Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$, their de Broglie wavelength in the frame of reference attached to their centre of mass is:
Answer: (B) $\lambda_{CM} = \dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$
Let the velocities be $\vec v_1$ and $\vec v_2$ ($\vec v_1 \perp \vec v_2$). For equal masses $\vec v_{CM} = \dfrac{\vec v_1 + \vec v_2}{2}$, so in the CM frame each electron moves with speed
$$\left|\vec v_1 - \vec v_{CM}\right| = \frac{|\vec v_1 - \vec v_2|}{2} = \frac{\sqrt{v_1^2 + v_2^2}}{2}$$
With $v_1 = \dfrac{h}{m\lambda_1}$ and $v_2 = \dfrac{h}{m\lambda_2}$:
$$\lambda_{CM} = \frac{h}{m\cdot\frac12\sqrt{v_1^2 + v_2^2}} = \frac{2}{\sqrt{\dfrac{1}{\lambda_1^2} + \dfrac{1}{\lambda_2^2}}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$$
Solution by Sreeraj P, M.Sc Physics