(a) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?

(b) The work function of aluminium is 4·2 eV. If two photons each of energy 2·5 eV are incident on its surface, will the emission of electrons take place? Justify your answer.

(c) The stopping potential in an experiment on the photoelectric effect is 1·5 V. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.

 

 

 
 
 
 
 

Answers (1)

(a) When a photon of the energy h\nu is absorbed by an electron in the photosensitive material, a part of the energy absorbed is used up in liberating it from the surface (the work function), 

\therefore The remaining energy appears as the kinetic energy of the photoelectron.

Such that,

K_{max}=h\nu -\phi _{o}\; \; \; \; (\therefore \phi _{o}is \; work\; function)

and if, h\nu > \phi _{o}

 the electron is emitted 

(b) If the two photons each of energy 2.5 eV are incident on the given surface, then the emission of electrons will not take place.

Such that,

energy h\nu, of a single photon, is less than the work function \phi _{o}

\because K_{max}=h\nu -\phi _{o}

if  h\nu < \phi _{o} so, hence, no emission will take place.

(c) Given; the stopping potential (\nu _{o})=1.5V

We know, K_{max}=eV_{o}                (\therefore K_{max}=Maximum kinetic energy )

K_{max}=1.6\times 10^{-19}\times 1.5=2.4\times 10^{-19}Joule

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