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. $\frac{1}{{\lambda }_{CM}}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$
• C. ${\lambda }_{CM}=\frac{2{\lambda }_1{\lambda }_2}{\sqrt{{{\lambda }_1}^2+{{\lambda }_2}^2}}$
• D. ${\lambda }_{CM}=\left(\frac{{\lambda }_1+{\lambda }_2}{2}\right)$

Answer 1

Answer: (B)

$\frac{1}{{\lambda }_{CM}}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$

Since $\lambda =\frac{h}{mv}$

$v=\frac{h}{m\lambda }$

Let $v_1$ and $v_2$ are the speeds of electrons

$v_{cm}=\frac{v_1+v_2}{2}$

$\frac{h}{2m{\lambda }_{cm}}=\frac{1}{2}(\frac{h}{m{\lambda }_1}+\frac{h}{m{\lambda }_2})$

$\frac{1}{{\lambda }_{cm}}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$