Crystal Structure Notes 2r2i11

UNIT IV

SOLID STATE PHYSICS

CHAPTER

CRYSTAL STRUCTURE 1.1 INTRODUCTION When two atoms are brought together, two kinds of forces: attraction and repulsion come into play. The force of attraction increases rapidly as the separation between atoms, decreases. At small distances the force of repulsion comes into being and increases more rapidly with further decrease in interatomic distance. At distance r = r0, attractive force becomes equal to the repulsive force and at this distance the interaction energy acquires minimum value and the atoms form a bound state. The minimum energy corresponds to the equilibrium-state. When a large number of atoms are brought together, the atoms in equilibrium-state, arrange themselves to form a closed pack structure, called solid. Thus, the forces of interaction between the structural particles (atoms, ions, group of atoms) are responsible for the existence of solids. The force of attraction between the particles prevents them from flying apart and the force of repulsion prevents them from merging.

Fig. 1.1.1 Variation of inter-atomic force and potential energy with distance

Solids occur in two forms: Crystalline and amorphous. A crystalline solid is one in which atoms or other structural units are arranged in an orderly repetitive array. That is, a crystalline solid exhibits long-range orders.

Amorphous (non-crystalline) solids do not show long range periodicity. An amorphous substance, such as glass, is actually regarded as a super-cooled liquid that differs from ordinary liquid in its physical properties. Usually viscosity is used to differentiate a liquid from amorphous solid. If the coefficient of viscosity is less than 1015 poise, it is called liquid, if it is greater than this, it is a solid. The transition from liquid to amorphous-state and vice-versa is gradual whereas the transition between a liquid and crystalline solid is sharp.

1.2

SPACE LATTICE OR CRYSTAL LATTICE

Space lattice is a geometrical abstraction. To have its meaning, imagine a network of straight lines constructed in such a way that it divides the entire space into identical volumes. The points of intersection of these lines are called lattice points. The network of points in three dimensions, in which the surrounding of each point is identical with the surrounding of other points, is called space lattice. By associating each lattice point with a single atom or group of atoms, called basis, crystal structure results. Thus, for every crystal there is a network of lattice points, which are occupied by either a single atom or group of atoms.

Fig. 1.2.1 Space lattice and fundamental vectors a, b and c

In 1848, a Russian mathematician Bravais showed that there are just 14 ways of arranging points in space lattice such that all the lattice points have exactly the same surroundings. Such lattices are called Bravais lattices. From these 14 space lattices, an unlimited number of different crystal structures can be made.

Fig. 1.2.2 Two dimensional space lattice spanned by basis vectors a and b

For any type of lattice, there exists three fundamental translation vectors a, b, and c, not lying in a plane, in of which any lattice point can be specified. A translation operation T is defined by ...(1.2.1) T = n1 a + n2 b + n3 c where n1, n2, n3 are integers. Thus means that by applying the operation T to a point r we reach another point r', which has the same environment as point r. r' = r + T = r + n1a + n2b + n3c

...(1.2.2)

The set of points r' given by Eqn. (2) for all possible values the integers n1, n2, and n3 defines the space lattice.

1.3

UNIT CELL

A parallelepiped formed by fundamental vectors a, b and c of space lattice is called unit cell. Each crystal is built up of a repetitive stacking of unit cells each identical in size, shape and orientation. The unit cell with basis is thus the fundamental building block of the crystal. A unit cell is always drawn with lattice point at each corner, but there may also be lattice points at the center of some of the faces or at the body center of the cell. Although there are infinite number of ways of choosing the unit cell, but the simplest way is conventionally accepted.

Primitive Cell A unit cell with lattice points at its corners only is called primitive cell. Thus, there is only one lattice point to each primitive cell. It is also defined as the smallest unit cell in volume of space lattice. Basis vectors: A set of linearly independent vectors a, b, c, which can be used to define a unit cell, are called basis vectors. Primitive basis vectors: A set of linearly independent vectors that define a primitive cell is called primitive basis vectors.

