1
Chapter Three Crystal binding
Why do atoms form crystals or solids?
Answer :
Interatomic forces
that bind atoms.
Atoms bind due to the Coulomb attractive forces
between
electrons and neighboring atomic ions
.
Contents:
☻
Types and strengths of binding forces
☻
Reason for crystal structure formation
☻
Mechanical properties of crystals
Cohesive energy U
≡
the energy that must be added to the crystal to separate its
components into neutral free atoms at rest
≡
Energy of free atoms
–
Crystal energy
Hence, U>0 to form a stable solid
• Magnitude ~ 1-10eV except for the inert gas crystals (0.02-0.2eV) • U ≤ Eion (Ionization energy = Binding energy of valence electrons) • U controls the melting temperature and bulk modulus
3
Types of bonds
(a) Van der Waals
(Molecular)
(b) Covalent
(c) Metallic
(d) Ionic
Electrons localized among atoms
Electrons shared by the neighboring atoms
Electrons free to move through sample
Electrons transferred to adjacent atoms
(a)
+
-+
-+
-+
-+
-+
-+
-(b)
+
+
+
+
+
(c)
+
+
+
+
+
(d)
+
+
+
+
-
-(a) Molecular bonding
Inert gas crystals : He, Ne, Ar, Kr, Xe, Rn
Transparent Insulators – completely filled outer electron shellsWeakly bonding – van der Waals bonding FCC structures except for He3 and He4
high ionization energies low melting temperatures
12.13 14.00
15.76 21.56
Ionization energy (eV)
161.4 115.8 83.81 24.56 Melting temperature (K) 0.16 0.12 0.08 0.02
Cohesive energy (eV/atom)
Xenon
Krypton
Argon
Neon
5 Phase diagrams of (a) 4He and (b) 3He.
6
Van der Waals –London Interaction
Consider two identical inert gas atoms
R
Neutral: positive nucleus + spherically symmetric distribution of electron charge
No interaction between atoms → No cohesion (NO solid)
?
Attractive interaction
Fluctuating dipole-dipole interaction
between the atoms
Reviews:
r
)
r
n(
r
d
P
r
3r
r
∫
=
r n(r) d -q +q←
=
qd
P
r
P r as r>>dElectric fields Attractive force Repulsive force
4 2 1
r
P
P
~
F
3 2 1r
P
P
~
U
3r
P
~
E
7
Inert gas solids
• On average spherically symmetric distribution of electron charge with the positive nucleus in the center
0
P
r
=
• But thermal fluctuations (finite T) cause instantaneous electric dipole moment
0
P
0
(t)
P
r
≠
→
2≠
-+
-+
P
v
≠
0
0
P
v
=
fluctuations
• On adjacent atoms if the dipoles are random there could be no net force (time average) • But dipole induces a dipole in neighboring atoms that always gives an
Model for inert gas solid –
two identical linear harmonic oscillators
R x2 -e +e x1 -e +ep
1and p
2are the momenta of these two oscillators
C is the force constant
Hamiltonian for the unperturbed system – no Coulomb interaction
2 2 2 2 2 1 2 1 o
Cx
2
1
2m
p
Cx
2
1
2m
p
H
=
+
+
+
Hamiltonian for Coulomb interaction energy of the system
3 2 1 2 R x , x
R
x
x
2e
2 1
−
→
<< 2 2 1 2 2 1 2 2 1x
R
e
x
R
e
x
x
R
e
R
e
H
−
−
+
−
−
+
+
=
9 Normal mode transformation -- symmetric (s) and anti-symmetric (a)
2
p
p
p
;
2
p
p
p
2
x
x
x
;
2
x
x
x
2 1 a 2 1 s 2 1 a 2 1 s−
≡
+
≡
−
≡
+
≡
Total Hamiltonian after the transformation
+
+
+
−
+
=
2 a 3 2 2 a 2 s 3 2 2 sx
R
2e
C
2
1
2m
p
x
R
2e
C
2
1
2m
p
H
2 2 2 2 2 1 2 1 oCx
2
1
2m
p
Cx
2
1
2m
p
H
=
+
+
