this differs from 03._crystalbindingandelasticconstants
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This differs from 03._CrystalBindingAndElasticConstants.ppt
only in the section “Analysis of Elastic Strain” in which a modified version of the Kittel narrative is used.
3. Crystal Binding and Elastic Constants
• Crystals of Inert Gases
• Ionic Crystals
• Covalent Crystals
• Metals
• Hydrogen Bonds
• Atomic Radii
• Analysis of Elastic Strains
• Elastic Compliance and Stiffness Constants
• Elastic Waves in Cubic Crystals
Introduction
Cohesive energy energy required to break up crystal into neutral free atoms.
Lattice energy (ionic crystals) energy required to break up crystal into free ions.
Kcal/mol = 0.0434 eV/molecule KJ/mol = 0.0104 eV/molecule
Crystals of Inert Gases
Atoms: •high ionization energy•outermost shell filled•charge distribution spherical
Crystal: •transparent insulators•weakly bonded•low melting point•closed packed (fcc, except He3 & He4).
Van der Waals – London Interaction
Van der Waals forces ＝ induced dipole – dipole interaction between neutral atoms/molecules.
Ref: A.Haug, “Theoretical Solid State Physics”, §30, Vol I, Pergamon Press (1972).
2 2 2 2
2 1 1 2
Q Q Q QV
R
R x x R x R x
Atom i charge +Q at Ri and charge –Q at Ri + xi.( center of charge distributions )
2 22
2 1 2 1 2 1ˆ ˆ ˆ ˆ ˆ2 R x x R x R x R x R x
2 1 R R R
1/22 212R a
R a
R a 2
2
2 2
ˆ3ˆ11
2 2
a
R R R R
R aR a ˆRR R
2 2 22 1 2 1 1 22 x x x x x x
2
2 1 1 23ˆ ˆ3
QV
R R x R x x x
2
1 2 1 2 1 232
Qx x y y z z
R ˆRR z
0H H V H0 = sum of atomic hamiltonians
2
0
0 0 00 0
j
j j
VE E V
E E
0 = antisymmetrized product of ground state atomic functions
1st order term vanishes if overlap of atomic functions negligible.2nd order term is negative & R6 (van der Waals binding).
Repulsive Interaction
Pauli exclusion principle (non-electrostatic) effective repulsion
Lennard-Jones potential: 12 6
4VR R
, determined from gas phase data
/Re
Alternative repulsive term:
Equilibrium Lattice Constants
Neglecting K.E. 12 6
14
2toti j i j i j
E U Np R p R
For a fcc lattice:
12
12
112.13188
i j i jp
6
6
114.45392
i j i jp
For a hcp lattice:
12 12.13229 6 14.45489
R n.n. dist
At equilibrium:
0dE
dR
12 6
12 613 7
14 12 6
2N
R R
1/6
0 12
6
2R
1.09 for fcc lattices
Experiment (Table 4):
Error due to zero point motion
Cohesive Energy
12 6
12 6
14
2totU R NR R
26
012
1
2totU R N
26
12
48
N
4 2.15N for fcc lattices
For low T, K.E. zero point motion.
For a particle bounded within length , p
2 2
2. .
2 2
pK E
m m
1
2
quantum correction is inversely proportional to the atomic mass:~ 28, 10, 6, & 4% for Ne, Ar, Kr, Xe.
Ionic Crystals
ions: closed outermost shells ~ spherical charge distribution
Cohesive/Binding energy = 7.9+3.615.14 = 6.4 eV
Electrostatic (Madelung) Energy
Interactions involving ith ion: i i jj i
U U
2/
2
. .R
i j
i j
qn ne
RU
qotherwise
p R
tot iU NUFor N pairs of ions:2
/R qN z e
R
z number of n.n.﹦ρ ~ .1 R0
j i i jp
﹦Madelung constant
At equilibrium: 0totdU
dR
2/
2Rz q
N eR
→ 0
2/2
0R q
R ez
2
00 0
1tot
N qU
R R
2
0
N qMadelung Energy
R
Evaluation of Madelung Constant
App. B: Ewald’s method j i i jp
1 1 12 1
2 3 4
2ln 2
KCl
i fixed
Kcal/mol = 0.0434 eV/molecule Prob 3.6
Covalent Crystals
• Electron pair localized midway of bond.• Tetrahedral: diamond, zinc-blende structures.• Low filling: 0.34 vs 0.74 for closed-packed.
