Quinn J.J., Yi K.-S. Solid State Physics: Principles and Modern Applications
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Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
28 1 Crystal Structures
It is convenient to define φ
i
, the potential energy of the i
th
atom as 527
φ
i
=
j
φ
ij
. (1.34)
Here, the prime on the sum implies that the term i = j is omitted. It is
528
apparent from symmetry considerations that φ
i
is independent of i for an 529
infinite lattice, so we can drop the subscript i. The total energy is then 530
U =
1
2
2Nφ = Nφ, (1.35)
where 2N is the number of atoms and N is the number of molecules.
531
It is convenient in evaluating φ to introduce a dimensionless parameter p
ij
532
defined by p
ij
= R
−1
r
ij
,whereR is the distance between nearest neighbors. 533
In terms of p
ij
, the expression for φ is given by 534
φ =
λ
R
n
j
p
−n
ij
−
e
2
R
j
(∓p
ij
)
−1
. (1.36)
Here, the primes on the summations denote omission of the term i = j.We
535
define the quantities 536
A
n
=
j
p
−n
ij
, (1.37)
and
537
α =
j
(∓p
ij
)
−1
. (1.38)
538
The α and A
n
are properties of the crystal structure; α is called the Madelung 539
constant. The internal energy of the crystal is given by Nφ,whereN is the 540
number of molecules. The internal energy is given by 541
U = N
λ
A
n
R
n
− α
e
2
R
. (1.39)
At the equilibrium separation R
0
,
∂U
∂R
R
0
must vanish. This gives the result 542
λ
A
n
R
n
0
= α
e
2
nR
0
. (1.40)
Therefore, the equilibrium value of the internal energy is
543
U
0
= Nφ
0
= −Nα
e
2
R
0
1 −
1
n
. (1.41)
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
1.6 Binding Energy of Ionic Crystals 29
Compressibility 544
The best value of the parameter n can be determined from experimental data 545
on the compressibility κ. κ is defined by the negative of the change in volume 546
per unit change in pressure at constant temperature divided by the volume. 547
κ = −
1
V
∂V
∂P
T
. (1.42)
548
The subscript T means holding temperature T constant, so that (1.42) is the 549
isothermal compressibility. We will show that at zero temperature 550
κ
−1
= V
∂
2
U
∂V
2
T =0
. (1.43)
Equation (1.43) comes from the thermodynamic relations
551
F = U − TS, (1.44)
and
552
dU = TdS − PdV. (1.45)
By taking the differential of (1.44) and making use of (1.45), one can see that
553
dF = −PdV − SdT. (1.46)
From (1.46) we have
554
P = −
∂F
∂V
T
. (1.47)
Equation (1.42) can be written as
555
κ
−1
= −V
∂P
∂V
T
= V
∂
2
F
∂V
2
T
. (1.48)
But at T =0,F = U so that
556
κ
−1
= V
∂
2
F
∂V
2
T =0
(1.49)
is the inverse of the isothermal compressibility at T =0.Wecanwritethe
557
volume V as 2NR
3
and use
∂
∂V
=
∂R
∂V
∂
∂R
=
1
6NR
2
∂
∂R
in (1.39) and (1.43). This 558
gives 559
κ
−1
T =0
=
αe
2
18R
4
0
(n − 1), (1.50)
560
or 561
n =1+
18R
4
0
αe
2
κ
. (1.51)
From the experimental data on NaCl, the best value for n turns out to be
562
∼9.4. 563
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
30 1 Crystal Structures
Evaluation of the Madelung Constant 564
For simplicity let us start with a linear chain. Each positive (+) atom has two 565
neighbors, which are negative (−)atoms,atp
01
= 1. Therefore, 566
α =
j
∓p
−1
ij
=2
1 −
1
2
+
1
3
−
1
4
+ ...
