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
[ stress ] = [ elastic constant ] x [ strain ]
1
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
2
Types of bonds
(a) Van der Waals Electrons localized among atoms
(Molecular)
- +
- - +
- -
(b) Covalent
Electrons shared by the neighboring atoms
(c) Metallic
Electrons free to move through sample
(d) Ionic
Electrons transferred to adjacent atoms
- +
- - +
- - +
- (a)
- +
- - +
- -
+
+
+
+
+
+
(b)
+
+
+
+
(c)
-
+
-
+
-
+
-
+
-
(d)
3
(a) Molecular bonding Inert gas crystals : He, Ne, Ar, Kr, Xe, Rn
Transparent Insulators – completely filled outer electron shells
Weakly bonding – van der Waals bonding
FCC structures except for He3 and He4
high ionization energies
low melting temperatures
Neon Argon Krypton Xenon
Cohesive energy (eV/atom)
0.02
0.08
0.12
0.16
Melting temperature (K)
24.56
83.81
115.8
161.4
Ionization energy (eV)
21.56
15.76
14.00
12.13
What holds an inert gas crystal ?
4
Phase diagrams of (a) 4He and (b) 3He.
5
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)
Fluctuating dipole-dipole interaction
Reviews:
+q
d
r
P
Attractive interaction
between the atoms
n(r)
-q
E ~
r
P = qd ←
P
as r>>d
r3
Electric fields
?
r
r
r r
P = ∫ d 3 r n( r ) r
P1P2
F~ 4
r
PP
U ~ 132
r
Attractive force
Repulsive force
6
Inert gas solids
• On average spherically symmetric distribution of electron charge with the
positive nucleus in the center
r
P =0
• But thermal fluctuations (finite T) cause instantaneous electric dipole moment
r
P (t) ≠ 0 → P 2 ≠ 0
v
P=0
-
+
-
fluctuations
+
- - -
v
P≠0
• 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
attractive force
7
Model for inert gas solid –
two identical linear harmonic oscillators
x2
+e
-e
R
+e
x1
-e
p1 and p2 are the momenta of these two oscillators
C is the force constant
Hamiltonian for the unperturbed system – no Coulomb interaction
2
p
1
p 12
1
Ho =
+ Cx 12 + 2 + Cx 22
2m 2
2m 2
Hamiltonian for Coulomb interaction energy of the system
2
2e x1x 2
e2
e2
e2
e2
H1 = +
−
−

→ −
x1 , x 2 << R
R R + x1 − x 2 R + x1 R − x 2
R3
8
Normal mode transformation -- symmetric (s) and anti-symmetric (a)
x1 + x 2
x1 − x 2
xs ≡
; xa ≡
2
2
p1 − p 2
p1 + p 2
; pa ≡
ps ≡
2
2
Total Hamiltonian after the transformation
 p s2 1 
2e 2  2   p a2 1 
2e 2  2 
+  C − 3  x s  + 
+  C + 3  x a 
H=
R    2m 2 
R  
 2m 2 
2
p 12
1
p
1
Ho =
+ Cx 12 + 2 + Cx 22
2m 2
2m 2
Two frequencies of the coupled oscillators
symmetric (s) and anti-symmetric (a)
C
H o , ωo =
m
Cs
C − 2e 2 / R 3
=
H , ωs =
m
m
Ca
C + 2e 2 / R 39
=
ωa =
m
m
The zero point energy
1
1
hω o + hω o
¾ The uncoupled oscillators
¾ The coupled oscillators
2
2
1
1
hω s + hω a
2
2
1 1


2
(
)
−
2
2
2
3
2


C - 2e / R
C
2e 
1  2e  2 2  2e 

 + K
1 −
 = ωo 1 − 
+
ωs =
=
3 
3 
3 
m
m  CR 
 2  CR  1× 2  CR 





1 1


2
1
/
2
−
(
)
2
2
2
3
2


C + 2e / R
C
2e 
1  2e  2 2  2e 
1 +
 = ωo 1 + 
+

 + K
ωa =
=
3 
3 
3 
m
m  CR 
 2  CR  1× 2  CR 





Therefore, the zero point energy of the coupled oscillators is
lowered from the uncoupled oscillators by
1/ 2
2
2


