1.

A proton and an alpha-particle are accelerated through the same potential difference. The ratio of de-Broglie wavelength of proton to the de-Broglie wavelength of alpha particle will be

Answer»

<P>`1 : 2`
`2 sqrt(2) : 1`
`2 : 1`
`1 : 1`

Solution :`lambda = (h)/(sqrt(2mqV))`
`(lambda_(p))/(lambda_(alpha)) = sqrt((2m_(alpha) q_(alpha) V)/(2m_(p)q_(p)V)) = sqrt((2 xx 4m_(p) xx 2E xx V)/(2m_(p) xx e xx V)) = 2 sqrt(2) : 1`


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