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
$\begin{array}{l}{v}_{cm}=\frac{v_1+v_2}{2}\\ \frac{h}{2m{\lambda }_{cm}}=\frac{1}{2}\left(\frac{h}{m{\lambda }_1}+\frac{h}{m{\lambda }_2}\right)\\ \frac{1}{{\lambda }_{cm}}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}\end{array}$