An electron of mass m and a photon have the same energy E in the range of a few eV.

(3) De-Broglie wavelength of electron $\left(\lambda_{e}\right)$ is given by

$\lambda_{e}=\frac{h}{p_{e}}=\frac{h}{\sqrt{2 m E}}$           $\ldots$ (i) $\quad(\because p=\sqrt{2 m E})$

Energy of photon $(E)$ is given by

$E=\frac{h c}{\lambda_{P}}$        (Here $\lambda p=$ wavelength of Photon)

$\Rightarrow \lambda_{P}=\frac{h c}{E}$ ...(ii)

On dividing (i) by (ii) we get

$\Rightarrow \quad \frac{\lambda_{e}}{\lambda_{P}}=\frac{h}{\sqrt{2 m E}} \frac{E}{h c}=\sqrt{\frac{E}{2 m}} \cdot \frac{1}{c}$