Two nuclei of mass number $3$ combine with another nucleus of mass number $4$ to yield a nucleus of mass number $10$. If the binding energy per nucleon for the mass numbers $3$, $4$ and $10$ are $5.6$ MeV, $7.4$ MeV and $6.1$ MeV, respectively, then in the process, $\Delta Mc^2 = $ _____ MeV.

The nuclear reaction is given by:

$2 X^{3} + Y^{4} \rightarrow Z^{10}$

The total binding energy of the reactants is the sum of the binding energies of the two nuclei of mass number $3$ and the nucleus of mass number $4$.

Binding energy of reactants $= 2 \times (3 \times 5.6) + (4 \times 7.4)$

$= 33.6 + 29.6 = 63.2$ MeV

The total binding energy of the product nucleus of mass number $10$ is:

Binding energy of product $= 10 \times 6.1 = 61.0$ MeV

The change in mass energy $\Delta M c^2$ is equal to the difference in the total binding energies of the reactants and the product.

$\Delta M c^2 = |63.2 - 61.0| = 2.2$ MeV

Answer: $2.2$