JEE Main 2024 — 27 January, Shift 1 — Question 52

Let: Mass number of the initial high mass nuclide, $A_{\text{initial}} = 236$. Binding energy per nucleon of the initial nuclide, $BE_{\text{initial}} = 7.6 \text{ MeV/Nucleon}$. Mass number of the middle mass nuclides (products), $A_{\text{final}} = 118$. Binding energy per nucleon of the middle mass nuclides, $BE_{\text{final}} = 8.6 \text{ MeV/Nucleon}$.

The fission process involves a high mass nuclide splitting into two middle mass nuclides. Therefore, the total mass number of the products is $2 \times 118 = 236$, which is consistent with the initial mass number.

The total binding energy of a nucleus is given by the product of its mass number ($A$) and its binding energy per nucleon ($BE_{\text{per nucleon}}$).

Total binding energy of the initial nuclide (reactant):

$$BE_{\text{total, initial}} = A_{\text{initial}} \times BE_{\text{initial}}$$ $$BE_{\text{total, initial}} = 236 \times 7.6 \text{ MeV}$$ $$BE_{\text{total, initial}} = 1793.6 \text{ MeV}$$

Total binding energy of the product nuclides (two fragments):

$$BE_{\text{total, final}} = 2 \times (A_{\text{final}} \times BE_{\text{final}})$$ $$BE_{\text{total, final}} = 2 \times (118 \times 8.6) \text{ MeV}$$ $$BE_{\text{total, final}} = 236 \times 8.6 \text{ MeV}$$ $$BE_{\text{total, final}} = 2030.16 \text{ MeV}$$

Energy released ($Q$) in the fission process: The energy released is the difference between the total binding energy of the products and the total binding energy of the reactants:

$$Q = BE_{\text{total, final}} - BE_{\text{total, initial}}$$ $$Q = 2030.16 \text{ MeV} - 1793.6 \text{ MeV}$$ $$Q = 236.56 \text{ MeV}$$

Rounding to one decimal place, the energy released is $236.6 \text{ MeV}$.

The final answer is $\boxed{236.6}$.