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(D) Let the mass of nucleus $A$ be $2m$,so the mass of each nucleus $B$ and $C$ is $m$. Let the initial velocity of $A$ be $v_0$. The initial momentum

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$\frac{\lambda_A}{2}, \lambda_A$

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(D) Let the mass of nucleus $A$ be $2m$,so the mass of each nucleus $B$ and $C$ is $m$. Let the initial velocity of $A$ be $v_0$. The initial momentum is $P_A = (2m)v_0$.According to the law of conservation of linear momentum,the final momentum must equal the initial momentum. Let $v$ be the velocity of $B$. Then the velocity of $C$ is $v/2$ in the opposite direction.$P_f = m v - m(v/2) = m v / 2$.Equating initial and final momentum: $2m v_0 = m v / 2$,which gives $v = 4v_0$.Now,calculate the momenta of $B$ and $C$:$P_B = m v = m(4v_0) = 4m v_0$.$P_C = m(v/2) = m(2v_0) = 2m v_0$.The de Broglie wavelength is given by $\lambda = h/P$.Given $\lambda_A = h / (2m v_0)$.$\lambda_B = h / P_B = h / (4m v_0) = \frac{1}{2} 	imes \frac{h}{2m v_0} = \frac{\lambda_A}{2}$.$\lambda_C = h / P_C = h / (2m v_0) = \lambda_A$.Thus,the wavelengths are $\lambda_B = \frac{\lambda_A}{2}$ and $\lambda_C = \lambda_A$.

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