1.4

PARAMETER OF A UNIT CELL

A unit cell is specified by six quantities: three edges a, b, c and three angles a, b, g defined in the figure. These quantities are termed the parameters of the unit cell. In crystallography, the distances are expressed in of a, b, c. The 14 space lattices with conventional unit cell, some by their primitive cells and others by non-primitive cells are shown in the Fig. (1.4.2). The 14 lattices may be grouped into 7 crystal systems, each of which has in common certain characteristic symmetry elements. Fig. 1.4.1 Parameters of a unit cell

Fig. 1.4.2 14 Bravais space lattices with conventional unit cell

Table 1.4.1: Seven Crystal Systems

System

Bravais lattice

Symbol

Unit cell characteristics

1 Triclinic

Simple

P

a ¹ b ¹ c, a ¹ b ¹ g ¹ 90°

2 Monoclinic

Simple Base centered

P C

a ¹ b ¹ c, a = b = 900 ¹ g

3 Orthorhombic

Simple Base centered Body centered Face centered

P C I F

a ¹ b ¹ c, a = b = g = 90°

4 Tetragonal

Simple Body centered

P I

a = b ¹ c, a = b = g = 90°

5 Cubic

Simple Body centered Face centered

P I F

a = b = c, a = b = g = 90°

6 Hexagonal

Simple

P

a = b ¹ c, a = 120°, b = g = 90°

7 Trigonal (Rhombohedral)

Simple

P

a = b = c, a = b = g ¹ 90°

Fig. 1.4.3 Primitive (rhombohedral) and non-primitive (cubic) unit cells for f.c.c. lattice

The crystallographic description of crystal requires specification of shape and size of unit cell and distribution of matter in it. The choice of unit cell is not unique. While choosing the unit cell, emphasis is given on the symmetry exhibited by it. If a non-primitive cell exhibits more symmetry

than a primitive cell, the former is preferred. For example, in F.C.C. lattice the simplest primitive cell is rhombohedral with three fundamental translation vectors, each of length L/Ö2, which make angles of 60° with respect to each other Fig. (1.4.3). A non-primitive cubic unit cell of length L exhibits strong symmetry and therefore, it preferred to primitive cell. The rhombohedral unit cell has volume (L3)/4 whereas the cubic unit cell has volume L3.

Fig. 1.4.4 Primitive and non-primitive (conventional) unit cell for b.c.c. lattice

Similarly for B.C.C. lattice, because of strong symmetry shown by cubic unit cell, it is conventionally accepted in comparison to primitive rhombohedral unit cell, which can be constructed with primitive vectors of length ( LÖ3)/2, The primitive cell has volume (L3)/2 and non-primitive cubic unit cell has volume L3.

1.5

SYMMETRY

An object is said to have symmetry if it coincides with itself as a result of some kinds of its spatial movement or rotation through some angle about an axis. The larger the number of ways by which the object can be made to coincide with itself, the more symmetric it is said to be. By symmetry element we mean an operation by which equivalent points of the object or space lattice are brought into coincidence. All the possible symmetry operations on an object can be expressed as a linear combination of following four symmetry elements. (i) n-fold rotation axis (Symmetry axes): If a rotation of a lattice about an axis through an angle 2p/n brings a lattice into a position indistinguishable from its initial position, then the axis is a n-fold rotation symmetry axis. The possible values of n are 1, 2, 3, 4 and 6. Five-fold rotational symmetry in a crystal lattice is impossible. Obviously any axis of a lattice is a one-fold symmetry axis.

