Photons with wavelength 220 nm are incident on an aluminium

Photons with wavelength 220 nm are incident on an aluminium plate with work funtion phi = 4.08 eV. What is the individual energy of the photons? What is the maximum energy of the electrons emitted? What would be the stopping potential?

Solution

Given:-

wavelength of photon=220nm= 220*10^-9 m

work function=4.08eV=4.08*1.6*10^-19 J=6.53*10^-19 J

a] Individual energy of the photon, E incident = h c/lambda= 6.6*10^-34* 3*10⁸/ 220*10^-9= 0.09*10^-17 J=9*10^-19 J

b] Maximum enery of the electron emitted is equal to kinetic energy of electrons

Applying the Einstein\'s phototelectric equation; we have E _incident= work function+ Ke _electron

Ke _electron= E _incident- work function=9*10^-19 J-6.53*10^-19 J=2.47*10^-19 J

c] stopping potential= Ke _electron

qV= 2.47*10^-19 J

V= 2.47*10^-19 J/ 1.6*10^-19= 1.54 V