crystal structure notes

8/13/2019 Crystal Structure Notes

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SOLID STATE PHYSICS

UNITIV


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CHAPTER

CRYSTALSTRUCTURE

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 distancesthe force of repulsion comes into being and increases more rapidly with further decrease in inter-atomic distance. At distance r= r0, attractive force becomes equal to the repulsive force and at thisdistance the interaction energy acquires minimum value and the atoms form a bound state. Theminimum energy corresponds to the equilibrium-state. When a large number of atoms are broughttogether, the atoms in equilibrium-state, arrange themselves to form a closed pack structure, calledsolid. 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. 1.1.1Variation of inter-atomic force and potential energy with distance

Solids occur in two forms: Crystallineand amorphous. A crystalline solid is one in which atomsor other structural units are arranged in an orderly repetitive array. That is, a crystalline solid exhibitslong-range orders.


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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 itsphysical properties. Usually viscosity is used to differentiate a liquid from amorphous solid. If thecoefficient of viscosity is less than 1015 poise, it is called liquid, if it is greater than this, it is asolid. The transition from liquid to amorphous-state and vice-versa is gradual whereas the transitionbetween 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 linesconstructed 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 whichthe 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 structureresults. Thus, for every crystal there is a network of lattice points, which are occupied by either asingle atom or group of atoms.

Fig. 1.2.1 Space lattice and fundamental vectors a, bandc

In 1848, a Russian mathematicianBravaisshowed that there are just 14 ways of arranging pointsin space lattice such that all the lattice points have exactly the same surroundings. Such lattices arecalledBravais lattices. From these 14 space lattices, an unlimited number of different crystal structurescan be made.

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


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For any type of lattice, there exists three fundamental translation vectors a, b, andc, not lyingin a plane, in terms of which any lattice point can be specified. A translation operation T is definedby

T = n1a+ n2b+ n3c ...(1.2.1)

where n1, n2, n3are integers. Thus means that by applying the operation T to a pointr we reachanother pointr', which has the same environment as pointr.

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 n3definesthe space lattice.

1.3 UNIT CELL

A parallelepiped formed by fundamental vectors a, band cof space lattice is called unit cell. Eachcrystal 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 alwaysdrawn with lattice point at each corner, but there may also be lattice points at the center of some ofthe faces or at the body center of the cell. Although there are infinite number of ways of choosingthe 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 latticepoint 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 unitcell, are called basis vectors.

Primitive basis vectors: A set of linearly independent vectors that define a primitive cell iscalledprimitive 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 defined in the figure. Thesequantities are termed the parameters of the unit cell. Incrystallography, the distances are expressed in terms ofa, b, c.

The 14 space lattices with conventional unit cell,some by their primitive cells and others by non-primitivecells are shown in the Fig. (1.4.2). The 14 lattices maybe grouped into 7 crystal systems, each of which has incommon certain characteristic symmetry elements.

Fig. 1.4.1Parameters of a unit cell


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Fig. 1.4.2 14 Bravais space lattices with conventional unit cell


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Table 1.4.1: Seven Crystal Systems

System Bravais lattice Symbol Unit cell characteristics

1 Triclinic Simple P a b c, 90

2 Monoclinic Simple P a b c, = 900Base centered C

3 Orthorhombic Simple P a b c, = = 90Base centered CBody centered IFace centered F

4 Tetragonal Simple P a = b c, = = 90Body centered I

5 Cubic Simple P a = b = c, = = 90Body centered IFace centered F

6 Hexagonal Simple P a = b c, = 120, = = 90

7 Trigonal Simple P a = b = c, = = 90 (Rhombohedral)

Fig. 1.4.3Primitive (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 celland 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


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than a primitive cell, the former is preferred. For example, in F.C.C. lattice the simplest primitivecell is rhombohedral with three fundamental translation vectors, each of length L/2, which makeangles of 60 with respect to each other Fig. (1.4.3). A non-primitive cubic unit cell of length Lexhibits strong symmetry and therefore, it preferred to primitive cell. The rhombohedral unit cellhas 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 isconventionally accepted in comparison to primitive rhombohedral unit cell, which can be constructedwith primitive vectors of length ( L3)/2, The primitive cell has volume (L3)/2 and non-primitivecubic 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 spatialmovement or rotation through some angle about an axis. The larger the number of ways by whichthe 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 spacelattice are brought into coincidence. All the possible symmetry operations on an object can be expressedas 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 anangle 2/n brings a lattice into a position indistinguishable from its initial position, thenthe 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 alattice is a one-fold symmetry axis.


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(ii) Plane of symmetry: This type of symmetry exists if one half of the lattice coincides withitself as a result of the mirror reflection of its points in a certain plane. This plane iscalled the symmetry plane of the lattice.

(iii) Inversion center (center of symmetry): A crystal lattice is said to possess the center ofsymmetry if it coincides with itself upon inversion through a certain point. This point iscalled the center of symmetry. By inversion operation the position vector r of a latticepoint is converted into r.

