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 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

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

The ratio of wavelength of a photon and that of an electron of same energy E will be

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

If the P.E. of an electron is -68 eV in hydrogen atom then find out K.E.,E of orbit where electron exist & radius of orbit.

Stopping potential of emitted photo electron is V when monochromatic light of wavelength 'lamda' incident on a metal surface. If wavelength of light incident becomes 'lambda/3' stopping potential of photoelectrons becomes V/4 then the threshold wavelength of metal is k'lambda' then k will be

An electron and a photon have same wavelength lambda , what is ratio of their kinetic energies ?

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.

If light of wavelength lambda_(1) is allowed to fall on a metal , then kinetic energy of photoelectrons emitted is E_(1) . If wavelength of light changes to lambda_(2) then kinetic energy of electrons changes to E_(2) . Then work function of the metal is