As shown, a metal cathode (emitter) is part of a circuit employing the photoelec

Question

As shown, a metal cathode (emitter) is part of a circuit employing the photoelectric effect in order to determine the work function of the metal cathode. The circuit contains an ammeter, variable emf, collector plate and a light source (which emits light of frequency 7.5 x 10^ 14 Hz. During the experiment, the voltage of the emf is gradually increased from 0.0 volts to 0.65 volts until it is observed that electric current in the circuit is zero at 0.65 volt.

(a) Calculate the wavelength ofthe photons which are incident on the cathode.

(b) Calculate the work function of the metal cathode (in electronvolts).

(c) Calculate the minimum frequency of the light necessary to cause electrons to be ejected from the metal cathode (i.e. the threshold frequency).

Explanation / Answer

Given that, Light source of frequency f= 7.5X 10^14 Hz

then, we know that, eV0= hf-hf0

a) incident light wave lenght= c/f= (3*10^8)/7.5*10^14 = 400nm

b) workfunction, hf0= hf-eV0 = 6.625*10^-34 Js x7.5*10^14 Hz- 1.6*10^-19 Cx0.65V= 3.91*10^-19 Joule=2.4 eV

c) threshold frequency, f0= hfo/h=0.59*10^14Hz

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