1
1  2e 
A
 × 2 = − 6
∆U = hωo − 
3 
2
R
 8  CR 

Attractive
interaction
10
The van der Waals interaction, the London interaction,
the induced dipole-dipole interaction
A
∆U = − 6 < 0
R
P1 fluctuation
R
P2
induced
e4
2
≡
h
A = hωo
ω
α
o
2C 2
where α is the electronic polarizability
r
r
“Polarizability” of the atom P = α E
r
r
P1
P2 = α E1 = α 3
R
P1P2
P12
U ~ − 3 ~ −α 6 < 0
R
R
11
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.
12
Empirical formula for such repulsive potential ∆U =
B
>0
12
R
The total potential for inert gas system
8
U(r) = UPauli + UvdW
U(R)/4ε
B
A
= 12 − 6
R
R
 σ 12  σ  6 
= 4ε   −   
 R  
 R 
6
4
2
U
0
-2
the Lennard-Jones potential
UPauli
0.85
UvdW
0.90
0.95
1.00
1.05
1.10
1.15
R/σ
where empirical parameters A=4εσ6 and B=4εσ12 are determined
from independent measurements made in the gas phase.
Values of ε[energy] and σ[length] are shown in Table 4.
13
1.20
N atoms in the crystal
U total
12
6






σ
σ
1
 −
 
= N(4ε ) ∑ 
p R  
i ≠ j  p ij R nn 
2


 ij nn  

where Rnn is nearest neighbor distance and
pijRnn is the distance between atom i and atom j
U total
 1  σ
1
= N(4ε )  ∑ 12 
2
 i ≠ j p ij  R nn
12
 1  σ

 −  ∑ 6 
 i≠ j p  R
ij 
nn





6



dimensionless
Both 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.
14
1
FCC structure, ∑ 12 = 12.13188 and
i ≠ j p ij
HCP structure,
1
= 14. 45392
∑
6
i ≠ j p ij
1
1
and
=
12
.
13229
= 14. 45489
∑
12
6
i ≠ j p ij
i ≠ j p ij
∑
Both structures have 12 nearest neighbors.
BCC structure,
1
∑ 12 = 9.11418
i ≠ j p ij
1
= 12.25330
6
i ≠ j p ij
and ∑
BCC structure has 8 nearest neighbors.
15
Cohesive energy of inert gas crystals at 0K
-- minimum Utotal (Equilibrium)

σ12
σ6 
(12.13)(−12) 13 − (14.45)(−6) 7 
R
R 

6
σ12  R  (12.13)(12) 
= 2Nε (14.45)(6) 13   −

R  σ 
(14.45)(6) 
dU total
= 2Nε
dR
=0
at R o = 1.09σ , U total (R o ) = −(2.15)(4Nε ) is a minimum
Neon
Argon Krypton Xenon
Ro (Å)
3.13
3.76
4.01
4.35
σ (Å)
2.74
3.40
3.65
3.98
Ro/σ
1.14
1.11
1.10
1.09
Deviation
FCC structure
Quantum corrections
16
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
1
6
i ≠ j p ij
1
α ≡ ∑ 12
i ≠ j p ij
β ≡∑
U tot /Nε
SC
BCC
HCP
FCC
8.4
12.25
14.45
14.45
6.2
9.11
12.13
12.13
-5.69
-8.24
-8.61
-8.62
FCC is favored.
17
(d) Ionic bonding
Alkali halides
Electron transfers between atoms to form two oppositely charged ions.
Strong electrostatic forces dominate.
Electron configuration : closed electronic shells
instead of
Li (1S22S)
For examples, LiF : Li+ (1S2)
F- (1S22S22p6) instead of
F (1S22S22p5)
Like inert gas atoms
w/. some distortion of charge distribution near
but
Charge distribution is
the region of contact with neighboring atoms
spherically symmetric.
Electron density distribution
in the base plane of NaCl
relative electron
concentration
18
Need to consider the ionization energies and electron affinities of atoms
Ionization energy I
energy that must be supplied in order to remove
an electron from a neutral atom
Electron affinity A
energy that is gained when an additional electron
is added to a neutral atom
Ionic bonding is produced whenever
an element w/. a relatively low ionization energy is combined with
an element w/. a high electron affinity.
19
e.g. NaCl crystal
e-
Na :
Na+
What is its cohesive energy ?
Valence electron loosely bound to ion
Ionization energy =5.14eV
(energy to remove electron from Na)
Na
Cl :
eCl
Cl
Na+
r=2.81Å
Ionic bound
-
Na+
+ e-
Seven valence electrons tightly bound
desire a filled outer shell
Electron affinity energy = 3.61eV
e- +
NaCl :
+ 5.14eV
Cl
Cl-
+ 3.61eV
Ion bind by electrostatic attraction
U~-e2/4πεr ~ -7.9eV
Na+ + Cl- = NaCl +7.9eV
Na + Cl = NaCl + 6.37eV
20
cohesive energy
N ions in the crystal and Uij is the interaction energy between ions i and j
rij
qiq j
ρ
rij
U ij = λ exp(− ) +
(i≠j)
CGS
long range electrostatic
short range Pauli repulsive
U ij = λ exp(−
R
ρ
)+
qiq j
p ijR
where R = nearest neighbor distance
U tot = ∑ U ij = Nz λ exp(−
i≠ j
R
ρ
)+∑
i≠ j
qiq j
p ijR

