An α particle and a proton are accelerated from rest by a potential...

$\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mqV}}$

$⇒ \dfrac{{\lambda}_{p}}{{\lambda}_{\alpha}} = \sqrt{\dfrac{{m}_{\alpha}{q}_{\alpha}}{{m}_{p}{q}_{p}}} = \sqrt{\dfrac{4{m}_{p} \times 2e}{{m}_{p} \times e}} = \sqrt{8} = 2\sqrt{2}$$= 2\sqrt{2}$

$⇒ \dfrac{{\lambda}_{p}}{{\lambda}_{\alpha}} = 2 \times 1 . 4 = 2 . 8$