(iv) Rotation-inversion axis:If a lattice remains unchanged upon rotation about an axis throughangle 2/n, n= 1, 2, 3, 4 and 6, followed by inversion about a lattice point through whichthe rotation axis passes, it is said to have rotation-inversion axis. This kind of symmetryis indicated in the figure.

The point A takes the position A' upon rotation through angle . Upon inversion about pointO, 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 terms 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 terms of fundamental vectors a,b

andc.(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 terms of

axial units, the intercepts are 4, 6, 3. The reciprocals of these numbers are1 1 1

, ,4 6 3

. By multiplying


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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, 3con crystallographic axes

If a plane intersects the crystal axes on the negative side of origin, the corresponding index isnegative. 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


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1.7 INTERPLANAR DISTANCE

To find the interplanar spacing for the set of planes with Miller indices (hkl), take the origin on anyone plane of this set and erect axes in the directions of crystal axes a,bandc.The interplanar spacingd for the planes of indices (hkl) is equal to the distance from origin to the nearest plane of the setunder question. The distance d is measured along the normal drawn from origin to the plane (hkl).Let the normal to the plane make angles 1, 2 and 3 with a-axis, b-axis and c-axis respectively.The plane (hkl) intersects a-axis at a/h, b-axis at b/kand c-axis at c/l.

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

From Fig. (1.7.1) we see that = = =1 2 3cos ,cos ,cos/ / /d d d

a h b k c l

For orthogonal axes,cos2 1+ cos

2 2+ cos2 3 = 1

Hence,

+ + =

2 2 22

2 2 21

h k ld

a b c


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=

+ +2 2 2

2 2 2

1d

h k l

a b c...(1.7.1)

For cubic system a = b = c

=+ +2 2 2

ad

h k l...(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 terms of axial units a, b, c) of the vector in that direction. For example, the direction of vectorha + 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 terms of its coordinates, which are expressed interms of unit cell edges. For example, [[324]] is reached by moving along a-axis a distance threetimes the length of vector a, then parallel to b-axis a distance twice the length of vector band finallyparallel to c-axis a distance four times the length of vector c. The indices of a point are writteninside a double square bracket. The indices of a point with coordinates x= 3a,y= 4b, z= 2care

written as [[ 342 ]].

1.8 SOME CHARACTERISTICS OF CUBIC SYSTEM

There are three possible cubic lattice: simple cubic(s.c.), body centeredcubic(b.c.c.) andface centeredcubic (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 adjoinat each corner. The share of each corner atom to a unit cell is 1/8 of an atom. Total number of atoms

per unit cell is =1

8 18

. Thus, simple cubic cell is a primitive cell. Since 6 equidistant nearest

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 2ris called the distance of nearest neighbour.The volume of unit cell V = a3

Volume of all atoms in a unit cell =

=3

3413 6

ar

Packing fractionf=

= = 0.52V 6


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Simple cubic structure is a loosely packed structure. The only one element polonium formssimple 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 thefigure. There are 8 atoms at the 8 corners, each shared by 8 unit cells and one atom at the bodycenter, which wholly belongs to the unit cell. Thus, the total number of atoms per unit cell is

+ =1

8 1 28

. The b.c.c. is not a primitive cell. The primitive cell is a rhombohedral in shape as

shown in the Fig. (1.4.4). The primitive cell contains only one atom.In close packing, the hard sphere radius is given by

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

=3

4

ar

Fig. 1.8.2Calculation of hard sphere radius rin terms of a


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Volume of unit cell V = a3

Volume of all atoms in unit cell = 34

23

r

=

34 3

23 4

a

Packing fractionf=

= =3

0.68V 8

Lithium, sodium, potassium, chromium crystallize in b.c.c. structures.

Face centered cubic structure: The conventional unit cell for face centered structure is shownin the figure. There are 8 atoms, each shared by 8 unit cells at 8 corners and 6 face-centered atoms,

each shared by 2 cells. The total number of atoms per unit cell is + =1 1

8 6 48 2

. Thus, the

conventional unit cell is not a primitive cell.In close packing, the hard sphere radius is given by

(4r)2= a2+ a2

=2

4

ar

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

Volume of cubic unit cell V= a3

Volume of all atoms of unit cell

= =3

34 243 6

ar

Packing fraction

= = =2

0.74V 6

f

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


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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 isequivalent to three primitive cells. If we consider an atom at the center of basal plane, we see that ithas 6 nearest neighbours in its own plane, 3 in a plane above and 3 in a plane below it. Thecoordination number is 12. The number of atoms per unit cell is 6.

OB = acos 30, OA = (2/3) OB =3

a

OC2= OA2 + AC2

= +

2 22

23

a c

a

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 2

223

a ca

Hence = =8

1.6333

c

a

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


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The packing fraction is given by

= = = =

343

2

6 20.74

V 63 3

2

rf

a c

Be, Mg, Zn, and Cd are some metals, which crystallize in hexagonal closed packed structure.

