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

Momentum of paricles `= (h)/(lambda_(1))` & `(h)/(lambda_(2))`
Since particles ar identical, mass of both particles are same.
`mv_(1) = (h)/(lambda_(1))` & `mv_(2) = (h)/(lambda_(2)), vec(v)_("COM") = (v_(1))/(2)hati + (v_(2))/(2)hatj`
`barv_(1COM) = (v_(1))/(2)hati - (v_(2))/(2)hatj rArr |vecv_(1COM)| = (sqrt(v_(1)^(2) + v_(2)^(2)))/(2) = vec(v)_("2COM")|`
`lambda = (h)/(mv) = (2h)/(msqrt(v_(1)^(2) + v_(2)^(2))) = (2)/(sqrt((mv_(1))^(2) + (mv_(2))^(2)))`
`= (2)/(sqrt(((h)/(lambda_(1)))^(2) + ((h)/(lambda_(2)))^(2))) = (2lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2) + lambda_(2)^(2)))`