free electron in_metal

UEEP2024 Solid State Physics

Topic 3 Free Electron in Metals

Drude’s classical theory of Electrical Conduction

• Drude assumed that a metal is composed of ions, which are stationary, and valence electrons, which are free to move.

• If no voltage applied to the metal then at each collision the electron is deflected in a different direction so that the overall motion is quite random.

• The valence electrons appears to be similar to the molecules in an idea gas.

• The velocity of the electrons with mass m at temperature T is given by the equation

where KB is the Boltzmann’s constant.

TKmv B23

21 2

Drude’s classical theory of Electrical Conduction

• The mean free path is the average distance that an electron travels between collisions.

• The relaxation time is the average time duration between collisions.

• The mean speed of the electron • The speed of electron at room temperature is

about 105 ms-1.• The mean free path is about 1 nm and the

relaxation time about 10-14 s.

.

v

Drude’s classical theory of Electrical Conduction

• When we apply an electric field to a sample, the electrons are attracted towards the positive end of the sample, a net flow of electrons in this direction.

• If the electric field is E, then the force on each electron is eE.

• Acceleration on each electron is .• Change in velocity • This quantity is called the drift velocity.

emeEa

.em

eEav

Drude’s classical theory of Electrical Conduction

• Electron mobility µ =e/m.

• The current density

where n is the number of valence electrons per unit volume,• Conductivity of the metal is given by • Drube’s model is a) consistent with ohm’s law. b) Explain the phenomenon of electrical resistance. c) Gives good values of conductivity.

.eEnAIJ

.en

Failures of Classical model• Conductivity should be directly proportional to the valence electron

concentration, not in good agreement with experimental data.• Electrical properties of alloys should be intermediate between the values of

corresponding pure materials. However many alloys have resistivity which are considerably larger than those of either of the pure constituents.

• The dependence of resistivity on temperature: Drube’s model predicted resistivity should be proportional to T1/2,

experimental measurement show that resistivity is actually proportional to T over a wide range of temperature.

• Drude ‘s model predict the molar specific heat capacity for monovalent metal of 9R/2, 6R for divalent metal and 15R/2 for trivalent metal. However, experiment results show that the molar specific heat capacity at room temperature is approximately 3R, regardless of the valency of the metal.

Example

Estimate the typical conductivity of a metal at 295 K assuming that the mean free path is about 1 nm and the number of valency electrons is about 1029 m-3.

Solution

• Thermal velocity

• Relaxation time

• Conductivity

1531

23

1016.11011.9

)295)(1038.1(33

ms

mTkv B

sv

155

9

1062.81016.1

101

117

31

15219292

1042.21011.9

1062.8106.110

mm

ne

Free Electron Fermi Gas• Atoms bounded by “free electrons”. Good

example is alkali metals (Li, K, Na, etc.)

Electron in a metal can move freely in a straight path over many atomic distances, undeflected by collisions with other conduction electron or by collisions with the atom cores

Free Electron Fermi Gas1. A conduction electron is not deflected by ion cores

arranged on a periodic lattice

2. A conduction electron is scattered only infrequently by other conduction electrons

Consequence of the Pauli exclusion principle

Free electron Fermi gas – A gas of free electrons subject to the Pauli principle

Free Electron Fermi Gas

),(ˆ),( trHtrt

i

For a general quantum system

(1)

For a single particle in three dimensions

),()(),(2

),( 22

trrVtrm

trt

i

(2)

Wave functionHamiltonian operator

Free Electron Fermi Gas(1-D)

nnn E

dxd

m

2

22

2

Consider a free electron gas in 1-D (electron of mass m is confined to a length L by infinite barriers), (2) can be expressed as

With boundary conditions

0)(0)0( Lnn

(3)Energy of the electron

Free Electron Fermi Gas(1-D)

LnwherexA nn

n

212sin

Lnkwhere

mkE nn

n

2

22

The solution of (3)

Energy En is given bywavevector

0 /L 2/L

K-space

k /L

Free Electron Fermi Gas3-D

)()(2

22

rErm kkk

Consider a free electron gas in 3-D (electron of mass m is confined to a cube of edge L), (2) can be expressed as

With boundary conditions

0)(0)0(0)(0)0(0)(0)0(

LzzLyyLxx

kk

kk

kk

(4)

Free Electron Fermi Gas(3-D)

LnLnLn zzyyxx 21,

21,

21

zyxA

zyxk

2sin2sin2sin

The solution of (4)

where

(5)

Free Electron Fermi Gas(3-D, periodic boundary conditions)

