A proton and an α -particle are accelerated from rest by 2   V and 4   V potentials,…

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

${\lambda }_p=\frac{h}{\sqrt{2m_p{q}_{p}V}} ...\left(1\right)$

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

${\lambda }_{\alpha }=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }V\text{'}}} ...\left(2\right)$

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

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{\sqrt{m_{\alpha }}}{\sqrt{m_p}}\times \sqrt{\frac{q_{\alpha }V\text{'}}{q_{p}V}} ...\left(3\right)$

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

$\begin{array}{rcl}\frac{{\lambda }_p}{{\lambda }_{\alpha }}& =& \sqrt{\frac{4}{1}}\times \sqrt{\frac{2\times 4}{1\times 2}}\\ & =& 4:1\end{array}$