(ii) Plane of symmetry: This type of symmetry exists if one half of the lattice coincides with itself as a result of the mirror reflection of its points in a certain plane. This plane is called the symmetry plane of the lattice. (iii) Inversion center (center of symmetry): A crystal lattice is said to possess the center of symmetry if it coincides with itself upon inversion through a certain point. This point is called the center of symmetry. By inversion operation the position vector r of a lattice point is converted into – r. (iv) Rotation-inversion axis: If a lattice remains unchanged upon rotation about an axis through angle 2p/n, n = 1, 2, 3, 4 and 6, followed by inversion about a lattice point through which the rotation axis es, it is said to have rotation-inversion axis. This kind of symmetry is indicated in the figure. The point A takes the position A' upon rotation through angle p. Upon inversion about point O, the point A' goes to A". Thus, A and A'' are related by two-fold rotation-inversion axis.

Fig. 1.5.1 Two-fold rotation-inversion axis

1.6

MILLER INDICES

A plane in a crystal is specified in of Miller indices. To find the Miller indices of a plane, following procedure is used. (i) Determine the intercepts of the plane on crystal axes in of fundamental vectors a, b and c. (ii) Take the reciprocal of these numbers in order. Reduce them to three smallest integers having the same ratio. (iii) Enclose these numbers in parentheses as (h k l). The Miller indices (h k l) denote a set of parallel planes. For example consider a plane with intercepts 4a, 6b and 3c on the crystal axes. In of 1 1 1 axial units, the intercepts are 4, 6, 3. The reciprocals of these numbers are , , . By multiplying 4 6 3

each of these numbers with lowest common multiple 12, these are reduced to smallest integers 3, 2, 4. The Miller indices of the plane are (324).

Fig. 1.6.1 Miller indices of a plane cutting intercepts 4a, 6b, 3c on crystallographic axes

If a plane intersects the crystal axes on the negative side of origin, the corresponding index is negative. By convention a minus sign is placed above the corresponding index. A plane intercepting the crystal axes at – 4a, 2b, – c, has Miller indices (12 4). Miller indices of some important planes of cubic lattices are shown in the Fig. (1.6.2).

Fig. 1.6.2 Miller indices of some important planes of cubic lattices

1.7

INTERPLANAR DISTANCE

To find the interplanar spacing for the set of planes with Miller indices (hkl), take the origin on any one plane of this set and erect axes in the directions of crystal axes a, b and c. The interplanar spacing d for the planes of indices (hkl) is equal to the distance from origin to the nearest plane of the set under question. The distance d is measured along the normal drawn from origin to the plane (hkl). Let the normal to the plane make angles d1, d2 and d3 with a-axis, b-axis and c-axis respectively. The plane (hkl) intersects a-axis at a/h, b-axis at b/k and c-axis at c/l.

Fig. 1.7.1 Interplanar spacing for set of planes (hkl )

From Fig. (1.7.1) we see that cos δ1 =

d d d ,cos δ2 = ,cos δ3 = a/ h b/ k c/l

For orthogonal axes, cos2 d1 + cos2 d2 + cos2 d3 = 1 Hence,

 h2 k 2 l 2  d2  2 + 2 + 2  = 1 b c   a

d=

1 h2 a2

+

k2 b2

+

l2

...(1.7.1)

c2

For cubic system a = b = c

d=

a h2 + k 2 + l 2

...(1.7.2)

Direction in a Crystal The direction in a crystal is expressed as a set of integers which has the same ratio as the components (in of axial units a, b, c) of the vector in that direction. For example, the direction of vector ha + kb + lc is expressed as [hkl]. In a cubic system the direction [hkl] is perpendicular to the plane (hkl) but this is not true for other systems.

Position of a Point in Unit Cell The position of a point in unit cell is specified in of its coordinates, which are expressed in of unit cell edges. For example, [[324]] is reached by moving along a-axis a distance three times the length of vector a, then parallel to b-axis a distance twice the length of vector b and finally parallel to c-axis a distance four times the length of vector c. The indices of a point are written inside a double square bracket. The indices of a point with coordinates x = –3a, y = – 4b, z = 2c are written as [[ 342 ]].

