In the fusion reaction ${}_1{}^{2}H+{}_1{}^{2}H\to {}_2{}^{2}He+{}_0{}^{1}n$, the masses of deuteron, helium and neutron expressed in amu are 2.015, 3.017 and 1.009 respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. $1 amu=931.5 MeV/c^2$ .

$∆m=2(2.015)-(3.017+1.009)=0.004 amu$

Energy released $=(0.004\times 931.5) MeV=3.726 MeV$

Energy released per deuteron $=\frac{3.726}{2}=1.863 MeV$

Number of deuterons in 1 kg $=\frac{6.02\times {10}^{26}}{2}=3.01\times {10}^{26}$

Energy released per kg of deuterium fusion $=(3.01\times {10}^{26}\times 1.863)=5.6\times {10}^{26} MeV$

                                                                      $=9.0\times {10}^{13}J$