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Calculate the amount of energy released when 1 kg of ""_(92)^(235)"U" undergoes fission reaction. |
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Answer» Solution :235 g of` ""_(92)^(235)"U"` has `6.02 xx 10^(23)` atoms. In one gram of`""_(92)^(235 "U"`, the number of atoms is equal to `(6.02 xx 10^(23))/(235) = 2.56 xx 10^(21)` So the number of atoms in 1 kg of `""_(92)^(235)"U" = 2.56 xx 10^(21)xx 1000 = 2.56 xx 10^(24)` Each `""_(92)^(235)"U"` is `Q = 2.56 xx 10^(24) xx 200 MeV = 5.12 xx 10^(26)MeV` By CONVERTING in terms of joules, `Q = 5.12 xx 10^(26) xx 1.6 xx 10^(-13)`J = `8.192 xx 10^(13)`J. In terms of Kilowatt HOUR `Q = (8.192 xx 10^(13))/(3.6 xx 10^(6))= 2.27 xx 10^(7)` kWh |
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