Two identical non-relativitic partcles A and B move at right angles to each othre, processing de Broglie wavelengths `lamda_1` and `lamda_2`, respectively. The de Broglie wavelength of each particle in their centre of mass frame of reference is

Options

A. `sqrt((lambda_(1)^(2)+lambda_(2)^(2))/(2))`
B. `(lambda_(1)+lambda_(2))/(2)`
C. `(2lambda_(1)lambda_(2))/(lambda_(1)+lambda_(2))`
D. `(2lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2)+lambda_(2)^(2)))`

Answer

D

Explanation

Let m be the mass of each particle and thus verticals are `v_(1)and v_(2)` respectively

`vecv_(cm(x))=(mv_(2))/(2m)=(v_(2))/(2)`
`vecv_(cm(y))=(mv_(1))/(2m)=(v_(1))/(2)`
`vecv_(2cm(x))=v_(2)-(v_(2))/(2)=(v_(2))/(2)`
`vecv_(2cm(y))=(v_(1))/(2)`
So magnitude of velocity of `2^(nd)`particle is C.O.M. fram becomes `sqrt(v_(2)^(2)/4+(v_(1)^(2))/(4))=sqrt(v_(1)^(2)+v_(2)^(2))/(2)`
`lambda=(h)/(mv_(1))rArrv_(1)=(h)/(mlambda_(1))`
`lambda_(2)=(h)/(mv_(2))rArrv_(2)=(h)/(mlambda_(2))`
So De broglie wavelength of `2^(nd)` paticle in COM frame is
`lambda=(h)/(msqrt((v_(1)^(2)+v_(1)^(2))/(2)))=(2h)/(msqrt(h^(2)/(m^(2)lambda_(1)^(2))+(h^(2))/(n_(2)^(2)lambda_(1)^(2))))=(2)/(sqrt((1)/(lambda_(1)^(2))+(1)/(lambda_(2)^(2))))`
`rArr(2lambda_(1)lambda_(2))/(sqrt(lambda_(1)^(2)+lambda_(2)^(2)))`