The work function for a given photosensitive surface is 2.5 eV. When light of frequency f falls on this surface, the emitted photoelectrons are completely stopped by applying a retarding potential of 4.1 V. Estimate the value of the frequency f of the light.

Dear student,
 Since the emitted photoelectrons are completely stopped by applying a retarding potential of 4.1 V, the maximum kinetic energy of any photoelectron would be 4.1eV.
If the incdent light has frequency f then the energy of any photon in it is hf, where 'h' is Plank's constant.
From Einstein's photoelectric effect equation       KEmax = hf - work function
                                                                             4.1ev   = hf - 2.5eV 
                                                                           f = 1.6 x 1015 Hz
Regards.

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