An `alpha`- particle and a proton are accelerated from rest by a potential difference of `100 V` After this their de Broglie wavelength are `lambda_(a) and lambda_(p)` respectively The ratio `(lambda_(p))/ lambda_(p)` , to the nearest integer is

Answer

`3`

Explanation

`P_(1) = sqrt(2m(100 eV))`
`lambda_(P) = (h)/(sqrt(2m(100eV))) rArr lambda_(alpha)=(h)/(sqrt(2(4m)2(100 eV)))`
`lambda_(P)/lambda_(alpha)=sqrt(8)`
rArr The ratio `lambda_(p)/lambda_(alpha)`, to the nearest integer, is equal to `3`.