• 3-D system• periodic boundary

conditions

,.....8,6,4,2,0,,LLLL

kkk zyx

),,(),,(),,(),,(),,(),,(

zyxLzyxzyxzLyxzyxzyLx

kk

kk

kk

The solution of (4)

rikArk exp)( the components of k satisfy

One allowed value of k per volume (2/L)3

Free Electron Fermi Gas(3-D, periodic boundary conditions)

2222

22

22 zyxk kkkm

km

E

Energy Ek is given by

K space

(6)

Free Electron Fermi Gas(density of state)

2/3

22

3

3

3 23/2

3/42

mEL

LkN

total number of allowed energy state

Density of states – number of allowed energy state per unit energy range

2/12/3

22

3 22

)( EmLdEdNED

(7)

(8)

Free Electron Fermi Gas(Fermi energy)

3/2

3

22 32

LN

mEF

The Fermi energy is defined as the energy of the topmost filled level in the ground state of the N electron system.

The state of the N electron system at absolute zero

From (7), we get

For N electron system

Free Electron Fermi Gas(Fermi-Dirac distribution)

• The Fermi-Dirac distribution gives the probability that an energy state with energy E is occupied by an electron

1exp

1)(

kTE

Ef

The electron density is given by

dEEfEDn )()(

Fermi level – the energy at which the probability of occupation is 1/2

Example

Show that the probability for an electron state at the Fermi energy is equal to 0.5 for all finite temperature.

Solution

• When E = EF,

.21

111

11

11

11)(

0

/)(/)(

e

eeEf TKEETKEE BFFBF

Example

Using the Fermi-Dirac distribution, determine the values of energy corresponding to f = 0.9 and f = 0.1 at a temperature of 300 K.

Solution

TKEETKEE

TKEEe

e

eEf

BF

BF

BF

TKEE

TKEE

TKEE

BF

BF

BF

20.220.2

)111.0ln(/)(111.01111.1

9.011

9.01

1)(

/)(

/)(

/)(

Fermi-Dirac Distribution

1.0

0.0

EF

Probability of occupationf(E)

Energy

Solution

TKEETKEE

TKEEe

e

eEf

BF

BF

BF

TKEE

TKEE

TKEE

BF

BF

BF

20.220.2

)9ln(/)(9110

1.011

1.01

1)(

/)(

/)(

/)(

Ohm’s Law

BvqF

qFmvmomentum

In an electric field and magnetic field B, the force on an electron of charge –q is

If B = 0,

mqv

thus

collision time

Ohm’s Law

mnqqnvJ

2

mnq

2

The electric current density is

If we define the electrical conductivity as

J

then

Ohm’s Law

Hall Effect

JBRBqnJ

qvBqF

H

0B

Hall coefficient

JBRBqnJ

qvBqF

H

Quantum StatisticsA single quantum mechanical system consists of N particles constrained to some volume V

assumption

1. weakly interacting particles

2. the particle density is low enough so that the energy of the system can be considered as the sum of the individual particle energies

1

ˆ

iii

sii

si

EnE

EHwavefunction of particle s

number of particles with energy Ei

Quantum Statistics• How to determine the most probable distribution of

finding n1 particles out of a total of N with energy E1, n2 with energy E2, and so on?

The most probable distribution (n1, n2 …,ns,….) is the one associated with the largest number of microscopically distinguishable arrangements

Thus, the plan of attack would be to derive an expression for P, the total number of microscopically distinct arrangement corresponding to a given arbitraty sequence (n1, n2 …,ns,….) and then find the particular sequence that maximizes P.

Three types of quantum particles

1. Identical but distinguishable particles• harmonic oscillators

2. Identical indistinguishable particles of half-odd integral spin (Fermions)• electrons• protons

3. Identical indistinguishable particles of integral spin (Bosons)• photons• phonons

The particles of concern to us fall into one of three categories

obey Pauli exclusion principle

Pauli exclusion principle not apply

N particle system

,.......,......,, 21 sEEE

Our model system is taken to contain N particles, each with a spectrum of allowed energy levels

We divide the single-particle energy spectrum into energy “bins”. Bin s, as an example, represents all the elementary quantum states whose energies lie within some arbitrarily chosen interval Es centered about Es. The number of quantum states in bin s is denoted by gs.

Identical but distinguishable particles

1

11

111

11 !!

!nN

nn

CgornNn

NgP

Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.

For bin 1, the total number of distinguishable choices is

For bin 2, the total number of distinguishable choices is

21

22

2212

122 !!

!nnN

nn

CgornnNn

nNgP

Identical but distinguishable particles

the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is

1

2121

!!

........,....,....,,

s s

ns

ss

ngN

PPPnnnPs

Identical indistinguishable particles of half-odd integral spin (Fermions)

11!!!

111

11 ng Cor

ngngP

Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.

For bin 1, the total number of distinguishable choices is

For bin 2, the total number of distinguishable choices is

22!!!

