Verified Solution • 2025–26 Session moderate

QJEEM24Phe29a: JEE Main 2024 Physics - Class 12 JEEM24Phe29a Solution: The explosive in a Hydrogen bomb is a mixture of and in some…

The explosive in a Hydrogen bomb is a mixture of 1H2,1H3{ }_1 \mathrm{H}^2,{ }_1 \mathrm{H}^3 and 3Li6{ }_3 \mathrm{Li}^6 in some condensed form. The chain reaction is given by

amp;3Li6+0n1→2He4+1H3amp;1H2+1H3→2He4+0n1\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 ; M(Li)=6.01690 amu,M(1H2)=2.01471 amu,M(2He4)=4.00388\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 amu\mathrm{amu} , and 1 amu=931.5 MeV]1 \mathrm{~amu}=931.5 \mathrm{~MeV}]

Multiple Choice Options

  • A 22.22 MeV
  • B 28.12 MeV
  • C 16.48 MeV
  • D 12.64 MeV

Step-by-Step Verified Solution

3Li6+0n1→2He4+1H31H2+1H3→2He4+0n13Li6+1H2→2(2He4)

Energy released in process

amp;Q=Δmc2amp;Q=[M(Li)+M(H2)−2×M(2He4)]×931.5 MeVamp;Q=[6.01690+2.01471−2×4.00388]×931.5 MeVamp;Q=22.216 MeVamp;Q=22.22 MeV\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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