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

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`