QJEEM24Ph984f: JEE Main 2024 Physics - Class 12 JEEM24Ph984f Solution: In a nuclear fission process, a high mass nuclide ( A ≈ 236 )…

To determine the energy released in a nuclear fission process, we use the difference in binding energy (BE) before and after the fission. The formula for energy released ($$Q$$ value) in the process is given by:

$$Q=(\text{Total BE of products})-(\text{Total BE of reactants})$$

In this case, the reactant is a high mass nuclide with atomic mass $A \approx 236$ and a binding energy of $$7.6\mathrm{MeV}/\mathrm{nucleon}$$. Each of the two middle mass nuclides formed as products has atomic mass $A \approx 118$ and a binding energy of $$8.6\mathrm{MeV}/\mathrm{nucleon}$$.

Therefore, we calculate the total binding energy of reactant and products as follows:

For reactant:

$${\text{BE}}_{\text{reactant}}=236\times 7.6\mathrm{MeV}$$

For products (since there are two identical products):

$${\text{BE}}_{\text{products}}=2\times (118\times 8.6)\mathrm{MeV}$$

Thus, the energy released ($$Q$$ value) is:

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

$$Q=2(118\times 8.6)-(236\times 7.6)$$

$$Q=236\times (8.6-7.6)$$

$$Q=236\times 1$$

$$Q=236\mathrm{MeV}$$

This calculation demonstrates how the difference in binding energy per nucleon before and after fission leads to the release of energy, consistent with the mass-energy equivalence principle.