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Two particles having de-Broglie waveleng...

Two particles having de-Broglie wavelengths `lambda_(1) ` and `lambda_(2)`, while moving along mutually perpendicular directions, undergo perfectly inelastic collision. The de-Broglie wavelength `lambda`, of the final particle is

A

`lamda=sqrt(lamda_1lamda_2)`

B

`2/lamda=1/lamda_1+1/lamda_2`

C

`1/lamda^2=1/lamda_1^2+1/lamda_2^2`

D

`lamda=(lamda_1+lamda_2)/2`

Text Solution

Verified by Experts

The correct Answer is:
C

Conserver momentum
along `x:(h)/(lamda_1)+0=h/lamdacostheta`
along y :`h/(lamda_2)+0=h/lamdasintheta`
Squaring and adding
`h^2(1/(lamda_1^2+lamda_2^2))=h^2/lamda_2implies1/lamda_2=1/lamda_1^2+1/lamda_2^2`
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