Application to metals 91region. Afullcalculation ofSorofFisfairlycomplicated,andwillnotbe attempted
here. (There is nopartition functionZto help on the way). The best route is to use
S=kkkBln,together withtheknownform off(ε).
Figure 8.3illustrates theschematicbehaviour ofthethermodynamicfunctions,
U,PandCV.
8 .2 APPLICATION TO METALS
Since this is a major topic of anybook on the physics of solids, this section will be
brief. The free-electron model is surprisingly successful in describing both trans-
port andequilibrium properties ofthe conduction electrons. The surpriseisthat
the verylarge electrostatic interactions, both between pairs of electrons and also
between an electron and the lattice ions, can for many purposes be neglected, or
atanyratebe treatedas small.The outcomeisthat the conduction electrons can
be modelled as an ideal FDgas, but the densityof states is somewhat modified to
take account of these interactions.See section 14.1 for a fuller discussion of these
interactions.
Thus the heat capacity of a metal has a contribution from the conduction electrons
whichis preciselyoftheform (8.11). SinceTTTFissolarge (typically 70 000 K, as
wehave seen) this termis verysmallat room temperature comparedwithNkB.And
the lattice contribution to the heat capacity of a solid (Chapters 3 and 9) is itself
oforder 3 NkB.Therefore, as a consequence ofFD statistics, theheat capacity ofa
metalis verysimilar to that ofa non-metal– a result confirmedbyexperiment. At
low temperatures, the lattice contribution is frozen out asT^3 in a crystalline solid
(Chapter 9). Hence thelinear electronic termbecomes readily measurable at around
1 K. Its magnitudeiswellunderstoodfrom equation (8.11). A transition metalhasa
large density of states (thed-electron states) at the Fermi energy compared to a ‘good’
metal like copper. So theelectronicheat capacity ofcopperis muchsmaller than that
of,say,platinum.
The contribution to the magnetic susceptibility of the conduction electrons is also
strongly influencedbyFD statistics. In section 3.1.4 wediscussedthemagnetization
ofaspin-^12 solidobeyingBoltzmann statistics. In weak fields, the result was
M=Nμ^2 B/kkkBT (8.12) and(3.10)(In this section we revert to the use ofμfor magnetic moment; we shalluse the
alternative symbolεFfor the Fermienergy.) When we come to consider themagnetic
contribution from aligning the conduction electron spins in an applied fieldB,the
situationisdifferent. Most ofthespins are unabletoalignbecause there are no
available emptystates.
The problem is illustrated in Fig. 8.4, relevant toT=0. The only spins which
realigninthefieldare thoseintheshadedstates, numberingμB×^12 g(εF)thefactor of
1
2 arisingsince only halfthe states areinthespin-downband. Eachofthese electrons