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(D) The initial momentum of the system is $P_{i} = 0$ because the neutron is slowly moving (momentum $\approx 0$) and the nucleus of mass $M$ is at re

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$\lambda$

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(D) The initial momentum of the system is $P_{i} = 0$ because the neutron is slowly moving (momentum $\approx 0$) and the nucleus of mass $M$ is at rest.According to the law of conservation of linear momentum,the final momentum $P_{f}$ must also be zero.Let $P_{1}$ be the momentum of the nucleus with mass $m_{1}$ and $P_{2}$ be the momentum of the nucleus with mass $5m_{1}$.Thus,$P_{1} + P_{2} = 0$,which implies $P_{1} = -P_{2}$.Taking the magnitude,$|P_{1}| = |P_{2}| = P$.The de-Broglie wavelength is given by $\lambda = \frac{h}{P}$.For the first nucleus,$\lambda_{1} = \frac{h}{P_{1}} = \lambda$.For the second nucleus,$\lambda_{2} = \frac{h}{P_{2}} = \frac{h}{P_{1}} = \lambda$.Therefore,the de-Broglie wavelength of the other nucleus is also $\lambda$.

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