In a photoelectric set up, a point source of light of power `3.2xx10^(-3)` W emits monoenergetic photons of energy 5.0 eV. The source is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius `8.0xx10^(-3)` m. The efficiency of photoelectron emission is one for every `10^(6)` incident photon. Assume that state is isolated and initially neutral and that the photoelectrons are instantly swept away after emission. Calculate the number of photoelectron emitted per sec, by sphere :

Options

A. `10^(2)`
B. `10^(3)`
C. `10^(4)`
D. `10^(5)`

Answer

D

Explanation

Power of the source `P=3.2xx10^(-3)W`
`P=2.2xx10^(-3)W`
Energy of one photon
`E=5.0eV=5xx1.6xx10^(-19)J=8.0xx10^(-19)J`
Number of photon emitted per sec by source,
`N=(P)/(E)=(3.2xx10^(-3))/(8.0xx10^(-19))=4xx10^(15)`
If `r=0.8 m` is the distance of the sphere from the source, then the number of photons incident on the sphere per sec unit area.
`=(N)/(4pir^(2))`
If `R=8.0xx10^(-3)m` is the radius of the sphere, the number of photons incident on sphere per sec.
`=(N)/(4pir^(2))xx4pi R^(2)=(NR^(2))/(4pir^(2))xx4`
`=(4xx10^(15)xx(8.0xx10^(-3))2)/(4xx(0.80)^(2))=4xx10^(11)`
As efficiency of photoelectron emission is 1 for every `106(6)` incident photons, therefore, the number of emitted photoelectrons per sec
`=4xx(10^(11))/(10^(6))=4xx10^(5)~~10^(5)`