Home
Class 12
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

Verified by Experts

The correct Answer is:
`1.2xx10^(-5)m`
Promotional Banner

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.