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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 ...


For more information, see NUREG/CR-5550 "Passive Nondestructive Assay of Nuclear ...

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