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 incident photons eject photoelectrons. find the number of photoelectrons emitted per second and their minimum and maximum 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 1.0 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 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% .

Ultraviolet light of wavelength 350 nm and intensity 1 W//m^2 is directed at a potassium surface having work function 2.2 eV (ii) if 0.5 percent of the incident photons produce photoelectric effect, how many photoelectrons per second are emitted from the potassium surface that has an area 1 cm^2 . E_("kmax") =1.3 eV, n= 8.8 xx 10^(11)("photo electron")/("second") or r =(Nhv)/t =nhv

Ultraviolet light of wavelength 66.26 nm and intensity 2 W//m^(2) falls on potassium surface by which photoelectrons are ejected out. If only 0.1% of the incident photons produce photoelectrons, and surface area of metal surface is 4 m^(2) , how many electrons are emitted per second?

A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency , then the number of photons emitted by the bulb in 20 seconds are : ( 1eV = 1.6 xx 10^(-19) J , hc = 12400 eV )

A beam of electromagnetic radiation of intensity 12.8 xx 10^(-5) W//cm^(2) is comprised of wavelength, lambda = 155 nm . It falls normally on a metal (work function phi = 2eV ) of surfacce area of the 1 cm^(2) . If one in 10^(3) photons ejects total number of electrons ejected in 1 s in 10^(x) . (hc = 1240 eVnm) 1eV = 1.6 xx 10^(-19) J =, then x is _____.

Light of wavelenght 5000 Å fall on a metal surface of work function 1.9 eV Find a. The energy of photon b. The kinetic energy of photoelectrons