Atoms & Nuclei

JEE Main level, single correct, medium

223 of 239

Question 223 of 239

The explosive in a Hydrogen bomb is a mixture of H21,H31H21,H31 and Li63Li63 in some condensed form. The chain reaction is given by Li63+n10→He42+H31Li63+n10\rightarrow He42+H31; H21+H31→He42+n10H21+H31\rightarrow He42+n10
During the explosion the energy released is approximately [Given : M(Li)=6.01690amu,M(H21)=2.01471amu,M(He42)=4.00388amuM(Li)=6.01690amu,M\left(H21\right)=2.01471amu,M\left(He42\right)=4.00388amu and 1amu=931.5MeV1amu=931.5MeV]

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Answer: (D) 22.22MeV22.22MeV

Combining the reactions, it can be written that
Li63+n10→He42+H31Li63+n10\rightarrow He42+H31
H21+H31→He42+n10H21+H31\rightarrow He42+n10
_____________________ Li63+H21\rightarrow 2\left(He42\right)
Hence, the energy released in process can be calculated as follows
Q=Δmc2=[M(Li)+M(H21)−2×M(He42)]×931.5MeV=[6.01690+2.01471−2×4.00388]×931.5MeV=22.22MeV\begin{matrix}Q & = & \Delta m{c}^{2} \\ & = & \left[M(Li)+M\left(H21\right)-2\times M\left(He42\right)\right]\times 931.5MeV \\ & = & [6.01690+2.01471-2\times 4.00388]\times 931.5MeV \\ & = & 22.22MeV\end{matrix}

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