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Energy released in fission

  1. May 28, 2007 #1
    1. The problem statement, all variables and given/known data
    Consider the following problem:
    1) Calculate the energy released in the fission reaction,
    Uranium(235,92) + neutron--->[Uranium(235,92)]*--->Neodymium(143,60) + Zirconium(90,40)+ 3 neutrons + 8 electrons +8 antineutrino

    2. Relevant equations

    3. The attempt at a solution

    If atomic rest masses are used in calculating the Q value, the term 8 e- may be dropped.Let M(U),M(Nd),M(Zr) and M(n) be the atomic masses of Uranium, Neodymium,Zirconium and neutron respectively.
    Q=[M(U) - M(Nd) - M(Zr) - {3-1}M(n)]c^2
    Q= 197.6 MeV
    This is the solution given in Schaums book on Modern Physics.
    Just because atomic rest masses are used, how can we neglect the 8 e- term?
  2. jcsd
  3. May 28, 2007 #2


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    Staff: Mentor

    In the fission process, the electrons are not annihilated, so the rest mass of electrons (92) never changes.

    does not have the correct balance.

    Fission does not produce protons - it simply transforms one nucleus into two nuclei.

    In the fission process, the sum of Z's of the new nuclei (Z1 and Z2) must equal Z(U) = 92, so:

    If Z1 = 40 (as in Zr), the other nuclei must have Z2 = 52 (Te).

    If Z2 = 60 (as in Nd), then the other nuclei must have Z1 = 32 (Ge).

    The number of electrons = Z (which for U = 92).

    What does change is the atomic binding energy of the electrons, but that is on the order of eV for valence electrons and increases toward the keV range for inner electrons of more massive nuclei.
    See - http://xray.uu.se/hypertext/XREmission.html
    Last edited: May 28, 2007
  4. May 29, 2007 #3
    I didn't get your point. Could u please explain it in detail?
  5. May 29, 2007 #4
    i think he's saying taht since the binding energy and rest energy of electrons is on the scale of eV and keV we need not care since the answer is on the scale of hundreds of MeV
  6. May 31, 2007 #5
    But the energy associated with 8 electrons is about 4 MeV and that doesn't seem to negligeble.
  7. Jun 1, 2007 #6


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    Homework Helper

    If you look at charge balance and assume the initial U was neutral, then you will see that the final Nd and Zi can't be neutral. They must be ionized with a net charge of +8. Hence if you use the neutral atomic masses for them, it already accounts for the 8 extra electrons.
  8. Jun 1, 2007 #7
    Thats cool!Thanx guys.
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