One milliwatt of light of wavelength `4560Å` is incident on a cesium surface. Calculate the photoelectric current liberated, assuming a quantum efficiency of 0.5%. Given work function for cesium = 1.93 eV :

Options

A. 1.856 mA
B. `1.856muA`
C. `18.56muA`
D. none of these.

Answer

B

Explanation

Energy of each photon of incident light is
`E=hv=(hv)/(lambda)=(6.62xx10^(-34)xx3xx10^(8))/(4560xx10^(-10))`
`=4.32xx10^(-19)J`
Power of source `=10^(3)W(=P)`
Number of photons liberarted per sec.
`N=(P)/(E)=(10^(-3))/(4.32xx10^(-19))=2.32xx10^(15)` per sec.
As quantum efficiency is `0.5%` of effective number of incident photons is only `0.5%` which ejecte photoelectron.
`:.` Number of electrons ejected as
`N_(e)= 2.32xx10^(15)xx(0.5)/(100)=1.16xx10^(13)//sec`
`:.` Corresponding photo electric current proudced,
`i=(q)/(t)=(n.N_(e))/(t)=1.6xx10^(-19)xx1.16xx10^(13)`
`=1.856xx10^(-6)A`
`u=1.856 muA`.