Transcript Document

Lecture VII
Why Solids?
 most elements are solid at room temperature
 atoms in ~fixed position
“simple” case - crystalline solid
 Crystal Structure
Why study crystal structures?
 description of solid
 comparison with other similar materials classification
 correlation with physical properties
Metals, insulators & semiconductors?
Pure metals: resistivity
increases rapidly with
increasing temperature.
Diamond
Resistivity (Ωm)
At low temperatures all
materials are insulators
or metals.
1020-
1010Germanium
100 Copper
10-100
100
200
Temperature (K)
300
Semiconductors: resistivity decreases rapidly with increasing
temperature.
Semiconductors have resistivities intermediate between metals
and insulators at room temperature.
Metals and insulators
Measured resistivities range over more than 30 orders of magnitude
Material
Resistivity
(Ωm) (295K)
Resistivity
(Ωm) (4K)
10-5
10-12
Copper
2  10-6
10-10
SemiConductors
Ge (pure)
5  102
1012
Insulators
Diamond
1014
1014
1020
1020
Potassium
“Pure”
Metals
Polytetrafluoroethylene
(P.T.F.E)
Early ideas
• Crystals are solid - but solids are not
necessarily crystalline
• Crystals have symmetry (Kepler) and long
range order
• Spheres and small shapes can be packed to
produce regular shapes (Hooke, Hauy)
?
Kepler wondered why snowflakes have 6 corners,
never 5 or 7. By considering the packing of
polygons in 2 dimensions, demonstrate why
pentagons and heptagons shouldn’t occur.
Empty space
not allowed
CRYSTAL TYPES
Three types of solids, classified according to atomic
arrangement:
(a) crystalline and (b) amorphous materials are
illustrated by microscopic views of the atoms, whereas
(c) polycrystalline structure is illustrated by a more
macroscopic view of adjacent single-crystalline regions,
such as (a).
Crystal structure
Amorphous structure
quartz
Definitions
1. The unit cell
“The smallest repeat unit of a crystal structure, in 3D,
which shows the full symmetry of the structure”
The unit cell is a
box with:
• 3 sides - a, b, c
• 3 angles - , , 
14 possible crystal structures (Bravais lattices)
3D crystal lattice
cubic
a=b=c
==
tetragonal
a=bc
 =  =  = 90o
monoclinic
abc
 =  = 90o
  90o
orthorhombic
abc
 =  =  = 90o
trigonal (rhombohedral)
a=b=c
 =  =   90o
hexagonal
triclinic
a=bc
abc
 =  = 90o;  = 120o       90o
Chemical bonding
Types:
Ionic bonding
Covalent bonding
Metallic bonding
Van der Walls bonding
+
-
+
-
Bonding in Solids
Melting
point
(K)
Molecular
crystals
Metals
Ionic
crystals
W(3683)
C(<3500)
BN(3270)
3000
Mo(2883)
Pt(2034)
Fe(1808)
2000
1000
organic
crystals
0
Covalent
crystals
Cu(1336)
Al(933)
Pb(600)
Na(371)
Hg(234)
SiO2(2001)
LiF(1143)
KCl(1063)
Si(1683)
Ge(1240)
Electrons in metals
P. Drude: 1900 kinetic gas theory of electrons, classical
Maxwell-Boltzmann distribution
independent electrons
free electrons
scattering from ion cores (relaxation time approx.)
A. Sommerfeld: 1928
Fermi-Dirac statistics
F. Bloch’s theorem: 1928
Bloch electrons
L.D. Landau: 1957
Interacting electrons (Fermi liquid theory)
Metallic bond
Atoms in group IA-IIB let electrons to roam in
a crystal. Free electrons glue the crystal
Attract
eAttract
Repel
Repel
Na+
Na+
Attract
Attract
e-
Additional binding due to interaction of partially filled d – electron shells
takes place in transitional metals: IIIB - VIIIB
Core and Valence Electrons
Most metals are formed from atoms with partially filled atomic orbitals.
e.g. Na, and Cu which have the electronic structure
Na 1s2 2s2 2p6 3s1
Cu 1s2 2s2 2p6 3s23p63d104s1
Insulators are formed from atoms with closed
(totally filled) shells e.g. Solid inert gases
He 1s2
Ne
1s2 2s2 2p6
Or form close shells by covalent bonding
i.e. Diamond
Note orbital filling in Cu
does not follow normal rule
Simple picture. Metal have CORE electrons that are bound to the
nuclei, and VALENCE electrons that can move through the metal.
Why are metals good conductors?
Consider a metallic Sodium crystal to comprise of a lattice of Na+
ions, containing the 10 electrons which occupy the 1s, 2s and 2p
shells, while the 3s valence electrons move throughout the crystal.
The valence electrons form a very dense ‘electron gas’.
+ _ + + +
_
_ _
+ _ + +_ + _
_
_
+ + _+ +
_
_
+ _ + + +
+
+
+
_
+
Na+ ions:
Nucleus plus
10 core
electrons
+ _ +_
+
+
We might expect the negatively charged electrons to interact very
strongly with the lattice of negative ions and with each other.
In fact the valence electrons interact weakly with each other &
electrons in a perfect lattice are not scattered by the positive ions.
Free classical electrons:Assumptions
We will first consider a gas of free classical electrons subject to
external electric and magnetic fields. Expressions obtained will be
useful when considering real conductors
(i)
FREE ELECTRONS: The valance electrons are not affected
by the electron-ion interaction. That is their dynamical behaviour is
as if they are not acted on by any forces internal to the conductor.
(ii)
NON-INTERACTING ELECTRONS: The valence electrons
from a `gas' of non-interacting electrons. They behave as
INDEPENDENT ELECTRONS; they do not show any `collective'
behaviour.
(iii)
ELECTRONS ARE CLASSICAL PARTICLES:
(iv)
ELECTRONS ARE SCATTERED BY DEFECTS IN THE
LATTICE: ‘Collisions’ with defects limit the electrical conductivity.
This is considered in the relaxation time approximation.
Ohms law and electron drift
n free electrons per m3 with charge –e ( e = +1.6x10-19 Coulombs )
L
V = E/L = IR
(Volts)
Area A
Resistance R = rL/A (Ohms)
Resistivity r = AR/L
(Ohm m)
E = V/L = rI/A = rj
(Volts m-1)
Electric field E
Conductivity s = 1/r (low magnetic field) Force on electron F
Drift velocity vd
j = sE
(Amps m-2)
I = dQ/dt
Current density j = I/A
(Coulomb s-1)
dx
Force on electrons F = -eE results in a
constant electron drift velocity, vd.
Area A
Charge in volume element dQ = -enAdx
vd
j=
1 dQ
A dt
=  en
dx
dt
=  env
d
Electrical Conductivity
In the absence of collisions, the average momentum of free
electrons subject to an electric field E would be given by
 dp

