Find the ratio of de Broglie wavelength of a proton and as `alpha`-particle which have been accelerated through same potential difference.

Options

A. `(1)/(sqrt(2))`
B. `(1)/(2sqrt(2))`
C. `(1)/(3sqrt(2))`
D. `2sqrt(2)`

Answer

4

Explanation

Kinetic energy gained by a charge q after being accelerated through a potential difference V volt,
`qV=(1)/(2)mv^(2)rArrv=sqrt((2qV)/(m))`
`therefore " " mv = sqrt(2mqV)`
de Broglie wavelenght =
`lambda=(h)/(mv)=(h)/(sqrt(2mqV))therefore (lambda_(p))/(lambda_(alpha))=sqrt((m_(alpha)q_(alpha)V_(alpha))/(m_(p)q_(p)V_(p)))`
putting `V_(alpha) = V_(p)`,
`(lambda_(p))/(lambda_(alpha))=sqrt((4xx2)/(1xx1))=2sqrt(2)`