You are given that Mass of ${}_3^7Li$ = 7.0160u,
Mass of ${}_2^4He$ = 4.0026u
and Mass of ${}_1^1H$ = 1.0079u.
When 20 g of ${}_3^7Li$ is converted into ${}_2^4He$ by proton capture, the energy liberated, (in kWh), is :
[Mass of nucleon = 1 GeV/c²]

${}_3^7Li$ + ${}_1^1H$ $\to$ 2(${}_2^4He$) <br><br>$\Delta$m = $\left[ {{M_{Li}} + {M_H}} \right] - 2\left[ {{M_{He}}} \right]$ <br><br>= (7.0160 + 1.0079) - 2 $\times$ 4.0003 <br><br>= 0.0187 <br><br>Energy released in 1 reaction = $\Delta$mc² <br><br>In use of 7.016 u Li energy is $\Delta$mc². <br><br>In use of 1 gm Li energy is ${{\Delta m{c^2}} \over {{m_{Li}}}}$. <br><br>In use of 20 gm energy is = ${{\Delta m{c^2}} \over {{m_{Li}}}} \times 20$ <br><br>= $${{0.087 \times 1.6 \times {{10}^{ - 19}} \times {{10}^9}} \over {7.016 \times 1.6 \times {{10}^{ - 24}}}} \times 20$$ J <br><br>= 0.05 $\times$ 10¹⁴ J <br><br>= 1.33 $\times$ 10⁶ kWh