1.8

SOME CHARACTERISTICS OF CUBIC SYSTEM

There are three possible cubic lattice: simple cubic (s.c.), body centered cubic (b.c.c.) and face centered cubic (f.c.c). Simple cubic structure: The cubic unit cell for simple cubic lattice is shown in the figure. There are 8 atoms at the 8 corners of the cell, each corner atom is shared by 8 unit cells that ad at each corner. The share of each corner atom to a unit cell is 1/8 of an atom. Total number of atoms

1 = 1 . Thus, simple cubic cell is a primitive cell. Since 6 equidistant nearest 8 neighbours surround the atom at one corner, the coordination number is 6. In a close packing with hard spheres, the hard sphere radius r is given by a = 2r. where a is length of side of cubic cell. The distance 2r is called the distance of nearest neighbour. The volume of unit cell V = a3

per unit cell is 8 ×

4 πa3 Volume of all atoms in a unit cell W = 1 × πr 3 = 3 6 Packing fraction f =

Ω V

=

π 6

= 0.52

Simple cubic structure is a loosely packed structure. The only one element – polonium forms simple cubic structure for a particular range of temperature.

Fig. 1.8.1 Simple cubic unit cell

Body centered cubic structure: The conventional unit cell for b.c.c. structure is shown in the figure. There are 8 atoms at the 8 corners, each shared by 8 unit cells and one atom at the body center, which wholly belongs to the unit cell. Thus, the total number of atoms per unit cell is 1 8 × + 1 = 2 . The b.c.c. is not a primitive cell. The primitive cell is a rhombohedral in shape as 8 shown in the Fig. (1.4.4). The primitive cell contains only one atom. In close packing, the hard sphere radius is given by

(4r)2 = a2 + ( 2 a)2 \

r=

3a 4

Fig. 1.8.2 Calculation of hard sphere radius r in of a

Volume of unit cell V = a3

4 Volume of all atoms in unit cell W = 2 × πr 3 3 4  3  a = 2 × π 3  4  Packing fraction f =

Ω V

=

3 8

3

π = 0.68

Lithium, sodium, potassium, chromium crystallize in b.c.c. structures. Face centered cubic structure: The conventional unit cell for face centered structure is shown in the figure. There are 8 atoms, each shared by 8 unit cells at 8 corners and 6 face-centered atoms, 1 1 each shared by 2 cells. The total number of atoms per unit cell is 8 × + 6 × = 4 . Thus, the 8 2 conventional unit cell is not a primitive cell. In close packing, the hard sphere radius is given by (4r)2 = a2 + a2

r=

\

2a 4

Fig. 1.8.3 Conventional unit cell for f.c.c. structure

Volume of cubic unit cell V = a3

4 2πa3 Volume of all atoms of unit cell Ω = 4 × πr 3 = 3 6 Packing fraction f =

2π Ω = = 0.74 V 6

F.C.C. is closest packed structure. Ag, Al, Au, Cu, Pb, Pt crystallize in f.c.c. structure.

1.9

HEXAGONAL CLOSE PACKED (H.C.P.) STRUCTURE

The conventional unit cell for hexagonal packed structure is shown in Fig. (1.9.1). This unit cell is equivalent to three primitive cells. If we consider an atom at the center of basal plane, we see that it has 6 nearest neighbours in its own plane, 3 in a plane above and 3 in a plane below it. The coordination number is 12. The number of atoms per unit cell is 6.

a

OB = a cos 30°, OA = (2/3) OB = 2

2

OC = OA + AC 2

3 2

 a  c a′2 =   +   3 2

2

Fig. 1.9.1 Unit cell for hexagonal packed structure

In an ideal close packing all the 12 neighbours are at the same distance i.e., a = a' 2

\

 a  c a2 =   +   3 2

Hence

c 8 = = 1.633 a 3

2

In close packing, hard sphere radius is given by 2r = a

The packing fraction is given by 3 2π Ω 6 × 43 πr = = = 0.74 2 V  3 3a c  6    2    Be, Mg, Zn, and Cd are some metals, which crystallize in hexagonal closed packed structure.

