JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 20)
The explosive in a Hydrogen bomb is a mixture of $${ }_1 \mathrm{H}^2,{ }_1 \mathrm{H}^3$$ and $${ }_3 \mathrm{Li}^6$$ in some condensed form. The chain reaction is given by
$$\begin{aligned} & { }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+{ }_1 \mathrm{H}^3 \\ & { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 \end{aligned}$$
During the explosion the energy released is approximately
[Given ; $$\mathrm{M}(\mathrm{Li})=6.01690 \mathrm{~amu}, \mathrm{M}\left({ }_1 \mathrm{H}^2\right)=2.01471 \mathrm{~amu}, \mathrm{M}\left({ }_2 \mathrm{He}^4\right)=4.00388$$ $$\mathrm{amu}$$, and $$1 \mathrm{~amu}=931.5 \mathrm{~MeV}]$$
Explanation
$$\begin{aligned} & { }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+{ }_1 \mathrm{H}^3 \\ & { }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^3 \rightarrow{ }_2 \mathrm{He}^4+{ }_0 \mathrm{n}^1 \\ & \hline{ }_3 \mathrm{Li}^6+{ }_1 \mathrm{H}^2 \rightarrow 2\left({ }_2 \mathrm{He}^4\right) \\ & \hline \end{aligned}$$
Energy released in process
$$\begin{aligned} & \mathrm{Q}=\Delta \mathrm{mc}^2 \\ & \mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left(\mathrm{H}^2\right)-2 \times \mathrm{M}\left({ }_2 \mathrm{He}^4\right)\right] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5 \mathrm{~MeV} \\ & \mathrm{Q}=22.216 \mathrm{~MeV} \\ & \mathrm{Q}=22.22 \mathrm{~MeV} \end{aligned}$$
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