Question

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 :

• A. $\lambda_{CM} = \lambda_1 = \lambda_2$
• B. $\lambda_{CM} = \frac{2 \lambda _{1} \lambda _{2}}{\sqrt{\lambda _{1}^{2} + \lambda _{2}^{2}}} $
• C. $\frac{1}{\lambda_{CM} } = \frac{1}{\lambda _{1}} + \frac{1}{\lambda _{2}} $
• D. $\lambda_{CM} = \left(\frac{\lambda_{1} + \lambda_{2}}{2} \right)$

Answer 1

Answer: (C)

$ \begin{array}{l} \text { Since } \lambda=\frac{ h }{ mv } \\ v =\frac{ h }{ m \lambda} \end{array} $
Let $v_{1}$ and $v_{2}$ are the speeds of electrons
$ \begin{array}{l} v _{ cm }=\frac{ v _{1}+ v _{2}}{2} \\ \frac{ h }{2 m \lambda_{ cm }}=\frac{1}{2}\left(\frac{ h }{ m \lambda_{1}}+\frac{ h }{ m \lambda_{2}}\right) \\ \frac{1}{\lambda_{ cm }}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}} \end{array} $