Solve this following

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: [Mass of nudeon $=1 \mathrm{GeV} / \mathrm{c}^{2}$ ]

 

  1. $8 \times 10^{6}$

  2. $1.33 \times 10^{6}$

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

  4. $4.5 \times 10^{5}$


Correct Option: , 2

Solution:

${ }_{3}^{7} \mathrm{Li}+{ }_{1}^{1} \mathrm{H} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$

$\Delta \mathrm{m} \Rightarrow\left[\mathrm{m}_{\mathrm{Li}}+\mathrm{m}_{\mathrm{H}}\right]-2\left[\mathrm{M}_{\mathrm{He}}\right]$

Energy released in 1 reaction $\Rightarrow \Delta \mathrm{mc}^{2}$.

In use of $7.016 \mathrm{u} \mathrm{Li}$ energy is $\Delta \mathrm{mc}^{2}$

In use of $1 \mathrm{gm}$ Li energy is $\frac{\Delta \mathrm{mc}^{2}}{\mathrm{~m}_{\mathrm{Li}}}$

In use of $20 \mathrm{gm}$ energy is $\Rightarrow \frac{\Delta \mathrm{mc}^{2}}{\mathrm{~m}_{\mathrm{Li}}} \times 20 \mathrm{gm}$

$\Rightarrow \frac{[(7.016+1.0079)-2 \times 4.0026] \mathrm{u} \times \mathrm{c}^{2}}{7.016 \times 1.6 \times 10^{-24} \mathrm{gm}} \times 20 \mathrm{gm}$

$\Rightarrow\left(\frac{0.0187 \times 1.6 \times 10^{-19} \times 10^{9}}{7.016 \times 1.6 \times 10^{-24} \mathrm{gm}} \times 20 \mathrm{gm}\right)$ Joule

$\Rightarrow 0.05 \times 10^{+14} \mathrm{~J}$

$\Rightarrow 1.4 \times 10^{+6} \mathrm{kwh}$

$\left[1 \mathrm{~J} \Rightarrow 2.778 \times 10^{-7} \mathrm{kwh}\right]$

Ans. (2)

 

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