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(B) The energy of a photon is given by $E = \frac{hc}{\lambda_{	ext{photon}}}$,which implies $\lambda_{	ext{photon}} = \frac{hc}{E}$.The de-Broglie

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$\frac{1}{c}\left(\frac{E}{2 m}\right)^{1 / 2}$

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(B) The energy of a photon is given by $E = \frac{hc}{\lambda_{	ext{photon}}}$,which implies $\lambda_{	ext{photon}} = \frac{hc}{E}$.The de-Broglie wavelength of an electron with energy $E$ is given by $\lambda_{	ext{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Taking the ratio of the two wavelengths:$\frac{\lambda_{	ext{electron}}}{\lambda_{	ext{photon}}} = \frac{h / \sqrt{2mE}}{hc / E} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc}$.Simplifying the expression:$\frac{\lambda_{	ext{electron}}}{\lambda_{	ext{photon}}} = \frac{1}{c} 	imes \frac{E}{\sqrt{2mE}} = \frac{1}{c} 	imes \sqrt{\frac{E^2}{2mE}} = \frac{1}{c} \left(\frac{E}{2m}ight)^{1/2}$.

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).

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