Let K _ 1 and K _ 2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ _ 1 and λ _ 2 , respectively are incident on a metallic surface. If λ _ 1 = 3 λ _ 2 then :

Correct answer: B. K _ 1 < K _ 2 3 According to Einstein's photoelectric equation (h c)/(λ) = ϕ _ 0 + KE _ max , where ϕ _ 0 is a work function of a metal ∴ K _ 1 = h c λ _ 1 - ϕ _ 0 (i) K _ 2 = h c λ _ 2 - ϕ _ 0 (i i) or K _ 1 - K _ 2 = hc[ 1 λ _ 1 - 1 λ _ 2 ] = hc[ 1 3λ _ 2 - 1 λ _ 2 ] = - 2hc 3 λ _ 2 (Given λ _ 1 = 3 λ _ 2 ) = - (2)/(3) ( K _ 2 + ϕ _ 0 ) (Using (ii)) or K _ 1 = K _ 2 - (2)/(3) K _ 2 - (2)/(3) ϕ _ 0 = K _ 2 3 - (2)/(3) ϕ _ 0 or K _ 1 < K _ 2 3

JEE Main 202228 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Evening Shift
Subject: Physics
Chapter: Dual Nature of Matter

JEE Main 2022 · 28 June · Evening Shift · Physics · Dual Nature of Matter

Let K _ 1 and K _ 2 be the maximum kinetic energies of...

Let

{K}_{1}

and

{K}_{2}

be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength

{\lambda}_{1}

and

{\lambda}_{2}

, respectively are incident on a metallic surface. If

{\lambda}_{1} = 3{\lambda}_{2}

then :

Options

A ${K}_{1} > \dfrac{{K}_{2}}{3}$
B ${K}_{1} < \dfrac{{K}_{2}}{3}$
C ${K}_{1} = \dfrac{{K}_{2}}{3}$
D ${K}_{2} = \dfrac{{K}_{1}}{3}$

Correct answer

B. ${K}_{1} < \dfrac{{K}_{2}}{3}$

Step-by-step solution

According to Einstein's photoelectric equation

\dfrac{h c}{\lambda} = {ϕ}_{0} + {\text{KE}}_{\text{max}}

, where

{ϕ}_{0}

is a work function of a metal

∴ {K }_{1} = \dfrac{h c}{{\lambda}_{1}} - {ϕ}_{0} \ldots \left(\text{i}\right)

{K }_{2} = \dfrac{h c}{{\lambda}_{2}} - {ϕ}_{0} \ldots \left(\text{i} \text{i}\right)

\text{or}{K}_{1} - {K}_{2} = hc\left[\dfrac{1}{{\lambda}_{1}} - \dfrac{1}{{\lambda}_{2}}\right]

= hc\left[\dfrac{1}{{3\lambda}_{2}} - \dfrac{1}{{\lambda}_{2}}\right] = - \dfrac{2hc}{3{\lambda}_{2}}\left(\text{Given}{\lambda}_{1} = 3{\lambda}_{2}\right)

= - \dfrac{2}{3} \left({K }_{2} + {ϕ}_{0}\right) \left(\text{Using (ii)}\right)

or {K }_{1} = {K }_{2} - \dfrac{2}{3} {K }_{2} - \dfrac{2}{3} {ϕ}_{0} = \dfrac{{K }_{2}}{3} - \dfrac{2}{3} {ϕ}_{0}

\text{or} {K }_{1} < \dfrac{{K }_{2}}{3}

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