Vo for photo electrons emitted from a surface by light of wavelength 4910 A is 0.71 V. When the incident wavelength is changed the Vo is 1.43...

The external photoelectric effect happens when an incoming
photon strike an electron into an atom on the surface of a metal. If the energy of the
photon is greater than the work function of the metal `W` (energy necessary to move the
electron from its energy level into the atom to infinite distance) than the electron
will be released from the atom having a certain kinetic energy. The kinetic energy of
the electron emitted is usually measured by stopping it into a reverse electric
potential. Hence the law of the photoelectric effect
is:

`E_(ph) =W +E_k`

`h*nu =W
+(m*v^2)/2`

`(h*c)/lambda = W +eU`

For initial photon from text we have `lambda =4910 A` and
`U_1=0.71 V` . The work function of the metal is

`W =
(h*c)/lambda-e*U_1 =(6.626*10^-34*3*10^8)/(4910*10^-10) -1.6*10^-19*0.71 = 2.91*10^-19 J
=1.82 eV`

For the second photon the stopping potential is
`U_2 =1.43 V`

`lambda = (h*c)/(W+eU_2)
=(6.626*10^-34*3*10^8)/((1.82+1.43)*1.6*10^-19)=3822.7 A`

The new photon wavelength is 3822.7
Angstrom.