A proton and an alpha-particle are accel...

A proton and an `alpha`-particle are accelerated through same potential difference. Find the ratio of their de-Brogile wavelength.

`sqrt(8) : 1`

`1 : sqrt(8)`

`1 : 2`

`1 : sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:

`lambda=(h)/(sqrt(2meV))rArrlambdaprop(1)/(sqrt(m))`
`(lambda_(p))/(lambda_(alpha))=sqrt((m_(alpha))/(m_(p)))=sqrt(4)=2`

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