You are given that Mass of ${ }_{3}^{7} \mathrm{Li}=7.0160 \mathrm{u}$,
Mass of ${ }_{2}^{4} \mathrm{He}=4.0026 \mathrm{u}$
and Mass of ${ }_{1}^{1} \mathrm{H}=1.0079 \mathrm{u}$.
When $20 \mathrm{~g}$ of ${ }_{3}^{7} \mathrm{Li}$ is converted into ${ }_{2}^{4} \mathrm{He}$ by
proton capture, the energy liberated, (in $\mathrm{kWh}$ ), is: [Mass of nudeon $=1 \mathrm{GeV} / \mathrm{c}^{2}$ ]
Correct Option: , 2
${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$
$\Delta \mathrm{m} \Rightarrow\left[\mathrm{m}_{\mathrm{Li}}+\mathrm{m}_{\mathrm{H}}\right]-2\left[\mathrm{M}_{\mathrm{He}}\right]$
Energy released in 1 reaction $\Rightarrow \Delta \mathrm{mc}^{2}$.
In use of $7.016 \mathrm{u} \mathrm{Li}$ energy is $\Delta \mathrm{mc}^{2}$
In use of $1 \mathrm{gm}$ Li energy is $\frac{\Delta \mathrm{mc}^{2}}{\mathrm{~m}_{\mathrm{Li}}}$
In use of $20 \mathrm{gm}$ energy is $\Rightarrow \frac{\Delta \mathrm{mc}^{2}}{\mathrm{~m}_{\mathrm{Li}}} \times 20 \mathrm{gm}$
$\Rightarrow \frac{[(7.016+1.0079)-2 \times 4.0026] \mathrm{u} \times \mathrm{c}^{2}}{7.016 \times 1.6 \times 10^{-24} \mathrm{gm}} \times 20 \mathrm{gm}$
$\Rightarrow\left(\frac{0.0187 \times 1.6 \times 10^{-19} \times 10^{9}}{7.016 \times 1.6 \times 10^{-24} \mathrm{gm}} \times 20 \mathrm{gm}\right)$ Joule
$\Rightarrow 0.05 \times 10^{+14} \mathrm{~J}$
$\Rightarrow 1.4 \times 10^{+6} \mathrm{kwh}$
$\left[1 \mathrm{~J} \Rightarrow 2.778 \times 10^{-7} \mathrm{kwh}\right]$
Ans. (2)
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