Question

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

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

Answer 1

The correct option is D $1.33\times {10}^6$

Given: ${}_3^{7}Li{+}_1^{1}H\to 2\left({}_2^{4}He\right)$
$\Delta m\to [m_{Li}+m_H]-2\left[M_{He}\right]$
Energy released $=\Delta m{c}^2$
In Use of $1gLi$ energy released $=\frac{\Delta m{c}^2}{m_{Li}}$
In Use of $20g$ energy released$=\frac{\Delta m{c}^2}{m_{Li}}\times 20g$
$=\frac{\left[\right(7.0160+1.0079)-2\times 4.0026]u\times c^2}{7.0160\times 1.6\times {10}^{-24}}\times 20g$

$=(\frac{0.0187\times 1.6\times {10}^{-19}\times {10}^9}{7.016\times 1.6\times {10}^{-24}}\times 20g)=480\times {10}^{10}\text{J}$

$\because 1\text{J}=2.778\times {10}^{-7}\text{kWh}$
$\therefore$ Energy released $=480\times {10}^{10}\times 2.778\times {10}^{-7}=1.33\times {10}^6\text{kWh}$