In a photo electric effects set-up, a point source of light of poewer `3.2xx10^(-3)` W emits mono energetic photons of energy 5.0 Ev. The source is located at a distance of 0.8m from the centre of a stationary metallic sphere of work function 3.0 Ev & of radius `8.0xx10^(6)`m. The efficiency of photo electric emission is one of every `10^(6)` incident photons. Assume that the sphere is isolated and initially neutral, and that photo electrons are instantly swept awy after emission.
(a) Calculate the number of photo electrons emitted.per second.
(b) Find the ratio of the wavelength of incident light to the de-Broglie wave length of the fastest photo electrons emitted. ltbr. (c) It is observed that the photo electron emission stops ata certain time tafter the light source is switched on. Why ?
(d) Evaluate the time t.

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.