+
m
C
ω
,
H
o o=
m
R
/
2e
C
m
C
ω
m
R
/
2e
C
m
C
ω
,
H
3 2 a a 3 2 s s+
=
=
−
=
=
Two frequencies of the coupled oscillators symmetric (s) and anti-symmetric (a)
+
×
−
+
+
=
+
=
+
=
+
×
−
+
−
=
−
=
=
K
K
2 3 2 3 2 o 2 / 1 3 2 3 2 a 2 3 2 3 2 o 2 / 1 3 2 3 2 sCR
2e
2
1
)
2
1
(
2
1
CR
2e
2
1
1
ω
CR
2e
1
m
C
m
R
/
2e
C
ω
CR
2e
2
1
)
2
1
(
2
1
CR
2e
2
1
1
ω
CR
2e
1
m
C
m
R
/
2e
-C
ω
Therefore, the zero point energy of the coupled oscillators is
lowered
from the uncoupled oscillators by
o o
2
1
2
1
ω
ω
h
h
+
a s2
1
2
1
ω
ω
h
h
+
The zero point energy
¾
The uncoupled oscillators¾
The coupled oscillators6 2 3 2 o
R
A
2
CR
2e
8
1
ω
2
1
U
=
−
×
−
=
∆
h
Attractive
interaction
11
The van der Waals interaction, the London interaction,
the induced dipole-dipole interaction
lity
polarizabi
electronic
the
is
re
whe
ω
2C
e
ω
A
2 o 2 4 oα
α
h
h
≡
=
0
R
A
U
=
−
6<
∆
P2 inducedP1 fluctuation
“Polarizability” of the atom
P
=
α
E
r
r
3 1 1 2R
P
E
P
r
=
α
r
=
α
6 2 1 3 2 1R
P
~
R
P
P
~
U
−
−
α
R< 0
What limits attraction ? --
Repulsive force (Pauli exclusion principle)
Two electrons cannot have all their quantum number s the same.
A B A B
Charge distributions overlap • When charge distributions of two atoms overlap, there is a tendency for electrons
from atom B to occupy in part states of atom A occupied by electrons of atom A, and vice versus.
• Pauli exclusion principle prevents multiple occupancy, and electron distribution of atoms with closed shells can overlap only if accompanies by the partial
promotion of electrons to unoccupied high energy state of the atom.
The electron overlap increases the total energy of the system and
gives a repulsive contribution to the interaction.
13 0 R B U= 12 > ∆
Empirical formula for such repulsive potential
R
R
ε
4
R
A
R
B
6 12 6 12
−
=
−
=
σ
σ
The total potential for inert gas system
the Lennard-Jones potential
where empirical parameters A=4
εσ
6 andB=4
εσ
12are determined
from independent measurements made in the gas phase.
Values of ε[energy] and σ[length] are shown in Table 4.
0.85 0.90 0.95 1.00 1.05 1.10 1.15 1.20 -2 0 2 4 6 8 UPauli UvdW U U( R) /4 ε R/σ
U(r)
=
U
Pauli+
U
vdW14 N atoms in the crystal
( )
∑
−
=
≠j i 6 nn ij 12 nn ij totalR
p
R
p
4
ε
N
2
1
U
σ
σ
where R
nnis nearest neighbor distance and
p
ijR
nnis the distance between atom i and atom j
( )
∑
−
∑
=
≠ ≠ 6 nn j i 6ij 12 nn j i 12ij totalR
p
1
R
p
1
4
ε
N
2
1
U
σ
σ
dimensionlessBoth lattice sums can be done for any structure.
Sum of 1/pn converges rapidly for large n.
More distant neighbors have more influence on the latter term than the former term.
15
FCC structure,
14
.
45392
p
1
13188
.
12
p
1
j i 6ij j i 12ij=
=
∑
∑
≠ ≠ andHCP structure,
14 .45489 p 1 13229 . 12 p 1 j i 6ij j i∑≠ 12ij = and ∑≠ =Both structures have 12 nearest neighbors.
25330 . 12 p 1 11418 . 9 p 1 j i ij6 j i 12ij = ∑ = ∑ ≠ ≠ and
BCC structure,
16
Cohesive energy of inert gas crystals at 0K
-- minimum Utotal (Equilibrium)
−
=
−
−
−
=
)
6
)(
45
.
14
(
)
12
)(
13
.
12
(
σ
R
R
σ
(14.45)(6)
2N
ε
R
σ
)
6
)(
45
.
14
(
R
σ
12)
(12.13)(
2N
ε
dR
dU
6 13 12 7 6 13 12 totalminimum
a
is
)
4N
)(
15
.
2
(
)
R
(
U
,
09
.
