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

`6.25xx10^(11)s^(-1) s^(-1), (0.)/(50) eV`

Explanation

Energy of photon, `E=10.6xx1.6xx10^(-19)`,
`I=2.0Wm^(-2), A=10^(-4)m^(2), phi_(0)=5.6 eV`,
`eta=0.53%`
No. of photoelectrons ejected per sec `=0.53/100xx`no. of photons falling per sec
`=0.53/100xx(IA)/E=0.53/100xx(2xx10^(-4))/(10.6xx1.6xx10^(-19))`
`=6.25xx10^(11)s^(-1)`
Minimum energy of ejected photoelectrons =zero K.E.
Maximum energy of ejected photoelectron `=10.6-5.6`
`=5.0 eV`