2- STRUCTURE OF HEXAGONAL CLOSED PACKED CRYSTALSshodhganga.inflibnet.ac.in/bitstream/10603/44200/7/07_chapter 2.pdf · 2.2 ATOMIC PACKING FRACTION:- In crystallography, atomic packing
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2- STRUCTURE OF HEXAGONAL CLOSED PACKED
CRYSTALS
2.1 INTRODUCTION:
In the present chapter we are confined to study the crystal
structure of H.C.P. metals (Be, Mg, Zn, Cd………), and represent the
detailed study of direct lattice vector, reciprocal lattice vector and
packing fraction of H.C.P. metals. Emphasis will be given to the study of
the first Brillouin zone of H.C.P. structure. The theoretical approach to
obtain the lattice contribution of thermal capacity of metallic crystal is
generally made by applying sampling techniques. For this purpose, the
phonon - frequencies have to be computed for a large number of point
within the first Brillouin zone. Here we have presented a brief
summary of the process to divide the 1st Brillouin zone of hexagonal
close packed structure into 1000 points, which later on reduce to 84
non-equivalent- representative points within the 1/24th part of the
zone if symmetry consideration is properly accounted for.
In a metal, the nuclei of the atoms are arranged in a periodical
spatial array, or crystal lattice, which may be thought of as constructed
from identical unit cells, each containing one or more atoms.
Conveniently position of the atom is taken to be at a corner of the cell.
Now, if the edges of the cells are specified by the vectors a1, a2, a3 any
atomic position, or lattice point may be obtained by a translation
vector Rn, given by
Rn = m1 a1 + m2 a2 + m3 a3 ……(2.1)
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Where m1, m2 and m3 are integers. A more complicated lattice
may be described as a Bravais lattice with a basis, that is to say, the unit
cell contains more than one atom and the basis gives the position of
similarly situated atoms within two unit cells differ by a translational
vector of the form (2.1).
The hexagonal close-packed structure is made by stacking close-
packed planes in a simple sequence. The lowest energy arrangement
for seven atoms in a plane is a hexagonal array in which six atoms
surround a central one. The atom in H.C.P. structure are arranged in
equidistant parallel planes, in each of which every atom has six
equidistant neighbors. A second plane of atoms similarly packed may
be placed over the first so as to fit as tightly as possible. If the centers of
the first layer are represented by points marked A, the second layer
may be represented by the points marked B, then each points B will be
at the apex of a regular tetrahedron of which three points marked A
form the base as shown in Fig. (2.1a). In the H.C.P. structure, let us
assume that the radius of each sphere be r, then the length of each side
of the tetrahedron will be 2r. Therefore, the standing height of the
tetrahedron is as 2r Sin 600 = √ r. The distance between the centroid
and midpoint of the any side of the basal triangle is
√ .r
√ .
Hence height of the tetrahedron is
√(√ )
√ =2(
)
If the spacing between first and third layer is represented by c, then
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=
(
)
= √
This is called the ideal axial ratio, but this ratio is not satisfied by
any of the H.C.P. metals. The actual ratio in H.C.P. metals deviates quite
appreciably from the ideal value of c/a=1.632. it is lowest for
beryllium (c/a=1.567) and highest for cadmium (c/a=1.885). The ratio
of c/a of some H.C.P. metals are given in table 2.1 as below,.
(Table- 2.1)
Metals a (A0) c (A0) c/a
Be 2.281 3.577 1.567
Mg 3.2028 5.1998 1.623
Zn 2.6648 4.9467 1.856
Cd 2.9788 5.6167 1.885
Co 2.5053 4.0886 1.632
Ho 3.577 5.616 1.57003
Sc 3.309 5.268 1.592
Tl 3.4566 5.5248 1.5983
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Consequently, six immediate neighbors of an atom in the basal
plane due to the deviation of axial ratio are either farther or nearer
than the other six ones, three above and three below the basal plane
depending upon the actual axial ratio being lower or higher than the
ideal value. Thus a sets of twelve nearest neighbors split into the two
sets of neighbors-nearest and next nearest, each containing six atoms.
Each atom of the hexagonal lattice has six nearest neighbors at a
distance (
)
, located on z=
planes. Six second neighbours
at a distance ‘a’ lying in the basal plane and six third neighbor at a
distance (
)
,lying on the z=
planes. This type of
distribution is found in these H.C.P. structures except in the case of zinc
and cadmium, which posses a high axial ratio. The neighbours, which
are first and second in the case of Be, interchange their labels in the
case of Zn and Cd and become the second and first respectively.
