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Transcript
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.