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CHAP. 33]                      PHYSICAL AND QUANTUM OPTICS                            415



        SOLVED PROBLEM 33.13
              A proton (m = 1.67 × 10 −27  kg) is accelerated through a potential difference of 200 V. (a) What is its
              kinetic energy in electronvolts? (b) What is its velocity?
              (a) Since the charge on the proton is +e, its kinetic energy is 200 eV.
              (b)                      KE = (200 eV)(1.6 × 10 −19  J/eV) = 3.2 × 10 −17  J
                            1
                               2
                  Since KE = mv ,
                            2


                                           2(KE)    (2)(3.2 × 10 −17  J)  5
                                      v =        =                = 2 × 10 m/s
                                             m       1.67 × 10 −27  kg
        SOLVED PROBLEM 33.14
              In the photoelectric effect, light directed at the surface of certain metals causes electrons to be emitted. In
              the case of potassium, 2 eV of work must be done to remove an electron from the surface. (a) If light of
              wavelength 5 × 10 −7  m falls on a potassium surface, what is the maximum energy of the photoelectrons
              that emerge? (b) If light of wavelength 4 × 10 −7  m falls on the same surface, will the photoelectrons
              have more or less energy?

              (a) Since c = f λ, the frequency of the light is
                                                        8
                                                c  3 × 10 m/s
                                                                    14
                                            f =  =           = 6 × 10 Hz
                                                λ   5 × 10 −7  m
                  The energy of each photon is therefore
                                                              14
                                    E = hf = (6.63 × 10 −34  J·s)(6 × 10 Hz) = 3.98 × 10 −19  J
                  Since 1 eV = 1.6 × 10 −19  J,

                                                  3.98 × 10 −19  J
                                             E =              = 2.49 eV
                                                 1.6 × 10 −19  J/eV
                  This is the maximum energy that can be given to an electron by a photon of this light. Because 2 eV is needed
                  to remove an electron, the maximum energy of the photoelectrons in this situation is 0.49 eV.
              (b) A shorter wavelength means a higher frequency and hence more energy to be imparted to the photoelectrons.




                                     Multiple-Choice Questions


         33.1. Which one or more of the following produce coherent electromagnetic waves?
               (a) two lasers of the same frequency
               (b) two antennas fed by the same radio transmitter
               (c)  a pinhole in a cover over a monochromatic light source and its reflection in a mirror
               (d) two pinholes in a cover over a monochromatic light source

         33.2. The resolving power of a lens can be improved by increasing which one or more of the following?
               (a)  the diameter of the lens  (c)  the wavelength of the light
               (b)  the object distance  (d)  the brightness of the light

         33.3. A beam of transverse waves whose vibrations occur in all directions perpendicular to their direction of motion is
               (a) polarized   (c) resolved
               (b) unpolarized  (d) diffracted
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