The ratio of de-Broglie wavelength of an α -particle and a proton...

As we know, de Broglie wavelength in terms of energy of particle is given by, $\lambda = \dfrac{1}{\sqrt{2mE}}$.

Energy gained by proton, $E = qV$.

Therefore,

$\dfrac{{\lambda}_{\alpha}}{{\lambda}_{p}} = \dfrac{\dfrac{h}{\sqrt{2{m}_{\alpha}{q}_{\alpha}V}}}{\dfrac{h}{\sqrt{2{m}_{p}{q}_{p}V}}}$

$⇒ \dfrac{{\lambda}_{\alpha}}{{\lambda}_{p}} = \sqrt{\dfrac{1}{8}}$

$⇒ m = 8$