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Neutron energy needed to produce fission of 208Pb

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Estimate the neutron energy needed to produce fission of 208Pb. Is it likely that such neutrons woould be released in the resulting fission?

    2. Relevant equations

    Eexcitation=Qexcitation+Tn
    Qexcitation=[m(208Pb)+mn-m(209Pb)]c2
    V=(e2/4πεo)(Z1Z2/r)
    r=1.2fm[(A1)1/3+(A2)1/3]




    3. The attempt at a solution

    I'm a little confused just on how to approach this problem. I know that in order to fission the excitation energy must be greater than the activation energy. But doesn't the activation energy depend on what the products of the fission reaction are, of which there is more than one possible combination?

    Anyways my best guess was to just come up with a fission reaction that could possibly happen and since Z=50 and N=82 are nuclear shell magic numbers I came up with the following

    208Pb + n → 132Sn + 75Ge + 2n

    using
    Q = [m(208Pb)+mn-m(132Sn) - m(75Ge)- 2mn]

    and
    Eactivation= V - Q

    I find the activation energy and then find the kinetic energy the neutron needs for the excitation energy to be greater than the activation energy.

    My main problem is I don't know if I'm alright assuming the fission products I did. And as I look around on the internet I get the sense that the activation energy is a set value for a particular isotope, which greatly confuses me. If anyone could just kinda push me in the right direction that would be greatly appreciated!
     
  2. jcsd
  3. Oct 12, 2014 #2
    It's correct that Eactivation= V - Q
    if Eactivation less than binding energy of last neutron in 209Bi
    then a neutron with zero kinetic energy , fission immediately occur
    but if Eactivation greater than binding energy of last neutron
    the difference between them is neutron kinetic energy
     
  4. Oct 12, 2014 #3
    also see John .R lamarsh- introduction to nuclear reactor theory chapter 3
     
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