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

(a) `P=n(hc)/(lambda) implies n=(P)/(hc//lambda) =(3.2xx10^(-3))/(5xx1.6xx10^(-19)) =4xx10^(15)`
where: n: number of photons/sec.
Number of photons/sec/`m^(2)` at sphere
`n'=(n)/(4pi d^(2))`
Number of photons/sec incident on sphere
`n''=(n)/(4pi d^(2)).pi r^(2) =(n)/(4)((r)/(d))^(2)`
`=(4xx10^(15))/(4)((8xx10^(-3))/(0.8))^(2) =10^(15)xx10^(-4) =10^(11)`
Since `10^(6)` photons emit one electron, hence the number of electrons emitted per second
`n_(0) =(10^(11))/(10^(6)) =10^(5)`
(b) `E =phi + K_(max) implies 5=3 +K_(max)`
`K_(max) =2 eV`
de Broglie wavelength, `lambda_(d) =(h)/(p) =(h)/(sqrt(2mK_(max)))`
Incident wavelength, `lambda =(hc)/(E)`
`(lambda)/(lambda_(d)) =(c sqrt(2mK_(max)))/(E)`
`(3xx10^(8)sqrt(2xx9.1xx10^(-31)xx2xx1.6xx10^(-19)))/(5xx1.6xx10^(-19))`
`=(3xx10^(8)xx2)/(5) sqrt((9.1xx10^(-31))/(1.6xx10^(-19)))`
`=1.2xx10^(8)xx(1)/(4)xx10^(-6) sqrt(91)`
`=286.18`
(c) As electrons are ejected from sphere, it becomes positively charged and its potential increases. When potential of sphere becomes equal to stopping potential, emission of electrons stopped.
(d) Let charge on sphere be q when its potential equals stopping potential
`V=V_(s)`
`(1)/(4pi in_(0)) (q)/(r)=V_(s)` `implies (9xx10^(9)q)/(8xx10^(-3)) =2`
`q=(16)/(9)xx10^(-12)C`
`q=n_(0)et`
`(16)/(9)xx10^(-12) =10^(5)xx1.6xx10^(-19)t`
`t =(10)/(9)xx100 =(1000)/(9) =111 sec`