Nuclear Physics Question 5
Question 5 - 2024 (29 Jan Shift 1)
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
${ }{5} \mathrm{Li}^{\dagger}+{ }{0} \mathrm{n}^{1} \rightarrow{ }{2} \mathrm{He}^{4}+{ }{1} \mathrm{H}^{3}$
${ }{1} \mathrm{H}^{2}+{ }{1} \mathrm{H}^{3} \rightarrow{ }{2} \mathrm{He}^{4}+{ }{0} \mathrm{n}^{1}$
During the explosion the energy released is approximately
[Given : $\mathrm{M}(\mathrm{Li})=6.01690 \mathrm{amu} . \mathrm{M}\left(\mathrm{H}^{2}\right)=2.01471 \mathrm{amu} . \mathrm{M}\left({ }_{2} \mathrm{He}^{4}\right)=4.00388 \mathrm{amu}$, and 1 $\mathrm{amu}=931.5$
$\mathrm{MeV}]$
(1) $28.12 \mathrm{MeV}$
(2) $12.64 \mathrm{MeV}$
(3) $16.48 \mathrm{MeV}$
(4) $22.22 \mathrm{MeV}$
Show Answer
Answer: (4)
Solution:
$$ \begin{aligned} & { }{3} \mathrm{Li}^{6}+{ }{0} \mathrm{n}^{1} \rightarrow{ }{2} \mathrm{He}^{4}+{ }{1} \mathrm{H}^{3} \ & \mathrm{H}^{2}+{ }{1} \mathrm{H}^{3} \rightarrow{ }{2} \mathrm{He}^{4}+{ }{0} \mathrm{n}^{1} \ & \hline{ }{3} \mathrm{Li}^{6}+{ }{1} \mathrm{H}^{2} \rightarrow 2\left({ }{2} \mathrm{He}^{4}\right) \end{aligned} $$
Energy released in process
$\mathrm{Q}=\Delta \mathrm{mc}^{2}$
$\mathrm{Q}=\left[\mathrm{M}(\mathrm{Li})+\mathrm{M}\left(\mathrm{H}^{2}\right)-2 \times \mathrm{M}\left(\mathrm{H}_{2} \mathrm{He}^{4}\right)\right] \times 931.5 \mathrm{MeV}$
$\mathrm{Q}=[6.01690+2.01471-2 \times 4.00388] \times 931.5 \mathrm{MeV}$
$\mathrm{Q}=22.216 \mathrm{MeV}$
$\mathrm{Q}=22.22 \mathrm{MeV}$
