Explanation

(c) If a charge particle of mass m and charge q is accelerated through a potential difference V and E is the energy acquired by the particle , then
$E = qV$
If v is velocity of particle, then
$E = \frac{1}{2}mv^{2}$
Or $v = \sqrt{\left( \frac{2E}{m} \right)}$
Now, de-Broglie wavelength of particle is
$\lambda = \frac{h}{mv} = \frac{h}{m\sqrt{(2E\ /m)}}$
Substituting the value of $E$, we get
$\lambda = \frac{h}{\sqrt{2mqV}}$
For electron, $\lambda_{e} = \frac{h}{\sqrt{2m_{e}eV}}$
For proton, $\lambda_{p} = \frac{h}{\sqrt{2m_{p}eV}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\lambda_{e}}{\lambda_{p}} = \sqrt{(\frac{m_{p}}{m_{e}})}$