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 1.0 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

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 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% .

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% .

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

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

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