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`.

Answer

`6.25 xx 10^(11)`, zero , `0.5 eV`

Explanation

Energy of incident photon,
`E_(1)=10.6 eV=10.6xx1.6xx10^(-19) J=16.96xx10^(-19) J`
Energy incident per unit area per unit (intensity)`=2J`
`:.` No. of photon incident on unit area in unit time
`=(2)/(16.96xx10^(-19))implies 1.18xx10^(14)`
Therefore, number of photon incident per unit time on given area `(1.0xx10^(-4) m^(2))`
`=(1.18xx10^(18))(1.0xx10^(-4))=1.18xx10^(14)`
but only `0.83 %` of incident photons emit photoelectrons
`:.` No. of photoelectrons emitted per second (n)
`n=(0.53/100) (1.18xx10^(14))implies n=6.25xx10^(11)`
`K_("min")=0` and `K_("max")=E_(1)-` work function
`=(10.6-5.6) eV=5.0 eV`
`:. K_("max")=5.0 eV`