A photon and an electron (mass $m$ ) have the same energy $E$. The ratio $\frac{{\lambda }_{\mathrm{photon}}}{{\lambda }_{\mathrm{electron}}}$ of their de Broglie wavelengths is: (c is the speed of light)

We are asked for the ratio of the de Broglie wavelengths of a photon and an electron when both have the same total energy $E$.

**1. Photon wavelength** For a photon, its energy and wavelength are related by

Re‑arranging gives

$\lambda$_photon=hc/E

**2. Electron wavelength** An electron’s de Broglie wavelength is

$\lambda$_electron=h/p

Its kinetic energy (the given energy $E$) is

$E=p{2}^2/2m$

Solving for the momentum $p$ gives

$p=\sqrt{2mE}$

Thus the electron wavelength is

$\lambda$_electron=h/2mE

**3. Form the ratio**

$\frac{\lambda }{}$_photon λ_electron = hc/E h/2mE

Cancel the Planck constant $h$ and simplify:

$\frac{c}{/}=c·\sqrt{\frac{2m}{E}}$

Therefore the required ratio is

$c\sqrt{\frac{2m}{E}}$

This matches the answer option given.