in an experimental set up to study the photoelectric effect a point source of light of power `3.2xx10^(-3)` W was taken. The source can emit mono energetic photons of energy 5eV and is located at a distance of 0.8 m from the center of a stationary metallic sphere of work-function 3.0 eV. The radius of the sphere is `r = 8xx10^(-3)` m. The efficiency of photoelectric emission is one for every `10^6` incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swept away after the emission).
Time after which photoelectric emission stops is

Explanation

(i) Energy of emitted photons `E_(1)=5.0eV = 8xx 10^(-19) J`,Energy emittedper second by the source `E_(2)=3.2xx10^(-3) `J ,No. of photons emitted per second =`E_(2)/E_(1)=4 XX10^(15)`,No. of photons incident on the sphere per second, `n=(4xx10^(15))/(4pi (0.8)^(2))xxpi (8xx10^(-3))^(2) approx 10^(11)` per second.No. of photoelectron emitted per second `n'=(10^(11))/(10^(6))=10^(5)` per second.(ii) Max. K.E. of photo electron `k_(max)`=Energy of incident photons -work function =`2eV=3.2xx10^(-19)J`,de-Broglie wavelength of these photo electrons. `lambda_(1)=(h)/sqrt ((2k^(max)m_(e))=8.68A` wavelength of incident light `lambda _(2) =(12375)/(5)=2475`, The desired ratio is `(lambda_(2))/(lambda_(1))=285.1`.(iii) Emission of photoelectrons is stopped when its potential when is equal to the stopping potential required for the fastest moving electrons..(iv) `k_(max)=2eV` stopping potential `v_(0)=2volt`, `implies 2= kq/r `, `implies q= 1.78 xx 10^(-12)`c, `implies t =(q)/(10^(5)xx1.6xx10^(-19))`, ` t approx 111 seconds.