24

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)

25

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

26

=

(

)

= √

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

27

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

28

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

29

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.

30

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

31

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

32

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)

33

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).

34

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,

35

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).

36

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.

37

(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.

38

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.

39

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

40

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

41

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

42

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

43

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.

**********