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

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`