In a photoelectric effect experiment the frequency of photon
In a photoelectric effect experiment, the frequency of photons bombarding the surface is increased until photoelectrons just start to leave the surface. If this occurs at a frequency of 6.0 1014 Hz, what is the work function of the surface?
Solution
W = hv0 ;
W = 6.626 e -34 * 6 e 14 ;
W = 3.9756 e -19 J <-----ans
W = 3.9756 e -19 / [ 1.6 e -19 ]
W = 2.484 eV <---------ans
