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 :
$\left[\right.$ Mass of nucleon $\left.=1 \mathrm{GeV} / \mathrm{c}^{2}\right]$
Correct Option: , 4
(4) ${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \longrightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$
$\Delta m \rightarrow\left[m_{\mathrm{Li}}+m_{\mathrm{H}}\right]-2\left[M_{\mathrm{He}}\right]$
Energy released $=\Delta m c^{2}$
In use of $1 \mathrm{~g}$ Li energy released $=\frac{\Delta m c^{2}}{m_{\mathrm{Li}}}$
In use of $20 \mathrm{~g}$ energy released $=\frac{\Delta m c^{2}}{m_{\mathrm{Li}}} \times 20 \mathrm{~g}$
$=\frac{[(7.016+1.0079)-2 \times 4.0026] u \times c^{2}}{7.016 \times 1.6 \times 10^{-24}} \times 20 \mathrm{~g}$
$=\left(\frac{0.0187 \times 1.6 \times 10^{-19} \times 10^{9}}{7.016 \times 1.6 \times 10^{-24}} \times 20\right)=480 \times 10^{10} \mathrm{~J}$
$\because 1 \mathrm{~J}=2.778 \times 10^{-7} \mathrm{kWh}$
$\therefore$ Energy released $=480 \times 10^{10} \times 2.778 \times 10^{-7}$
$=1.33 \times 10^{6} \mathrm{kWh}$
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