Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are `lambda_(1)` and `lambda_(2)` their the Brogile wavelength in the frame of reference attached to their centre of mass is :

Options

A. `lamda_1+lamda_2`
B. `(2lamda_1lamda_2)/((sqrt(lamda_1^2+lamda_2^2)))`
C. `(lamda_1lamda_2)/((sqrt(|lamda_1^2+lamda_2^2|)))`
D. `((lamda_1+lamda_2))/(2)`

Answer

B

Explanation

Let m be the mass each particle, then `lamda_1=((h)/(mv_1))` and `lamda_2=((h)/(mv_2))`, where `v_1` and `v_2` are the velocities of two particles as shown in the figure.
`vecvCM=(mvecv_1+vecv_2)/(2m)=(vecv_1+vecv_2)/(2)`
Velocity of A w.r.t. C frame is
`vecv_(1c)=vecv_1-vecv_(CM)=(vecv_1-vecv_2)/(2)`
`|vecv_(1C)|=sqrt((vecv_1-vecv_2)/(2))=|vecv_(2c)|`
So, required wavelength is
`lamda=(h)/(m|vecv_(1c)|)=(h)/(m)xx(2)/((h)/(m)sqrt((1)/(lamda_1^2)+(1)/(lamda_2^2)))`
`=(2lamda_1lamda_2)/(sqrt(lamda_1^2+lamda_2^2))`