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JEE Main 2024 (29 Jan, Shift 1) · Nuclear Fission & Fusion

Nuclear Fission & Fusion question from JEE Main 2024 (29 Jan, Shift 1)

Q11JEE Main 2024 · 29 JAN · SHIFT 1Medium

Nuclear Fission & Fusion

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

3Li6+0n1→2He4+1H3{}_{3}\text{Li}^{6} + {}_{0}\text{n}^{1} \rightarrow {}_{2}\text{He}^{4} + {}_{1}\text{H}^{3}

1H2+1H3→2He4+0n1{}_{1}\text{H}^{2} + {}_{1}\text{H}^{3} \rightarrow {}_{2}\text{He}^{4} + {}_{0}\text{n}^{1}

During the explosion the energy released is approximately

[Given; M(Li)=6.01690 amu,M(1H2)=2.01471 amu,M(2He4)=4.00388 amu\text{M(Li)} = 6.01690\text{ amu}, \text{M}\left({}_{1}\text{H}^{2}\right) = 2.01471\text{ amu}, \text{M}\left({}_{2}\text{He}^{4}\right) = 4.00388\text{ amu}, and 1 amu=931.5 MeV]1\text{ amu} = 931.5\text{ MeV}]

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Answer: (a)

  1. Adding the two reactions, nn and 3H{}^3\text{H} cancel: 6Li+2H→2 4He{}^6\text{Li} + {}^2\text{H} \to 2\,{}^4\text{He}.
  2. Δm=6.01690+2.01471−2×4.00388=8.03161−8.00776=0.02385 amu\Delta m = 6.01690 + 2.01471 - 2\times4.00388 = 8.03161 - 8.00776 = 0.02385\text{ amu}.
  3. E=0.02385×931.5≈22.22 MeVE = 0.02385\times931.5 \approx 22.22\text{ MeV}.
  4. So the answer is (a) 22.22 MeV22.22\text{ MeV}.
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