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

In a photoelectric effect experiment, the frequency of photons bombarding the surface is increased until photoelectrons just start to leave the surface. If this

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