electronic structure of metals. free electron model traditional model: explains qualitatively the...

Electronic structureof

Metals

Free electron model

Traditional model:Explains qualitatively the high electric conductivity.

EK=3/2 n k T heat capacity = 3/2 n k

Failure: no large heat capacities; no large susceptibility (expected for free charge carriers)

Quantized free electron

• Particle in a box• Fermi-Dirac

statistics• Work function• Photoelectric

effect• Explains low

magnetic susceptibility

)(8

23

22

212

2

nnnma

hE

2

2

2

2

2

22

,, 8 z

z

y

y

x

xnnn L

n

L

n

L

n

m

hE

zyx 222

2

2

,, 8 zyxnnn nnnLm

hE

zyx

Degeneracy

As the number of atoms becomes very large (10^20):

We need more levels to fill the electrons in.

Degeneracy increases (no. of orbitals with the same energy)

Better to work with the so-called wave vector k

kh

p

km

h

m

pmvE

km

hk

Ekkk

E

L

n

L

n

L

nEE

L

n

L

n

L

nk

L

nk

L

nk

L

nk

kkkk

kkkk

ozyx

o

z

z

y

y

x

xok

z

z

y

y

x

x

z

zz

y

y

y

x

xx

zyx

zyx

2

822

1

8

2

2

222

2

2

22

2

222

2

222

222

2

2222

•All degenerate combinations of nx, ny, nz have the

same energy with a specific value for the k-vector.

•these combinations are at the same distance (k) from

the origin.

k

•The highest energy level is called the Fermi level: EF

•The Fermi energy is determined by the number of electrons N in the system

•What is the relation between N and kF?

•Only positive values for kx, ky, kz possible!

•Count all energy states (all volume elements) in above volume. VOLUME:

2

2

2

8 FF km

hE

kF

3

3

4

8

1Fk

•Each volume element is

•Number of volume elements (energy states):

•What is the relation between N and kF?

Vdndndn

Vd

L

nd

L

nd

L

nddkdkdkd

zyx

z

z

y

y

x

xzyx

33

..

....

26

134

81

3

23

3

Nk

V

V

k

F

F

322

2

2312 3

8

3

V

N

m

hE

V

Nk FF

Spin degeneracy

•Fermi velocity is the velocity of electrons at Fermi level

•Few estimates for Na:

•bcc; a=4.2Å

•1 valence electron per atom

•Valence electron density=

312

2

322

2

2

3

2

2

13

8

V

N

m

hv

vmV

N

m

hEF

322

338

103

102.4

2

cmV

N

cm

electron

V

N

•Fermi enenrgy = 3.5 eV

•Fermi velocity ≈108 cm/s

Density of state DOSnumber of states (N) with energy E

i.e. have energy between E and E+dE322

2

2 3

8

V

N

m

hE E

Number of states up to the energy E

21

23

2

2

2

23

23

2

2

2

8

2)(

8

3

Eh

mV

Ed

NdEDDOS

Eh

mVN

E

E

FE

EdEDN0

)(

FE

EdEfEDN0

)()(

At non-zero temperature :The Fermi-distribution is included

The problem of heat capacity•In the kinetic gas theory: Each monoatomic gas particle has the energy

•Classical theory predicts this value for one mole of free electrons (1 mole of group AI metal)

•Value measured is only 0.01 of expected

•Solution: Pauli exclusion principle, Fermi distribution

TkE Bk 2

3

RNcc

kdT

dEc

avVmV

BV

2

3

2

3

,

•When we heat a metal at zero Kelvin, not every electron can gain the energy kBT .

•Only those electrons within the range EF,EF-kBT can become excited.

•The fraction of particles that become excited

•Number of excited electrons

•The total electronic thermal energy is Gives the right heat capacity values

F

B

E

Tk

NE

Tk

F

B

TkNE

TkEU B

F

Bk

• Fraction of electrons above EF in Ag at 300 K?