1.10 SOME CRYSTAL STRUCTURES

Sodium chloride (rock salt) NaCl structure: The sodium chloride structure results by arrangingNa+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 byone-half of the body diagonal of the unit cube. The positions of Cland Na+are:

Cl 0 0 0 0 0 0 Na+ 0 0 0 0 0 0

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

of Na+ per unit cell is + =1 1

8 6 48 2

. Similarly the number of Cl per unit cell is also 4. Thus

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 oftwo simple cubic lattices, the corner of one sub-lattice lies at the body center of the other. Cesiumions lie at the lattice points of one sub-lattice and chloride ions at those of the other sub-lattice. Eachion is at the center of the cube whose corners are occupied by ions of opposite kind. The coordination


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number is 8. The number of Cs+per unit cell = =18 18

. The number of Clper unit cell is also 1.

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 HCP

Volume of unit cell a3 a3 a3 33a2c/2

No. of atoms per unit cell 1 2 4 6

Coordination number 6 8 12 12

Nearest neighbour distance a a3/2 a2/2 a

(2r)

Atomic radius (r) a/2 a3/4 a2/4 a/2

Atomic packing fraction /6 = 0.52 3/8 = 0.68 2/6 = 0.74 2/6 = 0.74

SOLVEDEXAMPLES

Ex. 1.Show that the lattice constant for cubic cell is given by

=

1/ 3

A

M

N

na

where n = number of molecules per unit cell, M = molar mass, = density, NA= Avogadrosnumber.

Sol.Let a= length of side of cubic cell.


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Volume of unit cell v= a3

Molar volume of crystal VM=

M

This volume contains NAmolecules.

Volume available to a single molecule = A

M

Nmv

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

=

A

M

N

nv

This is equal to a3.Hence

=

1/ 3

A

M

N

na

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

=snd

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

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

=snd

abc

Sol.Consider a crystal region of volume V containing N lattice planes, each of area S. Let srepresent the number of lattice points per unit area of lattice plane. The number of lattice points inthe volume V is

N' = sS N ...(1)This number may also be calculated as follows.Let = volume of unit cell n= number of lattice points per unit cell.

The number of unit cells in volume V is

V

.

The number of lattice points in volume V is N'=V

n

Thus =V

SNs n

Whence =

V

SNsn

...(2)


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If dis interplanar spacing of the lattice planes then V= Sd N ...(3)Making use of (3) in (2), we get

=snd

...(4)

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

=snd

abc ...(5)

In cubic crystal = a3we have

=3s

nda

...(6)

Ex. 3.The rock salt (NaCl) has fcc structure and contains 4 molecules per unit cell. Calculate thelattice constant for the crystal.

Molecular weight of NaCl = 58.45 kg/kmolDensity = 2180 kg/m3

Avogadro Number NA= 6.02 1026kmol1

Sol.

= = =

1/ 2

26A

M 4 58.455.63

N 2180 6.02 10

na

Ex. 4.Find the ratio of intercepts on the crystal axes by plane (231) in a simple cubic lattice.Sol. Letx1,x2,x3be the intercepts by the plane on the axes. In terms of axial units the intercepts

are

= = =1 2 3, ,x x x

m n pa b c

where m, n, p are numbers.

Now =1 1 1

: : 2 : 3 : 1m n p

=

1 1 1

: : : :2 3 1m n p

= 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 cubiclattice.

Sol. =+ +2 2 2

hkl

ad

h k l


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= =+ +100 1 0 0

ad a

= =+ +

1101 1 0 2

a ad

= =+ +

1111 1 1 3

a ad

=100 110 1111 1

: : 1: :2 3

d d d

Ex. 6.In a tetragonal lattice a = b = 2.5 , c= 1.8. Calculate the lattice spacing between the

(111) planes.

Sol. =

+ +2 2 2

2 2 2

1hkld

h k l

a b c

= =

+ +111

2 2 2

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1 1 1

2.5 2.5 1.8

d

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

Sol. Density of lattice points = =3s

nd nd

abc a

For simple cubic lattice n = 1, d100= a = 2.5 1010m

= = = =

19 2

3 2 10 2

1 11.6 10 m

(2.5 10 m)s

a

a a

Ex. 8.Zinc has hcp structure. The height of unit cell is 4.9 , the nearest neighbour distance is2.7 . Calculate the volume of the unit cell.

Sol.Lattice constant a = 2r = 2.7 Volume of unit cell

=23 3

V2

a c=

10 2 103 3(2.7 10 m) (4.9 10 m)2

= 9.4 10 29 m3


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

=23 3

V ,2

a c where a = 2r= 3.2

and = =8

5.23

c a

V = 1.4 1028m3

QUESTIONSANDPROBLEMS

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, bodycentered and face centered cubic cells. Calculate the number of atoms per unit conventionalunit 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 cubiccrystal.

6. Obtain an expression for the lattice constant of a cubic crystal in terms of molar mass anddensity 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 cubicstructure of lattice constant 2.5 .

9. Copper has fcc structure. Radius of copper atom is 1.28 . Calculate the inter-planar distance

for plane (111).[Hint:a = 4r/2]

10. Magnesium has hcp structure. Atomic radius of Mg is 1.6 . Calculate the volume of unit cell.

crystal structure notes

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