A phototube is illuminated by light of a known wavelength. Electrons are ejected

Question

A phototube is illuminated by light of a known wavelength. Electrons are ejected from the photocathode with some kinetic energy K. They are collected as anode current unless a variable retarding potential V is large enough to stop the electrons. For a given potential V all electrons with K???eV will be stopped, and at some value V0 even the fastest electrons with a kinetic energy Kmax will be stopped when

Kmax = hf - W = e V0 ,

with f ?the frequency of the incident light, and W the work function of the cathode material. Derive and explain the meaning of this equation.

Explanation / Answer

The maximum kinetic energy Kmax of an ejected electron is given by - Kmax = hf - W where h is the Planck constant and f is the frequency of the incident photon. The term W is the work function, which gives the minimum energy required to remove a delocalised electron from the surface of the metal. For a given frequency of the incident radiation, the stopping potential Vo is related to the maximum kinetic energy of the photoelectron that is just stopped from reaching plate Q. If m is the mass and vmax is the maximum velocity of photoelectron emitted, then Kmax = (1/2) m vmax^2 If e is the charge on the electron and V0 is the stopping potential, then the work done by the retarding potential in stopping the electron = eV0, which gives (1/2) m vmax^2 = eV0 The above relation shows that the maximum velocity of the emitted photoelectron is independent of the intensity of the incident light. Hence, Kmax = eV0 Hence, Kmax = hf - W = e V0

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