Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below:
$$_1^2X + _1^2X = _2^4Y$$
The binding energies per nucleon of $$_1^2X$$ and $$_2^4Y$$ are $$1.1 \text{ MeV}$$ and $$7.6 \text{ MeV}$$ respectively. The energy released in this process is ______ MeV.

Given the nuclear fusion reaction $$ _1^2X + \, _1^2X \rightarrow \, _2^4Y$$ and the binding energy per nucleon of $$_1^2X = 1.1 \text{ MeV}$$ and of $$_2^4Y = 7.6 \text{ MeV}$$.

Since each $$_1^2X$$ has 2 nucleons, its total binding energy is $$2 \times 1.1 = 2.2 \text{ MeV}$$, so for two such nuclei $$\text{Total BE}_{reactants} = 2 \times 2.2 = 4.4 \text{ MeV}$$.

Meanwhile, $$_2^4Y$$ has 4 nucleons, so its total binding energy is $$\text{Total BE}_{product} = 4 \times 7.6 = 30.4 \text{ MeV}$$.

Using the relation Energy released = Total BE of products - Total BE of reactants gives $$= 30.4 - 4.4 = 26 \text{ MeV}$$.

The energy released in this process is $$\textbf{26}$$ MeV.