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 reaction is
$$\;^3X + \;^3X + \;^4Y \;\longrightarrow\; ^{10}Z$$

First write the total binding energy (BE) of each nucleus.
Binding energy of a nucleus = (binding energy per nucleon) $$\times$$ (mass number).

For mass number $$3$$:
Total BE = $$5.6\;\text{MeV per nucleon}\times 3 = 16.8\;\text{MeV}$$.
There are two such nuclei, so initial contribution = $$2 \times 16.8 = 33.6\;\text{MeV}$$.

For mass number $$4$$:
Total BE = $$7.4\;\text{MeV per nucleon}\times 4 = 29.6\;\text{MeV}$$.

Therefore, total initial binding energy is
$$E_{\text{initial}} = 33.6 + 29.6 = 63.2\;\text{MeV}$$.

For the product nucleus of mass number $$10$$:
Total BE = $$6.1\;\text{MeV per nucleon}\times 10 = 61.0\;\text{MeV}$$.

The change in binding energy (final minus initial) is
$$\Delta E = E_{\text{final}} - E_{\text{initial}} = 61.0 - 63.2 = -2.2\;\text{MeV}.$$

A negative value means the final nucleus is less tightly bound; energy of $$2.2\;\text{MeV}$$ must be supplied, which equals the increase in mass-energy of the system:
$$\Delta M c^2 = 2.2\;\text{MeV}.$$

Hence, the required value is $$2.2\;\text{MeV}$$.
Option C which is: $$2.2$$