Page 107 - Electrical Properties of Materials
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Thermionic emission                          89

                                                          ∗
            which should be evaluated as a function of temperature and differentiated.  ∗  See, for example. F. Seitz. Modern the-
            The result is                                                    ory of solids. McGraw-Hill, New York,
                                                                             1940, p. 146.
                                           2   2
                                          π k B
                                      c v =     T,                    (6.25)
                                          2 E F
            which agrees reasonably well with eqn (6.23) obtained by heuristic arguments.
            This electronic specific heat is vastly lower than the classical value (3/2)k B for
            any temperature at which a material can remain solid.


            6.4 The work function
            If the metal is heated, or light waves are incident upon it, then electrons may  Metal  Vacuum
            leave the metal. A more detailed experimental study would reveal that there is
                                                                                             φ
            a certain threshold energy the electrons should possess in order to be able to         Energy
                                                                               E
            escape. We call this energy (for historical reasons) the work function and denote  F
            it by φ. Thus, our model is as shown in Fig. 6.2. At absolute zero temperature
            all the states are filled up to E F , and there is an external potential barrier φ.     Distance
               It must be admitted that our new model is somewhat at variance with the old
            one. Not long ago, we calculated the energy levels of the electrons on the as-  Fig. 6.2
            sumption that the external potential barrier is infinitely large, and now I go back  Model for thermionic emission
            on my word and say that the potential barrier is finite after all. Is this permiss-  calculation. The potential barrier that
            ible? Strictly speaking, no. To be consistent, we should solve Schrödinger’s  keeps the electrons in the metal is
                                                                             above the Fermi energy level by an
            equation subject to the boundary conditions for a finite potential well. But
                                                                             energy φ.
            since the potential well is deep enough, and the number of electrons escap-
            ing is relatively small, there is no need to recalculate the energy levels. So I am
            cheating, but not excessively.


            6.5 Thermionic emission
            In this section we shall be concerned with the emission of electrons at high
            temperatures (hence the adjective thermionic). As we agreed before, the elec-
            tron needs at least E F + φ energy in order to escape from the metal, and all
            this, of course, should be available in the form of kinetic energy. Luckily, in
            the free-electron model, all energy is kinetic energy, since the potential energy
            is zero and the electrons do not interact; so the relationship between energy
            and momentum is simply

                                      1     2  2   2
                                  E =     p + p + p z  .              (6.26)
                                               y
                                           x
                                     2 m
               A further condition is that the electron, besides having the right amount of
            energy, must go in the right direction. Taking x as the coordinate perpendicular
            to the surface of the metal, the electron momentum must satisfy the inequality

                                    p 2 x  p 2 x0
                                       >     = E F + φ.               (6.27)
                                   2 m   2 m
               However, this is still not enough. An electron may not be able to scale the
            barrier even if it has the right energy in the right direction. According to the
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