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`

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 .

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 .

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 .

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 .

A stream of photons of energy 10.6 eV and intensity 2.0 W m^(-2) is incident on a platinum surface. Area of the surface is 1.0 xx 10^(-4) m^(2) and its work function is 5.6 eV. 0.53 % of incident photons emit photoelectrons. Find the number of photoelectrons emitted per second and maximum and minimum energies of the emitted photoelectrons in eV. (1eV = 1.6 xx 10^(-19)J)

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]

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]

A metal plate of area 1xx10^(-4)m^(2) is illuminated by a radiation 16 mW//m^(2) .The work function of the metal is 5eV.The energy of the incident photons is 10 eV and only 10% of it produces photo-electrons .The number of emitted photoelectrons per second and their maximum energy respectively will be

A light of intensity 16 mW and energy of each photon 10 eV incident on a metal plate of work function 5 eV and area 10^(-4)m^(2) then find the maximum kinetic energy of emitted electrons and the number of photoelectrons emitted per second if photon efficiency is 10% .