When electron or any charged particle is accelerated through potential difference V, then kinetic energy gained is given by $E=eV$ ...... (i)
$E=\frac{1}{2}m{v}_0^2=\frac{p^2}{2m}=\frac{h^2}{2m.{\lambda }^2}$ ...... (ii)
$\therefore eV=\frac{h^2}{2m.{\lambda }^2} \Rightarrow \lambda =\frac{h}{\sqrt{2meV}}$ ....... (iii)
When proton of mas $M$ is accelerated through a potential difference of $9V$, then the de - Broglie wavelength obtained is
${\lambda }^{\prime }=\frac{h}{\sqrt{2M e\left(9V\right)}}=\frac{h}{3\times \sqrt{2MeV}}\times \frac{\sqrt{m}}{\sqrt{m}}$
$\therefore {\lambda }^{\prime }=\frac{\lambda }{3}.\sqrt{\frac{m}{M}}$