222

22 ng Cor

ngngP

Identical indistinguishable particles of half-odd integral spin (Fermions)

the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is

1

2121

!!!

........,....,....,,

s sss

s

ss

nngg

PPPnnnP

Identical indistinguishable particles of integral spin (Bosons)

!1!

!1

11

111

gngnP

Determine P of N particles such that bin 1 contains n1 particles, bin 2 n2 particles, and so on.

For bin 1, the total number of distinguishable choices is

For bin 2, the total number of distinguishable choices is

!1!

!1

22

222

gngnP

Identical indistinguishable particles of integral spin (Bosons)

the total number of distinguishable choices in which bin 1 contains n1 particles, bin 2 n2 particles, and so on, is

1

2121

!!1!1

........,....,....,,

s ss

ss

ss

nggn

PPPnnnP

Lagrange method

constEnE

constNn

sss

ss

1

1

we wish to find the set of ns for which P is maximized subjected to conditions :

We constrain our solution using Lagrange multipliers forming the function:

1121 )ln(,.......,,

sss

sss nEEnNPnnnF

Lagrange method

sallfornF

s

0

0lnln sss Eng

For Identical but distinguishable particles

ss

s Egn

expMaxwell-Boltzmann

Lagrange method

0lnln ssss Enng

For Identical indistinguishable particles of half-odd integral spin (Fermions)

1exp

s

ss E

gn

Fermi-Dirac

Lagrange method

0lnln ssss Enng

For Identical indistinguishable particles of half-odd integral spin (Fermions)

1exp

s

ss E

gn

Bose-Einstein

Energy Band

Electron in a metal can move freely in a straight path over many atomic distances, undeflected by collisions with other conduction electron or by collisions with the atom cores

Energy Band

),(ˆ),( trHtrt

i

For a general quantum system

(1)

For a single particle in three dimensions

),()(),(2

),( 22

trrVtrm

trt

i

(2)

Wave functionHamiltonian operator

Energy Band

nnn E

dxd

m

2

22

2

free electron gas

(3)Energy of the electron

V(r)=0

Free Electron Model

Energy Band0)( rVif

Considering a periodic potential, V(r+T)=V(r)

)()()(2

)( 22

rrVrm

rE

(2) can be expressed as (4)

Energy Band

G

riGGeUrV )(

G

iGxGeUxV )(

The periodic potential V(r) may be expanded as a Fourier series in the reciprocal lattice vectors G

or

(5)

For 1-D system

Thus, (4) can be rewritten as

)()(2

)( 2

22

xeUdx

xdm

xEG

iGxG

(6)

Energy Band

G k

xGkikG

k

ikxk

eCUxxV

eCkmdx

xdm

)(

22

2

22

)()(

2)(

2

k

ikxkeCx)(

The wavefunction Ψ(x) may be expressed as a Fourier series summed over all values of the wavevector permitted by the boundary conditions

then

(7)

Energy Band

k

ikxk

G k

xGkikG

ikxk

k

eCEeCUeCkm

)(22

2

From (6), we get

mkwhere

CUCE

k

GGkGkk

2

0

22

Each Fourier component must have the same coefficient on both sides of the equation

(8)

central equation

Energy Band

G

xGkiGk

ikxkk eCeCx )()(

Once we determine The Ck from central equation, the Ψ(x) is given as

(9)

Rearrange (9), we get

)(

)()(

xueCwhere

exueeCx

kG

iGxGk

ikxk

ikx

G

iGxGkk

(10)

Energy Band (Bloch Theorem)

)()()exp()()(

Trururikrur

kk

kk

The solutions of the Schrodinger equation for a periodic potential must be of a special form

The eigenfunctions of the wave equation for a periodic potential are the product of a plane wave exp(ik•r) times a function uk(r) with the periodicity of the crystal lattice

(11)Translation vector

Bloch theorem

Energy Band (Kronig-Penney Model)

00

)(xbDeCe

axBeAex

QxQx

iKxiKx

a0-b

Uo

If V(x) is

The solutions for (2) are

wheremKEandmQEUo 2

222

22

Energy Band (Kronig-Penney Model)

)()(00

00

DCQBAikdxd

dxd

DCBA

xx

xx

The constants A,B,C,D are chosen so that Ψ and dΨ/dx are continuous at x=0 and x=a

)()0()( baikexbbaxa Bloch theorem

(12)

(13)

Energy Band (Kronig-Penney Model)

)(

)(

)(

)(

baikQbQbiKaiKa

baik

bxax

baikQbQbiKaiKa

baikbxax

eDeCeQBeAeiK

edxd

dxd

eDeCeBeAe

e

(14)

(15)

Energy Band (Kronig-Penney Model)

