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Calculate the energy released in the following nuclear reaction and hence calculate the energy released when 235 gram of uranium-235 undergoes fission. U_(92)^(235) + n_(0)^(1) to Kr_(36)^(92) + Ba_(56)^(141) + 3n_(0)^(1) Rest masses of U^(235), Ba^(141) ,Kr^(92) and neutron are 235.04390 amu, 140.91390 amu, 91.89730 amu and 1.00867 amu respectively. Avogadro number = 6.023 xx 10^(23). |
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Answer» SOLUTION :Sum of the masses of the reactants `=236.05257` AMU Sum of the masses of the products `=235.83721` amu Mass converted to energy per fission `=0.215236` amu Energy RELEASED per fission `=0.21536 xx 931` `=200.50016` MeV `200.5 xx 1.6 xx 10^(-13) = 3.208 xx 10^(-11)` J Energy released when 235 gram of `U^(235)` undergoes fission `=200.5 xx 6.023 xx 10^(23)` `=1.2076 xx 10^(26)` MeV or `1.932 xx 10^(13)` J |
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