You are given that Mass of \({ }_{3}^{7} Li =7.0160\, u\) Mass of \({ }_{2}^{4} He =4.0026\, u\) and 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: [Mass of nudeon \(\left.=1\, GeV / c ^{2}\right]\)

\(\Delta m \Rightarrow\left[ m _{ Li }+ m _{ H }\right]-2\left[ M _{ He }\right]\)

Energy released in \(1\) reaction \(\Rightarrow \Delta mc ^{2}\)

In use of \(7.016\, u\) \(Li\) energy is \(\Delta mc ^{2}\)

In use of \(1\, gm\) \(Li\) energy is \(\frac{\Delta mc ^{2}}{ m _{ Li }}\)

In use of \(20\, gm\) energy is \(\Rightarrow \frac{\Delta mc ^{2}}{ m _{ Li }} \times 20 gm\)

\(\Rightarrow \frac{[(7.016+1.0079)-2 \times 4.0026] u \times c ^{2}}{7.016 \times 1.6 \times 10^{-24} gm } \times 20 gm\)

\(\Rightarrow\left(\frac{0.0187 \times 1.6 \times 10^{-19} \times 10^{9}}{7.016 \times 1.6 \times 10^{-24} gm } \times 20 gm \right)\) Joule

\(\Rightarrow 0.05 \times 10^{+14} J\)

\(\Rightarrow 1.4 \times 10^{+6} kwh\)

\(\left[1 J \Rightarrow 2.778 \times 10^{-7} kwh \right]\)