Nucleus $$A$$ having $$Z = 17$$ and equal number of protons and neutrons has 1.2 MeV binding energy per nucleon. Another nucleus $$B$$ of $$Z = 12$$ has total 26 nucleons and 1.8 MeV binding energy per nucleons. The difference of binding energy of $$B$$ and $$A$$ will be ______ MeV.

The binding energy of a nucleus is obtained by multiplying its binding energy per nucleon with its total number of nucleons (mass number $$A$$).

For nucleus $$A$$: $$Z = 17$$ and the number of neutrons equals the number of protons.
Hence, mass number $$A_A = Z + N = 17 + 17 = 34$$.

Binding energy per nucleon of nucleus $$A$$ is $$1.2 \text{ MeV}$$, so the total binding energy is
$$BE_A = A_A \times (\text{binding energy per nucleon}) = 34 \times 1.2 = 40.8 \text{ MeV}$$.

For nucleus $$B$$: mass number $$A_B = 26$$ (given) and binding energy per nucleon is $$1.8 \text{ MeV}$$.
Therefore,
$$BE_B = A_B \times (\text{binding energy per nucleon}) = 26 \times 1.8 = 46.8 \text{ MeV}$$.

The required difference in binding energy is
$$BE_B - BE_A = 46.8 \text{ MeV} - 40.8 \text{ MeV} = 6.0 \text{ MeV}$$.

Hence, the difference of binding energy of nucleus $$B$$ and nucleus $$A$$ is $$6 \text{ MeV}$$.