Metals having H.C.P. structure may be described as two
interpenetrating simple hexagonal lattice. There are two atoms per
unit cell, with one atom positioned at the origin and the other removed
a distance.
r =
a1 +
a2 +
a3 ……… (2.2)
Where a1, a2 and a3 are primitive lattice basis vectors. The
primitive translation vector a1, a2 and a3 for a hexagonal lattice are so
chosen that the angle between a1 and a2 is 1200 and I a1 I = I a2 I= a and
I a3 I=c, where a and c are lattice constants. The unit cell of H.C.P.
structure is a rhombohedron of height ‘c’ with a rhombus of side ‘a’ as
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its base. The orientation of the cartesian axes with the hexagonal axes
is shown in Fig. (2.1b). The x and z axes lie along the positive directions
of a1 and a3 respectively. The primitive translation vectors can,
therefore, may be expressed as
a1 =a 1 …………. (2.3a)
a2=
1 +√
2 …………. (2.3b)
a3= c 3 ………… (2.3c)
Where the axes 1, 2 and 3 are the unit vector along three
Cartesian axes.
2.2 ATOMIC PACKING FRACTION:-
In crystallography, atomic packing factor (APF) or packing
fraction is the fraction of volume in a crystal structure that is occupied
by atoms i.e., the ratio of the volume occupied by the atoms to the
volume of the unit cell of the structure. It is dimensionless and always
less than unity. For practical purposes, the APF of a crystal structure is
determined by assuming that atoms are rigid spheres. For one-
component crystals (those contain only one type of atom), the APF is
represented mathematically by
Natoms. Vatoms
APF= Vunitcell
Where Natoms is the number of atoms in the unit cell, Vatom is the
volume of an atom, and Vunitcell is the Volume occupied by the unit cell.
Thus from above, the side length of the hexagon
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a= 2r and
Height of the hexagon c= √
(4r)
It is then possible to calculate the APF as follows:
Natoms. Vatoms
APF = Vunit cell
= (
)
√
= (
)
√ (√
)
= (
)
(
√ ) (√
)
=
√ ~ 0.74
For multiple- structure, the APF can exceed 0.74
2.3 RECIPROCAL LATTICE AND BRILLOUIN ZONE:-
The concept of reciprocal lattice space has been widely used in
X-ray crystallography and in quantum theory of metals. A special
variant of the unit cell of the reciprocal lattice known as Brillouin-
zone, forms one of the important concepts in the theory of solid state,
particularly in the theory of electronic energy bands.
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We define a set of vectors b1, b2, b3 perpendicular to the
coordinate planes of the oblique axes a1, a2, a3 by means of the
relations.
ai . bj =2 π ij (i, j=1,2,3) …… (2.4)
Where ij the kronecker delta, define by
ij = 1 for i=j
= 0 for i≠j
The vectors bj are called the reciprocal vectors and the set of
points whose position vectors are given by.
Kn = n1 b1 + n2 b2 + n3 b3 …… (2.5)
Where n1, n2, n3 take all integral values including zero, are called
reciprocal lattice points. The explicit expressions for the reciprocal
vectors are,
b1 = (a2 × a3) ………. (2.6 a)
b2 =
(a3 × a1) ……... (2.6 b)
and b3 = (a1 × a2) ……….(2.6 c)
Where V= a1. a2 × a3, is the volume of the unit cell of the crystal
lattice. These vectors define reciprocal space of crystal. Since the
phonon vector q (=2πk ) has same dimension as the reciprocal lattice
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vectors, the ‘ q ’ space is identified with the reciprocal space. The
orientation of the primitive reciprocal vectors is shown in Fig. (2.2a).
The angle included between b1 and b2 is 600 and they are
oriented at an angle of 300 from a1 and a2 respectively. Vector b3 lies in
the direction of a3. Obviously these vectors also define a hexagonal
lattice. Thus the hexagonal lattice is its own reciprocal. In
crystallographic notation the direction b3 and b2 are defined
respectively as the [0001], [01 0] directions. Thus in the [0001]
direction Iq1 I= I q2 I =0 and I q3 I = I q I and for the [01 0] direction
Iq1 I = I q3 I =0 and I q2 I = I q I, where q1, q2, q3 are the components of
the phonon wave vector q along three rectangular axes.
