Home
Class 11
CHEMISTRY
Photoelectrons are liberated by ultra li...

Photoelectrons are liberated by ultra light of wavelength `2000 Å` from a metallic surface for which the photoelectric threshold is 4000Å. Calculate the de Broglie wavelenth of electrons emitted with maximum kinetic energy.

Text Solution

AI Generated Solution

To solve the problem step by step, we will calculate the maximum kinetic energy of the emitted photoelectrons and then use that to find the de Broglie wavelength of the electrons. ### Step 1: Calculate the energy of the incident photons The energy of the incident photons can be calculated using the formula: \[ E = \frac{hc}{\lambda} \] where: ...
Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    ALLEN|Exercise Exercise - 01|38 Videos
  • ATOMIC STRUCTURE

    ALLEN|Exercise Exercise - 02|59 Videos
  • IUPAC NOMENCLATURE

    ALLEN|Exercise Exercise - 05(B)|7 Videos

Similar Questions

Explore conceptually related problems

Photo electrons are liberated by ultraviolet light of wavelength 3000 Å from a metalic surface for which the photoelectric threshold wavelength is 4000 Å . Calculate the de Broglie wavelength of electrons emitted with maximum kinetic energy.

Photoelectron are liberated by altra voilet light of wavelength 3000Å from a metallic surface for which the photoelectron thershold is 4000 Å calculate de broglic wavelength of electron with maximum kinetic energy

Calculate de - Broglie wavelength of an electron having kinetic energy 2.8xx10^(-23)J

Light of wavelength 2000 Å falls on a metallic surface whose work function is 4.21 eV. Calculate the stopping potential

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

Light of wavelength 200 nm incident on a metal surface of threshold wavelength 400 nm kinetic energy of fastest photoelectron will be

Light of wavelength 400 nm strikes a certain metal which has a photoelectric work function of 2.13eV. Find out the maximum kinetic energy of the photoelectrons

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.

A photon of frequency n causes photoelectric emmission from a surface with threshold frequency n_0 .The de Broglie wavelength lambda of the photoelectron emitted is given as

An electron is moving with a kinetic energy of 4.55 xx 10^(-25)J . Calculate its de-Broglie wavelength.