An electron of mass m and a photon have the same energy E. The...

The de-Broglie wavelength of an electron of mass $m$ and kinetic energy $E$ is given by:

$\lambda_e = \dfrac{h}{\sqrt{2mE}}$

The energy of a photon is given by $E = \dfrac{hc}{\lambda_p}$, where $\lambda_p$ is its wavelength.

Therefore, the wavelength of the photon is:

$\lambda_p = \dfrac{hc}{E}$

The ratio of their de-Broglie wavelengths is:

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}$

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{h}{\sqrt{2mE}} \times \dfrac{E}{hc}$

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{1}{c} \dfrac{E}{\sqrt{2mE}}$

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{1}{c} \sqrt{\dfrac{E^2}{2mE}}$

$\dfrac{\lambda_e}{\lambda_p} = \dfrac{1}{c} \sqrt{\dfrac{E}{2m}}$