A metal plate of area `1 xx 10^(-4) m^(2)` is illuminated by a radiation of intensity `16m W//m^(2)`. The work function of the metal is 5eV. The energy of the incident photons is 10eV. The energy of the incident photons is 10eV and 10% of it produces photo electrons. The number of emitted photo electrons per second their maximum energy, respectively, will be : `[1 eV = 1.6 xx 10^(-19)J]`
The energy of emitted photoelectrons from a metal is 0.9ev , The work function of the metal is 2.2eV . Then the energy of the incident photon is
The work function of a metal is 2.3 eV and the wavelength of incident photon is 4.8 xx 10 ^( -7) m. Find maximum kinetic energy of photo electrons
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 a mental is X eV when light of energy 2X eV is made to be incident on it then the maximum kinetic energy of emitted photo electron will be
A photon of wavelenth 3000 A strikes a metal surface, the work function of the metal being 2.20eV . Calculate (i) the energy of the photon in eV(ii) the kinetic energy of the emitted photo electron and (iii) the velocity of the photo electron.
When a beam of 10.6 eV photons of intensity 2.0 W //m^2 falls on a platinum surface of area 1.0xx10^(-4) m^2 and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. Find the number of photoelectrons emitted per second and their minimum and maximum energy (in eV). Take 1 eV = 1.6xx 10^(-19) J .
Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be
Find the kinetic energy of emitted electron, if in a photoelectric effect energy of incident Photon is 4 eV and work function is 2.4 eV.
The work function of metal is 1 eV . Light of wavelength 3000 Å is incident on this metal surface . The velocity of emitted photo - electrons will be
when a beam of 10.6 eV photons of intensity 2.0 W//m^(2) falls on a platinum surface of area 1.0 xx 10^(4) m^(2) and work function 5.6 eV , 0.53 % of the incidentphotons eject photoelectrons find the number of photoelectrons emited per second and their minimum energies (in eV)Take 1 eV= 1.6 xx 10^(-19) J
