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

Answer

(a) `10^(5)s^(-1)` (b) `286.18`
(d) `(1000)/(9)sec=111s`

Explanation

(A) Energy of emitted photons
`E_(1)= 5.03V=5.0xx1.6xx10^(-19)J.= 8.0xx10^(l-19)J`
Power the point source is `3.2xx10^(-3)W` or `3.2xx10^(-3)J//s`.
Therefore, energy emitted per second, `E_(2)= 3.2xx10^(-3)J`
Hence, number of photons emitted per second `n_(1)=(E_(2))/(E_(1))`
or `n_(1)=(3.2xx10^(-3))/(8.0xx10^(-19)) = n_(1)= 4.0xx10^(15) "photon"//s`
Number of photons on until area at a distance of `0.8 m` from the sources `S` will be
`n_(2)=(n_(1))/(4pi(0.8)^(2))=(4.0xx10^(15))/(4pi(0.64))~~5xx10^(14)"phton"//m^(2)`
The area of metallic sphere over which photons will fall is:
`A= pir^(2)=pi(8xx10^(-3))m^(2)~~2.01xx10^(-4)m^(2)`
Thereforem number of photons incident on the sphere per second are
`n_(3)=n_(2)A=(5.0xx10^(14)xx2.01xx10^(-4))~~10^(11)//s`.
But Since, one photoelectron is emitted for every `10^(6)` photons, hence number of photoelectrons emitted per second.
`n=(n_(3)).(10^(6))=(10^(11))/(10^(6))=10^(5)//s or n=10^(5)//s`
(B) Maximum kinetic energy of photoelectrons `K_(max)=` Energy of incident photons-work function.
`=(5.0-3.0)eV = 2.0 eV= 2.0xx1.6xx10^(-19)H`
`K_(max)=3.2xx10^(-19)J`
The de-Broglie wavelength of these phtoelectrons will be
`lambda_(1)=(h)/(P)=(h)/(sqrt(2K_(max)m))`
Hence `h`= Planck's constant and `m`= mass of electron.
`lambda_(1)=(6.63xx10^(-34))/(sqrt(2xx3.2xx10^(-19)xx9.1xx10^(11)))`
`8.68xx10^(-10)m=8.68Å`
Wavelength of incident light `=(12375)/(E_(1)("in"eC))`
or `lambda_(2)=(12375)/(5)Å=2475Å`
Therefore, the desired ratio is:
`(lambda_(2))/(lambda_(1))=(2475)/(8.68)=285.1`
(C ) As soon as electron are emitted from the metal sphere, it gets positively charged and acquired positive potential gradually increases as more and more photoelectrons are emitted from its surface. Emsission of photoelectrons is stopped when its potential is equal to the stopping potential required for fastest moving
(D) A discussed in part `(C )` emission of photoelectrons is stopped when potential on the metal spehre is equal to stopping potential of fastest moving electrons.
Since, `K_(max)=2.0eV`.
Thereforem stopping potential `V_(0)=2V`. Let `q` be the chrage required for the potential on the sphere to be equal to stopping potential or `2V`. Then
`2=(1)/(4piepsilon_(0)).(q)/(r )=(9.0xx10^(9))(q)/(8.0xx10^(-3))`
`q = 1.78xx10^(-12)C`.
photoelectrons emitted per second `=10^(5)` [Part a]
or change emitted per second `=(1.6xx10^(-19))xx10^(5)C`
`=(1.6xx10^(-14))C`.
Therefore, time required to acquire the charge `q` will be
`t=(q)/(1.6xx10^(-14))=(1.78xx10^(-12))/(1.6xx10^(-14))s`.
or `t~~111s`