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 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 ?

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 energy of a photon of frequency 4 xx 10^(15) Hz is ?

663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5 xx 10^(9) incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is "_________________" .

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

The work function of a surface is 3.1 eV. A photon of frequency 1x 10^(15) Hz. Is incident on it. Calculate the incident wavelength is photoelectric emission occur or not.

A photon of frequency v is incident on a metal surface whose threshold frequency is v_(0) . The kinetic energy of the emitted photoelectrons will be