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Find the amount of energy released when `1` atom of Uranium `._(92)U^(235)(235.0439 "amu")` undergoes fission by slow neturon `(1.0087 "amu")` and is splitted into Krypton `._(36)Kr^(92)(91.8973 "amu")` and Barium `._(56)Br^(141)(140.9139 "amu")` assuming no energy is lost. Hence find the enrgy in `kWh`, when `1g` of it undergoes fission. |
Answer» `._(92)^(235)U+._(0)^(235)nrarr._(56)^(141)Ba+._(36)^(92)Kr+3(._(0)^(1)n)` `Deltam = [{235.0439+1.0087}-{140.9139+91.8973+3xx1.0087}]` `= 0.2135amu` Equilvalent enrgy `= 0.2153xx931 = 200MeV` Number of atoms/fission in `1g` of atom is `(1)/(235) xx 6.023xx10^(23) = 2.56xx10^(21)` Energy released in fission of `1g` of `._(92)^(235)U` `= 2.56xx10^(21) xx 200MeV` `= 2.56 xx 10^(21) xx 200 xx 1.6 xx 10^(-13)J` `= (2.56xx10^(21)xx200xx1.6xx10^(-13))/(3.6xx10^(6))kWh` `= 2.28 xx 10^(4) kWh` |
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