1
R
at
o=
σ
total o=
−
ε
=0
1.09 1.10 1.11 1.14 Ro/σ 3.98 3.65 3.40 2.74 σ (Å) 4.35 4.01 3.76 3.13 Ro (Å)Xenon
Krypton
Argon
Neon
FCC structureDeviation
Quantum corrections
17 Expect structure to form crystals which have lowest energy,
largest cohesive energy
Gibbs free energy : G=U – TS + PV
Assuming T=0, P=0, and no kinetic energy of atomic motion
-8.62 -8.61 -8.24 -5.69 12.13 12.13 9.11 6.2 14.45 14.45 12.25 8.4
FCC
HCP
BCC
SC
∑ ≡ ≠j i p12ij 1 αε
/N
U
tot ∑ ≡ ≠j i p6ij 1 β FCC is favored.(d) Ionic bonding
Alkali halidesElectron transfers between atoms to form two oppositely charged ions. Strong electrostatic forces dominate.
Electron configuration : closed electronic shells
For examples, LiF : Li+(1S2) instead of Li (1S22S)
F- (1S22S22p6) instead of F (1S22S22p5)
Like inert gas atoms w/. some distortion of charge distribution near
the region of contact with neighboring atoms
but
Charge distribution is spherically symmetric.
Electron density distribution
19 Need to consider the ionization energies and electron affinities of atoms
energy that must be supplied in order to remove an electron from a neutral atom
Ionization energy I
energy that is gained when an additional electron is added to a neutral atom
Electron affinity A
Ionic bonding is produced whenever
an element w/. a relatively low ionization energy is combined with
an element w/. a high electron affinity.
20 e.g. NaCl crystal What is its cohesive energy ?
Valence electron loosely bound to ion Ionization energy =5.14eV
(energy to remove electron from Na) Na+
e -Na :
Cl
e
-Cl : Seven valence electrons tightly bound desire a filled outer shell
Electron affinity energy = 3.61eV
+ 3.61eV e- + Cl Cl -Na + 5.14eV Na+ + e -NaCl : Na+ Cl -r=2.81Å
Ion bind by electrostatic attraction U~-e2/4πεr ~ -7.9eV
Na+ + Cl- = NaCl +7.9eV
Ionic bound Na + Cl = NaCl + 6.37eV
21
N ions in the crystal and Uij is the interaction energy between ions i and j ( i ≠ j )
r
q
q
)
r
exp(
λ
U
ij j i ij ij=
−
ρ
+
long range electrostatic short range Pauli repulsive
CGS
R
p
q
q
)
R
exp(
λ
U
ij j iij
=
−
ρ
+
where R = nearest neighbor distance
−
−
=
+
−
=
=
∑
∑
≠ ≠R
α
q
)
R
exp(
z
λ
N
R
p
q
q
)
R
exp(
λ
Nz
U
U
2 j i ij j i j i ij totρ
ρ
p
α
j ij '∑
±
≡
where z = number of nearest Madelung constant neighbors of any ion
minimum U
total(Equilibrium)
z
λ
ρα
q
)
ρ
R
(
exp
R
0
R
q
N
α
)
ρ
R
exp(
ρ
Nz
λ
dR
dU
2 o 2 o 2 2 tot=
−
=
+
−
−
=
At equilibrium R
o,
−
−
=
o o 2 totR
ρ
1
R
q
N
α
U
Madelung energy
Short range repulsive
ρ = 0.1Ro
Madelung constant
α
: geometric sum
depends on relative distance, number, and sign of neighboring atoms --- crystal structures and basis
23
One dimension : line of ions of alternating signs
+
-
+
+
r R+
+
-
-
-
p
α
ij ij '∑
±
=
(
)
•
•
•
+
−
+
−
=
+
−
+
−
+
•
•
•
=
⇒
−
+
−
+
•
•
•
=
±
=
∑
4
x
3
x
2
x
x
x
1
n
and
4
1
3
1
2
1
1
2
4R
1
3R
1
2R
1
R
1
2
r
R
α
4 3 2 ij ij 'l
α
24
In three dimensions, it is more complicated to calculate
α
.
very slowly convergent
very long range electrostatic forces
Special mathematical tricks are used to calculate Madelung constant.
1.641
4
ZnS (Wurtzite)
1.6381
4
GaAs (Zinc blende)
1.7627
8
CsCl (BCC)
1.7476
6
NaCl (FCC)
α
Coordinate No
structure
Higher coordination number gives larger Madelung constant.