. (1.52)
If you remember that the power series expansion for ln(1 + x)isgivenby
567
−
∞
n=1
(−x)
n
n
= x −
x
2
2
+
x
3
3
−
x
4
4
+ ··· and is convergent for x ≤ 1, it is 568
apparent that 569
α =2ln2. (1.53)
If we attempt the same approach for NaCl, we obtain
570
α =
6
1
−
12
√
2
+
8
√
3
−
6
2
+ ··· . (1.54)
This is taking six opposite charge nearest neighbors at a separation of one
571
nearest neighbor distance, 12 same charge next nearest neighbors at
√
2times 572
that distance, etc. It is clear that the series in (1.54) converges very poorly. 573
The convergence can be greatly improved by using a different counting pro- 574
cedure in which one works with groups of ions which form a more or less 575
neutral array. The motivation is that the potential of a neutral assembly of 576
charges falls off much more quickly with distance than that of a charged 577
assembly. 578
Evjen’s Method 579
We will illustrate Evjen’s method
2
by considering a simple square lattice in 580
two dimensions with two atoms per unit cell, one at (0, 0) and one at (
1
2
,
1
2
). 581
The crystal structure is illustrated in Fig. 1.25. The calculation is carried out 582
as follows: 583
1. One considers the charges associated with different shells where the first 584
shell is everything inside the first square, the second is everything outside 585
the first but inside the second square, etc. 586
2. An ion on a face is considered to be half inside and half outside the square 587
defined by that face; a corner atom is one quarter inside and three quarters 588
outside. 589
3. The total Madelung constant is given by α = α
1
+ α
2
+ α
3
+ ···,where 590
α
j
is the contribution from the ith shell. 591
As an example, let us evaluate the total charge on the first few shells. The 592
first shell has four atoms on faces, all with the opposite charge to the atom 593
2
H.M. Evjen, Phys. Rev. 39, 675 (1932).
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
1.6 Binding Energy of Ionic Crystals 31
PRIMITIVE
UNIT CELL
Fig. 1.25. Evjen method for a simple square lattice in two dimensions
at the origin and four corner atoms all with the same charge as the atom at 594
the origin. Therefore, the charge of shell number one is 595
Q
1
=4
1
2
− 4
1
4
=1. (1.55)
Doing the same for the second shell gives
596
Q
2
=4
1
2
− 4
3
4
− 4
1
2
+8
1
2
− 4
1
4
=0. (1.56)
Here the first two terms come from the remainder of the atoms on the outside
597
of the first square; the next three terms come from the atoms on the inside of 598
the second square. To get α
1
and α
2
we simply divide the individual charges 599
by their separations from the origin. This gives 600
α
1
=
4(
1
2
)
1
−
4(
1
4
)
√
2
1.293, (1.57)
601
α
2
=
4(
1
2
)
1
−
4(
3
4
)
√
2
−
4(
1
2
)
2
+
8(
1
2
)
√
5
+
4(
1
4
)
2
√
2
0.314. (1.58)
This gives α α
1
+ α
2
∼ 1.607. The readers should be able to evaluate α
3
602
for themselves. 603
Madelung Constant for Three-Dimensional Lattices 604
For a three-dimensional crystal, Evjen’s method is essentially the same with 605
the exception that 606
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
32 1 Crystal Structures
Fig. 1.26. Central atom and the first cube of the Evjen method for the NaCl
structure
1. The squares are replaced by cubes. 607
2. Atoms on the face of a cube are considered to be half inside and half outside 608
the cube; atoms on the edge are
1
4
inside and
3
4
outside, and corner atoms 609
are
1
8
inside and
7
8
outside. 610
We illustrate the case of the NaCl structure as an example in the three 611
dimensions (see Fig. 1.26). 612
For α
1
we obtain 613
α
1
=
6(
1
2
)
1
−
12 (
1
4
)
√
2
+
8(
1
8
)
√
3
1.456. (1.59)
For α
2
we have the following contributions: 614
1. remainder of the contributions from the atoms on the first cube 615
=
6(
1
2
)
1
−
12 (
3
4
)
√
2
+
8(
7
8
)
√
3
616
2. Atoms on the interior of faces of the second cube 617
= −
6(
1
2
)
2
+
6(4) (
1
2
)
√
5
−
6(4) (
1
2
)
√
6
618
3. Atoms on the interior of edges of the second cube 619
= −
12 (
1
4
)
√
8
+
12(2) (
1
4
)
√
9
620
4. Atoms on the interior of the coners of the second cube 621
= −
8(
1
8
)
√
12
622
Adding them together gives 623
α
2
=
3 −
9
√
2
+
7
√
3
+
−
3
2
+
12
√
5
−
12
√
6
+
−
3
√
8
+
6
3
−
1
√
12
0.295.
