X-rays of wavelength `lambda` fall on photosensitive-surface emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Bfoglie wave-length of eleectrons emitted will be `root``((hlambda)/(2mc))`.What is threshold frequency?

Light rays of wavelength lambda and lambda/2 are incident on a a photosensitive metal surface, IF the maximum kinetic energy of the emitted photoelectrons from the metal surface in 2nd case be 3 times the maximum kinetic energy of emitted photoelectrons in the 1st case then determine the work function of the metal.

A 5 W point source emits monochromatic light of wave length 500Å What should be the work function of the meal from whose surface this light can liberate photoelectrons?

Ratio of the work functions of two metal surface is 1 : 2 . If threshold wavelength of photoelectric effect for the 1st metal is 6000Å , what is the corresponding value for the 2nd metal surface?

An electromagnetic wave of wavelength lamda is incident on a photoelectrons emitted from this surface have the de Broglie wavelength lamda_(1) prove that lamda = ((2mc)/(h)) lamda_(1)^(2)

A photon of wavelength 4x10^-7 m strikes a metal surface work function of the metal is 2.13eV calculate the kinetic energy of the emitted photoelectrons in eV

Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2eV, for the incident radiation of wavelength 300nm.

The work function of a metal surface for electron emission is W. What will be the threshold frequency of incident radiation for photoelectric emission?

Work function of a metal surface is 2eV. Maximum kinetic energy of photoelectrons emitted from the surface for incidence of light of wavelength 4140 Å is 1eV. What is the threshold wavelength of radiation for that surface?

Threshold wavelength for photoelectric emission from a metal surface is 3800Å , Utraviolet light of wave-length 2600Å is incident on the metal surface. Find the work function of the metal and