Explanation:
\(\Delta m = 2(2.015) - (3.017 + 1.009)\)
\(\,\,\,\,\,\,\,\, = 0.004\,{\text{amu}}\)
\(\because \) Energy released
\(\,\,\,\,\,\,\,\, = (0.004 \times 931.5)MeV\)
\(\,\,\,\,\,\,\,\, = 3.726MeV\)
Energy released per deuterium,
\(\,\,\,\,\,\,\, = \frac{{3.726}}{2} = 1.863MeV\)
Number of deuterons in \(1\,kg\)
\(\,\,\,\,\,\,\, = \frac{{6.02 \times {{10}^{26}}}}{2} = 3.01 \times {10^{26}}\)
\(\because \) Energy released per \(kg\) of deuterium fusion
\(\,\,\,\,\,\, = \left( {3.01 \times {{10}^{26}} \times 1.863} \right)\)
\(\,\,\,\,\,\, = 5.6 \times {10^2}MeV\)
\(\,\,\,\,\,\, = 9 \times {10^{13}}\;J\)