(1.60)
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
1.6 Binding Energy of Ionic Crystals 33
Thus, to the approximation α α
1
+ α
2
we find that α 1.752. The exact 624
result for NaCl is α =1.747558 ..., so Evjen’s method gives a surprisingly 625
accurate result after only two shells. 626
Results of rather detailed evaluations of α for several different crystal struc- 627
tures are α(NaCl) = 1.74756, α(CsCl) = 1.76267, α(zincblende) = 1.63806, 628
α(wurtzite) = 1.64132. The NaCl structure occurs much more frequently than 629
the CsCl structure. This may seem a bit surprising since α(CsCl) is about 1% 630
larger than α(NaCl). However, core repulsion accounts for about 10% of the 631
binding energy (see (1.41)). In the CsCl structure, each atom has eight nearest 632
neighbors instead of the six in NaCl. This should increase the core repulsion 633
by something of the order of 25% in CsCl. Thus, we expect about 2.5% larger 634
contribution (from core repulsion) to the binding energy of CsCl. This negative 635
contribution to the binding energy more than compensates the 1% difference 636
in Madelung constants. 637
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
34 1 Crystal Structures
Problems 638
1.1. Demonstrate that 639
(a) The reciprocal lattice of a simple cubic lattice is simple cubic. 640
(b) The reciprocal lattice of a body centered cubic lattice is a face centered 641
cubic lattice. 642
(c) The reciprocal lattice of a hexagonal lattice is hexagonal. 643
1.2. Determine the packing fraction of 644
(a) A simple cubic lattice 645
(b) A face centered lattice 646
(c) A body centered lattice 647
(d) The diamond structure 648
(e) A hexagonal close packed lattice with an ideal
c
a
ratio 649
(f) The graphite structure with an ideal
c
a
. 650
1.3. The Bravais lattice of the diamond structure is fcc with two carbon atoms 651
per primitive unit cell. If one of the two basis atoms is at (0,0,0), then the 652
other is at (
1
4
,
1
4
,
1
4
). 653
(a) Illustrate that a reflection through the (100) plane followed by a non- 654
primitive translation through [
1
4
,
1
4
,
1
4
] is a glide-plane operation for the 655
diamond structure. 656
(b) Illustrate that a fourfold rotation about an axis in diamond parallel to 657
the x-axis passing through the point (1,
1
4
, 0) (the screw axis) followed 658
by the translation [
1
4
, 0, 0] parallel to the screw axis is a screw operation 659
for the diamond structure. 660
1.4. Determine the group multiplication table of the point group of an 661
equilateral triangle. 662
1.5. CsCl can be thought of as a simple cubic lattice with two different atoms 663
[at (0, 0, 0) and (
1
2
,
1
2
,
1
2
)] in the cubic unit cell. Let f
+
and f
−
be the atomic 664
scattering factors of the two constituents. 665
(a) What is the structure amplitude F (h
1
,h
2
,h
3
) for this crystal? 666
(b) An X-ray source has a continuous spectrum with wave numbers k 667
satisfying: k is parallel to the [110] direction and 2
−1/2
2π
a
≤|k|≤
668
3 ×2
1/2
2π
a
,wherea is the edge distance of the simple cube. Use the
669
Ewald construction for a plane that contains the direction of incidence 670
to show which reciprocal lattice points display diffraction maxima. 671
(c) If f
+
= f
−
, which of these maxima disappear? 672
1.6. A simple cubic structure is constructed in which two planes of A atoms 673
followed by two planes of B atoms alternate in the [100] direction. 674
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
1.6 Binding Energy of Ionic Crystals 35
1. What is the crystal structure (viewed as a non-Bravais lattice with four 675
atoms per unit cell)? 676
2. What are the primitive translation vectors of the reciprocal lattice? 677
3. Determine the structure amplitude F (h
1
,h
2
,h
3
) for this non-Bravais lat- 678
tice. 679
1.7. Powder patterns of three cubic crystals are found to have their first four 680
diffraction rings at the values given below:
t4.2 φ
A
30
◦
35
◦
50
◦
60
◦
t4.3 φ
B
21
◦
29
◦
36
◦
42
◦
t4.4 φ
◦
30
◦
50
◦
60
◦
74
◦
681
The crystals are monatomic, and the observer believes that one is body 682
centered, one face centered, and one is a diamond structure. 683
1. What structures are the crystals A, B, and C? 684
2. The wave length λ of the incident X-ray is 0.95
˚
A. What is the length of 685
the cube edge for the cubic unit cell in A, B, and C, respectively? 686
1.8. Determine the ground state atomic configurations of C(6), O(8), Al(13), 687
Si(16), Zn(30), Ga(31), and Sb(51). 688
1.9. Consider 2N ions in a linear chain with alternating ±e charges and a 689
repulsive potential AR
−n
between nearest neighbors. 690
1. Show that, at the equilibrium separation R
0
, the internal energy becomes 691
U(R
0
)=−2ln2×
Ne
2
R
0
1 −
1
n
.
