A proton and an α -particle are accelerated from rest by 2V and 4V...

The formula to calculate the de Broglie wavelength of the proton is given by

${\lambda}_{p} = \dfrac{h}{\sqrt{2{m}_{p}{q}_{p}V}} . . . \left(1\right)$

The formula to calculate the de Broglie wavelength of the alpha particle is given by

${\lambda}_{\alpha} = \dfrac{h}{\sqrt{2{m}_{\alpha}{q}_{\alpha}V '}} . . . \left(2\right)$

Divide equation (1) by equation (2) and simplify to obtain the required ratio.

$\dfrac{{\lambda}_{p}}{{\lambda}_{\alpha}} = \dfrac{\sqrt{{m}_{\alpha}}}{\sqrt{{m}_{p}}} \times \sqrt{\dfrac{{q}_{\alpha}V '}{{q}_{p}V}} . . . \left(3\right)$

Substitute the values of the known parameters into equation (3) to calculate the required ratio.

$\begin{cases}\dfrac{{\lambda}_{p}}{{\lambda}_{\alpha}} & = & \sqrt{\dfrac{4}{1}} \times \sqrt{\dfrac{2 \times 4}{1 \times 2}} \\ & = & 4: 1\end{cases}$