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(C) For an electron with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.The de-Broglie wavelength

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$\frac{1}{c} \sqrt{\frac{E}{2m}}$

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(C) For an electron with kinetic energy $E$,the momentum $p$ is given by $E = \frac{p^2}{2m}$,which implies $p = \sqrt{2mE}$.The de-Broglie wavelength of the electron is $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.For a photon,the energy $E$ is related to its wavelength $\lambda_p$ by $E = \frac{hc}{\lambda_p}$,which implies $\lambda_p = \frac{hc}{E}$.The ratio of the wavelengths is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc} = \frac{1}{c} \sqrt{\frac{E}{2m}}$.

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelength of the electron to the wavelength of the photon is ($c =$ velocity of light).

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