Kinematics and Energetics of Binary Nuclear Fission

To determine the correct statements, we analyze the energetics and kinematics of the binary nuclear fission reaction $N \rightarrow P + Q$, where the heavy parent nucleus $N$ is initially at rest.

1. Conservation of Energy

The total energy before the fission process is equal to the total energy after the process. $\text{Total initial energy} = M_N c^2$ $\text{Total final energy} = M_P c^2 + M_Q c^2 + E_P + E_Q$

Applying the law of conservation of energy: $M_N c^2 = M_P c^2 + M_Q c^2 + E_P + E_Q$

Rearranging the terms in terms of the mass defect $\delta = M_N - M_P - M_Q$: $E_P + E_Q = (M_N - M_P - M_Q) c^2 = \delta c^2$

This confirms that Statement (A) is correct.

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2. Conservation of Linear Momentum

Since the parent nucleus $N$ is initially at rest, the total initial momentum of the system is zero. By conservation of linear momentum: $\vec{p}_P + \vec{p}_Q = 0 \implies \vec{p}_P = -\vec{p}_Q$

Taking the magnitude of momentum for both daughter nuclei: $|\vec{p}_P| = |\vec{p}_Q| = p$

Using the non-relativistic expression for momentum ($p = M v$): $M_P v_P = M_Q v_Q \implies \frac{v_P}{v_Q} = \frac{M_Q}{M_P}$

This confirms that Statement (C) is correct.

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3. Kinetic Energies and Reduced Mass

The kinetic energy of each daughter nucleus in terms of momentum $p$ is given by: $E_P = \frac{p^2}{2 M_P} \quad \text{and} \quad E_Q = \frac{p^2}{2 M_Q}$

The total kinetic energy is: $E_P + E_Q = \frac{p^2}{2 M_P} + \frac{p^2}{2 M_Q} = \frac{p^2}{2} \left( \frac{1}{M_P} + \frac{1}{M_Q} \right) = \frac{p^2}{2} \left( \frac{M_P + M_Q}{M_P M_Q} \right)$

Defining the reduced mass $\mu = \frac{M_P M_Q}{M_P + M_Q}$, we have: $E_P + E_Q = \frac{p^2}{2\mu}$

Equating this to the total energy released $\delta c^2$: $\frac{p^2}{2\mu} = \delta c^2 \implies p^2 = 2 \mu \delta c^2 \implies p = c \sqrt{2 \mu \delta}$

This confirms that Statement (D) is correct.

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4. Kinetic Energy of $P$

Using the magnitude of momentum derived above, the kinetic energy of fragment $P$ is: $E_P = \frac{p^2}{2 M_P} = \frac{2 \mu \delta c^2}{2 M_P} = \frac{\mu}{M_P} \delta c^2 = \left( \frac{M_Q}{M_P + M_Q} \right) \delta c^2$

Therefore, Statement (B) is incorrect because it has $M_P$ instead of $M_Q$ in the numerator.

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Conclusion

The correct options are A, C, and D.