In the photoelectric effect, an electromagnetic wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is $2.14 \text{ eV}$ and the stopping potential is $2 \text{ V}$, what is the wavelength of the electromagnetic wave (in $\text{ nm}$)? (Given $hc = 1242 \text{ eV nm}$, where $h$ is Planck's constant and $c$ is the speed of light in vacuum.)

  • A
    $400$
  • B
    $600$
  • C
    $200$
  • D
    $300$

Explore More

Similar Questions

Statement $-1$: When ultraviolet light is incident on a photocell, its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X$-rays, both $V_0$ and $K_{max}$ increase.
Statement $-2$: Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

The photoelectric work function for a metal is $2.4 \ eV$. Among the four wavelengths,the wavelength of light for which photoemission does not take place is: (in $nm$)

The stopping potential $(V_s)$ as a function of the frequency $(\nu)$ of the incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graphs show that the work function of $A$ is

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

Difficult
View Solution

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

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