K.E of photon electrons is `E`. When light of wavelength,`lambda` falls on them.If `K.E` of photon electrons reduces to half i.e.`E/2`, then,wavelength of incident light will be ..

The graph between the energy of photoelectrons E and the wavelength of incident light (lamda ) is

The kinetic energy of an electron is E when the incident wavelength is lamda To increase ti KE of the electron to 2E, the incident wavelength must be

The wavelength lambda_(e ) of an photon of same energy E are related by

K.E. of photo electron is E when incident frequency is lambda_(1) . It is 2E when incident frequency is lambda_(2) . The relation between the wavelength is

The KE of the photoelectrons is E when the incident wavelength is (lamda)/(2) . The KE becomes 2E when the incident wavelength is (lamda)/(3) . The work function of the metal is

Assertion: if wavelength of light is doubled, energy and momentum of photons are reduced to half. Reason: By increasing the wavelength, speed of photons will decrease.

Light of wavelength 4000 Å falls on a photosensitive plate with photoelectric work function 1.96 eV. The K.E. of photoelectrons emitted will be :

Electrons are emitted from the cathode of a photocell of negligible work function, when photons of wavelength lamda are incident on it. Derive the expression for the de Broglie wavelength of the electrons emitted in terms of the wavelength of the incident light

Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is lambda . If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is