JEE Main 2022 — Physics Modern Physics

2022mcqeasy

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 :

Official previous-year question

Held on 28 Jun 2022 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

According to Einstein's photoelectric equation

$\frac{h⁡c}{\lambda }={\phi }_{0}+{\text{KE}}_{\text{max}}$ , where ${\phi }_{0}$ is a work function of a metal

$\therefore { K⁡ }_{1} = \frac{ h⁡ c }{ {\lambda }_{1} } - {\phi }_{0} \ldots ( \text{i} )$

${ K⁡ }_{2} = \frac{ h⁡ c }{ {\lambda }_{2} } - {\phi }_{0} \ldots ( \text{i} \text{i} )$

$\text{or} {K}_{1}-{K}_{2}=hc[\frac{1}{{\lambda }_{1}}-\frac{1}{{\lambda }_{2}}]$

$=hc[\frac{1}{{3\lambda }_{2}}-\frac{1}{{\lambda }_{2}}]=-\frac{2hc}{3{\lambda }_{2}} (\text{Given} {\lambda }_{1}=3{\lambda }_{2})$

$= - \frac{ 2 }{ 3 } ( { K⁡ }_{2} + {\phi }_{0} ) ( \text{Using (ii)} )$

$\mathrm{or}⁡ { K⁡ }_{1} = { K⁡ }_{2} - \frac{ 2 }{ 3 } { K⁡ }_{2} - \frac{ 2 }{ 3 } {\phi }_{0} = \frac{ { K⁡ }_{2} }{ 3 } - \frac{ 2 }{ 3 } {\phi }_{0}$

$\text{or} { K⁡ }_{1} < \frac{ { K⁡ }_{2} }{ 3 }$

Your note

JEE Main Modern Physics in other years

Modern Physics 2026 Modern Physics 2025 Modern Physics 2024 Modern Physics 2023 Modern Physics 2021 Modern Physics 2020 Modern Physics 2019 Modern Physics 2018 Modern Physics 2017 Modern Physics 2016 Modern Physics 2015 Modern Physics 2014

Keep practicing Modern Physics

Work through every JEE Main Modern Physics PYQ, year by year.

Browse all Physics topics

Solution

According to Einstein's photoelectric equation

$\frac{h⁡c}{\lambda }={\phi }_{0}+{\text{KE}}_{\text{max}}$ , where ${\phi }_{0}$ is a work function of a metal

$\therefore { K⁡ }_{1} = \frac{ h⁡ c }{ {\lambda }_{1} } - {\phi }_{0} \ldots ( \text{i} )$

${ K⁡ }_{2} = \frac{ h⁡ c }{ {\lambda }_{2} } - {\phi }_{0} \ldots ( \text{i} \text{i} )$

$\text{or} {K}_{1}-{K}_{2}=hc[\frac{1}{{\lambda }_{1}}-\frac{1}{{\lambda }_{2}}]$

$=hc[\frac{1}{{3\lambda }_{2}}-\frac{1}{{\lambda }_{2}}]=-\frac{2hc}{3{\lambda }_{2}} (\text{Given} {\lambda }_{1}=3{\lambda }_{2})$

$= - \frac{ 2 }{ 3 } ( { K⁡ }_{2} + {\phi }_{0} ) ( \text{Using (ii)} )$

$\mathrm{or}⁡ { K⁡ }_{1} = { K⁡ }_{2} - \frac{ 2 }{ 3 } { K⁡ }_{2} - \frac{ 2 }{ 3 } {\phi }_{0} = \frac{ { K⁡ }_{2} }{ 3 } - \frac{ 2 }{ 3 } {\phi }_{0}$

$\text{or} { K⁡ }_{1} < \frac{ { K⁡ }_{2} }{ 3 }$

JEE Main 2022 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics

JEE Main 2022 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics