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`

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.