At 300 K, thermal energy kT = 4 x 10-21 J = 0.025 eVOnly electrons with this energy below EF will be

affected Fraction is roughly kT/EF = 0.46% (Exact value is

9kT/16EF)

Only small fraction contributes to conduction

Fermi-Dirac Statistics

distribution of carriers on the available energy levels satisfies three conditions:

• The Pauli exclusion principle:

• The equilibrium condition: Minimum free Enthalpy G

• The conservation of particles (or charge) condition; i.e.constant number of carriers regardless of the distribution

TF=EF / k

T at which a gas of classical particles would have to be heated to have the average energy per particle equal

to the fermi energy at 0 K.

MetalElectron concentr. X10-28 m-3

EF

eV

TF

K

Na2.653.2337500

K1.402.1224600

Cu8.497.0581200

Ag5.855.4863600

Au5.95.5364100

Still not satisfactory:• Can not answer why are there metals, semiconductors and

non metals? why is the mean free path for some electron

states very large?

• Oversimplification Potential in crystal not constant but periodic

Solution of Schroedinger equation yields:

k: wave vector;│k│= 2(Describes electron wave)

When Bragg’s condition is met (at /a for 1-dim. crystal, certain energy value are not obtainable (gap)

MO-Approach

all states between ± 2 (J ± 2K)

Na

Why is Be metallic?

Insulators and Semiconductors

Positive and negative charge carriers

n-type semiconductor

p-type semiconductor

Pure diamond:colorlessB added Diamond: blueN added Diamond: yellow

Inorganic solids• GaP, GaAs: semiconductors, Isoelectronic with Si• KCl: insulator Isoelectronic with Si• Band model applicable???

NaCl

Na+:1s2 2s2 2p6

empty 3s, 3p band

Cl-: 1s2 2s2 2p6 3s2 3p6

full 3p band

E

C-band

V-band

gap

Transition corresponds to e-transfer from Cl- to Na+

Correlation between electronegativity difference and gap energies

222 CEE hg

Eh homopolar band gapEg actual gapC charge transfer term

2

2

,g

i E

CfIonicity

C related to

Color of some band-gap semiconductors

SubstanceMineral

namePigment

name

Bandgap(eV)Color

CDiamond-5.4Colorless

ZnOZinciteZinc white3.0Colorless

CdSGreenockiteCadmium yellow2.6Yellow

CdS1-xSex-Cadmium orange2.3Orange

HgSCinnabarVermillion2.0Red

HgSMetacinnabar-1.6Black

Si-

1.1  Black

PbSGalena-0.4Black

Yellow cadmium sulfide CdS, (E, = 2.6 eV), and black cadmium selenide CdSe (Eg = 1 .6 eV), which have the same structure and form a solid-solution series. Plate IX illustrates the yellow-orange-red-black sequence of these mixed crystals as the band-gap energy decreases. Mixed crystals such as Cd4SSe3 form the painter's pigment cadmium orange and are also used to color glass and plastic.

Transition metals

pale green (d-d transition)= 10-14-1cm-1

t2g banddxy, dxz, dyz overlap = 103 -1cm-1

Guidelines• Good d-overlap occurs if:

1) Formal charge on cation is smallTiO: metallic, TiO2: insulator; Cu2O, MoO2:semi; CuO, MoO3 insulator

2) Cation from early transition metalsTiO: metallic, NiO, CuO: poor semiconductors Li-spinels

3) Cation in second or third transition seriesCr2O3 poor conductor, Mo and W oxides: good conductors

4) Anion reasonably electropositiveNiO: poor conductor, NiS, NiSe, NiTe: good conductors

Effects 1, 2 and 3: keep d-orbitals spread out Effect 4: reduction of ionicity and band gap

Effect of crystal structure

• Fe3O4: inverse spinel [Fe3+]tetr[Fe2+,Fe3+]octO4

almost metallic conductivity

Oct. sites close to each other (edge sharing) charge migration (Fe2+ Fe3+)

• Mn3O4: normal spinel [Mn2+]tetr[Mn3+2]octO4

tetr. and oct. sites far from each other,

no charge transfer

• Li-spinels: LiMn2O4 [Li+]tetr[Mn3+,Mn4+]octO4 semiconductor

LiV2O4 [Li+]tetr[V3+,V4+]octO4 metallic

Molecular crystals

VdW Forces