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Lecture 2 Applications of the free electron gas model In this lecture we will look at some materials properties which can be described within the FEG model. 2.1 Specific heat We want to calculate the contribution of the electrons to the specific heat of a metal. The specific heat at constant volume CV is given by CV = ∂ETOT , ∂T (2.1) where ETOT is the total energy of the system. The total energy of the FEG can be obtained by adding up the energies of each occupied electronic state (times 2 for the spin): Z ∞ ETOT = 2 dE E n(E)fT (E), (2.2) ∂fT (E) . ∂T (2.3) 0 therefore the specific heat is: Z CV = 2 ∞ dE E n(E) 0 It is convenient to define x = (E − EF )/kB T . With this definition we find ∂fT (E) x ex = . ∂T T (1 + ex )2 (2.4) A plot of the r.h.s. (Fig. 2.1) shows that this function is nonzero only within |x| / 5, meaning that the integrand in Eq. (2.2) is non-vanishing only within 5 kB T from the Fermi energy (about 125 meV at room temperature, much smaller than the Fermi 16 2 Applications of the free electron gas model energy). This implies that we can take the density of states outside of the integral and set it equal to its value at the Fermi energy nF = n(EF ): Z ∞ x ex dE E CV = 2nF T (1 + ex )2 0 Z ∞ xex dx (x + EF /kB T ) = 2nF kB2 T (1 + ex )2 −E /k T Z ∞F B 2 x xe dx = 2nF kB2 T (1 + ex )2 −∞ 2π 2 = nF kB2 T. (2.5) 3 From Eqs. (1.26) and (1.37) we have the simple result nF = 3ρ/4EF , therefore we can rewrite the specific heat as follows: π2 kB T ρ kB . (2.6) 2 EF We see that the electron contribution to the specific heat is proportional to the temperature T . The lattice contribution goes instead as T 3 and is much larger, unless we perform measurements at very low temperature. We can compare the result in Eq. (2.6) CV = Figure 2.1: The function F (x) = xex (1 + ex )−2 . This function is nonnegligible only for |x| / 5. with the specific heat of a classical gas (law of Dulong and Petit): 3 CVclass = ρ kB . 2 If we take the ratio between the previous two equations we obtain: CV π 2 kB T = , 3 EF CVclass (2.7) (2.8) therefore in the FEG the specific heat is reduced from the classical value by the ratio kB T /EF . This means that only the fraction of electrons with energy within kB T from the Fermi energy can contribute to the specific heat. Electrical conductivity 2.2 17 Electrical conductivity We now discuss briefly the Drude theory of the electrical conductivity. In this case we do not make use of quantum mechanics, instead we simply treat the electrons as a classical gas. The results are only qualitative but still very useful because they allow us to introduce some important concepts. Suppose we apply a DC electric field to a metal. Our free electron gas starts moving because every electron is accelerated by the electric field. The equation of motion of each electron can be written as me me v dv = −eE − . dt τ (2.9) The first term on the rhs is the force due to the electric field. The second term is a “loss of momentum” due to collisions between the electron and the vibrations of the lattice or between the electron and the defects in the sample. This equation is telling us that an electron is accelerated uniformly by the electric field for a time τ , before it gets scattered by something and loses its momentum in the collision. The approximation given by Eq. (2.9) is not quite accurate but is the simplest possible description of electron motion in metals. In steady state the l.h.s. vanishes and we have: v=− eτ E. me (2.10) τ is called the relaxation time and it is very difficult to calculate. The electrical current density is obtained as j = −eρv, (2.11) since the number of electrons going through the surface A in the time dt is dN = ρAvdt, and jA = dN/dt. If we use Eq. (2.10) we obtain j= ρe2 τ E. me (2.12) If we now introduce the electrical conductivity σ through j = σel E and the resistivity ρel = 1/σel we obtain Drude’s law: ρe2 τ , me me = . ρe2 τ σel = (2.13) ρel (2.14) In practice the huge complexity of our initial problem is now hidden inside the relaxation time τ . The best conductivities in elemental metals (at room temperature) are found for Ag (6.2·105 ohm−1 cm−1 ), Cu (5.9·105 ohm−1 cm−1 ), Au (4.6·105 ohm−1 cm−1 ), Al (3.7·105 18 2 Applications of the free electron gas model ohm−1 cm−1 ). All other elemental metals show conductivity values below 2·105 ohm−1 cm−1 . This explains why Cu and Al are used as electrical cables. Measurements indicate that different scattering mechanisms add up following the Matthiessen’s rule: 1 1 1 = + + ··· (2.15) τ τel−ph τel−imp where “el-ph” means electron-phonon scattering, “el-imp” means scattering with the impurities of the lattice and so on. For example, in pure Cu at 4 K the relaxation time is τ ∼ 1 ns. From this time we can calculate how far an electron at the Fermi level can travel (in average) between two subsequent collisions: λ = vF τ ∼ 0.3 cm. λ is called the mean free path. At room temperature (T = 300 K) due to collisions with phonons, the mean free path in Cu in drastically reduced to λ ∼ 300 Å. Figure 2.2: Schematic plot of resistivity vs. temperature. Electron-impurity scattering gives rise to a constant contribution to the resistivity, while electronphonon scattering contributes a temperature-dependent resistivity. 