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 $_{3}Li^{6}+_{0}n^{1}\rightarrow_{2}He^{4}+_{1}H^{3}$ $_{1}H^{2}+_{1}H^{3}\rightarrow_{2}He^{4}+_{0}n^{1}$ During the explosion the energy released is approximately [Given : $M(Li)=6.01690$ amu. $M(_{1}H^{2})=2.01471$ amu. $M(_{2}He^{4})=4.00388$ amu, and $1$ amu$=931.5$ MeV]

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Modern Physics

The explosive in a Hydrogen bomb is a mixture of

_{1} H^{2}

_{1} H^{3}

and

_{3} L i^{6}

in some condensed form. The chain reaction is given by

_{3} L i^{6} +_{0} n^{1} \to_{2} H e^{4} +_{1} H^{3}

_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} +_{0} n^{1}

During the explosion the energy released is approximately
[Given :

M (L i) = 6.01690

amu.

M (_{1} H^{2}) = 2.01471

amu.

M (_{2} H e^{4}) = 4.00388

amu, and

1

amu

= 931.5

MeV]

Answer: (D)

Net reaction:

_{3} Li^{6} +_{1} H^{2} \to 2_{2} He^{4}

(adding the two given steps and cancelling the neutron and

_{1} H^{3}

).

Mass defect:

Δ m = M (Li) + M (_{1} H^{2}) - 2 M (_{2} He^{4}) = 6.01690 + 2.01471 - 2 (4.00388)

= 8.03161 - 8.00776 = 0.02385 amu

Energy released:

Q = 0.02385 \times 931.5 = 22.22 MeV

Key confirmed: D.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option