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Calculate Energy during Nuclear Fission

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Another possible form of the fission of U-235 is:
    [tex]\begin{array}{cc}235&92\[\tex] U + [tex]\begin{array}{cc}1&0\[\tex] n [tex]/rightarrow[/tex] [tex]\begin{array}{cc}141&56\[\tex] Ba + [tex]\begin{array}{cc}92&36\[\tex] Kr + 3 [tex]\begin{array}{cc}1&0\[/tex] n

    a) Given the masses of the particles in the table below, calculate the amount of energy released in the fission of a U-235 nucleus.

    U = 234. 993 u
    1 n = 1.008 u
    Ba = 140.883 u
    Kr = 91.905 u


    2. Relevant equations



    3. The attempt at a solution
    Before i start, im not sure what to do with the last product formed from the U-235 reaction. I'mm not sure if its 3 protons or 2 protons and 1 neutron. anyways i just assumed it was 3 neutrons though.

    Mass or reactants :
    234.993 u + 1.008 u = 236.001 u
    Mass of products:
    140.883 u + 91.905 u + 1.008 u(3) = 235.812 u

    Mass difference:
    236.001 u
    - 235.812 u
    = 0.189 u
    Energy released:

    m = (0.189 u)(1.6605 x 10^-27 kg/u)
    =3.138345 x 10^-28 kg

    E = mc^2
    = (3.138345 x 10^-28 kg)(2.998 x 10^8 m/s)^2
    = 2.82 x 10^-11 J

    Im honestly not sure that this is correct.
     
  2. jcsd
  3. Oct 13, 2009 #2
    i couldn't do the laTeX image, but its 235/92 U + 1/0 n = 141/56 Ba + 92/63 Kr + 3 1/0 n
     
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