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IN a nuclear reactor, fission is produced in 1 g of `.^(235)U` `(235.0349 am u)`. In assuming that `._(53)^(92)Kr (91.8673 am u)` and `._(36)^(141)Ba (140.9139 am u)` are produced in all reactions and no energy is lost, calculate the total energy produced in killowatt. Given: `1 am u =931 MeV`. |
Answer» Correct Answer - `2.28 xx 10^(4)kWh` The nuclear fission reaction in the above process is `._(92)^(235)U +._(0) + ._(0)^(1)n rarr._(56)^(141)Ba +._(36)^(92)Kr+3 (._(0)^(1)n` Thesum of the mass before reaction is `235.0439 +1.007=236.0526 a.m. u.` The sum of the masses after reaction is `140.9139 +91.8973 +(1.0087) =235.8375 a.m.u.` Mass defect, `Delta m =236.0526 -235. 8373 =0.2175 a.m.n` Energy released in the fission of `.^(235)U` nulceus is given by `DeltaE=Deltam xx 931.2 MeV` `=0.2153 xx 931.2 =200 MeV` Number of atoms in `1 g` of `.^(235)U` `N=(6.02 xx10^(23))/(235)=2.56 xx10^(21)` Energy released in fission of `1g` of `.^(235)U`, `E=200 xx 2.56 xx 10^(21) MeV` `=5.12 xx10^(23) MeV =(5.12 xx 10^(23)) xx (1.6 xx10^(-13))` `=8.2 xx 10^(10)J` `(8.2 xx 10^(10))/(3.6 xx 10^(6) kWh =2.28 xx10^(4) kWh` |
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