In a nuclear fission process, a high mass nuclide $(A \approx 236)$ with binding energy $7.6 \mathrm{~MeV} /$ Nucleon dissociated into middle mass nuclides $(\mathrm{A} \approx 118)$, having binding energy of $8.6 \mathrm{~MeV} / \mathrm{Nucleon}$. The energy released in the process would be ______ $\mathrm{MeV}$.

<p>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:</p><p>$Q = (\text{Total BE of products}) - (\text{Total BE of reactants})$</p><p>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 $8.6 \mathrm{MeV}/\mathrm{nucleon}$.</p><p>Therefore, we calculate the total binding energy of reactant and products as follows:</p><p>For reactant:</p><p>$\text{BE}_{\text{reactant}} = 236 \times 7.6 \mathrm{MeV}$</p><p>For products (since there are two identical products):</p><p>$\text{BE}_{\text{products}} = 2 \times (118 \times 8.6) \mathrm{MeV}$</p><p>Thus, the energy released ($Q$ value) is:</p><p>$Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactant}}$</p><p>$Q = 2(118 \times 8.6) - (236 \times 7.6)$</p><p>$Q = 236 \times (8.6 - 7.6)$</p><p>$Q = 236 \times 1$</p><p>$Q = 236 \mathrm{MeV}$</p><p>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.</p>