A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity $v$ is given by

$\lambda =\frac{h}{mv}$       $(\because p=mv)$

de-Broglie wavelength of a proton of mass $m_1$ and kinetic energy $k$ is given by

${\lambda }_1=\frac{h}{\sqrt{2m_{1}k}}$           $(\because p=\sqrt{2mk})$

${\lambda }_1=\frac{h}{\sqrt{2m_{1}qV}}....(i)$        $[\because k=qV]$

For an alpha particle mass $m_2$ carrying charge $q_0$ is accelerated through potential $V$, then

${\lambda }_2=\frac{h}{\sqrt{2m_2{q}_{0}V}}$

$\because$ For $\alpha -particle{(}_2^{4}He)$ :  $q_0=2q$ and $m_2=4m_1$

$\therefore {\lambda }_2=\frac{h}{\sqrt{2\times 4m_1\times 2q\times V}}....(ii)$

The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get
$\frac{{\lambda }_1}{{\lambda }_2}=\frac{h}{\sqrt{2m_{1}qV}}\times \frac{\sqrt{2\times m_1\times 4\times 2qV}}{h}=\frac{4}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

We get  $\frac{{\lambda }_1}{{\lambda }_2}=2\sqrt{2}$