Page 37 - Modern physical chemistry
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26                       Structure in Molecules and Atoms

             where A. is the wavelength. It has the direction of the beam.
                Particles with a  definite momentum may be obtained by passing a  heterogeneous
             beam through a velocity selector. This may be a mechanical device consisting of a series
             of rotating discs containing openings that line up for the desired velocity. Or,  when the
             particles are charged, the selector may employ a crossed electric and magnetic field.
                Alternatively, charged particles may be accelerated from approximate rest by a mea-
             sured voltage increment V. If+e is the charge on the particle, the loss of potential energy,
             and the gain in kinetic energy,  during the acceleration is ±eV.  Parameter e may be the
             magnitude of the charge on an electron or a small multiple of it. So a convenient unit of
             energy is the kinetic energy acquired by an electron on moving freely through a voltage
             rise of one volt, the electron volt (eV).
                When m is the mass of the particle and p its momentum, the gain in kinetic energy is
                                                 2
                                               L=±eV,                                [2.18]
                                               2m

             as long as Newtonian mechanics is applicable.

             Example 2. 1
                Relate the wavelength of a homogeneous electron beam to the voltage accelerating
             the electrons.
                From equation (2.18), we have

                                                       112
                                             p=  2meV  )  .
                                                (
             Substituting this expression into the de Broglie equation
                                                    h
                                                 p=-
                                                     A
             obtained from (2.16) and (2.17), and solving for the wavelength yields
                                                    h
                                             A=----
                                                (2mevf/2


                Now introduce accepted values of the fundamental constants to get
              A =          6.6261 x 10- 34   J s     1  = 12.264 x 10-10  V- 1I2  m = 12.264 V- I12  A.
                 [ (             )(            )J 1I2  V1l2
                  2 9.1094 x 10- 31  kg  1.6022 x 10- 19  C


             This can be rewritten in the form
                                                       112
                                            A =  15~41  ]  A.
                                               [



             2.4 Diffraction by Randomly Oriented Molecules
                Bond distances and bond angles can be obtained from diffraction data. In a common
             setup, a homogeneous beam of particles passes through a low pressure jet of the material.
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