the amount of energy (in kJ/mol) released during this process is (given: $U=235.0439$ amu, $Ba=141.9164$ amu, $^{91}Kr=90.9234$ amu, neutron $=1.00866$ amu)
(a)$3.12 \times 10^{12}$
(b)$2.8 \times 10^{11}$
(c)$1.0 \times 10^{9}$
(d)$1.68 \times 10^{10}$
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.
Explanation
${}^{235}_{92}U+{}^{1}_{0}n \rightarrow {}^{142}_{56}Ba+{}^{91}_{36}Kr+3\,{}^{1}_{0}n$Energy released per atom $(E)=(\Delta m)c^{2}$.$E=\left[(235.0435+1.00866)-(141.9164+90.9234+3\times1.00866)\right]\times1.67\times10^{-27}\times(3\times10^{8})^{2}\,\mathrm{J}$Therefore, $E(\text{per atom})=2.8\times10^{-11}\,\mathrm{J}$ [$1\,\mathrm{amu}=1.67\times10^{-27}\,\mathrm{kg}$, $c=3\times10^{8}\,\mathrm{m/s}$].Energy released from 1 mole disintegration of $^{235}_{92}U$ = $E(\text{per mole})=E(\text{per atom})\times N_{A}=2.8\times10^{-11}\times6.023\times10^{23}$$=1.68\times10^{13}\,\mathrm{J}=1.68\times10^{10}\,\mathrm{kJ}$