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(A) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda_{ph}}$,where $\lambda_{ph}$ is the wavelength of the photon.Thus,$\lambda_{ph} =

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

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(A) The energy of a photon is given by $E = h
u = \frac{hc}{\lambda_{ph}}$,where $\lambda_{ph}$ is the wavelength of the photon.Thus,$\lambda_{ph} = \frac{hc}{E}$.The de Broglie wavelength of an electron with kinetic energy $E$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.Now,the ratio of the wavelength of the photon to the wavelength of the electron is:$\frac{\lambda_{ph}}{\lambda_e} = \left( \frac{hc}{E} ight) 	imes \left( \frac{\sqrt{2mE}}{h} ight)$$= \frac{c}{E} 	imes \sqrt{2mE} = c \sqrt{\frac{2mE}{E^2}} = c \sqrt{\frac{2m}{E}}$.

If $m$ is the mass of an electron and $c$ is the speed of light,then the ratio of the wavelength of a photon of energy $E$ to the wavelength of an electron of the same energy is:

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