In the case of the hexagonal crystal, we have following expressions
from equations (2.3) and (2.6)
b1 =
( 1 +
√ 2 ) …………(2.7a)
b2 = √
( 2 ) ………….(2.7b)
b3 =
( 3 ) …………..(2.7 c)
Thus the volume of the primitive cell of the reciprocal lattice is
b1 . b2 × b3 =
√ ………….(2.8)
The vector Kn in general with the help of equation (2.5) and (2.7) is
expressed by
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Kn =
n1 1 +
√ (n1+2n2) 2 +
n3 3 …… (2.9)
The shortest non-zero ‘Kn’ are the following eight vectors
√
2 ,
( 1 -
√ 2),
( 1 +
√
2 ) , 3
In terms of bj, these vector are
b1, ( b1 - b2), b2, b3 ……… (2.10)
The equation of the planes perpendicular to the vectors given by (2.9)
is
2 q . Kn + I Kn I2 = 0 ………….(2.11)
Where q is the phonon-wave vectors defined by q =
, and Kn is the
vector bisected by the plane.
In order to obtain first Brillouin-zone, we draw planes which are
perpendicular bisectors of the vector (non-zero shortest) defined by
equation (2.10) and the equation of these planes, obtaind from
equation(2.11) are.
I q1 I √
I q2 I =
…… (2.12 a)
I q2 I =
√ …… (2.12 b)
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I q3 I =
…… (2.12 c)
The boundary planes enclose a hexagonal prism of height
and base side
. Thus the first Brillouin-zone is a simple hexagonal
lattice as shown in Fig(2.3).
The following symbols have been used according to the method
developed by Raghavacharyulu1 in a study of the diamond lattice and
later on used by Iyengar et. al.2 in the theoretical study of H.C.P. metals.
(i) Г- stands for the point q=0
(ii) M- stands for the zone boundary along [0110] directions.
(iii) A- stands for the zone boundary along [0001] directions.
(iv) K- stands for the zone boundary along [1120] direction.
(v) Σ- Represents the symmetry direction [0110]
(vi) ∆- Represents the symmetry direction [0001]
(vii) T,T’- Represents the symmetry direction [1120]
Phonon dispersion curves are drawn along these symmetry
directions.( ∆, Σ,T and T’ directions)
2.4 CHOICE OF ALLOWED WAVE VECTORS:-
The allowed wave-vectors are determined by Born’s3 cyclic
condition according to which the first Brillouin zone is divided into
2n × 2n × 2n=(2n)3 miniature cells of the basic vectors
(j=1,2,3).
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A vector of this miniature cell is a possible wave vector q of the
permitted vibrations. The vector q may accordingly be written as,
q = n1
+n2
+ n3
With the help of the expressions (2.7) we get
q =
(
) 1 +
√
2 +
3 …(2.13)
Where n1, n2, n3 are integers which can assume values from –n to
+n. Phonon wave vector q in terms of its components along three
rectangular axes may be written as
q = I q1 I 1 + I q2 I 2 + I q3 I 3 ...... (2.14)
Where
I q1 I =
…….(2.15 a)
I q2 I =
√
…….(2.15 b)
I q3 I =
…….(2.15 c)
In practice, due to symmetry consideration we need to consider
only the 1/24th part of the first Brillouin-zone, which is further
irreducible under symmetry operations. The irreducible segment of the
Brillouin-zone is a triangular prism of height
. The base of this
segment, which is obtained by projecting the segment in the basal
plane, is shown by ABC in Fig. (2.2b). The sides of this base are,
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AB=
, BC=
, CA=
√
In the present work we have taken the irreducible segment of
the first Brillouin-Zone to lie in a positive octant such that CA coincides
with the y-axis, BC is equally inclined with b1 and b2. The points lying
in this segment are given by values of n1,n2,n3 which are all positive
subject to the condition,
n1 + 2n2 < 2n
n3 < n
n2 < n
n1 < n2
The first Brillouin-zone is divided into 1000 miniature cells4,5
taking 2n to be 10. This gives a total of 84 points including origin in the
irreducible segment. Each of the representative point is equivalent to
many other similar points. From symmetry considerations, however an
account is taken of all these paints to get the total number of points
exactly 1000. In order to get the correct number of similar points and
hence the statistical weight, one has to fill the whole space by
polyhendra of the first Brillouin-zone. The observed number of points
on the faces, the edges and the corners of the zone must be divided by
the number of polyhendra sharing these points. The procedure for the
hexagonal lattice is described below. The representative points in the
first Brillouin Zone, are given in table (2.2) and the components of
representative wave vectors within 1/24th part of first Brillouin Zone
are shown table (2.3).
