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