 dt

= F =  eE

 Field
The rate of change of the momentum due to collisions is
 dp

 dt
At equilibrium
Now
 dp

 dt


 Collisions
= p / t p

 dp 




dt
 Field

 Collisions
=0
So
p = - etΕ
j = -nevd = -nep/me = (ne2tp /me) E
So the conductivity is s = j/E = ne2tp /me
The electron mobility, m, is defined as the drift velocity per unit
applied electric field
m = vd / E = etp /me (units m2V-1s-1)
Examples of ionic bonding
• Metal atoms with 1 electron to lose can form
ionic bonds with non-metal atoms which need
to gain 1 electron:
– Eg. sodium reacts with fluorine to form sodium
fluoride:
So the formula
for sodium
fluoride is NaF
sodium
sodiumion
atom
(Na+)
-)
fluoride
ion
(F
fluorine atom
(Na)
(F)
Examples of ionic bonding
C sC l C ry stal S tructure
• C h lo rid e io n s fo rm
sim p le cub es w ith cesiu m
io n s in th e cen ter
Examples of ionic bonding:NaCl
•Each sodium atom is surrounded by its
six nearest neighbor chlorine atoms (and
vice versa)
•Electronically – sodium has one electron
in its outer shell: [Ne]3s1 and Chlorine
has 7 (out of 8 “available” electron
positions filled in its outer shell)
[Ne]3s23p5
•Sodium “gives up” one of its electrons to
the chlorine atom to fill the shells
resulting in [Ne] [Ar] cores with Na+ and
Cl- ions
•Coulombic attraction with tightly bound
electron cores
NaCl
• Potential energy:
U = 
1
e
2
4  0 r

 - Madelung constant, m – integer number

for ro, the equilibrium position between the ions:
U0 = 

4  0

B
r
m
2
e 
1
1  
r0 
m
U0 is the cohesive energy, i.e. the energy per ion to remove the ion out of the
crystal.
Ionic bonding
Repulsive potential  1/rm
Total potential
Attractive potential  -1/r
Properties of the ionic crystals
• medium cohesive energy (2-4 eV/ atom).
– low melting and boiling temp. .
• Low electrical conductivity.
– (the lack of the free electrons).
• Transparent for VIS light
– ( energy separation between neighbouring levels > 3 eV)
• Easily dissolved in water.
– Electrical dipoles of water molecules attract the ions
Covalent bonding: molecular orbitals
Consider an electron in the ground, 1s, state of a hydrogen atom
i.e.  (r) =
1

