To find the ratio of the de Broglie wavelengths of a photon and an electron when they both have the same energy $E$, follow these steps:

Photon Wavelength:

For a photon, the energy is given by:

$ E = \frac{h c}{\lambda_{\mathrm{Ph}}} $

Where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda_{\mathrm{Ph}}$ is the wavelength of the photon. Rearranging this equation gives:

$ \lambda_{\mathrm{Ph}} = \frac{h c}{E} $

Electron Wavelength:

For an electron, the energy related to its momentum is:

$ E = \frac{p^2}{2m} $

Where $m$ is the mass of the electron and $p$ is its momentum. Using the de Broglie wavelength expression $p = \frac{h}{\lambda_e}$, we can write:

$ E = \left( \frac{h}{\lambda_e} \right)^2 \times \frac{1}{2m} $

Solving for the electron's wavelength $\lambda_e$:

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

Ratio of Wavelengths:

Now, calculate the ratio of the wavelengths $\frac{\lambda_{\mathrm{Ph}}}{\lambda_e}$ as follows:

$ \frac{\lambda_{\mathrm{Ph}}}{\lambda_e} = \frac{\frac{h c}{E}}{\frac{h}{\sqrt{2mE}}} = c \sqrt{\frac{2m}{E}} $

This ratio equation defines the relation between the photon and electron wavelengths when both have the same energy $E$.