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

Energy incident on surface per second
`P = 1A`
`= 2.0 xx 1.0 xx 10^(-4) = 2 xx 10^(-4) J`
Energy of each photon
`= 10.6 e V = 10.6 xx 1.6 xx 10^(-19)J`
Number of photon incident on the surface
` = (2 xx 10^(-4))/(10.6 xx 1.6 xx 10^(-19))`
`Number of photon emitted
`= (0.53)/(100) xx (2 xx 10^(-4))/(10.6 xx 1.6 xx 10^(-19)) = 6.25 xx 10^(11)`
According to Einstein's photoelectric equation, maximum `KE` of photoelectrons
`E_(k)` = epsilon - W = 10.6 e V - 5.6 e V = 5 e V`
Minimum kinetic energy of photoelectrons = Zero.