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(D) The energy of a photon is given by $E = \frac{hc}{\lambda_p}$.Therefore,the wavelength of the photon is $\lambda_p = \frac{hc}{E}$.For an electron

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The energy of a photon is given by $E = \frac{hc}{\lambda_p}$.Therefore,the wavelength of the photon is $\lambda_p = \frac{hc}{E}$.For an electron,the de Broglie wavelength is $\lambda_e = \frac{h}{p}$,where $p$ is the momentum.Since the kinetic energy of the electron is $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Thus,the wavelength of the electron is $\lambda_e = \frac{h}{\sqrt{2mE}}$.Now,the ratio of the wavelengths is $\frac{\lambda_p}{\lambda_e} = \left( \frac{hc}{E} ight) 	imes \left( \frac{\sqrt{2mE}}{h} ight)$.Simplifying this,we get $\frac{\lambda_p}{\lambda_e} = c \sqrt{\frac{2mE}{E^2}} = c \sqrt{\frac{2m}{E}}$.

List $I$	List $II$
$A$. Planck's constant $(h)$	$I$. $[M^1 L^2 T^{-2}]$
$B$. Stopping potential $(V_s)$	$II$. $[M^1 L^1 T^{-1}]$
$C$. Work function $(\phi)$	$III$. $[M^1 L^2 T^{-1}]$
$D$. Momentum $(p)$	$IV$. $[M^1 L^2 T^{-3} A^{-1}]$

The ratio of the wavelength of a photon of energy $E$ to that of an electron of the same energy is ($m=$ mass of an electron,$c=$ speed of light,$h=$ Planck's constant).

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Match List $I$ with List $II$:

List $I$ List $II$

$A$. Planck's constant $(h)$ $I$. $[M^1 L^2 T^{-2}]$

$B$. Stopping potential $(V_s)$ $II$. $[M^1 L^1 T^{-1}]$

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