Nucleus $$A$$ is having mass number $$220$$ and its binding energy per nucleon is $$5.6$$ MeV. It splits in two fragments $$B$$ and $$C$$ of mass numbers $$105$$ and $$115$$. The binding energy of nucleons in $$B$$ and $$C$$ is $$6.4$$ MeV per nucleon. The energy $$Q$$ released per fission will be:

We are given a nucleus $$A$$ with mass number $$220$$ and binding energy per nucleon $$5.6$$ MeV. It splits into fragments $$B$$ (mass number $$105$$) and $$C$$ (mass number $$115$$), each with binding energy per nucleon $$6.4$$ MeV.

Formula: The energy released in fission is the difference between the total binding energy of the products and the total binding energy of the parent nucleus:

$$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{parent}}$$

Calculate the total binding energy of the parent nucleus $$A$$:: $$\text{BE}_A = 220 \times 5.6 = 1232 \text{ MeV}$$

Calculate the total binding energy of fragment $$B$$:: $$\text{BE}_B = 105 \times 6.4 = 672 \text{ MeV}$$

Calculate the total binding energy of fragment $$C$$:: $$\text{BE}_C = 115 \times 6.4 = 736 \text{ MeV}$$

Calculate the total binding energy of products:: $$\text{BE}_{\text{products}} = 672 + 736 = 1408 \text{ MeV}$$

Calculate the energy released:: $$Q = 1408 - 1232 = 176 \text{ MeV}$$

The correct answer is Option D: 176 MeV.