you have to take into account that the initial particle can be divided into two etc, but basically it is the same principle.
    Last edited: Sep 23, 2007
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