3/2
- r/ ao
ao e
The Hamiltonian is H =
where
- 2 2
2m
a o is the Bohr Radius
-

where
r
=
e
2
4 o
The expectation value of the electron energy is
< E > =   (r) H  (r)dV
This gives <E> = E1s = -13.6eV
F(r)
E1s
V(r)
+
Hydrogen Molecular Ion
Consider the H2+ molecular ion in which
one electron experiences the potential
of two protons. The Hamiltonian is
H =
- 2 2
+ U( r ) =
2m
- 2 2
2m
-

-
e-
r

p+
R
r |r - R|
We approximate the electron wavefunctions as

 ( r ) = A [  ( r ) +  (| r - R |)]  A[  1 +  2 ]
and
  ( r ) = B [  ( r )   (| r - R |)]  B[ 1   2 ]
p+
Bonding and
anti-bonding states
1 .2
1 .0

 (r)
0 .8
0 .6
0 .4
0 .2
Expectation values of the
energy are:

E = E1s – (R) for
 (r)

 (r)
- 0 .2
- 0 .4
- 0 .6
- 0 .8
- 1 .0
V(r)
- 1 .2

E = E1s + (R) for
0 .0
 (r)
- 1 .4
-6
-4
0
2
4
6
2
4
6
r
1 .2
(R) - a positive function
-2
 (r)
1 .0
0 .8
2
0 .6
Two atoms: original 1s state
leads to two allowed electron
states in molecule.
0 .4
0 .2
0 .0
- 0 .2
- 0 .4
- 0 .6
- 0 .8
Find for N atoms in a solid
have N allowed energy states
- 1 .0
- 1 .2
- 1 .4
-6
-4
-2
0
r
covalent bonding – H2 molecule
Anti-bonding
2s
bonding
Anti-bonding
1s
bonding
•
8
energy(eV)
6
4
parallel spin
2
0
-2
antiparallel spin
-4
-6
R0
0.1
0.2
0.3
nuclear separation (nm)
system energy (H2)
0.4
Covalent bonding
Crystals: C, Si, Ge
Covalent bond is formed by two electrons, one
from each atom, localised in the region
between the atoms (spins of electrons are
anti-parallel )
Example: Carbon 1S2 2S2 2p2
C
C
3D
Diamond:
tetrahedron,
cohesive energy 7.3eV
2D
Covalent Bonding in Silicon
•Silicon [Ne]3s23p2 has four
electrons in its outermost shell
•Outer electrons are shared with
the surrounding nearest neighbor
atoms in a silicon crystalline lattice
•Sharing results from quantum
mechanical bonding – same QM
state except for paired, opposite
spins
(+/- ½ ħ)
Covalent bond
Atoms in group III, IV,V,&VI tend to form
covalent bond
Filling factor
T. :0.34
F.C.C :0.74
ionic – covalent mixed
diamond lattice
zinc blend crystals (ZnS, GaAs)
As
Properties of the covalent crystals
• Strong, localized bonding.
• High cohesive energy
(4-7 eV/atom).
– High melting and boiling temperature.
• Low conductivity.
The Hall Effect
Ex, jx
Ey
Bz
vd = vx
An electric field Ex causes a current jx to flow.
A magnetic field Bz produces a Lorentz force in the y-direction on
the electrons. Electrons accumulate on one face and positive
charge on the other producing a field Ey .
F = -e (E + v  B). In equilibrium jy = 0 so Fy = -e (Ey - vxBz) = 0
Therefore Ey = +vxBz
jx = -nevx
so
The Hall resistivity is rH
For a
general vx.
vx+ve or -ve
Ey = -jxBz/ne
= Ey/jx
= -B/ne
The Hall coefficient is RH = Ey/jxBz = -1/ne
j
E
The Hall Effect
Bz
The Hall coefficient RH = Ey/jxBz = -1/ne
j=jx
vd = vx
Ey
The Hall angle is given by tan  = Ey/Ex = rH/r
Ex
For many metals RH is quiet well described by this expression
which is useful for obtaining the electron density, in some cases.
However, the value of n obtained differs from the number of
valence electrons in most cases and in some cases the Hall
coefficient of ordinary metals, like Pb and Zn, is positive seeming
to indicate conduction by positive particles!
This is totally inexplicable within the free electron model.
The (Quantum)Free Electron model:
Assumptions
(i) FREE ELECTRONS: The valance electrons are not affected by
the electron-ion interaction. That is their dynamical behaviour is as
if they are not acted on by any forces internal to the conductor.
(ii) NON-INTERACTING ELECTRONS: The valence electron from a
`gas' of non-interacting electrons. That is they behave as
INDEPENDENT ELECTRONS that do not show any `collective'
behaviour.
(iii) ELECTRONS ARE FERMIONS: The electrons obey Fermi-Dirac
statistics.
(iv) ‘Collisions’ with imperfections in the lattice limit the electrical
conductivity. This is considered in the relaxation time
approximation.
Free electron approximation
U(r)
U(r)
Neglect periodic potential & scattering (Pauli)
Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb)
Eigenstates & energies
  2 2

