Question

The work function for lead is 4.14 eV.

(a) Find the cutoff wavelength for lead.
nm

(b) What is the lowest frequency of light incident on lead that releases photoelectrons from its surface?
Hz

(c) If photons of energy 5.80 eV are incident on lead, what is the maximum kinetic energy of the ejected photoelectrons?
eV

Answer 1

We know from the photo electric-efect that , but the kinectic energy K=0 when the electrones are removed with the minimun energy possible for doing so, given by the work funtion W. So we know that for items "a" and "b", the condition for this particular photo-electric effect is the following:

, but we know that and that , so we can write:

from where

a) and this is the cutoff wavelength for lead.

b) Again we take our equation , BUT THIS TIME WE SOLVE FOR , THAT IS, the lowest frequency of light incident on lead that releases photoelectrons from its surface. So we get:

from where

b) and this is the lowest frequency of light incident on lead that releases photoelectrons from its surface.

c) For part c we need to use the whole first equation where E=5.8eV, W=4.14Ev and K is teh kinetic energy we need to solve for:

and this is the maximum kinetic energy of the ejected photoelectrons, as the incident energy E exceeds the work function, the electrones will end up with a leftover kinetic energy of 1.66eV.