f =

1.10

SOME CRYSTAL STRUCTURES

Sodium chloride (rock salt) NaCl structure: The sodium chloride structure results by arranging Na+ and Cl– alternately at the lattice points of a cubic lattice. The crystal structure can be thought of as superposition of two f.c.c. sub-lattices, the lattice points of one sub-lattice are occupied by Na+ and those of other are occupied by Cl–. The two f.c.c. sub-lattices are displaced from each other by one-half of the body diagonal of the unit cube. The positions of Cl– and Na+ are: Cl– Na+

000 ½½½

½½0 00½

½0½ 0½0

0½½ ½00

Fig. 1.10.1 Conventional unit cell of sodium chloride structure

Six neighbours of opposite kinds surround each ion. The coordination number is 6. The number 1 1 of Na+ per unit cell is 8 × + 6 × = 4 . Similarly the number of Cl– per unit cell is also 4. Thus 8 2 there are four sodium chloride molecules in a unit cell. Cesium chloride structure: In cesium chloride crystal, the Cs+ lie at the corners of a cubic cell and Cl– at the body centered position. The CsCl crystal may be thought of as superposition of two simple cubic lattices, the corner of one sub-lattice lies at the body center of the other. Cesium ions lie at the lattice points of one sub-lattice and chloride ions at those of the other sub-lattice. Each ion is at the center of the cube whose corners are occupied by ions of opposite kind. The coordination

number is 8. The number of Cs+ per unit cell = 8 ×

1

= 1 . The number of Cl– per unit cell is also 1. 8 So there is one cesium chloride molecule per unit cell.

Fig. 1.10.2 Conventional unit cell of cesium chloride

Summary of properties of some crystal structures Property

SC

BCC

FCC

H

Volume of unit cell

a3

a3

a3

3Ö3a2c/2

No. of atoms per unit cell

1

2

4

6

Coordination number

6

8

12

12

Nearest neighbour distance

a

aÖ3/2

aÖ2/2

a

a/2

aÖ3/4

aÖ2/4

a/2

p/6 = 0.52

Ö3p/8 = 0.68

Ö2p/6 = 0.74

Ö2p/6 = 0.74

(2r) Atomic radius (r) Atomic packing fraction

SOLVED EXAMPLES Ex. 1. Show that the lattice constant for cubic cell is given by 1/ 3

 nM  a=   ρNA 

where n = number of molecules per unit cell, M = molar mass, r = density, NA = Avogadro’s number. Sol. Let a = length of side of cubic cell.

Volume of unit cell v = a3 M

Molar volume of crystal VM =

ρ

This volume contains NA molecules. Volume available to a single molecule vm =

M ρ NA

If there are n molecules in a unit cell, then the volume of unit cell is

v=

nM ρ NA

3

This is equal to a . Hence 1/ 3

 nM  a=   ρNA 

Ex. 2. Show that the number of lattice points per unit area of a lattice plane with interplanar spacing d is given by

ρs =

nd Ω

where n = number of lattice point per unit cell and W is volume of unit cell.

For crystal with orthogonal axes, Ω = abc , the above formula reduces to

ρs =

nd abc

Sol. Consider a crystal region of volume V containing N lattice planes, each of area S. Let rs represent the number of lattice points per unit area of lattice plane. The number of lattice points in the volume V is ...(1) N' = rs S N This number may also be calculated as follows. Let W = volume of unit cell n = number of lattice points per unit cell. V . The number of unit cells in volume V is Ω The number of lattice points in volume V is N' =

Thus

Whence

ρs SN = ρs =

V Ω

V Ω

n

n

nV ΩSN

...(2)

If d is interplanar spacing of the lattice planes then V=SdN Making use of (3) in (2), we get nd

ρs =

...(3)

...(4)

Ω

For crystals with orthogonal axes (cubic, tetragonal, orthorhombic) W = abc. We have

ρs =

nd

...(5)

abc

In cubic crystal W = a3 we have

ρs =

nd

...(6)

a3

Ex. 3. The rock salt (NaCl) has fcc structure and contains 4 molecules per unit cell. Calculate the lattice constant for the crystal. Molecular weight of NaCl = 58.45 kg/kmol Density = 2180 kg/m3 Avogadro Number NA = 6.02 ´ 1026 kmol –1 1/ 2

 nM  a=   ρNA 

Sol.

