Home
Class 11
CHEMISTRY
Light of wavelenght 4000Å falls on the s...

Light of wavelenght 4000Å falls on the surface of cesium . Calculate the maximum kinetic energy of the photoelectron emiited . The critical wavelenght for photoelectric effect in cesium is 6600Å.

Text Solution

Verified by Experts

The correct Answer is:
`1.95xx10^(-19)` Joule

`N//A`
Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    BANSAL|Exercise Exercise.1|48 Videos
  • ATOMIC STRUCTURE

    BANSAL|Exercise Exercise.2|21 Videos
  • ATOMIC STRUCTURE

    BANSAL|Exercise Solved Examples|17 Videos
  • GENERAL ORGANIC CHEMISTRY

    BANSAL|Exercise Exercise 3|9 Videos

Similar Questions

Explore conceptually related problems

Light of wavelength 2000Å is incident on a metal surface of work function 3.0 eV. Find the minimum and maximum kinetic energy of the photoelectrons.

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 ?

Find the maximum kinetic energy of the photoelectrons ejected when light of wavelength 350mn is incident on a cesium surface. Work function of cesium = 1.9 e V.

If light of wavelength 6200Å falls on a photosensitive surface of work function 2 eV, the kinetic energy of the most energetic photoelectron will be

Light of wavelength 5000 Å falls on a sensitive plate with photoelectric work function of 1.9 eV . The kinetic energy of the photoelectron emitted will be

If the frequency of light in a photoelectric experiment is double then maximum kinetic energy of photoelectron

Ligth of wavelength 4000 Å is incident on a metal plate whose work function is 2 eV . What is maximum kinetic enegy of emitted photoelectron ?