)(coscoscoshsinsinh2

22

bakKaQbKaQbQKKQ

oUband limlim

0

From (12)-(15), we can get

If the potential is a periodic delta function

kaKaKaKaP coscossin)/(

Then, (16) can reduce to

(16)

(17)P=Q2ba/2

Energy Band (Kronig-Penney Model)

-6 -4 -2 0 2 4 6

-2

-1

0

1

2

3

4

5

6

P=3/2

(P/K

a)si

n(K

a)+c

os(K

a)

Ka ()

Not allowed

Energy Band (Kronig-Penney Model)

Energy gap

(Ka that is not allowed)

Band Structure

Actual band structures are usually exhibited as plots of energy versus wavevector in the first Brillouin zone. This is helpful in visualization and economical of graph paper.

2/a

Band Structure

/a 2/a 3/ak

E

If the band structure of A is :

/ak

E

Band Structure

k

E

Eg

k

E

Eg

k

E

Eg

Direct Indirect Overlap

(negative bandgap)

Bandgap – the difference in energy between the lowest point of the conduction band the highest point of the valence band.

valence band

conduction band

Band Structure

band structure of Si

Energy Band(conductor and insulator)

A band is filled from low energy to high energy

Energy Band (conductor and insulator)

A full band cannot conduct electricity. A full band is always a full band no matter what the external field are

An empty band cannot conduct electricity because it does not have charge carrier.

Only a partially filled band can conduct electricity. The occupied states are not “balance” under an external field. This unbalance causes the current flow.

Energy Band(Equation of motion and effective mass)

2

2

2

11dkEd

me

The equation of motion of an electron in an energy band is :

dtdkEdm

dkd

dtdm

dtdvm

dtdvm

dtdkF

eee

e

2

From (1), we can get

(1)

effective mass

)( BvEqdtdkF

Energy Band(Equation of motion and effective mass)

)(2

22

0 AmkEE

Approximate solution near a zone boundary

(2)

From (1), we get

Ammthus

mA

dkEd

e 2

22

2

2

11 (3)

An electron of mass m when put into a crystal respond to applied fields as if the mass were me (effective mass)

Electron rest mass

Example

By determine the number of grid points contained beneath the surface of radius nmax, show that the Fermi energy (i.e. the energy of the highest occupied state at T = 0K ) is given by

.32

3/222

VN

mEF

Each allowed quantum state can be represented by a grid point (nx, ny, nz).Grid points with the same energy are connected by a

surface of constant radius. In order to determine the Fermi energy, find the values of the radius nmax, just contains sufficient states to accommodate all the valence electrons in the crystal

ny

nx

nz

nmax

Solution

• The volume contained beneath the surface corresponds to one-eight of a sphere of radius nmax, the volume is

• Each grid point corresponding to a cube of unit volume, this expression also gives the number of grid points beneath the surface.

• As each grid point corresponding to a state which can accommodate two electrons for a crystal containing N electrons we require N/2 states.

•

634

81 3

max3max

nn

3/1

max

3max 362

NnnN

Solution

• By writing

• Energy

2/1222max zyx nnnn

3/2223/2

3

32

3/2222

max

22

2222222

32

32

322

22

VN

mh

LN

mh

NLm

hnLm

h

nnnLm

hmkhE zyx

Thermal Conductivity in Metals

• Thermal conductivity of a Fermi gas is

where n is the electron concentration k is the Boltzmann constant T is the temperature m is them mass of electron and is the collision time.

mTnklv

mvTnkK FF

el 33

22

2

22

Thermal Conductivity in Metals

• Do the electrons or phonons carry the greater part of the heat current in a metal?

• In pure metals the electronic contribution is dominant at all temperatures.

• In impure metals or in disordered alloys, the electron mean free path is reduced by collisions with impurities, and phonon contribution may be comparable with the electronic contribution.

Wiedemann-Franz law

• The Wisdemann-Franz law states that for metals at not too low temperatures the ratio of the thermal conductivity to the electrical conductivity is directly proportional to temperature, with the value of the constant of proportionality independent of the particular metal.

Tek

mne

mTnkK 22

2

22

33

Lorenz number L

• The Lorenz number L is defined as

• The value of

• This remarkable result involves neither n nor m.• Experimental values of L at 0oC and 100oC are in

good agreement.

TKL

.ohm/deg- watt1045.23

2822

ekL

Experimental Lorenz numbersL ×108 watt-ohm/deg2

Metal 0oC 100oC

Ag 2.31 2.37

Au 2.35 2.40

Cd 2.42 2.43

Cu 2.23 2.33

Mo 2.61 2.79

Pb 2.47 2.56

Pt 2.51 2.60

Su 2.52 2.49

W 3.04 3.20

Zn 2.31 2.33