next up previous
Next: White-dwarf stars Up: Quantum statistics Previous: The Stefan-Boltzmann law

Conduction electrons in a metal

The conduction electrons in a metal are non-localized (i.e., they are not tied to any particular atoms). In conventional metals, each atom contributes a single such electron. To a first approximation, it is possible to neglect the mutual interaction of the conduction electrons, since this interaction is largely shielded out by the stationary atoms. The conduction electrons can, therefore, be treated as an ideal gas. However, the concentration of such electrons in a metal far exceeds the concentration of particles in a conventional gas. It is, therefore, not surprising that conduction electrons cannot normally be analyzed using classical statistics: in fact, they are subject to Fermi-Dirac statistics (since electrons are fermions).

Recall, from Sect. 8.5, that the mean number of particles occupying state $s$ (energy $\epsilon_s$) is given by

\bar{n}_s = \frac{1}{{\rm e}^{\,\beta\,(\epsilon_s-\mu)} + 1},
\end{displaymath} (662)

according to the Fermi-Dirac distribution. Here,
\mu \equiv -k\,T\,\,\alpha
\end{displaymath} (663)

is termed the Fermi energy of the system. This energy is determined by the condition that
\sum_r \bar{n}_r = \sum_r\frac{1}{{\rm e}^{\,\beta\,(\epsilon_r-\mu)} + 1}
\end{displaymath} (664)

where $N$ is the total number of particles contained in the volume $V$. It is clear, from the above equation, that the Fermi energy $\mu$ is generally a function of the temperature $T$.

Let us investigate the behaviour of the Fermi function

F(\epsilon) = \frac{1}{{\rm e}^{\,\beta\,(\epsilon-\mu)} + 1}
\end{displaymath} (665)

as $\epsilon$ varies. Here, the energy is measured from its lowest possible value $\epsilon=0$. If the Fermi energy $\mu$ is such that $\beta\,\mu\ll 1$ then $\beta\,(\epsilon-\mu)\gg 1$, and $F$ reduces to the Maxwell-Boltzmann distribution. However, for the case of conduction electrons in a metal we are interested in the opposite limit, where
\beta\,\mu \equiv \frac{\mu}{k\,T} \gg 1.
\end{displaymath} (666)

In this limit, if $\epsilon\ll \mu$ then $\beta\,(\epsilon-\mu)\ll 1$, so that $F(\epsilon)=1$. On the other hand, if $\epsilon\gg \mu$ then $\beta\,(\epsilon-\mu)\gg 1$, so that $F(\epsilon)=
\exp[-\beta\,(\epsilon-\mu)]$ falls off exponentially with increasing $\epsilon$, just like a classical Boltzmann distribution. Note that $F=1/2$ when $\epsilon=\mu$. The transition region in which $F$ goes from a value close to unity to a value close to zero corresponds to an energy interval of order $k\,T$, centred on $\epsilon=\mu$. This is illustrated in Fig. 10.

Figure 10: The Fermi function.
\epsfysize =3in

In the limit as $T\rightarrow 0$, the transition region becomes infinitesimally narrow. In this case, $F=1$ for $\epsilon<\mu$ and $F=0$ for $\epsilon>\mu$, as illustrated in Fig. 10. This is an obvious result, since when $T=0$ the conduction electrons attain their lowest energy, or ground-state, configuration. Since the Pauli exclusion principle requires that there be no more than one electron per single-particle quantum state, the lowest energy configuration is obtained by piling electrons into the lowest available unoccupied states until all of the electrons are used up. Thus, the last electron added to the pile has quite a considerable energy, $\epsilon=\mu$, since all of the lower energy states are already occupied. Clearly, the exclusion principle implies that a Fermi-Dirac gas possesses a large mean energy, even at absolute zero.

Let us calculate the Fermi energy $\mu=\mu_0$ of a Fermi-Dirac gas at $T=0$. The energy of each particle is related to its momentum ${\bf p} =\hbar\,{\bf k}$ via

\epsilon = \frac{p^2}{2\,m} =\frac{\hbar^2\,k^2}{2\,m},
\end{displaymath} (667)

where ${\bf k}$ is the de Broglie wave-vector. At $T=0$ all quantum states whose energy is less than the Fermi energy $\mu_0$ are filled. The Fermi energy corresponds to a Fermi momentum $p_F=\hbar\,k_F$ which is such that
\mu_0 = \frac{p_F^{\,2}}{2\,m} = \frac{\hbar^2\,k_F^{\,2}}{2\,m}.
\end{displaymath} (668)

Thus, at $T=0$ all quantum states with $k<k_F$ are filled, and all those with $k>k_F$ are empty.

Now, we know, by analogy with Eq. (514), that there are $(2\pi)^{-3}\,V$ allowable translational states per unit volume of ${\bf k}$-space. The volume of the sphere of radius $k_F$ in ${\bf k}$-space is $(4/3)\,\pi\,k_F^{\,3}$. It follows that the Fermi sphere of radius $k_F$ contains $(4/3)\,\pi\,k_F^{\,3}\,(2\pi)^{-3}\,V$ translational states. The number of quantum states inside the sphere is twice this, because electrons possess two possible spin states for every possible translational state. Since the total number of occupied states (i.e., the total number of quantum states inside the Fermi sphere) must equal the total number of particles in the gas, it follows that

2\,\frac{V}{(2\pi)^3}\left(\frac{4}{3}\,\pi\,k_F^{\,3}\right) = N.
\end{displaymath} (669)

The above expression can be rearranged to give

k_F = \left(3\,\pi^2\,\frac{N}{V}\right)^{1/3}.
\end{displaymath} (670)

\lambda_F \equiv \frac{2\pi}{k_F} =\frac{2\pi}{(3\pi^2)^{1/3}}
\end{displaymath} (671)

which implies that the de Broglie wavelength $\lambda_F$ corresponding to the Fermi energy is of order the mean separation between particles $(V/N)^{1/3}$. All quantum states with de Broglie wavelengths $\lambda\equiv 2\pi/k > \lambda_F$ are occupied at $T=0$, whereas all those with $\lambda<\lambda_F$ are empty.

