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Nuclear Physics(Semi Empirical Mass Formula)

  1. Aug 7, 2013 #1
    1. Question : Use the Semi Empirical Mass Formula to estimate the energy released in the spontaneous fission reaction;

    235/92 Uranium -----> 87/35 Bromine + 145/57 Lanthanum + 3n




    2. Equations used :

    (i). Δm = m(reactants) - m(products)
    (ii). E = mc^2
    (iii). u = atomic mass unit



    3. Attempt to solution:

    atomic mass for 235/92 U = 235.043929 u
    " " " 87/35 Br = 86.920711 u
    " " " 145/57 La = 144.921765 u
    " " " 1/0 n = 1.009 u

    thus, Δm = m(reactants) - m(products)
    = 235.043929 - (86.920711 + 144.921765 + (3*1.009))
    = 2.19245u

    apply E = mc^2 = 2.19245u * (931.5 Mev/c^2)/u = 2043MeV




    4. Problem:

    i'm not sure if i'm answering the question right since it says "use the SEMF" , but this is how i am calculating the energy so far. Is this how i use the SEMF to calculate the energy, if not, how should the SEMF appear in the calculations? please kindly help......
     
    Last edited: Aug 7, 2013
  2. jcsd
  3. Aug 7, 2013 #2

    mfb

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    Staff: Mentor

    I think you are supposed to estimate the nucleon masses with the SEMF instead of looking them up.

    I get a different result for 235.043929 - (86.920711 + 144.921765 + (3*1.009)).
     
  4. Aug 8, 2013 #3
    thank you.
     
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