d


 = i


U
 2m

dt


 k (r, t ) =  0 e

i( t k r )
2 
2

Ek =
k
2m

k = 2(n x / L x , n y / L y , nz / L z )
Ek
Unit volume in k-space:
1/(23)
|k|
Density of states
G(E) - this is the number of allowed states within a band:
g(E)dE=2g(k)dk
with g(k) equal to the density of states within the k-space. 2 is due to two possible spin
values.
2 g (k )
g (E ) =
dE / dk
g(k) is equal to the number of states within the space between two spheras of radii k
and k+dk, which is equal to the number of states per unit volume (1/(2)3) multiplied by
the volume between the spheras (4k2dk). Thus :
g ( k ) dk =
1
( 2 )
2
3
4 k dk  g ( k ) =
k
2
2
2
g (k ) =
k
2
2
2 g (k )
g (E ) =
2
=
dE / dk
For:
2
E =
 k
k
2
1

2
dE / dk
2
2m
g (E ) =
k
2

2
1
dE / dk
=
k
2

2
2m
2
2 k
=
mk
 
2
2
g(E)
E
=
m
 
2
3
2 mE
Statistics & DOS
2k T
B
1
Fermi-Dirac statistics:
T=0.1m
1
fFD (E) =
e
E m
kT
0.5 1 2
1
m=EF
Density of states:
g(E)
dk
m
g ( E ) = 2  g (k ) 
= 2 3 2mE
dE  
E
Occupation of states
D(E)
2kBT
|kF|
Fermi
Surface
EF
E
Bound States in atoms
F6
F7
F8
F9
0
Electrons in isolated 0
atoms occupy discrete
allowed energy levels -1
E0, E1, E2 etc. .
Increasing
Binding
E1
Energy
E0
-2
The potential energy of
an electron a distance r -3
from a positively charge
nucleus of charge q is -4
V (r) =
 qe
2
4 o r
V(r)
E2
-5
-8
-6
-4
-2
0
r
r
2
4
6
8
Bound and “free” states in solids
The 1D potential energy
of an electron due to an 0
0
array of nuclei of charge
q separated by a distance
-1
R is
2
 qe
V (r) = 
-2
n 4   o r  nR
Where n = 0, +/-1, +/-2 etc. -3
V(r)
E2
E1
E0
V(r)
Solid
-3
This is shown as the
black line in the figure.
-4
V(r) lower in solid (work
function).
-5
-8
-6
-4
-2
Naive picture: lowest
binding energy states can +
+
become free to move
Nuclear positions
throughout crystal
00
r2
r
+r
4
+
R
6
8
+
Energy Levels and Bands
In solids the electron states of tightly bound (high binding
energy) electrons are very similar to those of the isolated atoms.
Lower binding electron states become bands of allowed states.
We will find that only partial filled band conduct
Band of allowed energy states.
E
+
position
+
+
+
Electron level similar to
that of an isolated atom
+
Energy band
theory
2 atoms
6 atoms
Solid state
N~1023 atoms/cm3
Metal – energy band theory
Insulator -energy band theory
diamond
semiconductors
Intrinsic conductivity
ln(s)
1/T
s s = s 0se
 E g / 2 kT
Extrinsic conductivity – n – type semiconductor
ln(s)
s d = s 0d e
1/T
 E d / kT
Extrinsic conductivity – p – type semiconductor
Conductivity vs temperature
ln(s)
s s = s 0se
 E g / 2 kT
s d = s 0d e
1/T
 E d / kT