Find the energy released in the following fissionreaction n + 235 U --> 141 Ba +

Question

Find the energy released in the following fissionreaction     n + 235U  -->       141Ba +     92Kr +3 n, where    mn   = 1.008665u,                  mu = 235.043915   u,               mBa=   140.9139u,                 mKr = 91.8973 and           1u = 1.66 x 10-12 kg. Convert the unit of the result to MeV Find the energy released in the following fissionreaction     n + 235U  -->       141Ba +     92Kr +3 n, where    mn   = 1.008665u,                  mu = 235.043915   u,               mBa=   140.9139u,                 mKr = 91.8973 and           1u = 1.66 x 10-12 kg. Convert the unit of the result to MeV

Explanation / Answer

  for the given problem let us take that    mp as mass of proton    me as mass of electron    mn as mass of nuetron    M is the mass of nucleus    Z is atomic number and    A is the mass number    then the mass defect will be    m = [Z mp + (A - Z)mn] - M    then the binding energy will be    BE = m c2    

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