R αq 2 

= N  zλ exp(− ) −
ρ
R 

where z = number of nearest
neighbors of any ion
α≡∑
j
'
±
p ij
Madelung constant
21
minimum Utotal (Equilibrium)
dU tot
Nzλ
R Nα q 2
exp(− ) +
=−
=0
2
dR
ρ
ρ
R
2
R
ραq
At equilibrium Ro, R o2 exp(− o ) =
ρ
zλ
U tot
Nα q 2
=−
Ro

ρ 

1 −
 Ro 
Short range repulsive
ρ = 0.1Ro
Madelung energy
Madelung constant α : geometric sum
depends on relative distance, number, and sign of neighboring atoms
----- crystal structures and basis
22
One dimension : line of ions of alternating signs
R
+
α=∑
ij
'
±
p ij
-
+
r
-
+
-
+
-
+
-
1
1
1
α

1
' ±
= ∑ = 2 −
+
−
+ • • •
R
rij

 R 2R 3R 4R
ij

 1 1 1
⇒ α = 21 − + − + • • • 

 2 3 4


x 2 x3 x 4

and ln (1 + x ) =  x − + − + • • • 
2
3
4


α = 2 ln(2) =1.386 for one dimensional chain
23
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.
Coordinate No
α
NaCl (FCC)
6
1.7476
CsCl (BCC)
8
1.7627
GaAs (Zinc blende)
4
1.6381
ZnS (Wurtzite)
4
1.641
structure
Higher coordination number gives larger Madelung constant.
24
It depends on the structure of the crystal but not unit cell dimensions.
Madelung energy
1 Nα q 2
EM = −
4πε R o
in a binary ionic crystal
w/. 2N ions in the crystal
Sometimes,
the core-core repulsive energy in other general form is considered.
E total
1 Nα q 2 NA
=−
+ n
4πε R o
Ro
Equilibrium
dE = − PdV + TdS
The first law of thermodynamics
dE
At T=0, the equilibrium sample volume is determined by
=0
dV
dE
=0
dR o
25
The equilibrium nearest-neighbor distance
 4πε nA 

R = −
2
 αq 
eq
o
1
( n −1)
E total
Then, the total energy
1 Nα q 2  1 
1− 
=−
eq 
4πε R o  n 
crystal
LiF
LiCl
NaF
NaCl
KCl
CsCl
R eq
o (Å)
2.01
2.57
3.32
2.82
3.17
3.57
n
6.20
7.30
6.41
8.38
8.55
10.65
A(J mn)
2.6×10-79 2.3×10-89 5.0×10-88 1.8×10-99 1.0×10-100 3.4×10-120
1
The isothermal compressibility κ ≡ −
V
d 2E
=V 2
κ
dV
1
1
Teq
κ
=
dP
dV T
α q (n − 1)
V=Na3=NCRo3
2
( )
36πε C R
eq 4
o
or
( )
eq 4
o
36πε C R
n = 1+
α q 2κ 26
(b) Covalent bonding
C, Si, Ge
Tetrahedral bond
C
C
Si
Si
Ge
Ge
}
Organic chemistry / diamond
7.3eV/atom
4.6eV/atom
Semiconductor
3.9eV/atom
4 atoms in the valence band bond to 4 neighboring atoms
diamond
Tetrahedral bonding
Nature of chemical bonds in a
diamond or zinc blende structure
27
(0,1/2,1/2)
Tetrahedral sp3 bond
(1/2,0,1/2)
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.
(1/4,1/4,1/4)
(1/2,1/2,0)
(0,0,0)
High electron concentration
ƒThe bond is usually formed from two electrons, one from
each atom participating in the bond.
ƒ Electron forming the bond tend to be partially localized in the
region between two atoms joined by the bond.
ƒ The spins of two electrons in the bond are antiparallel.
spin dependent coulomb energy
distortion of electron cloud
around atoms
28
Calculated valence electron concentration in Ge.
29
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
ψ
↑↑
↑↓
r
↑↓
r
↑↑
Pauli exclusion principle forbids two electrons with the same states.
↑↑ same spins: electrons must stay apart
30
↑↓ opposite spins: electrons can occupy the same place
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
31
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-
HF2- is stabilized by a hydrogen bond.
H+
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.
The hydrogen bond connects only two atoms.
32
(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, …
33
Mechanical properties of solid
Bond --- harmonic oscillation
r
r r
F = − k ( r − ro )
Crystal --- A collection of harmonic oscillators
a homogeneous continuous medium rather than a periodic array of atoms
Apply forces
stress
1D
Elastic regime
displacements of atoms
strain ε
F ku kl u
=
=
A A
A l
stress [Nt/m2]
dimensionless
elastic constant [Nt/m2]
σ=Cε
Elastic behavior is the fundamental distinction between solids and fluids.
Elasticity describes the dimensional change under external stresses.
34