In the fusion reaction, ${}_1^{2}H+{}_1^{2}H\to {}_2^{3}He+{}_0^{1}n$. The masses of deuterons, helium and neutron expressed in amu are $2.015,3.017$ and $1.009$ respectively. If $1\text{kg}$ of deuterium undergoes complete fusion. Find the amount of total energy released,
( $1\text{amu}=931.5{\text{MeV/C}}^2$)

The correct option is A $9\times {10}^{13}\text{J}$
Given:
${}_1^{2}H+{}_1^{2}H\to {}_2^{3}He+{}_0^{1}n$

Mass defect, $\Delta m=2\left(m_H\right)-m_{He}-m_n$
$=2\left(2.015\right)-(3.017+1.009)$
$=0.004\text{amu}$

Also, $1\text{amu}=931.5{\text{MeV/C}}^2$
So, $E=0.004\text{amu}=0.004\times 931.5=3.726\text{MeV}$
$E=3.726\times 1.6\times {10}^{-13}\text{J}=5.96\times {10}^{-13}\text{J}$
For $1\text{kg}$ of Deuterium available, moles $=\frac{1000\text{g}}{2\text{g}}=500$
So, $N=500N_A=3.01\times {10}^{26}$
Energy released $=\frac{N}{2}\times 5.96\times {10}^{-13}\approx 9\times {10}^{13}\text{J}$
Hence, option $\left(C\right)$ is correct.