# You are given that mass

Question:

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

1. (1) $4.5 \times 10^{5}$

2. (2) $8 \times 10^{6}$

3. (3) $6.82 \times 10^{5}$

4. (4) $1.33 \times 10^{6}$

Correct Option: , 4

Solution:

(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 $=480 \times 10^{10} \times 2.778 \times 10^{-7}$

$=1.33 \times 10^{6} \mathrm{kWh}$