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B Difference in binding energy is equal to the energy released

  1. Aug 19, 2016 #1
    Can someone please explain why is the difference in binding energy is equal to the energy released in a nuclear fission/fusion.

    From my understanding of binding energy, it is the energy needed to break bond between neutron and proton or energy released when forming into a nucleus. However, what I dont understand is that during nuclear fission/fusion, the energy released is equivalent to the difference in the products binding energy and the intial binding energy.

    In order to cause nuclear fission to happen, energy must be supplied onto the system in order for bond breaking but why would the bonding state would go from low to high?
  2. jcsd
  3. Aug 19, 2016 #2
    The total energy of a nucleus is equal to the energy of the nucleon masses plus the binding energy (which is negative). A large nucleus can be split into two nuclei. If energy is to be conserved then the energy associated with the large nucleus must be equal to the sum of the energies associated with the smaller nuclei. But the number of nucleons (nucleon mass) of the large nucleus is equal to that of the two smaller nuclei. After compensating for the nucleon masses the binding energy of the large nucleus should equal to the sum of the binding energies of the smaller nuclei. But it does not If you plot the measured binding energy per nucleon vs mass number you see that this value decreases monotonically from a maximum value of about 8.7 MeV/nucleon for A ≅ 60 to about 7.5 MeV/nucleon at A ≅ 240. To make it true we must add another term to the sum of the binding energies of the smaller nuclei. This term then is the difference between the binding energy of the large nucleus and the sum of the binding energies of the smaller nuclei or the energy (positive) released.

    Why is the binding energy per nucleon of the large nucleus less than that of the sum of the two smaller nuclei? That answer lies in the interactions of the nucleons, i.e., the nuclear force. I'm not a nuclear theorist so this is as far as I can go.
  4. Aug 19, 2016 #3
    To be able split the larger nucleus into smaller nuclei, don't you need to supply external work onto the system? And my curiosity is since binding energy is the energy needed to break the bond of the deuteron apart, therefore external work must be applied onto the system, so my question is since energy is conserved, does all the external work is being covertered into unbinding energy?
  5. Aug 20, 2016 #4


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    The energy released in fission and fusion reactions is equated to the mass defect :
    In fission the reactant ( a heavy element ) produces two lighter elements.
    The combined masses of the products are less then the mass of the reactant.
    In fusion the masses of the reactants are greater that the mass of the product.
    For simplicity : H + H --.> H2
    In both cases the energy from the mass defect equals mc2
    In terms of binding energy the nuclei in products of fission and fusion are more stable than nuclei of reactants
    I had this topic in first year chemistry
  6. Aug 20, 2016 #5


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    Edit. H + H = He
  7. Aug 20, 2016 #6
    Usually in order to produce large numbers of fissions you do need extra energy.. But what is known as spontaneous fission does exist and this happens without extra energy. The decay of a nucleus by alpha particle emission is a special case of fission. The fact that the final state has less total mass than the initial state is why this can happens. Spontaneous fission has been observed in heavy elements as Uranium Thorium and Americium. The escape or separation of the fission products from one another is inhibited by the coulomb barrier but not prevented. The alpha particle because of its small charge can tunnel out.more easily than larger nuclei. For larger fission fragments you do need the extra energy to expedite the process but even so spontaneous fission does occur albeit very slowly.

    You might think of nucleus as a turbulent interaction of nucleons forming various preferred combinations that for short periods of time momentarily existing at least as a semi stable nuclei. If this combination occurs often enough then it has an increased chance of separating itself. What is actually going on is determined by the complexities of the nuclear interaction.
  8. Aug 20, 2016 #7
    so for example, in an artificial nuclear fission; let's take Uranium 235 as an example. As U235 being bombarded by a moving neutron, it splits out two smaller nuclei Br+Kr+3neutrons and energy is released in the process. So it is right to think that K.E of the neutron is being totally converted into energy needed to break the bond of the Uranium235 in order for fission to happen?
    I am having difficulty in thinking that why isn't KE energy is being included in to the nuclear equation since E=mc2?
  9. Aug 20, 2016 #8
    I'm not sure about your use of the term bond. But yes the energy of the neutron totally goes into the dissociation of the nucleus It is a little simplistic to say the nucleon have formed bonds with one another. I think the fact that radioactive nuclei usually have several decay modes suggest something more complex. I do not think the fission is like radiolysys where an x-ray dissociates a molecule into two parts.

    But remember:

    mc2 = KE + PE + m02
  10. Aug 20, 2016 #9
    I got it. Thank you for the great explanation! :smile::thumbup:
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