Question

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are \(\lambda_1\) and \(\lambda_2\), their de Broglie wavelength in the frame of reference attached to their center of masses :

• 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. \(\lambda_{CM}=\left(\frac{\lambda_1+\lambda_2}{2}\right)\)

Hint:

In the centre-of-mass frame two equal masses have equal and opposite velocities, each of magnitude half of the relative speed \(|\vec v_1-\vec v_2|\).

Answer 1

The Correct Option is B

Both electrons have mass \(m\). Let \(\vec v_1\perp\vec v_2\) with \(v_1=\frac{h}{m\lambda_1}\), \(v_2=\frac{h}{m\lambda_2}\).
Velocity of the centre of mass: \(\vec v_{cm}=\frac{\vec v_1+\vec v_2}{2}\).
Velocity of electron 1 in the CM frame: \(\vec v_1-\vec v_{cm}=\frac{\vec v_1-\vec v_2}{2}\); electron 2 has the opposite velocity, so both have the same speed \(v'=\frac{|\vec v_1-\vec v_2|}{2}=\frac{\sqrt{v_1^2+v_2^2}}{2}\) (since the velocities are perpendicular).
\(\lambda_{CM}=\frac{h}{mv'}=\frac{2h}{m\sqrt{v_1^2+v_2^2}}=\frac{2}{\sqrt{\frac{1}{\lambda_1^2}+\frac{1}{\lambda_2^2}}}=\frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}\).
Answer: (2) \(\lambda_{CM}=\frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}\)