The work function of a metal is 2.31 eV. Photoelectric emission occurs when light of frequency `6.4 xx 10^(14)` Hz is incident on the metal surface. Calculate the energy of the incident radiation .

A photon of frequency 5 xx 10^(14)Hz is incident on the surface of a metal. What is the energy of the incident photon in eV?

The work function of caesium is 2.14 eV. When light of frequency 6xx10^(14)Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons. (b) stopping potential and (c) maximum speed of the emitted photoelectrons. given , h=6.63xx10^(-34)Js, 1eV=1.6xx10^(-19)J, c=3xx10^(8)m//s .

The given graph shows the variation of KE with frequency of incident radiations for two surfaces A and B. (a) Which of the metals has greater work function ? For which of the metals will stopping potential be more for the same frequency of incident radiation ? (b) The work function caesium metal is 2.14 eV. When light of frequency 6xx10^(14) Hz. incident on the metal surface , what is the maximum KE of the photoelectrons and stopping potential ?

The threshold frequency for a certain metal for photoelectric effect is 1.7x10^(15)Hz . When light of frequency 2.2xx10^(15)Hz is incident on the metal surface, the kinetic energy of the emitted photoelectrons is 3.3xx10^(-19)J . What is the value of the Planck's constant ?

Work function of potassium metal is 2.30 eV . When light of frequency 8xx 10^(14) Hz is incident on the metal surface, photoemission of electrons occurs. The stopping potential of the electrons will be equal to

Work function for caesium metal is 2.14 eV. Let a beam of light of frequency 6xx10^(14)Hz incident over the metal surface. Now, match the following columns and choose the correct option from code given.