THE FREE ELECTRON GAS : A MODEL FOR METALS

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.

• Therefore, the total increase in the thermal energy, DE

• The electronic heat capacity, cel, becomes

which is directly proportional to T in agreement with experimental results.

• Based on Sommerfeld’s approach,

DE = NmT

TF

æ

èç

ö

ø÷kBT

cel =¶DE

¶T» NmkB

T

TF

cel =1

2p 2NmkB

T

TF