1.

A photon and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is :

Answer»

<P>2
`sqrt8`
`(1)/(sqrt8)`
1

Solution :Kinetic energy of charged, particle, accelerated through a potential DIFFERENCE V,
i.e., `K=qV`
`:.` Momentum of the particle,
`p=SQRT(2mK)= sqrt(2mqV)`
de-Broglie wavelenth of the particle
`lambda=(h)/(p)=(h)/(sqrt(2mqV))`
`:.m_(alpha)=4m_(p) and q_(2)=2q_(1)`
`:. (lambda_(p))/(lambda_(alpha))=sqrt((m_(2)q_(2))/(m_(1)q_(1)))= sqrt((4xx2)/(1xx1))= sqrt(8)`


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