Energy of fission reaction

  1. The binding energy (BE) per nucleon for 235U is 7.6 Mev. The 235U undergoes a nuclear fission to produce two fragments both having a BE of 8.5Mev. What is the energy released from a complete fission of 1kg of 235U (joules)?

    Here I assumed that it breaks into two 118X element. Therefore
    BE(product) per atom is 2*118*8.5=2006
    BE(reactant) per atom is 236*7.6=1793.6
    Energy released per atom= 212.4 Mev
    Number of atoms in 1 kg U=1/.235*6.022*[itex]10^{23}[/itex] = 2.5625*[itex]10^{24}[/itex]
    Total energy = 2.5625*[itex]10^{24}[/itex] *212.4
    = 5.4428*[itex]10^{26}[/itex] Mev
    = 8.7*[itex]10^{13}[/itex] joules
    I want know if my answer is correct ?
  2. jcsd
  3. Seems about right (up to two sig. figs.)
    1 person likes this.
  4. mfb

    Staff: Mentor

    You have U235, not 236. You don't have to (and you should not) assume specific decay products, as you know the total number of nucleons both in the initial nucleus (235) and as the sum of all decay products (has to be the same).
  5. Yes but U235 is stable and usual fission occurs only by bombarding with a neutron , thus making it U236 which is unstable.Also the question says it breaks up into two same products which means they should have the same number of nucleons. Therefore having an even number of nucleons is the only way it can happen which supports the stability argument .
  6. QuantumPion

    QuantumPion 880
    Science Advisor
    Gold Member

    Fission does not fragment into two equal parts, you get one heavy and one light fission product and 2-3 free neutrons, e.g. I-135 + Tc99 + 2n.
  7. mfb

    Staff: Mentor

    U235 is not stable. While fission reactions in commercial applications are usual induced by other neutrons, this is not always the case (and the problem statement, as you wrote it, does not mention that). If you want to consider induced fission, you have to take the energy released in the U236 formation into account.

    Where? I just see that their binding energies per nucleon can be assumed to be the same (a reasonable approximation for typical fission products).
    1 person likes this.
  8. If I can't make the assumption that they are the same product how else can I solve it, and I have taken U236 for calculating the energy.
  9. mfb

    Staff: Mentor

    You know the binding energy per nucleon, and you know the total number of nucleons.

    If an apple costs 30 Cents and you have to buy 100 apples by going to the shop 2 times, it does not matter how many apples you buy each time (like 50+50 or 20+80 or whatever), the total cost is independent of that.

    Which is wrong.
    1 person likes this.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?

Draft saved Draft deleted