Given that the mass of ${ }_{3}^{7} Li = 7.0160 \, u$,the mass of ${ }_{2}^{4} He = 4.0026 \, u$,and the mass of ${ }_{1}^{1} H = 1.0079 \, u$. When $20 \, g$ of ${ }_{3}^{7} Li$ is converted into ${ }_{2}^{4} He$ by proton capture,the energy liberated (in $kWh$) is: [Take $1 \, u = 931.5 \, MeV/c^2$ and $1 \, kWh = 3.6 \times 10^6 \, J$]

• A
  $8 \times 10^{6}$
• B
  $1.33 \times 10^{6}$
• C
  $6.82 \times 10^{5}$
• D
  $4.5 \times 10^{5}$