Pauli exclusion → exchange interaction
H2
Ar : Filled outermost shell → van der Waal interaction (3.76A)Cl2 : Unfilled outermost shell → covalent bond (2A)
s2 p2 → s p3 → tetrahedral bonds
Metals
Metallic bonding: • Non-directional, long-ranged.• Strength: vdW < metallic < ionic < covalent• Structure: closed packed (fcc, hcp, bcc)• Transition metals: extra binding of d-electrons.
Hydrogen Bonds
• Energy ~ 0.1 eV• Largely ionic ( between most electronegative atoms like O & N ).• Responsible (together with the dipoles) for characteristics of H2O.• Important in ferroelectric crystals & DNA.
Atomic Radii
Na+ = 0.97AF = 1.36ANaF = 2.33Aobs = 2.32A
Standard ionic radii ~ cubic (N=6)
Bond lengths:F2 = 1.417ANa –Na = 3.716A NaF = 2.57A
Tetrahedral:C = 0.77ASi = 1.17ASiC = 1.94AObs: 1.89A
Ref: CRC Handbook of Chemistry & Physics
Ionic Crystal Radii
E.g. BaTiO3 : a = 4.004ABa++ – O– – : D12 = 1.35 + 1.40 + 0.19 = 2.94A → a = 4.16ATi++++ – O – – : D6 = 0.68 + 1.40 = 2.08A → a = 4.16ABonding has some covalent character.
Analysis of Elastic Strains
ˆ ixLet be the Cartesian axes of the unstrained state
ix be the the axes of the stained state
Using Einstein’s summation notation, we have
ˆi j i j j xˆ ˆi i i j j x x x
1 1 11 1 11 2 k k x x 2 2 211 11 12 131 2
ˆi irr x
i ir r x
R r r ˆi i j jr x
Position of atom in unstrained lattice:
Its position in the strained lattice is defined as
Displacement due to deformation:
i j j iu r
iii ii
i
ue
x
jii j i j j i
j i
uue
x x
ˆi i ir x x ˆi iu x
Define ( Einstein notation suspended ):
i j
Dilation
1 1 2 2 3 3ˆ ˆ ˆi i i j j j k k ka b c x x x
1 1 2 2 3 3i i j j k k i j kV
1 2 3ˆ ˆ ˆV a b c abc x x x
1 2 3V x x x
211 22 33
V VO
V
211 22 331V O
1 even permutation of 123
1 for odd permutation of 123
0 otherwisei j k ijk
2123 1 23 2 1 3 3i i j j k i jkV V O
2Tr O
where
Stress Components
Xy = fx on plane normal to y-axis = σ12 .
