Question

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

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

Answer 1

The correct option is D

$1.33\times {10}^6$

Step 1: Given Data

Mass of lithium, ${\mathrm{m}}_{\mathrm{Li}}=7.0160\mathrm{u}$

Mass of helium, ${\mathrm{m}}_{\mathrm{He}}=4.0026\mathrm{u}$

Mass of hydrogen, ${\mathrm{m}}_{\mathrm{H}}=1.0079\mathrm{u}$

Mass of nucleon $=\frac{GeV}{c^2}$

Step 2: To find

Energy liberated in $\mathrm{kWh}$ when $20\mathrm{g}$ of $\mathrm{Li}$ is converted to $\mathrm{He}$

Step 3: Formulae and applied concept:

Reaction ${\mathrm{Li}}_3^7+{\mathrm{H}}_1^1\to 2{\mathrm{He}}_2^4$

Change in mass,

$$\begin{gathered} ∆\mathrm{m}=\left({\mathrm{m}}_{\mathrm{Li}}+{\mathrm{m}}_{\mathrm{H}}\right)-2{\mathrm{m}}_{\mathrm{He}} \\ \Rightarrow ∆\mathrm{m}=\left(7.0160+1.0079\right)-2\left(4.0003\right) \\ \Rightarrow ∆\mathrm{m}=0.0187\mathrm{u} \end{gathered}$$

Energy liberated in reaction$=∆{\mathrm{mc}}^2$

For $7.0160\mathrm{u}\mathrm{Li}$, energy liberated $=∆{\mathrm{mc}}^2$

$1\mathrm{g}\mathrm{Li}$ liberates $=\frac{∆{\mathrm{mc}}^2}{{\mathrm{m}}_{\mathrm{Li}}}$

$20\mathrm{g}\mathrm{Li}$ liberates$=\frac{∆{\mathrm{mc}}^2}{{\mathrm{m}}_{\mathrm{Li}}}\times 20$

$=\frac{0.0187\times 931.5\times {10}^6\times 1.6\times {10}^{-19}\times (20/7)\times 6.023\times {10}^{23}}{36\times {10}^5}$

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

Therefore, the energy liberated is $1.33\times {10}^6\mathrm{kWh}$

Hence, the correct answer is option D.