The radiations of 1 eV and 2.5 eV are incident on a metal having a work function of 0.5 eV . The ratio of their maximum velocities is

Photons of energy 1.5 eV and 2.5 eV are incident on a metal surface of work function 0.5 eV. What is the ratio of the maximum kinetic energy of the photoelectrons ?

The light photons of energy 1.5 eV and 2.5 eV are incident on a metallic surface of work function 0.5 eV, the ratio of the maximum kinetic energy of the photoelectrons is

Two photons of energy 2.5 eV and 3.5 eV fall on a metal surface on work function 1.5 eV. The ratio of the maxium velocities of the photoelectrons emitted from the metal surface is

Two differenct photons of energies, 1 eV and 2.5 eV, fall on two identical metal plates having work function 0.5 eV, Then the ratio of maximum KE of the electrons emitted from the two surface is -

A photon energy 3.4 eV is incident on a metal having work function 2 eV . The maximum K.E. of photoelectrons is equal to

Two radiations of photons energies 1 eV and 2.5 eV, successively illuminate a photosensitive metallic surface of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons is

A light of 4 eV incident on metal surface of work function phi eV . Another light of 2.5 eV incident on another metal surface of work function phi_2 eV . Find the ratio of maximum velocity of photo electrons (v_1/v_2) .