4 × 58.45   = = 5.63 Å 26   2180 × 6.02 × 10 

Ex. 4. Find the ratio of intercepts on the crystal axes by plane (231) in a simple cubic lattice. Sol. Let x1, x2, x3 be the intercepts by the plane on the axes. In of axial units the intercepts are

x1 a

x2

= m,

= n,

b

x3 c

=p

where m, n, p are numbers. Now

\

1 m

:

1 n

:

1 p

= 2:3:1

m : n : p=

1 2

:

1 3

:

1 1

=3:2:6 x1 : x2 : x3 = 3 : 2 : 6 Ex. 5. Find the ratio of interplanar distances of planes (100), (110) and (111) in a simple cubic lattice. Sol.

dhkl =

a h2 + k 2 + l 2

a

d100 =

1+ 0 + 0 a

d110 =

a 1 2

2 a

=

1+ 1+ 1

d100 : d110 : d111 = 1:

a

=

1+1+ 0

d111 =

=a

3

1

:

3

Ex. 6. In a tetragonal lattice a = b = 2.5 Å, c = 1.8 Å. Calculate the lattice spacing between the (111) planes. Sol.

1

dhkl =

h2 a2

k2

+

+

b2

l2 c2

1

d111 =

1

+

2.52

1 2.52

+

1

= 1.26 Å

1.82

Ex. 7. Calculate the density of lattice points in the lattice plane (100) in a simple cubic lattice of lattice constant a = 2.5 Å.

ρs =

Sol. Density of lattice points

nd nd = abc a3

For simple cubic lattice n = 1, d100 = a = 2.5 × 10–10 m \

ρs =

a a

3

=

1 2

a

=

1 (2.5 × 10−10 m)2

= 1.6 × 1019 m − 2

Ex. 8. Zinc has h structure. The height of unit cell is 4.9 Å, the nearest neighbour distance is 2.7 Å. Calculate the volume of the unit cell. Sol. Lattice constant a = 2r = 2.7 Å Volume of unit cell

V=

3 3a2 c 2

=

= 9.4 × 10

3 3(2.7 × 10 −10 m)2 (4.9 × 10−10 m)

– 29

2 3

m

Ex. 9. Calculate the volume of unit cell for a hexagonal close packed structure in which the atom occupying the lattice points has radius of 1.6 Å. Sol. Volume of the unit cell

V=

3 3a2 c , 2

where a = 2r = 3.2 Å

8 a = 5.2 Å 3

and

c=

\

V = 1.4 × 10–28 m3

QUESTIONS

AND

PROBLEMS

1. Distinguish between amorphous and crystalline solids. Define the following: (a) Space lattice (b) Unit cell (c) Primitive cell (d) Basis 2. Describe the seven system of crystals with examples. Give diagrams of simple cubic, body centered and face centered cubic cells. Calculate the number of atoms per unit conventional unit cell. 3. What are Miller indices? How are they specified? 4. Derive an expression for inter-planar spacing for orthogonal systems. 5. Obtain an expression for the number of lattice points per unit area of a lattice plane in a cubic crystal. 6. Obtain an expression for the lattice constant of a cubic crystal in of molar mass and density of the crystal. 7. With diagram describe the structure of sodium chloride and cesium chloride structure. 8. Calculate the number of atoms per unit area in the plane (100), (110), (111) in simple cubic structure of lattice constant 2.5 Å. 9. Copper has fcc structure. Radius of copper atom is 1.28 Å. Calculate the inter-planar distance for plane (111). [Hint: a = 4r/Ö2] 10. Magnesium has h structure. Atomic radius of Mg is 1.6 Å. Calculate the volume of unit cell.