To solve this problem, we need to find the de Broglie wavelengths for both the electron and the photon and then find their ratio.$\\$The de Broglie wavelength $\lambda$ is given by:$\\ \lambda = \frac{h}{p} \\$where $h$ is Planck's constant and $p$ is the momentum.$\\$For an electron:$\\$The energy $E$ of the electron is given by the relation:$\\ E = \frac{p^2}{2m} \\$Solving for momentum $p: \\ p = \sqrt{2mE} \\$Thus, the de Broglie wavelength for the electron is:$\\ \lambda_e = \frac{h}{\sqrt{2mE}} \\$For a photon:$\\$The energy $E$ of the photon is given by:$\\ E = pc \\$Solving for momentum $p: \\ p = \frac{E}{c} \\$Thus, the de Broglie wavelength for the photon is:$\\ \lambda_p = \frac{h}{\frac{E}{c}} = \frac{hc}{E} \\$Now, we find the ratio of the de Broglie wavelengths:$\\ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}} \\$Simplifying the expression:$\\ \frac{\lambda_e}{\lambda_p} = \frac{h \cdot E}{hc \cdot \sqrt{2mE}} \\$Cancel out $h$ from the numerator and denominator:$\\ \frac{\lambda_e}{\lambda_p} = \frac{E}{c \cdot \sqrt{2mE}} \\$Further simplifying:$\\ \frac{\lambda_e}{\lambda_p} = \frac{1}{c} \left( \frac{E}{2m} \right)^{1/2} \\$Therefore, the correct option is Option 3: $\frac{1}{c} \left( \frac{E}{2m} \right)^{1/2} \\$