Key Point The de-Broglie wavelength of a particle of mass $m$ and moving with velocity $v$ is given by
$\lambda=\frac{h}{m v} (\because p=m v)$
de-Broglie wavelength of a proton of mass $m_{1}$ and kinetic energy $k$ is given by
$\lambda_{1} =\frac{h}{\sqrt{2 m_{1} k}} (\because p=\sqrt{2 m k})$
$=\frac{h}{\sqrt{2 m_{1} q v}} \ldots \text { (i) }[\because k=q V]$
For an alpha particle mass $m_{2}$ carrying charge $q_{0}$ is accelerated through potential $V$, then
$\lambda_{2}=\frac{h}{\sqrt{2 m_{2} q_{0} V}}$
$\because$ For $\alpha$ -particle $\left({ }_{2}^{4} He \right)$
$\therefore q_{0}=2 q$ and $m_{2}=4 m_{1}$
$\therefore \lambda_{2}=\frac{h}{\sqrt{2 \times 4 m_{1} \times 2 q \times V}}$ ...(ii)
The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
$\frac{\lambda_{1}}{\lambda_{2}} =\frac{h}{\sqrt{2 m_{1} q V}} \times \frac{\sqrt{2 \times m_{1} \times 4 \times 2 q V}}{h}= \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$=2 \sqrt{2}$