the free electron gas : a model for metals · 2018-04-16 · classical free electron gas...
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CLASSICAL FREE ELECTRON GAS
• CLASSICAL FREE ELECTRON GAS– Ohmic law
V=voltage, I=current, R=resistance, r=resistivity, s=conductivity, and E=electric field. L and A are, respectively the length and cross-sectional area of the metal.– The three-dimensional thermal conduction in metals
JQ is the thermal current density, K is the thermal conductivity and T is absolute temperature.
V = IR
I(= JA) =V = EL( ) / R(=rL
A)
or
J =sE where s =1
r
JQ = -KÑT
Drude’s Model (1900):On Electrical Conduction in Metals
• “Valency”, ZV, is the number of electrons in the outermost atomic shell (outside the “core”). The core to refers to the system composed of an atom's non-valence electrons and its nucleus.
• In metals, or solids in general, atoms (cores plus valence electrons) number densities are in the range 1022 - 1023 atoms/cm3.
• Atomic cores in a solid are typically separated by distances ~ 0.2 to 0.5 x 10-9 m, or equivalently, 0.2 - 0.5 nanometers.
• In a metal electrons in the outermost atomic shell (ZV) can move around freely.• In the presence of an electric field their motion is organized and an average drift
velocity vD and a net current, J, are attained as
• Nm, the free electron density, is independent of the electric field and varies between ~ 0.9x1023 cm-3 in cesium to ~ 2.5x1023 cm-3 in beryllium.
J = NmevD
• vD is proportional to the applied electric field, E.
• m is a measure of the ease with which the electron moves in a solid and is called the electron mobility. m is expressed in units of velocity per electric field.
• As a first approximation one can assume the electron to be accelerated by a force of magnitude eE over a collision free time t, which represents the average time between collisions. For an electronic mass, m, the drift velocity becomes
• The conductivity, s, and the mobility relations are
vD = mE
vD =eEt
m
J = NmevD = NmeeEt
m= sE or s =
Nme2t
m
and
m =et
mor s = Nmem
Drude’s Model (1900):On Thermal Conduction in Metals
• For a thermal conduction process along the x-direction
• Since half of the electron density Nm at x comes from the high temperature side and half of it comes from the low temperature side the thermal current density can be written as
• Assuming that the variation in temperature over the average free path, lx = vxt, (JQ)x may be expanded about x
• Generalizing to a three-dimensional heat flow and using
cel is the electronic specific heat (electronic heat capacity per unit volume).
JQ( )x
= -K x
dT
dx
æ
èç
ö
ø÷
JQ( )x
=1
2Nmvx E T x- vxt( )éë ùû-E T x+ vxt( )éë ùû{ }
JQ( )x
= Nmvx2tdE
dT-dT
dx
æ
èç
ö
ø÷
JQ =1
3v
2td NmE( )dT
-ÑT( ) =1
3v
2tcel -ÑT( )
where
cel =d NmE( )dT
vx2 = vy
2 = vz2 =
1
3v
2
• Using the Maxwell-Boltzmann statistics to describe the free electron gas in Drude’s metal one finds that the energy of the electron is
• With this value for the electron energy gives for cel
• Unfortunately for Drude his estimate of cel is two orders of magnitude higher than what is observed experimentally.
• However the major success of Drude’s theory is in explaining the observation of Wiedemann and Franz for the ratio K/s.
• From equations above it can be easily shown that
• which is the experimentally observed Wiedemann-Franz law for metals.
E=1
2mv
2
=3
2kBT
K
s=
3
2
kB
e
æ
èç
ö
ø÷T
cel =3kBNm
2
Solved Example
The Weidemann-Franz constant K/sT = 2.4x10-8 J W K-2 s-1. For a composite material made of alternating thin lamellae of metal A (r=1.0x10-6 W cm) and metal B (r = 4.0x10-5 W cm) of equal thickness, calculate the room-temperature thermal conductivities of the two separate phases, and of the composite material parallel to and perpendicular to the lamellae.
SolutionWe can use
K
sT=Kr
T= 2.4´10-8 or K =
2.4´10-8T
r
For metal AKA =
2.4´10-8( ) 300( )
1.0´10-8( )= 720 JK -1m-1s-1 Ans.
For metal BKB =
2.4´10-8( ) 300( )
4.0´10-7( )=18 JK -1m-1s-1 Ans.
For the composite we use Fig a for current flowing parallel to the lamellae. In this case
1
r//
=1
rA 1.0( )0.5´1.0( )
é
ëêê
ù
ûúú
+1
rB 1.0( )0.5´1.0( )
é
ëêê
ù
ûúú
=1
2rA+
1
2rB
or
r// =1
2 1.0´10-6( )+
1
2 4.0´10-5( )
é
ë
êê
ù
û
úú
-1
=1.95 JK -1m-1s-1
Using this last equation in the first one gets
K// =2.4´10-8( ) 300( )
1.95´10-8( )= 369 JK -1m-1s-1
In Fig. b we show current flowing perpendicular to the lamellae and we, hence, have
r^ =rA 0.5( )1.0´1.0( )
+rB 0.5( )1.0´1.0( )
=1
2rA +
1
2rB
or
r^ =1
21.0´10-6( ) +
1
24.0´10-5( ) = 2.05´10-5 Wcm
Using this last equation in the first one gets
K^ =2.4´10-8( ) 300( )
2.05´10-7( )= 35.12 JK -1m-1s-1
Sommerfeld’s Model for Metals
• Consider a metal of a free electron density Nm.
• In the ground state of this free electron gas, electrons will fill up Nm states of lowest energy.
• Thus all states are filled up to an energy EF, known as the Fermi energy, determined by the requirement that the number of states with E < EF, obtained by integrating the density of states between 0 and EF, should equal Nm ; i. e.,
or
• The occupied states are those inside the Fermi sphere in k-space.
• The surface of this sphere is called the Fermi surface and the radius is the wave-number kF. kF may be written as
• It is implicit in the above analysis that all energy states below the Fermi level are occupied, whereas above the Fermi level all energy states are empty.
• This is the case at T = 0 K which is termed the ground state.
• At T > 0 electrons can be thermally excited to higher states : i. e., some energy states above the Fermi level may be occupied and, therefore, some states below the Fermi level may be vacant.
• Therefore we need to use the Fermi-Dirac occupation probability of the energy states accessible to the electrons in the gas.
Nm = f E( )E=0
¥
ò ge E( )dE
EF, kF, and TF for selected metals
• We define the Fermi temperature TF by EF = kBTF.• Only at a temperatures ~ TF that the particles in a classical gas attain kinetic
energies as high as EF.• Only at temperatures > TF will the free electron gas behave like a classical gas.• In practice metals vaporize before the temperature TF is reached.
• As the temperature of the metal is increased not every electron gains an energy ~ kBT as expected classically, but only those electrons in states within an energy range kBTof the Fermi level are excited thermally; these electrons gain an energy which is itself of the order of kBT.
• For the electron gas with the electron density number of Nm, only a fraction of T/TF of this total number is expected to be within an energy range ~ kBT of the Fermi level at any temperature T.