Given a photon and an electron with identical energy \( E \), we aim to find the ratio \( \frac{\lambda_p}{\lambda_e} \) of the photon's wavelength \( \lambda_p \) to the electron's de Broglie wavelength \( \lambda_e \).
The photon energy is expressed as:
\[ E = \frac{hc}{\lambda_p} \]
Rearranging for \( \lambda_p \):
\[ \lambda_p = \frac{hc}{E} \]
The de Broglie wavelength of an electron is defined by:
\[ \lambda_e = \frac{h}{p} \]
For an electron with kinetic energy \( E \), its momentum \( p \) is:
\[ p = \sqrt{2mE} \]
Substituting this momentum into the de Broglie wavelength equation:
\[ \lambda_e = \frac{h}{\sqrt{2mE}} \]
The ratio \( \frac{\lambda_p}{\lambda_e} \) is calculated as:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\frac{hc}{E}}{\frac{h}{\sqrt{2mE}}} \]
Simplifying this expression yields:
\[ \frac{\lambda_p}{\lambda_e} = \frac{hc}{E} \times \frac{\sqrt{2mE}}{h} = \frac{c\sqrt{2mE}}{E} \]
Assuming \( E \) represents kinetic energy, the simplified ratio is:
\[ \frac{\lambda_p}{\lambda_e} = \frac{\sqrt{2mE}}{c} \]
Therefore, the determined ratio is:
\(\frac{\sqrt{2mE}}{c}\).