The explosive in a Hydrogen bomb is a mixture of $$_1H^2$$, $$_1H^3$$ and $$_3Li^6$$ in some condensed form. The chain reaction is given by $$_3Li^6 + _0n^1 \rightarrow _2He^4 + _1H^3$$; $$_1H^2 + _1H^3 \rightarrow _2He^4 + _0n^1$$
During the explosion the energy released is approximately [Given: $$M(Li) = 6.01690$$ amu, $$M(_1H^2) = 2.01471$$ amu, $$M(_2He^4) = 4.00388$$ amu and $$1$$ amu $$= 931.5$$ MeV]

We need to find the total energy released in the hydrogen bomb chain reaction.

Reaction 1: $$_3Li^6 + _0n^1 \rightarrow \, _2He^4 + \, _1H^3$$

Reaction 2: $$_1H^2 + \, _1H^3 \rightarrow \, _2He^4 + \, _0n^1$$

Adding both reactions (the $$_1H^3$$ and $$_0n^1$$ cancel as they appear on opposite sides):

$$_3Li^6 + \, _1H^2 \rightarrow 2 \, _2He^4$$

Mass of reactants:

$$m_{reactants} = M(Li^6) + M(H^2) = 6.01690 + 2.01471 = 8.03161 \text{ amu}$$

Mass of products:

$$m_{products} = 2 \times M(He^4) = 2 \times 4.00388 = 8.00776 \text{ amu}$$

Mass defect:

$$\Delta m = m_{reactants} - m_{products} = 8.03161 - 8.00776 = 0.02385 \text{ amu}$$

Using the conversion factor $$1 \text{ amu} = 931.5$$ MeV:

$$E = \Delta m \times 931.5 = 0.02385 \times 931.5 = 22.22 \text{ MeV}$$

The correct answer is Option (4): 22.22 MeV.