Kinetic Energy and Particle Conservation in Nuclear Fission of Uranium

To determine the correct choice, we analyze the fission reaction using conservation laws (atomic number $Z$, mass number $A$, and momentum) and energy considerations.

1. Conservation of Mass Number ($A$) and Atomic Number ($Z$)

The reaction is represented as: $^{236}_{92}\text{U} \rightarrow ^{140}_{54}\text{Xe} + ^{94}_{38}\text{Sr} + x + y$

• Conservation of Mass Number ($A$): $236 = 140 + 94 + A_x + A_y \implies A_x + A_y = 236 - 234 = 2$

• Conservation of Atomic Number ($Z$): $92 = 54 + 38 + Z_x + Z_y \implies Z_x + Z_y = 92 - 92 = 0$

Since $A_x + A_y = 2$ and $Z_x + Z_y = 0$, the particles $x$ and $y$ must both be neutrons ($n$), which have $A = 1$ and $Z = 0$. Thus, $x = n$ and $y = n$.

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2. $Q$-value of the Reaction and Total Kinetic Energy

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

• $\text{BE}(^{236}_{92}\text{U}) = 236 \times 7.5\text{ MeV} = 1770\text{ MeV}$
• $\text{BE}(^{140}_{54}\text{Xe}) = 140 \times 8.5\text{ MeV} = 1190\text{ MeV}$
• $\text{BE}(^{94}_{38}\text{Sr}) = 94 \times 8.5\text{ MeV} = 799\text{ MeV}$
• $\text{BE}(\text{neutrons}) = 0$

The total energy released ($Q$-value) in the reaction is the difference in binding energies: $Q = \text{BE}_{\text{products}} - \text{BE}_{\text{reactants}}$ $Q = (1190 + 799) - 1770 = 1989 - 1770 = 219\text{ MeV}$

The total kinetic energy of all emitted products equals $Q$: $K_{\text{Xe}} + K_{\text{Sr}} + K_x + K_y = Q$

Given $K_x = 2\text{ MeV}$ and $K_y = 2\text{ MeV}$: $K_{\text{Xe}} + K_{\text{Sr}} + 2 + 2 = 219$ $K_{\text{Xe}} + K_{\text{Sr}} = 215\text{ MeV}$

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3. Distribution of Kinetic Energy

Assuming the two emitted neutrons have opposite momenta such that their net momentum is zero (or negligible compared to the total momentum transferred to the fragments), conservation of linear momentum gives: $|\vec{p}_{\text{Xe}}| \approx |\vec{p}_{\text{Sr}}| = p$

Since kinetic energy $K = \frac{p^2}{2m}$, the ratio of kinetic energies of the two fragments is inversely proportional to their mass numbers: $\frac{K_{\text{Sr}}}{K_{\text{Xe}}} = \frac{A_{\text{Xe}}}{A_{\text{Sr}}} = \frac{140}{94}$

Using $K_{\text{Xe}} + K_{\text{Sr}} = 215\text{ MeV}$: $K_{\text{Sr}} = 215 \times \frac{140}{140 + 94} = 215 \times \frac{140}{234} \approx 128.63\text{ MeV} \approx 129\text{ MeV}$ $K_{\text{Xe}} = 215 \times \frac{94}{140 + 94} = 215 \times \frac{94}{234} \approx 86.37\text{ MeV} \approx 86\text{ MeV}$

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Conclusion

• $x = n$, $y = n$
• $K_{\text{Sr}} = 129\text{ MeV}$
• $K_{\text{Xe}} = 86\text{ MeV}$

This corresponds to Option A.