An electron of mass m and a photon have the s | vedclass

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(B) The de Broglie wavelength of an electron with energy $E$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.The wavelength of a photon w

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The de Broglie wavelength of an electron with energy $E$ is given by $\lambda_e = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.The wavelength of a photon with energy $E$ is given by $\lambda_p = \frac{hc}{E}$.The ratio of the wavelength of the electron to that of the photon is $\frac{\lambda_e}{\lambda_p} = \frac{h}{\sqrt{2mE}} 	imes \frac{E}{hc}$.Simplifying this,we get $\frac{\lambda_e}{\lambda_p} = \frac{1}{c} \frac{E}{\sqrt{2mE}} = \frac{1}{c} \sqrt{\frac{E}{2m}} = \frac{1}{c} \left(\frac{E}{2m}ight)^{1/2}$.

List-$I$	List-$II$
$A$. $E = hv$	$I$. de Broglie wavelength
$B$. Diffraction and Interference	$II$. Particle nature of light
$C$. $\lambda = h/p$	$III$. Wave nature of light
$D$. Compton effect	$IV$. Energy of photon

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the wavelength of the electron to that of the photon is: ($c$ being the velocity of light)

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