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The work function of caesium metal is 2.14 eV. When light of frequency 6 *10^{14} Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
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6.4: The Compton Effect - Physics LibreTexts


Nov 5, 2020 ... The Compton effect is the term used for an unusual result observed when ... Describe how experiments with X-rays confirm the particle nature of radiation ... Here the photon's energy Ef is the same as that of a light quantum of ... The wave relation that connects frequency f with wavelength λ and speed c ...


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NUREG/CR-5550 "Passive Nondestructive Assay of Nuclear ...


Michael C. Miller (Chapter 10) ... 5.1.6 Determination of Peak Position by the First Moment Method 105 ... 7.3.1 One-Component Example (Uranium Metal) ... electrons. The colors of the emitted light are characteristic of. the radiating elements ... The electron volt (eV) is a unit of energy equal to the kinetic energy gained by an ...


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Teja
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Given that  

Caesium metal work function = 2.14 eV

Incident light frequency \upsilon = 6*10^{14} Hz.


a) Maximum kinetic energy of the emitted electrons

        Photon's energy = energy needed for the emission of an electron (work function) + Maximum kinetic energy of the electron

                      E = W + K.E

                      K.E = h \upsilon - W

                        

                      E = \frac{ 6.63 * 10^{-34} * 6*10^{14}}{1.6*10^{-19}} - 2.14 eV

                      K.E = 2.485 - 2.140

                      K.E = 0.345 eV


          Hence, Maximum kinetic energy of the emitted electrons   K.E = 0.345 eV


b) Stopping potential    

              By conservation of energy, kinetic energy has to be equal to the change in potential energy.

                            K.E = eV

                             V = \frac{K.E}{e}               

                             V = \frac{0.345 eV}{1.6*10^{-19}} = \frac{0.345 * 1.6*10^{-19} V}{1.6*10^{-19}}

                             V = 0.345 V


c) Maximum speed of the emitted photoelectrons?  

        Mass of the electron m = 9.1*10^{-31}

        Kinetic energy K.E = \frac{1}{2} mv^2

                       v^2 = \frac{2K.E}{m}       

                       v^2 = \frac{2*0.345 eV}{9.1*10^{-31}}    

                       v^2 = \frac{2*0.345 * 6.1*10^{-19}}{9.1*10^{-31}}

                       v = \sqrt{\frac{2*0.345 * 6.1*10^{-19}}{9.1*10^{-31}}}

                       v = 3.323 * 10^{5} m/s

                       v = 332.3 km/s


          Hence, Maximum speed of the emitted photoelectrons v = 332.3 km/s