Home
Class 12
PHYSICS
If light of wavelength lambda(1) is allo...

If light of wavelength `lambda_(1)` is allowed to fall on a metal , then kinetic energy of photoelectrons emitted is `E_(1)`. If wavelength of light changes to `lambda_(2)` then kinetic energy of electrons changes to `E_(2)`. Then work function of the metal is

A

`(E_(1)E_(2)(lamda_(1)-lamda_(2)))/(lamda_(1)lamda_(2))`

B

`(E_(1)lamda_(1)-E_(2)lamda_(2))/((lamda_(1)-lamda_(2)))`

C

`(E_(1)lamda_(1)-E_(2)lamda_(2))/((lamda_(2)-lamda_(1)))`

D

`(lamda_(1)lamda_(2)E_(1)E_(2))/((lamda_(2)-lamda_(1)))`

Text Solution

Verified by Experts

The correct Answer is:
C
Promotional Banner

Topper's Solved these Questions

  • ELECTRON, PHOTON, PHOTOELECTRIC EFFECT AND X-RAYS

    ERRORLESS |Exercise Cathode Rays and Positive Rays|11 Videos

  • ELECTRON, PHOTON, PHOTOELECTRIC EFFECT AND X-RAYS

    ERRORLESS |Exercise Photon and Photoelectric Effect|4 Videos

  • ELECTRO MAGNETIC INDUCTION

    ERRORLESS |Exercise SET|20 Videos

  • ELECTRONICS

    ERRORLESS |Exercise Selv Evaluation Test|23 Videos

Similar Questions

Explore conceptually related problems

Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

In a photo-emissive cell, with exciting wavelength lambda_1 , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to lambda_2 , the maximum kinetic energy of electron is 2K then

When monochromatic light of wavelength 620 nm is used to illuminate a metallic surface, the maximum kinetic energy of photoelectrons emitted is 1 eV. Find the maximum kinetic energy of photoelectron emitted (in eV) if a wavelength of 155 nm is used on the same metallic surface. (hc = 1240 eV-nm)

The kinetic energy of most energetic electrons emitted from a metallic surface is doubled when the wavelength lamda of the incident radiation is changed from 400 nm to 310 nm. The work function of the metal is

Light of wavelength lambda strikes a photo - sensitive surface and electrons are ejected with kinetic energy is to be increased to 2 E , the wavelength must be changed to lambda' where

The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from lamda_1 " to " lamda_2 The work function of the metal is

In a photo-emissive cell, with exciting wavelength lambda , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to ( 3 lambda)/( 4) the kinetic energy of the fastest emitted electron will be :

The de Broglie wavelength (lambda) of a particle is related to its kinetic energy E as

Light of wavelength 4000 Å is incident on a metal surface. The maximum kinetic energy of emitted photoelectron is 2 eV. What is the work function of the metal surface ?