Question

You are given that Mass of ${}_3^{7}Li=7.0160u$
Mass of ${}_2^{4}He=4.0026u$
and Mass of ${}_1^{1}H=1.0079u$
When $20g$ of ${}_3^{7}Li$ is converted into ${}_2^{4}He$ by proton capture, the energy liberated, (in kWh), is: [Mass of nudeon $=1GeV/c^2]$

• 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)

${}_3^{7}Li+{}_1^{1}H\to 2\left({}_2^{4}He\right)$
$\Delta m\Rightarrow \left[m_{Li}+m_H\right]-2\left[M_{He}\right]$
Energy released in 1 reaction $\Rightarrow \Delta m{c}^2$
In use of $7.016uLi$ energy is $\Delta m{c}^2$
In use of $\mathrm{lgm}Li$ energy is $\frac{\Delta m{c}^2}{m_{Li}}$
In use of $20gm$ energy is $\Rightarrow \frac{\Delta m{c}^2}{m_{Li}}\times 20gm$
$\Rightarrow \frac{[(7.016+1.0079)-2\times 4.0026]u\times c^2}{7.016\times 1.6\times 10^{-24}gm}\times 20gm$
$\Rightarrow \left(\frac{0.0187\times 1.6\times 10^{-19}\times 10^9}{7.016\times 1.6\times 10^{-24}gm}\times 20gm\right)$ Joule
$\Rightarrow 0.05\times 10^{+14}J$
$\Rightarrow 1.4\times 10^{+6}kwh$
$\left[1J\Rightarrow 2.778\times 10^{-7}kwh\right]$