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A thermal neutron strikes U_(92)^(235) nucleus to produce fission. The nuclear reaction is as given below : n_(0)^(1) + U_(92)^(235) to Ba_(56)^(141) + Kr_(36)^(92) +3n_(0)^(1) + E Calculate the energy released in MeV. Hence calculate the total energy released in the fission of 1 Kg of U_(92)^(235). Given mass of U_(92)^(235) = 235.043933 amu Mass of neutron n_(0)^(1) =1.008665 amu Mass of Ba_(56)^(141)=140.917700 amu Mass of Kr_(36)^(92)=91.895400 amu |
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Answer» Solution :Mass of the REACTANTS `=235.043933+1.008665` `=236.052598` Mass of the PRODUCT `=140.917700 +91.895400+3(1.008665)` `=235.839095` Energy E `= Delta m XX 931 "MeV"` `=198.77` MeV 235 grams contain `6.0233 xx 10^(23)` atoms of `U^(235)` 1000 gram contains N number of atoms `N=(6.0233 xx 10^(23) xx 1000)/( 235)` `=2.5629 xx 10^(24)` atoms Total energy released `=198.77 xx 2.5629 xx 10^(24)` `=509.43 xx 10^(24)` MeV or `E=8.15 xx 10^(13)` J |
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