According to the de-Broglie hypothesis, the r | vedclass

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(B) For a photon, the energy $E$ is given by $E = h
u = \frac{hc}{\lambda_p}$. Therefore, the wavelength of the photon is $\lambda_p = \frac{hc}{E}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For a photon, the energy $E$ is given by $E = h
u = \frac{hc}{\lambda_p}$. Therefore, the wavelength of the photon is $\lambda_p = \frac{hc}{E}$.For an electron, the de-Broglie wavelength is $\lambda_e = \frac{h}{p}$. Since the kinetic energy $E = \frac{p^2}{2m}$, we have $p = \sqrt{2mE}$. Thus, $\lambda_e = \frac{h}{\sqrt{2mE}}$.The ratio of the wavelength of the photon to that of the electron is $\frac{\lambda_p}{\lambda_e} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{c}{E} 	imes \sqrt{2mE} = c \sqrt{\frac{2mE}{E^2}} = c \sqrt{\frac{2m}{E}}$.

According to the de-Broglie hypothesis, the ratio of the wavelength of a photon and that of an electron having the same energy $E$ is $(m = \text{mass of electron}, c = \text{velocity of light})$

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