Nuclear Physics Question 5
Question 5 - 2024 (29 Jan Shift 1)
The explosive in a Hydrogen bomb is a mixture of ${ } _1 H^{2},{ } _1 H^{3}$ and ${ } _3 Li^{6}$ in some condensed form. The chain reaction is given by
${ } _5 Li^{\dagger}+{ } _0 n^{1} \rightarrow{ } _2 He^{4}+{ } _1 H^{3}$
${ } _1 H^{2}+{ } _1 H^{3} \rightarrow{ } _2 He^{4}+{ } _0 n^{1}$
During the explosion the energy released is approximately
$[\text{Given} : M(Li)=6.01690 amu . M\left(H^{2}\right)=2.01471 amu . M\left({ } _2 He^{4}\right)=4.00388 amu$, and 1 $amu=931.5 MeV]$
(1) $28.12 MeV$
(2) $12.64 MeV$
(3) $16.48 MeV$
(4) $22.22 MeV$
Show Answer
Answer: (4)
Solution:
$$ \begin{aligned} & { } _3 Li^{6}+{ } _0 n^{1} \rightarrow{ } _2 He^{4}+{ } _1 H^{3} \\ & H^{2}+{ } _1 H^{3} \rightarrow{ } _2 He^{4}+{ } _0 n^{1} \\ & \hline{ } _3 Li^{6}+{ } _1 H^{2} \rightarrow 2\left({ } _2 He^{4}\right) \end{aligned} $$
Energy released in process
$Q=\Delta mc^{2}$
$Q=\left[M(Li)+M\left(H^{2}\right)-2 \times M\left(H _2 He^{4}\right)\right] \times 931.5 MeV$
$Q=[6.01690+2.01471-2 \times 4.00388] \times 931.5 MeV$
$Q=22.216 MeV$
$Q=22.22 MeV$