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The combined non equivalent 84 representative points lying in
1/24th part of first Brillouin Zone are given in table (2.4) along with
their statistical weights. For the computation of specific heats of
metals, phonon frequencies have to be calculated at these points.
2.5 POINTS LYING IN THE INTERIOR OF THE ZONE:-
For the interior of the zone the number of similar points
corresponding to the possible set of n1, n2, and n3 are given below.
Possible set Number of Similar points
(n1 n2 n3) 24
(n2 n2 n3) 12
(n2 n2 0) 6
(0 n2 n3) 12
(0 0 n3) 2
(n1 n2 0) 12
(0 n2 0) 6
Other points on the faces, edges and corners of the Brillouin-zone
are, as follows:
(a) Points lying on the faces like C:
These points lie on the lateral-surface of the Brillouin zone and
satisfy the condition n1+2n2=2n. Each of these points is shared by
two polyhedra and should therefore be counted as one-half of the
point.
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(b)Points lying on the faces like B:
These points lie on the top and the bottom surfaces and satisfy
the condition n3=n. Such points are also shared by two polyhedra.
Each point has therefore to be counted again as one-half of a point.
(c)Points lying on the edges like NN’:
These points satisfy the condition n1 + 2n2 =2n and n3 = n.
These edges are shared by four polyhedra. Such points should be
treated as one – fourth of a point.
(d) Points lying on the edges like EE’:
These points satisfy the condition n1 = n2 =
. Such points
are shared by three polyhedra and hence each such point has to be
counted as one-third of a point.
(e)Points lying on the corners:
These points satisfy the condition n1 = n2 =
and n3 = n and
are shared by six polyhedra, therefore each paint has to be counted
as one-sixth of a point.
Since in the present work we have taken 2n to be 10. Under such
condition there are no points on the edges like EE’ or at the corners.
There are points on the surfaces of the zone, each of which has to be
counted as half of a paint only, and on the edges like NN’ at which a
point has to be counted as one-fourth.
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Table-2.2
Representative points in the Brillouin Zone of HCP structure-
S.No. n1 n2 n3
1 0 0 x ; x can take integer value from 1 to
5 including 0(zero)
2 1 1 0,1,2,3,4,5.
3 0 1 0,1,2,3,4,5.
4 2 2 0,1,2,3,4,5.
5 1 2 0,1,2,3,4,5.
6 0 2 0,1,2,3,4,5.
7 3 3 0,1,2,3,4,5.
8 2 3 0,1,2,3,4,5.
9 1 3 0,1,2,3,4,5.
10 0 3 0,1,2,3,4,5.
11 2 4 0,1,2,3,4,5.
12 1 4 0,1,2,3,4,5.
13 0 4 0,1,2,3,4,5.
14 0 5 0,1,2,3,4,5.
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Table 2.3
Components of representative wave-vector within 1/24th part of
first Brillouin Zone.
S.No. q1
(in
)
q2
(in
√ )
q3
(in
)
1 0 0
; x can take integer value
from 1 to 5 including 0(zero)
2 0.1 0.3 0,0.2,0.4,0.6,0.8,1.0
3 0 0.2 0,0.2,0.4,0.6,0.8,1.0
4 0.2 0.6 0,0.2,0.4,0.6,0.8,1.0
5 0.1 0.5 0,0.2,0.4,0.6,0.8,1.0
6 0 0.4 0,0.2,0.4,0.6,0.8,1.0
7 0.3 0.9 0,0.2,0.4,0.6,0.8,1.0
8 0.2 0.8 0,0.2,0.4,0.6,0.8,1.0
9 0.1 0.7 0,0.2,0.4,0.6,0.8,1.0
10 0 0.6 0,0.2,0.4,0.6,0.8,1.0
11 0.2 1.0 0,0.2,0.4,0.6,0.8,1.0
12 0.1 0.9 0,0.2,0.4,0.6,0.8,1.0
13 0 0.8 0,0.2,0.4,0.6,0.8,1.0
14 0 1.0 0,0.2,0.4,0.6,0.8,1.0
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Table 2.4
Representative Points of the Brillouin – Zone
S.