(Static equilibrium → Torqueless) i j j i
y xX Y
Elastic Compliance & Stiffness Constants
i j i j k l k le S
1 11 xX 1 1 1
2 2 2
3 3 3
2 3 4
3 1 5
1 2 6
i j
e S
4 23 32 z yY Z
C e
S = elastic compliance tensor
Contracted indices
C = elastic stiffness tensor
Elastic Energy Density1
2U C e e
1
2C e C e U
e
C e 1
2C C e
C 1
2C C C
Let
then
1
2 i j k l i j k lU C u u
1
2C u u
Landau’s notations:
1
2ji
i jj i
uuu
x x
1
2
ii
i j
i jefor
ei j
1,2,3for
4,5,6ii
i j j i
uu
u u
u e
C
1
2C e e
Elastic Stiffness Constants for Cubic Crystals
Invariance under reflections xi → –xi C with odd numbers of like indices vanishes
Invariance under C3 , i.e.,
1111iiiiC C
x y z x x z y x
x z y x x y z x
All C i j k l = 0 except for (summation notation suspended):
1122ii k kC C 1212i k ikC C
2 2 2 2 2 211 1 2 3 12 1 2 2 3 3 1 44 4 5 6
1 1
2 2U C e e e C e e e e e e C e e e
1 111 12 12
2 212 11 12
3 312 12 11
4 444
5 544
6 644
0 0 0
0 0 0
0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
eC C C
eC C C
eC C C
eC
eC
eC
11C C 12C C 44C C , 1, 2,3 4,5,6
1
11 12 12 11 12 12
12 11 12 12 11 12
12 12 11 12 12 11
44 44
44 44
44 44
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
C C C S S S
C C C S S S
C C C S S S
C S
C S
C S
where 11 12
1111 12 11 122
C CS
C C C C
12
1211 12 11 122
CS
C C C C
4444
1S
C
1
11 12 11 12S S C C 1
11 12 11 122 2S S C C
Bulk Modulus & Compressibility
2 2 2 2 2 211 1 2 3 12 1 2 2 3 3 1 44 4 5 6
1 1
2 2U C e e e C e e e e e e C e e e
Uniform dilation:
211 12
12
6U C C
1 2 3 3e e e
4 5 6 0e e e
δ = Tr eik = fractional volume change
21
2B B = Bulk modulus
1 V
V p
11 12
12
3B C C = 1/κ κ = compressibility
See table 3 for values of B & κ .
Up
2
2
UB
p
pV
V
Elastic Waves in Cubic Crystals
Newton’s 2nd law:2
2
i ki
k
u
t x
don’t confuse ui with uα
i k ik j l j lC u →2
2
jlii k j l
k
uuC
t x
221
2jl
i k j lk j k l
uuC
x x x x
2l
i k j lk j
uC
x x
2 22 2 2 2 2 23 31 1 2 2 1 1
1111 1122 1133 1212 1221 1313 13312 2 2 21 1 2 1 3 2 1 2 3 1 3
u uu u u u u uC C C C C C C
t x x x x x x x x x x x
2 22 2 2 2 23 31 2 2 1 1
1111 1122 12122 2 21 1 2 1 3 2 1 2 3 1 3
u uu u u u uC C C
x x x x x x x x x x x
22 2 2 2 2
31 1 2 1 111 12 44 442 2 2 2
1 1 2 1 3 2 3
uu u u u uC C C C
t x x x x x x x
Similarly 22 2 2 2 2
32 2 1 2 211 12 44 442 2 2 2
2 2 3 2 1 1 3
uu u u u uC C C C
t x x x x x x x
2 2 2 22 2
3 3 3 32 111 12 44 442 2 2 2
3 3 2 3 1 2 1
u u u uu uC C C C
t x x x x x x x
Dispersion Equation2 2
2i l
i k j lk j
u uC
t x x
0
i ti iu u e k r
→2
0 0i i k j l k j lu C k k u
20 0il i k j l k j lC k k u
2 0i l i k j l k jC k k dispersion equation
2 0I kC i j imn j m nC k kkC
Waves in the [100] direction
2 0I kC i j imn j m nC k kkC
1,0,0kk → 211i j i jC kkC
1111 1112 1113
22111 2112 2113
3111 3112 3113
C C C
k C C C
C C C
kC11
244
44
0 0
0 0
0 0
C
k C
C
11L
Ck
0 1,0,0u Longitudinal
44T
Ck
0 0,1,0uTransverse, degenerate 0 0,0,1u
11112
2112
3113
0 0
0 0
0 0
C
k C
C
11 16 152
61 66 65
51 56 55
C C C
k C C C
C C C
Waves in the [110] direction
2 0I kC i j imn j m nC k kkC
1,1,02
kk →
2
11 12 21 22 2i j i j i j i j i j
kC C C C kC
1111 1221 1122 12122
2121 2211 2112 2222
3113 3223
0
02
0 0
C C C Ck
C C C C
C C
kC11 44 12 442
12 44 11 44
44
0
02
0 0 2
C C C Ck
C C C C
C
11 12 44
12
2L C C C k
0 1,1,0u Lonitudinal
442T
Ck
0 0,0,1u
Transverse 1 11 12
1
2T C C k
0 1, 1,0 u
Transverse
Prob 3.10
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