Page 130 - Electrical Properties of Materials
P. 130

112                           The band theory of solids

                                   Thus, in the three-dimensional case, the energy band extends from the
                                   minimum energy

                                                       E min = E 1 –2(A x + A y + A z )     (7.35)

                                   to the maximum energy
                                                      E max = E 1 +2(A x + A y + A z ).     (7.36)


                                   7.5  The effective mass

     The mass of an electron in a crys-  It has been known for a long time that an electron has a well-defined mass,
     tal appears, in general, different  and when accelerated by an electric field it obeys Newtonian mechanics. What
     from the free-electron mass, and is  happens when the electron to be accelerated happens to be inside a crystal?
     usually referred to as the effective  How will it react to an electric field? We have already given away the secret
     mass.                         when talking about cyclotron resonance.
                                     We shall obtain the answer by using a semi-classical picture, which, as the
                                   name implies, is 50% classical and 50% quantum-mechanical. The quantum-
                                   mechanical part describes the velocity of the electron in a one-dimensional
                                   lattice by its group velocity,

                                                                  1 ∂E
                                                             v g =    ,                     (7.37)
                                                                   ∂k
                                   which depends on the actual E–k curve. The classical part expresses dE as the
                                   work done by a classical particle travelling a distance, v g dt, under the influence
                                   of a force eE yielding

                                                           dE = eE v g dt
                                                                  1 ∂E
                                                              = eE    dt.                   (7.38)
                                                                    ∂k
                                     We may obtain the acceleration by differentiating eqn (7.37) as follows:

                                                                        2
                                                       dv g  1 d ∂E  1 ∂ E dk
                                                          =        =        .               (7.39)
                                                                         2
                                                       dt     dt ∂k    ∂k dt
                                     Expressing now dk/dt from eqn (7.38) and substituting it into eqn (7.39) we
                                   get
                                                                    2
                                                           dv g  1 ∂ E
                                                              =       eE .                  (7.40)
                                                                 2
                                                           dt     ∂k 2
                                     Comparing this formula with that for a free, classical particle
                                                               dv
                                                             m    = eE ,                    (7.41)
                                                               dt
                                   we may define

                                                                   ∂ E
                                                                    2    –1
                                                                2
                                                            ∗
                                                          m =                               (7.42)
                                                                   ∂k 2
   125   126   127   128   129   130   131   132   133   134   135