An electromagnetic wave of wavelength lambda | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Since the work function is negligible,the entire energy of the photon is co

Vedclass · Accepted Answer

$\lambda \propto \lambda_{1}^{2}$

Vedclass · Answer

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Since the work function is negligible,the entire energy of the photon is converted into the kinetic energy $(K)$ of the photoelectron: $K = \frac{hc}{\lambda}$.The de-Broglie wavelength $\lambda_{1}$ of the photoelectron is related to its momentum $(p)$ by $\lambda_{1} = \frac{h}{p}$.We know that kinetic energy $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.Substituting $K$ in the momentum equation: $p = \sqrt{2m \left( \frac{hc}{\lambda} \right)}$.Now,substitute $p$ into the de-Broglie wavelength formula: $\lambda_{1} = \frac{h}{\sqrt{2mhc/\lambda}} = \sqrt{\frac{h^2 \lambda}{2mhc}} = \sqrt{\frac{h \lambda}{2mc}}$.Squaring both sides: $\lambda_{1}^2 = \frac{h \lambda}{2mc}$.Since $h, m, c$ are constants,we get $\lambda_{1}^2 \propto \lambda$,which implies $\lambda \propto \lambda_{1}^2$.

List-$I$	List-$II$
$a$. Franck-Hertz Experiment	$i$. Particle nature of light
$b$. Photo-electric Experiment	$ii$. Discrete energy levels of atom
$c$. Davisson-Germer Experiment	$iii$. Wave nature of electron
	$iv$. Structure of atom

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de-Broglie wavelength $\lambda_{1}$,then:

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