2. Let the crystal be compressed so that R
0
→ R
0
(1 −δ). Show that the work 692
done per unit length in compressing the crystal can be written
1
2
Cδ
2
,and 693
determine the expression for C. 694
1.10. For a BCC and for an FCC lattice, determine the separations between 695
nearest neighbors, next nearest neighbors, ... down to the 5th nearest neigh- 696
bors. Also determine the separations between nth nearest neighbors (n = 697
1, 2, 3, 4, 5) in units of the cube edge a ofthesimplecube. 698
Uncorrected Proof
BookID 160928 ChapID 01 Proof# 1 - 29/07/09
36 1 Crystal Structures
Summary 699
In this chapter first we have introduced basic geometrical concepts useful in 700
describing periodic arrays of objects and crystal structures both in real and 701
reciprocal spaces assuming that the atoms sit at lattice sites. 702
A lattice is an infinite array of points obtained from three primitive 703
translation vectors a
1
, a
2
, a
3
. Any point on the lattice is given by 704
n = n
1
a
1
+ n
2
a
2
+ n
3
a
3
.
Any pair of lattice points can be connected by a vector of the form
705
T
n
1
n
2
n
3
= n
1
a
1
+ n
2
a
2
+ n
3
a
3
.
Well defined crystal structure is an arrangement of atoms in a lattice such
706
that the atomic arrangement looks absolutely identical when viewed from two 707
different points that are separated by a lattice translation vector. Allowed 708
types of Bravais lattices are discussed in terms of symmetry operations 709
both in two and three dimensions. Because of the requirement of transla- 710
tional invariance under operations of the lattice translation, the rotations of 711
60
◦
, 90
◦
, 120
◦
, 180
◦
, and 360
◦
are allowed. 712
If there is only one atom associated with each lattice point, the lattice 713
of the crystal structure is called Bravais lattice. If more than one atom is 714
associated with each lattice point, the lattice is called a lattice with a basis. 715
If a
1
, a
2
, a
3
are the primitive translations of some lattice, one can define a set 716
of primitive translation vectors b
1
, b
2
, b
3
by the condition 717
a
i
·b
j
=2πδ
ij
,
where δ
ij
=0ifi is not equal to j and δ
ii
= 1. It is easy to see that 718
b
i
=2π
a
j
× a
k
a
i
· (a
j
× a
k
)
,
where i, j,andk are different. The lattice formed by the primitive transla-
719
tion vectors b
1
, b
2
, b
3
is called the reciprocal lattice (reciprocal to the lattice 720
formed by a
1
, a
2
, a
3
), and a reciprocal lattice vector is given by 721
G
h
1
h
2
h
3
= h
1
b
1
+ h
2
b
2
+ h
3
b
3
.
Simple crystal structures and principles of commonly used experimental
722
methods of wave diffraction are also reviewed briefly. Connection of Laue 723
equations and Bragg’s law is shown. Classification of crystalline solids are 724
then discussed according to configuration of valence electrons of the elements 725
forming the solid. 726
Uncorrected Proof
BookID 160928 ChapID 02 Proof# 1 - 29/07/09
2 1
Lattice Vibrations 2
2.1 Monatomic Linear Chain 3
So far, in our discussion of the crystalline nature of solids we have assumed 4
that the atoms sat at lattice sites. This is not actually the case; even at the 5
lowest temperatures the atoms perform small vibrations about their equilib- 6
rium positions. In this chapter we shall investigate the vibrations of the atoms 7
in solids. Many of the significant features of lattice vibrations can be under- 8
stood on the basis of a simple one-dimensional model, a monatomic linear 9
chain. For that reason we shall first study the linear chain in some detail. 10
We consider a linear chain composed of N identical atoms of mass M 11
(see Fig. 2.1). Let the positions of the atoms be denoted by the parameters 12
R
i
, i =1, 2,...,N. Here, we assume an infinite crystal of vanishing surface 13
to volume ratio, and apply periodic boundary conditions. That is, the chain 14
contains N atoms and the Nth atom is connected to the first atom so that 15
R
i+N
= R
i
. (2.1)
16
The atoms interact with one another (e.g., through electrostatic forces, core 17
repulsion, etc.). The potential energy of the array of atoms will obviously be 18
a function of the parameters R
i
, i.e., 19
U = U(R
1
,R
2
,...,R
N
). (2.2)
We shall assume that U has a minimum U
R
0
1
,R
0
2
,...,R
0
N
for some partic-
20
ular set of values
R
0
1
,R
0
2
,...,R
0
N
, corresponding to the equilibrium state of
21
the linear chain. Define u
i
= R
i
−R
0
i
to be the deviation of the ith atom from 22
its equilibrium position. Now expand U about its equilibrium value to obtain 23
U (R
1
,R
2
,...,R
N
)=U
R
0
1
,R
0
2
,...,R
0
N
+
i
∂U
∂R
i
0
u
i
+
1
2!
i,j
∂
2
U
∂R
i
∂R
j
0
u
i
u
j
+
1
3!
i,j,k
∂
3
U
∂R
i
∂R
j
∂R
k
0
u
i
u
j
u
k
+ ··· . (2.3)