No.
Representative Points No. of Similar
Points
Statistical
weight
Remark
n1 n2 n3
1 0 5 5 3 3/1000 Edge
2 0 4 5 6 6/1000 Surface
3 1 4 5 12 12/1000 Surface
4 2 4 5 6 6/1000 Surface
5 0 3 5 6 6/1000 Surface
6 1 3 5 12 12/1000 Surface
7 2 3 5 12 12/1000 Surface
8 3 3 5 6 6/1000 Surface
9 0 2 5 6 6/1000 Surface
10 1 2 5 12 12/1000 Surface
11 2 1 5 6 6/1000 Surface
12 3 1 5 6 6/1000 Surface
13 1 0 5 6 6/1000 Surface
14 0 5 5 1 1/1000 Surface
15 0 4 4 6 6/1000 Surface
16 0 4 4 12 12/1000
17 1 4 4 24 24/1000
18 2 3 4 12 12/1000
19 0 3 4 12 12/1000
20 1 3 4 24 24/1000
21 0 3 4 24 24/1000
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22 3 3 4 12 12/1000
23 0 2 4 12 12/1000
24 1 2 4 24 24/1000
25 2 2 4 12 12/1000
26 0 1 4 12 12/1000
27 1 1 4 12 12/1000
28 0 0 4 2 2/1000
29 0 5 3 6 6/1000 Surface
30 0 4 3 12 12/1000
31 1 4 3 24 24/1000
32 2 4 3 12 12/1000
33 0 3 3 12 12/1000
34 1 3 3 24 24/1000
35 2 3 3 24 24/1000
36 3 3 3 12 12/1000
37 0 2 3 12 12/1000
38 1 2 3 24 24/1000
39 2 2 3 12 12/1000
40 0 1 3 12 12/1000
41 1 1 3 12 12/1000
42 0 0 3 2 2/1000
43 0 5 2 6 6/1000 Surface
44 0 4 2 12 12/1000
45 1 4 2 24 24/1000
46 2 4 2 12 12/1000
47 0 3 2 12 12/1000
48 1 3 2 24 24/1000
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49 2 3 2 24 24/1000
50 3 3 2 12 12/1000
51 0 2 2 12 12/1000
52 1 2 2 24 24/1000
53 2 2 2 12 12/1000
54 0 1 2 2 2/1000
55 1 1 2 12 12/1000
56 0 0 2 2 2/1000
57 0 5 1 6 6/1000 Surface
58 0 4 1 12 12/1000
59 1 4 1 24 24/1000
60 2 4 1 12 12/1000
61 0 3 1 12 12/1000
62 1 3 1 24 24/1000
63 2 3 1 24 24/1000
64 3 3 1 12 12/1000
65 0 2 1 12 12/1000
66 1 2 1 24 24/1000
67 2 2 1 12 12/1000
68 0 1 1 12 12/1000
69 1 1 1 12 12/1000
70 0 0 1 2 2/1000
71 0 5 0 3 3/1000 Surface
72 0 4 0 6 6/1000
73 1 4 0 12 12/1000
74 2 4 0 6 6/1000
75 0 3 0 6 6/1000
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76 1 3 0 12 12/1000
77 2 3 0 12 12/1000
78 3 2 0 6 6/1000
79 0 2 0 6 6/1000
80 1 2 0 12 12/1000
81 2 2 0 6 6/1000
82 0 1 0 6 6/1000
83 1 1 0 6 6/1000
84 0 0 0 1 1/1000
44
REFERENCES
1- Raghavacharayulu, I.V.V., Chand.J.Phys. 39, 1704 (1961).
2- Iyenger, P.K., Venkataraman, G., Vijayaraghavan, P.R. and Roy, A.P.,
Proc. At the symposium on inelastic scattering of neutrons, Mumbai
(India), Vol.1 .152(1964).
3- Born, M., “Atom. Theories des festen zustandes” J.B. Tenbner (1923).
4- Maurya J.R., Ph.D. Thesis, BHU, India (1975)
5- Maurya, J.R.; Singh, S.P. and Kushwaha, S.S. J.Sci. R. BHU, India Vol.
XXVIII (1), 1978.
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