X-rays of wavelength `lambda` fall on an phtot senstive surface,emitting electrons.Assuming that the work function of the surface can be netglected,prove that the de-Broglie wavelength of electrons emited will be `sqrt(h lambda//(2 m c))`.

Show that the de-Broglie wavelength lambda of electrons of energy E is givne by the relation: lambda= h /(sqrt(2 m E)) .

An electromagnetic wave of length lambda is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from the surface have the same de-Broglie wavelength lambda , prove that lambda=(2mc)/h lambda_1^2

Work function of a metl surface is 4.2 eV. The maximum wavelength which can eject electrons from this metal surface is:

A metal has work function 5eV. Will this metal emit electrons, when light of wavelength 400 nm falls on it?

Light of wavelength 500 nm falls on a metal whose work function is 1.9 eV. find the kinetic energy of the photo electrons emitted.

Consider a 20 W bulb emitting light of wavelength 5000 oversetcircA and shining on a metal surface kept at a distance 2m . Assume that the metal surface work function of 2eV and that each atom on the metal surface can be treated as a cirular disc of radius 1.5 A. will there be photoelectric emission?