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Consider one of fission reactions of `^(235)U` by thermal neutrons `._(92)^(235)U +n rarr ._(38)^(94)Sr +._(54)^(140)Xe+2n` . The fission fragments are however unstable and they undergo successive `beta`-decay until `._(38)^(94)Sr` becomes `._(40)^(94)Zr` and `._(54)^(140)Xe` becomes `._(58)^(140)Ce`. The energy released in this process is Given: `m(.^(235)U) =235.439u,m(n)=1.00866 u, m(.^(94)Zr)=93.9064 u, m(.^(140)Ce) =139.9055 u,1u=931 MeV]` .A. `156 MeV`B. `208 MeV`C. `465 MeV`D. cannot be computed |
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Answer» Correct Answer - b The complete fission is `._(92)^(235)U + n rarr ._(40)^(94)Zr +._(58)^(140)Ce + 2n+6e^(-1)` `Q =[m(.^(235)U) - m(.^(94)Zr) - m(.^(140)Ce) -m(n)]c^(2)` `=208 MeV`. |
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