According to Eq. (668), the Fermi energy at $T=0$ takes the form

\mu_0 = \frac{\hbar^2}{2\,m}\left(3\,\pi^2\,\frac{N}{V}\right)^{2/3}.
\end{displaymath} (672)

It is easily demonstrated that $\mu_0\gg k\,T$ for conventional metals at room temperature.

The majority of the conduction electrons in a metal occupy a band of completely filled states with energies far below the Fermi energy. In many cases, such electrons have very little effect on the macroscopic properties of the metal. Consider, for example, the contribution of the conduction electrons to the specific heat of the metal. The heat capacity $C_V$ at constant volume of these electrons can be calculated from a knowledge of their mean energy $\bar{E}(T)$ as a function of $T$: i.e.,

C_V = \left(\frac{\partial \bar{E}}{\partial T}\right)_V.
\end{displaymath} (673)

If the electrons obeyed classical Maxwell-Boltzmann statistics, so that $F\propto \exp(-\beta\,\epsilon)$ for all electrons, then the equipartition theorem would give
$\displaystyle \bar{E}$ $\textstyle =$ $\displaystyle \frac{3}{2}\,N\,k\,T,$ (674)
$\displaystyle C_V$ $\textstyle =$ $\displaystyle \frac{3}{2}\,N\,k.$ (675)

However, the actual situation, in which $F$ has the form shown in Fig. 10, is very different. A small change in $T$ does not affect the mean energies of the majority of the electrons, with $\epsilon\ll \mu$, since these electrons lie in states which are completely filled, and remain so when the temperature is changed. It follows that these electrons contribute nothing whatsoever to the heat capacity. On the other hand, the relatively small number of electrons $N_{\rm eff}$ in the energy range of order $k\,T$, centred on the Fermi energy, in which $F$ is significantly different from 0 and 1, do contribute to the specific heat. In the tail end of this region $F\propto \exp(-\beta\,\epsilon)$, so the distribution reverts to a Maxwell-Boltzmann distribution. Hence, from Eq. (675), we expect each electron in this region to contribute roughly an amount $(3/2)\,k$ to the heat capacity. Hence, the heat capacity can be written
C_V\simeq \frac{3}{2}\,N_{\rm eff}\,k.
\end{displaymath} (676)

However, since only a fraction $k\,T/\mu$ of the total conduction electrons lie in the tail region of the Fermi-Dirac distribution, we expect
N_{\rm eff} \simeq \frac{k\,T}{\mu}\,N.
\end{displaymath} (677)

It follows that
C_V \simeq \frac{3}{2}\,N\,k\,\frac{k\,T}{\mu}.
\end{displaymath} (678)

Since $k\,T\ll \mu$ in conventional metals, the molar specific heat of the conduction electrons is clearly very much less than the classical value $(3/2)\,R$. This accounts for the fact that the molar specific heat capacities of metals at room temperature are about the same as those of insulators. Before the advent of quantum mechanics, the classical theory predicted incorrectly that the presence of conduction electrons should raise the heat capacities of metals by 50 percent [i.e., $(3/2)\,R$] compared to those of insulators.

Note that the specific heat (678) is not temperature independent. In fact, using the superscript $e$ to denote the electronic specific heat, the molar specific heat can be written

c_V^{(e)} = \gamma\,T,
\end{displaymath} (679)

where $\gamma$ is a (positive) constant of proportionality. At room temperature $c_V^{(e)}$ is completely masked by the much larger specific heat $c_V^{(L)}$ due to lattice vibrations. However, at very low temperatures $c_V^{(L)}=A\,T^3$, where $A$ is a (positive) constant of proportionality (see Sect. 7.12). Clearly, at low temperatures $c_V^{(L)}=A\,T^3$ approaches zero far more rapidly that the electronic specific heat, as $T$ is reduced. Hence, it should be possible to measure the electronic contribution to the molar specific heat at low temperatures.

The total molar specific heat of a metal at low temperatures takes the form

c_V = c_V^{(e)} + c_V^{(L)} = \gamma\,T + A\,T^3.
\end{displaymath} (680)

\frac{c_V}{T} = \gamma + A\,T^2.
\end{displaymath} (681)

If follows that a plot of $c_V/T$ versus $T^2$ should yield a straight line whose intercept on the vertical axis gives the coefficient $\gamma$. Figure 11 shows such a plot. The fact that a good straight line is obtained verifies that the temperature dependence of the heat capacity predicted by Eq. (680) is indeed correct.

Figure 11: The low temperature heat capacity of potassium, plotted as $C_V/T$ versus $T^2$. From C. Kittel, and H. Kroemer, Themal physics (W.H. Freeman & co., New York NY, 1980).
\epsfysize =2in

next up previous
Next: White-dwarf stars Up: Quantum statistics Previous: The Stefan-Boltzmann law
Richard Fitzpatrick 2006-02-02