An electron (of mass $$m$$) and a photon have the same energy $$E$$ in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

We begin by recalling the two basic relations that connect energy with wavelength for an electron and for a photon.

For a material particle such as an electron, the de-Broglie relation is stated as $$\lambda_e=\frac{h}{p},$$ where $$\lambda_e$$ is the de-Broglie wavelength, $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the electron.

Because the energy given in the problem is only a few electron-volts, it is extremely small when compared with the electron’s rest-mass energy (about $$0.511\ \text{MeV}$$). Hence the electron moves non-relativistically and its kinetic energy $$E$$ is related to its momentum by the classical formula $$E=\frac{p^{2}}{2m},$$ where $$m$$ is the electron’s mass.

Solving this formula for the momentum, we get $$p=\sqrt{2mE}.$$ Substituting this value of $$p$$ into the de-Broglie relation, we obtain $$\lambda_e=\frac{h}{\sqrt{2mE}}.$$

Now, for a photon the connection between its energy and its wavelength is given by the Planck-Einstein relation $$E=\frac{hc}{\lambda_{\gamma}},$$ where $$\lambda_{\gamma}$$ is the photon’s wavelength and $$c$$ is the speed of light in vacuum.

Rearranging this for $$\lambda_{\gamma}$$ gives $$\lambda_{\gamma}=\frac{hc}{E}.$$

We are required to find the ratio of the electron’s de-Broglie wavelength to the photon’s wavelength, namely $$\frac{\lambda_e}{\lambda_{\gamma}} =\frac{\dfrac{h}{\sqrt{2mE}}}{\dfrac{hc}{E}}.$$

Dividing the two fractions, the factor $$h$$ cancels out and we get $$\frac{\lambda_e}{\lambda_{\gamma}} =\frac{E}{c\sqrt{2mE}}.$$

The numerator contains $$E$$ while the denominator contains $$\sqrt{E}$$, so we simplify by writing $$E=\sqrt{E}\times\sqrt{E}$$:

$$ \frac{\lambda_e}{\lambda_{\gamma}} =\frac{\sqrt{E}\,\sqrt{E}}{c\sqrt{2mE}} =\frac{\sqrt{E}}{c\sqrt{2m}}.$$ Combining the square-root terms in the denominator, we finally reach $$ \frac{\lambda_e}{\lambda_{\gamma}} =\frac{1}{c}\sqrt{\frac{E}{2m}}.$$ This matches option C.

Hence, the correct answer is Option C.