25
Madelung energy
o 2 MR
q
N
α
4
1
E
πε
−
=
in a binary ionic crystalw/. 2N ions in the crystal
Sometimes,
the core-core repulsive energy in other general form is considered.
n o o 2 total
R
NA
R
q
N
α
4
1
E
=
−
+
πε
EquilibriumTdS
PdV
dE
=
−
+
The first law of thermodynamics0
dV
dE
=
At T=0, the equilibrium sample volume is determined by
0
dR
dE
o
26 The equilibrium nearest-neighbor distance
(n 1) 1 2 eq o
q
α
nA
4
R
−
−
=
πε
−
−
=
n
1
1
R
q
N
α
4
1
E
eq o 2 totalπε
Then, the total energy
3.4×10-120 10.65 3.57
CsCl
1.0×10-100 8.55 3.17KCl
1.8×10-99 5.0×10-88 2.3×10-89 2.6×10-79 A(J mn) 8.38 6.41 7.30 6.20 n 2.82 3.32 2.57 2.01 (Å)NaCl
NaF
LiCl
LiF
crystal eq o R TdV
dP
V
1
−
≡
κ
V=Na
3=NCR
o3The isothermal compressibility
( )
κ
α
πε
2 4 eq oq
R
C
36
1
n
=
+
(
)
( )
eq 4 o 2R
C
36
1
n
q
1
πε
α
κ
−
=
eq T 2 2dV
E
d
V
1
=
κ
or27
(b) Covalent bonding
C, Si, GeTetrahedral bond
C C Organic chemistry / diamond Si Si
Ge Ge
4 atoms in the valence band bond to 4 neighboring atoms
Semiconductor
}
7.3eV/atom 4.6eV/atom 3.9eV/atomdiamond
Tetrahedral bondingNature of chemical bonds in a diamond or zinc blende structure
(1/4,1/4,1/4)
(0,0,0)
(1/2,1/2,0) (0,1/2,1/2)
(1/2,0,1/2)
High electron concentration Tetrahedral sp3 bond
Four lobes emanate from an atom at the center of a cube. Other atoms are at the ends of the dotted lines and lobes point from them toward the cube center.
The bond is usually formed from two electrons, one from
each atom participating in the bond.
Electron forming the bond tend to be partially localized in the
region between two atoms joined by the bond
.
The spins of two electrons in the bond are antiparallel.
distortion of electron cloud
around atoms
29
30
Consider simple covalent bond : H - H
Both hydrogen atoms would like to form a filled outer shell -- share electrons
Two cases :
↑↑
(same spins on electrons)↑↓
(opposite spins on electrons)ψ2 r ψ
↑↓
↑↑
r↑↓
↑↑
Pauli exclusion principle forbids two electrons with the same states.
↑↑
same spins: electrons must stay apart
31
Pauli exclusion principle modifies the distribution
of charge according to spin orientation.
Energy is lower when
electrons spend time
between nuclei
--attractive Coulomb
interaction from both
Spin-dependent
Coulomb energy
Exchange interaction
Neutral H has only one electron
→
covalent bonding with one other atom
But, there would be
a hydrogen bond
between them under certain conditions
~ 0.1eV
being formed only between the most electronegative atoms,
such as F, O, and N.
.
F
-F
-H
+HF
2-is stabilized by a hydrogen bond.
In the extreme ionic form of the hydrogen bond,
the hydrogen atom loses its electron to another atom in the molecule;
the bare proton forms the hydrogen bond.
33
(c) metallic bonding
most metals
High electrical conductivity : a large number of electrons in a
metal are free to move .
conduction electrons
Outer electrons of atoms that form metals are loosely bound.
The potential energy barrier between atoms is reduced, the electron energy may be well above the potential energy maximum and their wave functions are then nearly plane waves in regions between atoms.
Weak binding, 1~5eV/atom
Metals tend to crystallize in relatively closed packed structures : hcp, fcc, bcc, …
Mechanical properties of solid
Bond --- harmonic oscillation
(
r
r
o)
k
F
r
=
−
r
−
r
Crystal --- A collection of harmonic oscillators
a homogeneous continuous medium rather than a periodic array of atoms
Apply forces displacements of atoms
stress strain
ε
l
l
u
A
k
A
ku
A
F
=
=
dimensionless elastic constant [Nt/m2]σ
=
C
ε
1 D Elastic regime
stress [Nt/m2]Elastic behavior is the fundamental distinction between solids and fluids.
