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.

We need to find the difference of binding energy of nucleus $$B$$ and nucleus $$A$$.

Nucleus $$A$$ has $$Z = 17$$ and equal number of protons and neutrons. So the number of neutrons $$N = 17$$ and the mass number $$A_A = Z + N = 17 + 17 = 34$$.

Binding energy per nucleon = 1.2 MeV.

$$BE_A = 34 \times 1.2 = 40.8 \text{ MeV}$$

Nucleus $$B$$ has $$Z = 12$$ and total 26 nucleons. Binding energy per nucleon = 1.8 MeV.

$$BE_B = 26 \times 1.8 = 46.8 \text{ MeV}$$

$$BE_B - BE_A = 46.8 - 40.8 = 6.0 \text{ MeV}$$

The correct answer is $$6$$ MeV.