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Atoms and Nuclei question

2024 · 29 Jan · Shift 1 · Q80
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Atoms and Nuclei question

2024 · 29 Jan · Shift 1 · Q80

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
The explosive in a Hydrogen bomb is a mixture of 1H2,1H3{ }_1 \mathrm{H}^2,{ }_1 \mathrm{H}^31​H2,1​H3 and 3Li6{ }_3 \mathrm{Li}^63​Li6 in some condensed form. The chain reaction is given by 3Li6+0n1→2He4+1H31H2+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}​3​Li6+0​n1→2​He4+1​H31​H2+1​H3→2​He4+0​n1​ 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)=4.00388M(Li)=6.01690 amu,M(1​H2)=2.01471 amu,M(2​He4)=4.00388 amu\mathrm{amu}amu, and 1 amu=931.5 MeV]1 \mathrm{~amu}=931.5 \mathrm{~MeV}]1 amu=931.5 MeV]
  1. A
    22.22 MeV
  2. B
    28.12 MeV
  3. C
    16.48 MeV
  4. D
    12.64 MeV
View written solutionFree

Correct answer: A

  1. Find the net reaction

Given chain reactions:

3Li6+0n1→2He4+1H3{}_3\mathrm{Li}^6+{}_0\mathrm{n}^1 \rightarrow {}_2\mathrm{He}^4+{}_1\mathrm{H}^33​Li6+0​n1→2​He4+1​H3 1H2+1H3→2He4+0n1{}_1\mathrm{H}^2+{}_1\mathrm{H}^3 \rightarrow {}_2\mathrm{He}^4+{}_0\mathrm{n}^11​H2+1​H3→2​He4+0​n1

Add the two equations. The neutron (0n1)({}_0n^1)(0​n1) and tritium (1H3)({}_1H^3)(1​H3) cancel from both sides.

So the net reaction is:

3Li6+1H2→2 2He4{}_3\mathrm{Li}^6+{}_1\mathrm{H}^2 \rightarrow 2\,{}_2\mathrm{He}^43​Li6+1​H2→22​He4
  1. Use mass defect

Initial mass:

Mi=M(3Li6)+M(1H2)=6.01690+2.01471=8.03161 amuM_i=M({}_3\mathrm{Li}^6)+M({}_1\mathrm{H}^2)=6.01690+2.01471=8.03161\,\text{amu}Mi​=M(3​Li6)+M(1​H2)=6.01690+2.01471=8.03161amu

Final mass:

Mf=2×M(2He4)=2×4.00388=8.00776 amuM_f=2\times M({}_2\mathrm{He}^4)=2\times 4.00388=8.00776\,\text{amu}Mf​=2×M(2​He4)=2×4.00388=8.00776amu

Mass defect:

Δm=Mi−Mf=8.03161−8.00776=0.02385 amu\Delta m=M_i-M_f=8.03161-8.00776=0.02385\,\text{amu}Δm=Mi​−Mf​=8.03161−8.00776=0.02385amu
  1. Convert mass defect into energy

Given:

1 amu=931.5 MeV1\,\text{amu}=931.5\,\text{MeV}1amu=931.5MeV

Hence,

Q=Δm×931.5=0.02385×931.5 MeVQ=\Delta m\times 931.5=0.02385\times 931.5\,\text{MeV}Q=Δm×931.5=0.02385×931.5MeV Q≈22.22 MeVQ\approx 22.22\,\text{MeV}Q≈22.22MeV
  1. Match with options
22.22 MeV\boxed{22.22\,\text{MeV}}22.22MeV​

So the correct option is:

A

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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