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 V.What is the new wavelength?
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.