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I Energy released when 238U is divided into two 119Pd.

  1. Apr 14, 2017 #1
    Hi! I have a question about nuclear fission. Here is the problem:

    238_U (B/A = 7.6 MeV/n) is divided into two 119_Pd ( B/A = 8.5 MeV/n). How much energy will be released?

    I was thinking since the 119_Pd has larger B/A (binding energy per nucleus), energy is needed for this reaction to happen. But in my notes, the teacher has made this calculation 238*(8.5-7.6) = 214 MeV and states that this is released from the reaction. I do not undertand.

    Best regards
    ulrik
     
  2. jcsd
  3. Apr 14, 2017 #2

    jtbell

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    Staff: Mentor

    Binding energy is actually negative. That is, you must subtract it from the sum of the mass-energies of the individual nucleons, in order to get the total mass-energy of a nucleus. So if you think in terms of "average mass-energy per nucleon", this is smaller for 119Pd than for 238U.
     
  4. Apr 14, 2017 #3

    phyzguy

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    Science Advisor

    Although binding energy is defined in such a way that the numbers come out positive, it is actually a negative energy, since it represents how much energy is released when the nucleus is assembled from its constituents. so we can write:

    E(U238) = E(238 separate nuclei) - 238 * B/A(U238); E(Pd119) = E(119 separate nuclei) - 119 * B/A(Pd119)

    So if you subtract the final energy from the initial energy, you get:

    E(238 separate nucleu) - 238 * 7.6 - 2*(E(119 separate nuclei) - 119 * 8.5) = 238*(8.5 - 7.6)
     
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