The work function of potassium is 2.2 eV. UV light of wavelength 3000 `Å` and intensity 2 `Wm^(-2)` is incident on the potassium surface. (i) Determine the maximum kinetic energy of the photo electrons (ii) If 40% of incident photons produce photo electrons, how many electrons are emitted per second if the area of the potassium surface is 2 `cm^(2)` ?

Explanation

(i) The energy of the photon is `E = (hc)/(lambda) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(3000 xx 10^(-10))`
`E = 6.626 xx 10^(-19)`J = 4.14 eV br> Maximum KE of the photoelectrons is
`K_(max) = h upsilon - phi_(0) = 4.14 - 2.2 = 1.94` eV
(ii) The number of photons reaching the surface per second is
`n_(P) = (P)/(E) xx A`
`= (2)/(6.626 xx 10^(-19)) xx 2 xx 10^(-4) = 6.04 xx 10^(14)` photons/sec
The rate of emission of photoelectrons is
`= (0.40) n_(P) = 0.4 xx 6.04 xx 10^(14)`
`= 2.415 xx 10^(14)` photoelectrons/sec