Consider the fission reaction with atomic masses , , ,

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Problem

Consider the fission reaction 92235U+01n  →  3692Kr+56141Ba+3 01n^{235}_{92}\text{U}+^{1}_{0}\text{n}\;\rightarrow\;^{92}_{36}\text{Kr}+^{141}_{56}\text{Ba}+3\,^{1}_{0}\text{n} with atomic masses 92235U=235.0439 u^{235}_{92}\text{U}=235.0439\,\text{u}, 3692Kr=91.9262 u^{92}_{36}\text{Kr}=91.9262\,\text{u}, 56141Ba=140.9144 u^{141}_{56}\text{Ba}=140.9144\,\text{u}, 01n=1.0087 u^{1}_{0}\text{n}=1.0087\,\text{u}. Calculate the total energy released when 1.00 g1.00\,\text{g} of 92235U^{235}_{92}\text{U} is completely fissioned. Use NA=6.022×1023 mol−1N_{A}=6.022\times10^{23}\,\text{mol}^{-1}, 1 u=931.5 MeV/c21\,\text{u}=931.5\,\text{MeV}/c^{2} and 1 eV=1.6×10−19 J1\,\text{eV}=1.6\times10^{-19}\,\text{J}. Choose the closest value. (a) 7.1×106 MJ7.1\times10^{6}\,\text{MJ} (b) 8.2×104 MJ8.2\times10^{4}\,\text{MJ} (c) 7.1×104 MJ7.1\times10^{4}\,\text{MJ} (d) 7.1×102 MJ7.1\times10^{2}\,\text{MJ}

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