Two particles A and B of de-broglie wavelength `lambda_(1) and lambda_(2)` combine to from a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one dimensional).

Options

A. `lambda_(A)+lambda_(B)`
B. `lambda_(A)-lambda_(B)`
C. `(lambda_(A)lambda_(B))/(lambda_(A)+lambda_(B))`
D. `(lambda_(A)lambda_(B))/(lambda_(A)-lambda_(B))`

Answer

D

Explanation

For one dimensional motion,`vecp_(C)=vecp_(A)+vecp_(B)`
If `p_(A),p_(B) gt 0` or `p_(A),p_(B) lt 0`, i.e., `(p_(A)` and `p_(B)` are in same direction).
`p_(C)=p_(A)+p_(B)`
`h/lambda_(C)=h/lambda_(A)+h/lambda_(B)=h((lambda_(A)+lambda_(B))/(lambda_(A)lambda_(B)))`
`lambda_(C)=(lambda_(A)lambda_(B))/(lambda_(A)+lambda_(B))`
If `p_(A)gt0,p_(B)lt0` or `p_(A)lt0,p_(B)gt 0`
`(p_(A)` and `p_(B)` are in opposite direction)
`p_(C)|p_(A)-p_(B)|`
`h/lambda_(C)=|h/lambda_(A)-h/lambda_(B)|rArr(h|lambda_(A)-lambda_(B)|)/(lambda_(A)lambda_(B))`
`lambda_(C)=(lambda_(A)lambda_(B))/|(lambda_(A)-lambda_(B))|`