In a photoelectric effect set up, a point source of light of power `3.2 xx10^(-3) W` emits monochromatic photons of energy `5eV`.The source is located at a distance of a stationary metallic sphere of work function `3eV` and radius `8xx10^(-3)m`.The efficiency of photoelectron emission is one for every `10^(6)` incident photons.Assume that the sphere is isolated and initially neutral and the photoelectrons are initially swept away after emission.
Find the number of photons emitted per second

Explanation

Energy of a single phton `E = 5.0 eV = 8 xx 10^(-19) J`
Power of source `P = 3.2 xx 10^(-3) W`
`:.` number of photons emitted per second `n = (P)/(E) = (3.2 xx 10^(3))/(8 xx 10^(-19)) = 4 xx 10^(15)//s`
The number of photons incident per second on metal surface is `n_(0) = (n)/(4piR^(2)) xx pir^(2)`
`n_(0) = (4 xx 10^(15))/(4pi(0.8)^(2)) xx pi(8 xx 10^(-3)) = 1.0 xx 10^(11) "photon"//s`

Number of electrons emitted `= (1.0 xx 10^(11))/(10^(6)) = 10^(5)//s`
`KE_(max) = hv - phi= 5.0 -3.0 = 2.0 eV`
The photonelectron emission stops, when the metalli sphere acquires stopping potential
As `KE_(max) - 2.0 eV rArr "Stoppng potential" V_(0) = 2V rArr 2 = (q)/(4piepsi_(0)r) rArr q = 1.78 xx 10^(12) C`
Now charge `q = ("number of electron"//"second") xx l xx e rArr t = (1.78 xx 10^(-12))/(10^(5) xx 1.6 xx 10^(-19)) = 111s`