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

Options

A. `1 : 1`
B. `1 : 2`
C. `2 : 1`
D. `2sqrt(2) : 1`

Answer

4

Explanation

`qV=(1)/(2)mv^(2)ormv=sqrt(2qVm),`
`So lambda=(h)/(mv)=(h)/(sqrt(2qVm))i.e.lambdaprop(1)/(sqrt(q)m),`
`so(lambda_(p))/(lambda_(alpha))=sqrt((q_(alpha)m_(alpha))/(q_(p)m_(p)))=sqrt(2xx4)=2sqrt(2)`