The explosive in a Hydrogen bomb is a mixture of ₁ H², ₁ H³ and ₃ Li⁶ in some condensed form. The chain reaction is given by
₃ Li⁶+ ₀ n¹ → ₂ He⁴+ ₁ H³ ; ₁ H²+ ₁ H³ → ₂ He⁴+ ₀ n¹
During the explosion the energy released is approximately
[Given ; M(Li)=6.01690 amu, M( ₁ H²)=2.01471 amu, M( ₂ He⁴)=4.00388 amu, and 1 amu=931.5 MeV]

Solution

Physics sir · 4 steps

Add the two reactions so the neutron and tritium cancel; the net change is Li-6 plus deuterium to two He-4.

1. Add the reactions. The neutron used in the first reaction is given back by the second. The tritium made in the first is used up in the second. Both cancel, so their masses are never needed.

   ₃Li⁶ + ₁H² → 2 ₂He⁴

2. Mass defect. Mass before is 6.01690 + 2.01471 = 8.03161 amu. Mass after is 2 × 4.00388 = 8.00776 amu. So Δ m = 8.03161 - 8.00776 = 0.02385 amu.
3. Convert to energy. E = 0.02385 × 931.5 MeV = 22.216 MeV.
4. So. The energy released is about 22.22 MeV, which is option (a).

The idea to keep. In a chain of nuclear reactions, add them like chemical equations first. Particles that appear on both sides cancel, and the mass defect of the net reaction gives the total energy.

Answer from the NTA final answer key. Solution by PrepWiser.