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`.

Explanation

`P = IA =2xx10^(-4) W`
`P=n(hc)/(lambda)`
`2xx10^(-4) =nxx10.6xx1.6xx10^(-19)`
`n= (2xx10^(-4))/(10.6xx1.6xx10^(-19))`
`("number of electrons"//"sec")/(n) =53% =0.53`
`"Number of electrons"//"sec" =0.53 n`
`=(0.53xx2xx10^(-4))/(10.6xx1.6xx10^(-19))`
`=6.25xx10^(11)`
`E=phi + K_(max)`
`10.6=5.6 + K_(max) implies K_(max) =5 eV`
If electrons are emitted with zero velocity
`K_(min) =0`