A proton and deuteron are accelerated by same potential difference.Find the ratio of their de-Broglie wavelengths.

Explanation

The de Broglie wavelength `lambda` associated with same potential V is `lambda = (h)/(sqrt(2 meV))`
`therefore " " lambda alpha (1)/(sqrtm)` [For same potential]
Since alpha particle is `""_(2)^(4)He` and proton is `_(1)^(1)H`
`therefore` mass of alpha particle = 4 times mass of proton
`therefore " " alpha _("proton") alpha (1)/(sqrtM)` and `lambda_("alpha") alpha (1)/(sqrt4m)`
`therefore (lambda_("proton"))/(lambda_("alpha")) = (1)/(sqrtm) xx (sqrt(4m))/(1) = (2)/(1) therefore lambda_("proton") : alpha_("alpha") = 2 : 1`