1. Let the mass and initial momentum of nucleus $A$ be: $M \text{ and } p_A$ Since its de-Broglie wavelength is $\lambda_A$, $\lambda_A = \frac{h}{p_A}$

2. After fission: The nucleus splits into two equal masses, so each fragment has mass $m_B = m_C = \frac{M}{2}$

   Let the velocity of $B$ be $v$ in the same direction as $A$. Then $C$ moves in the opposite direction with velocity equal to half of $B$: $v_C = -\frac{v}{2}$

3. **Write momenta of $B$ and $C$: ** $p_B = \frac{M}{2}v$ $p_C = \frac{M}{2}\left(-\frac{v}{2}\right) = -\frac{Mv}{4}$

4. Apply conservation of momentum: Taking the direction of $A$ as positive, $p_A = p_B + p_C$ $p_A = \frac{Mv}{2} - \frac{Mv}{4} = \frac{Mv}{4}$

5. **Compare $p_B$ and $p_C$ with $p_A$: ** $p_B = \frac{Mv}{2} = 2\left(\frac{Mv}{4}\right) = 2p_A$ Magnitude of momentum of $C$: $|p_C| = \frac{Mv}{4} = p_A$

6. Use de-Broglie relation $\lambda = \dfrac{h}{p}$: For $B$, $\lambda_B = \frac{h}{p_B} = \frac{h}{2p_A} = \frac{\lambda_A}{2}$

   For $C$, $\lambda_C = \frac{h}{|p_C|} = \frac{h}{p_A} = \lambda_A$

7. Final result: $\lambda_B = \frac{\lambda_A}{2}, \qquad \lambda_C = \lambda_A$

So the correct option is D.