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 :

Answer

`lambda=(2lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2)+lambda_(2)^(2)))`

Explanation

Let `barp_(1)` and `barp_(2)` be the momenta of the two paritlces.
The velocity of the centre of mass

Hence, momentum of paritcles `1` in c.m. frame
`vecp_(1)=m(vecv_(1)-vecv_(cm))`
`= mvecv_(1)=m(vecp_(1)+vecp_(2))/(2m)`
`=(barp_(1)+barp_(2))/(2)`
`|vec(p)_(1)|=(1)/(2)sqrt(P_(1)^(2)+P_(2)^(2)-2P_(1)P_(2) cos 90^(@))`
`(h)/(lambda)=(1)/(2)sqrt((h^(2))/(lambda_(1)^(2)+lambda_(2)^(2)))`