2.3 Thermal conductivity The thermal conductivity κ is the proportionality coefficient between the thermal current density jth and the temperature gradient ∇T [∇ = (∂/∂x, ∂/∂y, ∂/∂z)]: jth = −κ∇T. (2.16) Using the standard kinetic theory of gases for the FEG it is possible to obtain the following expression for the thermal conductivity: κ= π 2 kB2 ρ τ T. 3me (2.17) In this expression the relaxation time τ and the electron density ρ are exactly the same that we used for the electrical conductivity in Sec. 2.2. Thermionic emission 19 Once again we have the problem that the relaxation time is difficult to evaluate within the theory. However, we can combine Eqs. (2.13) and (2.17) to obtain: κ π 2 kB2 = ∼ 2.4 · 10−8 W ohm K−2 . 2 σel T 3e (2.18) This relation (Wiedemann-Franz law) is useful because the relaxation time τ does not appear any more. In fact, since the relaxation time is the major source of uncertainty in the theory, a quantity which does not depend on τ should be reasonably described by the FEG theory. In pure metals it is found experimentally that the ratio κ/σel T is close to the theoretical value of Eq. (2.18): Cu 2.2 · 10−8 W ohm K−2 , Ag 2.3 · 10−8 W ohm K−2 , Au 2.4 · 10−8 W ohm K−2 . 2.4 Thermionic emission The FEG model can also be used to calculate the thermionic emission from a sample. The thermionic current arises from electrons which are able to escape from the sample due to thermal excitations. In this section we simplify the problem by asking what would be the total spherically-integrated thermionic current density: Z ∞ jTE = 2e v n(v)fT (v)dv, (2.19) vmin where 2 is the spin factor and v is the velocity of an electron in the eigenstate with energy E = me v 2 /2. The integration in Eq. (2.19) is performed over all the states with 2 enough energy Emin = me vmin /2 to escape from the potential well of the metal. With reference to Fig. 2.3, we define φ as the difference between the depth of the potential well V0 and the Fermi level EF : V0 = φ+EF . φ is called the work function of a metal and is the minimum energy required to extract electrons. With this notation the minimum energy required to escape from the metal is Emin = φ + EF . By combining Eqs. (1.37), (1.40), and (2.19) it is possible to arrive at the following equation (Dushman-Richardson equation): φ 2 . (2.20) jTE ∝ T exp − kB T This result indicates that the thermionic current exhibits an exponential dependence on the temperature. Equation (2.20) can be used to determine experimentally the work function. Typical work functions are 4.45 eV (Cu), 4.46 eV (Ag), 4.9 eV (Au), 4.2 eV (Al). 2.5 Hall effect This is our last application of the free electron gas model. The study of the Hall effect is useful to understand that the FEG theory has some limitations and we need to improve our description of electrons in solids. 20 2 Applications of the free electron gas model Figure 2.3: Illustration of the concept of work function. The electronic states are occupied up to EF . The work function φ is the energy difference between the vacuum level and the Fermi level. We consider the setup illustrated in Fig. 2.4. Let us call ux , uy , and uz the unit vectors along x, y, z respectively. There is an external magnetic field perpendicular to the Hall bar Bz uz , and an external longitudinal electric field Ex ux (along the x direction). The longitudinal electric field determines an electric current jx ux = −eρvx ux . Due to the Lorentz force evx Bz uy , the electrons are also pushed along the transverse direction (y axis). The steady state is reached when the charges accumulated on the edges establish a compensating electric field Ey uy . The equilibrium condition along the transverse direction can be written as: evx Bz uy − eEy uy = 0. (2.21) From this equation we obtain the electron velocity vx and rewrite the longitudinal current as: jx = −eρvx = −eρEy /Bz . (2.22) The ratio RH = Ey /jx Bz = − 1 ρe (2.23) is called Hall coefficient and can be measured. Figure 2.4: Schematic of Hall experiment. The longitudinal electric field Ex drives an electronic current jx . At the same time a perpendicular magnetic field Bz deflects the electron trajectories via the Lorentz force. The net result is that the electrons accumulate on one edge, leaving a region of net positive charge on the other edge. This charge redistribution gives rise to a transverse electric field Ey which can be measured. Hall effect 21 The interesting point in this story is that when measuring the Hall coefficients some metals such as Na exhibit a negative value [as we expect from Eq. (2.23)], but other metals, such as Al, have a positive RH value. Therefore it would appear that in some materials electrons are positively charged, contrary to what we expect. In order to resolve this difficulty we need to develop a better theory of electrons in metals. This will be our goal for the next few lectures.