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## Q. 9.8

Use the Fermi theory to compute the electronic contribution to the molar heat capacity of (a) copper and (b) silver, each at temperature T = 293 K. Express the results as a function of the molar gas constant R.

Strategy From the Fermi theory the molar heat capacity is $c_{ V }=2 \alpha R T / T_{ F }$. Recall that we can use $\alpha=\pi^{2} / 4$ at room temperature (T = 293 K), so

$c_{ V }=2\left(\frac{\pi^{2}}{4}\right) R \frac{T}{T_{ F }}=\frac{\pi^{2} R T}{2 T_{ F }}$

## Verified Solution

(a) Inserting numerical values (from Table 9.4, $T_{ F }=8.16 \times 10^{4}$ K for copper),

$c_{ V }=\frac{\pi^{2}(293 K )}{2\left(8.16 \times 10^{4} K \right)} R=0.0177 R$

for copper, which rounds to the value 0.02R quoted in the text.

(b) For silver, we simply replace the Fermi temperature with $6.38 \times 10^{4}$ K (Table 9.4):

$c_{ V }=\frac{\pi^{2}(293 K )}{2\left(6.38 \times 10^{4} K \right)} R=0.0227 R$

We see that the electronic contribution to the molar heat capacity is the same order of magnitude for these two good conductors, and in each case it is small compared with the lattice contribution (Section 9.3).

 Table 9.4 Fermi Energies (T = 300 K), Fermi Temperatures, and Fermi Velocities for Selected Metals Element $E_{ F }( eV )$ $T_{ F }\left(\times 10^{4} K \right)$ $u_{ F }\left(\times 10^{6} m / s \right)$ Element $E_{ F }( eV )$ $T_{ F }\left(\times 10^{4} K \right)$ $u_{ F }\left(\times 10^{6} m / s \right)$ Li 4.74 5.51 1.29 Fe 11.1 13 1.98 Na 3.24 3.77 1.07 Mn 10.9 12.7 1.96 K 2.12 2.46 0.86 Zn 9.47 11 1.83 Rb 1.85 2.15 0.81 Cd 7.17 8.68 1.62 Cs 1.59 1.84 0.75 Hg 7.13 8.29 1.58 Cu 7 8.16 1.57 Al 11.7 13.6 2.03 Ag 5.49 6.38 1.39 Ga 10.4 12.1 1.92 Au 5.53 6.42 1.4 In 8.63 10 1.74 Be 14.3 16.6 2.25 Tl 8.15 9.46 1.69 Mg 7.08 8.23 1.58 Sn 10.2 11.8 1.9 Ca 4.69 5.44 1.28 Pb 9.47 11 1.83 Sr 3.93 4.57 1.18 Bi 9.9 11.5 1.87 Ba 3.64 4.23 1.13 Sb 10.9 12.7 1.96 Nb 5.32 6.18 1.37 From N. W. Ashcroft and N. D. Mermin, Solid State Physics, Philadelphia: Saunders College (1976).