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(D) The initial momentum of the system is $P_i \approx 0$ because the neutron is moving slowly.According to the law of conservation of linear momentum

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$\lambda$

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(D) The initial momentum of the system is $P_i \approx 0$ because the neutron is moving slowly.According to the law of conservation of linear momentum,the final momentum of the system must also be zero $(P_f = 0)$.Let the momenta of the two product nuclei be $P_1$ and $P_2$. Since the total momentum is zero,we have $P_1 + P_2 = 0$,which implies $P_1 = -P_2$.In terms of magnitude,$|P_1| = |P_2| = P$.The de Broglie wavelength is given by the formula $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum.Since both nuclei have the same magnitude of momentum $P$,their de Broglie wavelengths will be $\lambda_1 = \frac{h}{P}$ and $\lambda_2 = \frac{h}{P}$.Therefore,$\lambda_1 = \lambda_2 = \lambda$.

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