Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Source of Energy in a Fission Reaction

  1. mass defect (mass converted to energy via E = mc^2)

    0 vote(s)
  2. decrease in nuclear binding energy (release of neutrons)

    0 vote(s)
  3. Neither is the correct explanation. (Please explain)

    0 vote(s)
  4. Both are different ways of explaining the same thing.

    1 vote(s)
  1. Mar 24, 2014 #1
    Hello. My teacher and I have had this very intense debate for quite some time now and we need your opinion on this issue. This question is, where does the energy released in a nuclear fission reaction come from? (Please use the terms "nuclear binding energy" and "mass defect")

    My take on this issue is that when Uranium 236 splits into Krypton, Barium and 3 neutrons, the nuclear binding energy on a per nucleon basis increases for this fission event. This is because Krypton and Barium have higher nuclear binding energy than Uranium, per nucleon. This, in turn, means that the products of the fission reaction (Krypton, Barium and 3 neutrons) have a smaller mass than Uranium does. This missing mass is converted to energy.

    In short, my belief is that the products of a fission reaction has more nuclear binding energy than the reactants does, therefore the products experience greater mass defect and thus possess less mass. This mass is converted to energy via E=mc^2.

    My teacher, however, has a different opinion. My teacher believes that although nuclear binding energy increases on a per nucleon basis in a fission reaction, the fact is that 3 neutrons have been released. This means that overall nuclear binding energy has decreased during the fission reaction. In short, my teacher's opinion is that the source of energy in a fission reaction is the nuclear binding energy that has been holding the 3 neutrons.

    Please indicate which theory is more correct and why. Thank you!
  2. jcsd
  3. Mar 24, 2014 #2


    User Avatar
    Gold Member

    I'm going with "both"....
    In fact in both cases you need "lesser" binding energy in order to have energy released... (it's just conservation of energy). But in order to calculate them, you are making use of E=mc^2...

    You can't, on the other hand, just drop things out of your calculations and say for example I'll consider only 1 and 2 and forget about the existence of 3 and 4... you are taking all the initial things together and all the final things together.
    So? either you say the one or the other, it's the same thing.
  4. Mar 24, 2014 #3
    Yes, but my teacher is saying the reactant side has more nuclear binding energy than the product side does, and I'm saying that the product side has more nuclear binding energy.

    Nuclear binding energy is the result of the mass defect when individual nucleons are combined together, so when uranium undergoes a fission reaction to produce krypton and barium, both of which have a higher nuclear binding energy on a per nucleon basis than uranium does, it is the loss in mass that results in the release of energy, right?
  5. Mar 25, 2014 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    This article has a lot of detail on the energy production from fission reactions:


    It would be pointless to repeat it in this forum verbatim. However, if there is a lack of clarity about some aspect, feel free to submit a more detailed query.
  6. Mar 25, 2014 #5

    No. In the nucleus the neutrons had a mass defect. Free neutrons have zero mass defect. You don't get energy from disappearing mass defect. You get energy from disappearing mass.
  7. Mar 25, 2014 #6
    You are right, not your teacher: binding is increasing, mass of reaction products goes down, the difference is released as (initially) kinetic energy of particles, and gammas.

    Whether "product side has more nuclear binding energy" depends on how you define sign of "binding energy". If you define it as positive quantity, then yes.

    But binding energy is in fact a *deficit* of energy. For example, deutherium nucleus's binding energy of 2.224 MeV can be said to be a difference of -2.224 MeV as compared to unbound prototon + neutron. If you define it like this, then more tightly bound nucleus has numerically smaller binding energy, since -3 < -2 etc.
  8. Mar 25, 2014 #7
    Thank you for your answers everyone, you explained this topic very well!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook