Question

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 $\left.=1 GeV / c ^{2}\right]$

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

Answer 1

Answer: (B)

${ }_{3}^{7} Li +{ }_{1}^{1} H \rightarrow 2\left({ }_{2}^{4} He \right)$
$\Delta m \Rightarrow \left[ m _{ Li }+ m _{ H }\right]-2\left[ M _{ He }\right]$
Energy released in 1 reaction $\Rightarrow \Delta mc ^{2}$
In use of $7.016 u Li $ energy is $\Delta m c^{2}$
In use of $\operatorname{lgm} Li$ 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
$\Rightarrow 0.05 \times 10^{+14} J$
$\Rightarrow 1.4 \times 10^{+6} kwh$
$\left[1 J \Rightarrow 2.778 \times 10^{-7} kwh \right]$