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 photons eject photoelectrons. Find the number of photoelectrons emitted per second

Explanation

Number of photons incident on the metal per second `=("Total energy incident on the metal per second")/("Energy of each photon")`
`=("Intensity x Area")/("Energy of each photon")`
Number of electrons emitted per second `=((0.53)/(100))` Number of photons incident on the metal surface
Number of electrons emitted per second `=((0.53)/(100))("Intensity x Area")/("Energy of each photon")`
`=((2.0)(1.0xx10^(-4)))/((10.6xx1.6xx10^(-19)))xx(0.53)/(100)=6.25xx10^(11)`
Minimum kinetic energy of photo electrons = 0
Maximum kinetic energy is, `K_(max)=E-W_(0)`
`=10.6-5.6=5eV`,