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: \(2.2\)