(A) The Einstein's photoelectric equation is given by $KE_{\max} = eV_S = \frac{hc}{\lambda} - \phi$.For $\lambda_1 = 280 \, nm$,the energy of the inc

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The Einstein's photoelectric equation is given by $KE_{\max} = eV_S = \frac{hc}{\lambda} - \phi$.For $\lambda_1 = 280 \, nm$,the energy of the incident photon is $E_1 = \frac{1240 \, eV \cdot nm}{280 \, nm} \approx 4.43 \, eV$.The stopping potential $V_{S1}$ is $V_{S1} = \frac{E_1 - \phi}{e} = 4.43 - 2.5 = 1.93 \, V$.For $\lambda_2 = 400 \, nm$,the energy of the incident photon is $E_2 = \frac{1240 \, eV \cdot nm}{400 \, nm} = 3.10 \, eV$.The stopping potential $V_{S2}$ is $V_{S2} = \frac{E_2 - \phi}{e} = 3.10 - 2.5 = 0.60 \, V$.The change in stopping potential is $\Delta V = V_{S1} - V_{S2} = 1.93 - 0.60 = 1.33 \, V \approx 1.3 \, V$.

Class 12 Physics Dual Nature of Radiation and Matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

Difficult

In a photoelectric experiment,ultraviolet light of wavelength $280 \, nm$ is used with a lithium cathode having a work function $\phi = 2.5 \, eV$. If the wavelength of incident light is switched to $400 \, nm$,find out the change in the stopping potential (in $V$).
$(h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s)$

JEE MAIN 2021 All JEE MAIN

A
$1.3$
B
$1.1$
C
$1.9$
D
$0.6$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Dual Nature of Radiation and Matter

Subtopic

Einstein's Photoelectric Equation and Energy Quantum of Radiation

Previous Year

JEE MAIN 2021 Paper All JEE MAIN years →

The slope of the graph between the frequency of incident light and the stopping potential for a given surface is:

Medium

View Solution

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?

DifficultMHT CET 2026

View Solution

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

Medium

View Solution

$A$ metal surface is illuminated by light of wavelength $400 \ nm$. The kinetic energy of the emitted photoelectrons is found to be $1.68 \ eV$. The work function of the metal is ......... $eV$. $(hc = 1240 \ eV \cdot nm)$

Difficult

View Solution

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for the photoelectric current is $V_s$. When the same surface is illuminated by a light of frequency $v/2$,the stopping potential is $V_s/4$. The threshold frequency of photoelectric emission is

EasyMHT CET 2022

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

The slope of the graph between the frequency of incident light and the stopping potential for a given surface is:

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

$A$ metal surface is illuminated by light of wavelength $400 \ nm$. The kinetic energy of the emitted photoelectrons is found to be $1.68 \ eV$. The work function of the metal is ......... $eV$. $(hc = 1240 \ eV \cdot nm)$

When a certain metal surface is illuminated with light of frequency $v$,the stopping potential for the photoelectric current is $V_s$. When the same surface is illuminated by a light of frequency $v/2$,the stopping potential is $V_s/4$. The threshold frequency of photoelectric emission is

Vedclass Test Series

Exam Paper Generator

Online Exam Module