Question

A 40 W ultraviolet light source of wavelength `2480 A`. Illuminates a magnesium (mg) surface placed 2 m away. Determine the number of photons emitted from the surface per second and the number incident on unit area of Mg surface per second . The photoelectric work function for Mg is `3.68eV`. Calculate the kinetic energy of the fastest electrons ejected from the surface. Determine the maximum wavelength for which the photoelectric effects can be observed with Mg surface.

Answer 1

Energy of a photon,
`E==(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/(2480xx10^(-10))`
`8xx10^(-19)J`
(a) Number of photons emitted per second by the source, `n=P/E=40/(8xx10^(-19))=5xx10^(19)`
As the surface is at a distance 2m from the point source, the number of photons incident per second per unit area on the surface will be
`n_(i)=n/(area of surface)=n/(4pir^(2))`
`=(5xx10^(19))/(4xx(22//7)xx(2)^(2))=1xx10^(18)`
(b) KE of the fastest photoelectrons is
`K_(max)=(hc)/lambda =phi_(0)=(8xx10^(-19))/(1.6xx10^(-19))-3.68`
`=5-3.68=1.32eV`
(c) Photoelectric effect will take place if the incident light is more than threshold wavelength `(lambda_(0))`.
So, `lambda_(0)=(hc)/(phi_(0))==(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/(3.68xx.16xx10^(-19))`
`3373xx10^(-10)m=3373Å`