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. `1/lamda_(cm)=1/lamda_1+1/lamda_2`
B. `lamda_(cm)=(2lamda_1lamda_2)/(sqrt(lamda_1^2+lamda_2^2))`
C. `lamda_(cm)=lamda_1=lamda_2`
D. `lamda_(cm)=((lamda_1lamda_2))/2`

Answer

B

Explanation

Momentum of each electron in the direction of X - axis and Y - axis `h/lamda_1 & h/lamda_2 hatj` respectively.
De Broglie wavelength `lamda=h/(mv)`
Velocities of first and second electrons are
`vecv_1=h/(mlamda_1)hati and vecv_2=h/(mlamda_2)hatj`
Velocity of centre of mass can be written as :
`vecv_(cm)=(mvecv_1+mvecv_2)/(m+m)=(vecv_1+vecv_2)/2`
Substituting values of `v_1 amd v_2` we get the following :
Velocity of centre of mass
`vecv_(cm)=h/(2mlamda_1)hati+h/(2mlamda_2)hatj`
Velocity of 1st particle with respect to centre of mass
`vecV_(1cm)=vecv_1-vecv_(cm)`
Substituting values we get the following
`vecV_(1cm)=h/(2mlamda_(1))hati-h/(2mlamda_2)hatj`
Linear momentum of the first electron with respect to centre of mass
`vecp_1=mvecV_(1cm)=h/(2mlamda_1)hati-h/(2mlamda_2)hatj`
`p_1=sqrt(h^2/(4lamda_1^2)+h^2/(4lamda_2^2))`
`lamda_(1cm)=h/(p_1)`
`lamda_(cm)=h/(sqrt(h^2/(4lamda_1^2)+h^2/(4lamda_2^2)))=(2lamda_1lamda_2)/(sqrt(lamda_1^2+lamda_2^2))`