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Crystalline Solids :-In Crystalline Solids the atoms are arranged in some regular periodic geometrical pattern in three dimensions- long range order Eg.

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Presentation on theme: "Crystalline Solids :-In Crystalline Solids the atoms are arranged in some regular periodic geometrical pattern in three dimensions- long range order Eg."— Presentation transcript:

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2 Crystalline Solids :-In Crystalline Solids the atoms are arranged in some regular periodic geometrical pattern in three dimensions- long range order Eg :- NaCl, CuSo 4,CsCl, ZnS, etc Amorphous Solids :- In an amorphous solids the atoms are not arranged in regular periodic geometrical pattern –short range order Eg :- Boran trioxide (B 2 O 3 ),Lamp Soot, glass etc Unit Cell :- A Unit cell is the volume of a solid from which the entire crystal can be constructed by a translation repetition in three directions in sp

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4 Crystal lattice :- An array of points in the space (2D,3D) in which every point has the same environment with respect to all other points is called space lattice Basis :- Atoms or molecules attached to each lattice point in the crystal system is called basis. Lattice+ Basis = Crystal

5 CsCl structure Lattice n-dimensional, infinite, periodic array of points, each of which has identical surroundings. Lattice n-dimensional, infinite, periodic array of points, each of which has identical surroundings. use this as test for lattice points lattice points

6 Crystallographic axes

7 Crystallographic axes:- The lines drawn parallel to the line of intersection of any three faces of the unit Cell which do not lie in the same plane are called Crystallographic axes X,Yand Z

8 Lattice Parameters Interfacial angles :- The angles between three Crystallographic axes represented by α,β, Ƴ are called interfacial angles * Primitives :- The intercepts a,b and c, which define the dimensions of the unit cell on the respective Crystallographic axes are called as primitives of the Unit Cell

9 An array of points such that every point has identical surroundings  In Euclidean space  infinite arrayEuclidean space  We can have 1D, 2D or 3D arrays (lattices) Space Lattice Translationally periodic arrangement of points in space is called a lattice or A lattice is also called a Space Lattice Note: points are drawn with finite size for clarity  in reality they are 0D (zero dimensional)

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11 Crystals are grouped into seven crystal systems, according to characteristic symmetry of their unit cell.symmetry The characteristic symmetry of a crystal is a combination of one or more rotations and inversions.

12 Crystal Systems There are Seven Basic Crystal Systems Cubic Tetragonal Orthorhombic Monoclinic Triclinic Rhombohedral (Trigonal) Hexagonal

13 Shape of UC Used as UC for crystal:Lattice Parameters Cube Cubic (a = b = c,  =  =  = 90  ) Square Prism Tetragonal (a = b  c,  =  =  = 90  ) Rectangular Prism Orthorhombic (a  b  c,  =  =  = 90  ) Parallelogram PrismMonoclinic (a  b  c,  =  = 90    ) Parallelepiped (general)Triclinic (a  b  c,      ) Parallelepiped (Equilateral, Equiangular) Rhombohedral (Trigonal) (a = b = c,  =  =   90  ) 120  Rhombic PrismHexagonal (a = b  c,  =  = 90 ,  = 120  )

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15 Latti In 1848, Auguste Bravais demonstrated that in a 3-dimensional system there are fourteen possible lattices A Bravais lattice is an infinite array of discrete points with identical environment seven crystal systems + four lattice centering types = 14 Bravais lattices Lattices are characterized by translation symmetry Auguste Bravais (1811-1863)

16 Bravais showed that there are only 14 independent ways of arranging points in space so that the environment looks the same from each point. These lattices are called Bravais lattices

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18 Cubic space lattices

19 Summary: Fourteen Bravais Lattices in Three Dimensions

20 Fourteen Bravais Lattices …

21 1CubicCube  PIFC Lattice point P I F Symmetry of Cubic lattices

22 PIFC 2TetragonalSquare Prism (general height)  I P Symmetry of Tetragonal lattices

23 PIFC 3OrthorhombicRectangular Prism (general height)  P I F C Symmetry of Orthorhombic lattices Note the position of ‘a’ and ‘b’ One convention Why is Orthorhombic called Ortho-’Rhombic’? Is there a alternate possible set of unit cells for OR?

24 PIFC 4Hexagonal 120  Rhombic Prism  What about the HCP? (Does it not have an additional atom somewhere in the middle?) A single unit cell (marked in blue) along with a 3-unit cells forming a hexagonal prism Note: there is only one type of hexagonal lattice (the primitive one) Symmetry of Hexagonal lattices

25 PIFC 5TrigonalParallelepiped (Equilateral, Equiangular)  Symmetry of Trigonal lattices Rhombohedra l Note the position of the origin and of ‘a’, ‘b’ & ‘c’

26 PIFC 6MonoclinicParallogramic Prism  Symmetry of Monoclinic lattices Note the position of ‘a’, ‘b’ & ‘c’ One convention

27 PIFC 7TriclinicParallelepiped (general)  Symmetry of Triclinic lattices

28 Arrangement of lattice points in the Unit Cell & No. of Lattice points / Cell Position of lattice pointsEffective number of Lattice points / cell 1P8 Corners = [8  (1/8)] = 1 2I 8 Corners + 1 body centre = [1 (for corners)] + [1 (BC)] = 2 3F 8 Corners + 6 face centres = [1 (for corners)] + [6  (1/2)] = 4 4 A/ B/ C 8 corners + 2 centres of opposite faces = [1 (for corners)] + [2  (1/2)] = 2

29 PH 0101 UNIT 4 LECTURE 229 MILLER INDICES DIFFERENT LATTICE PLANES

30 PH 0101 UNIT 4 LECTURE 230 MILLER INDICES The orientation of planes or faces in a crystal can be described in terms of their intercepts on the three axes. Miller introduced a system to designate a plane in a crystal. He introduced a set of three numbers to specify a plane in a crystal. This set of three numbers is known as ‘Miller Indices’ of the concerned plane.

31 PH 0101 UNIT 4 LECTURE 231 MILLER INDICES The orientation of planes or faces in a crystal can be described in terms of their intercepts on the three axes. Miller introduced a system to designate a plane in a crystal. He introduced a set of three numbers to specify a plane in a crystal. This set of three numbers is known as ‘Miller Indices’ of the concerned plane.

32 PH 0101 UNIT 4 LECTURE 232 MILLER INDICES Procedure for finding Miller Indices Step 1: Determine the intercepts of the plane along the axes X,Y and Z in terms of the lattice constants a,b and c. Step 2: Determine the reciprocals of these numbers.

33 PH 0101 UNIT 4 LECTURE 233 Step 3: Find the least common denominator (lcd) and multiply each by this lcd. Step 4:The result is written in paranthesis.This is called the `Miller Indices’ of the plane in the form (h k l). This is called the `Miller Indices’ of the plane in the form (h k l). MILLER INDICES

34 PH 0101 UNIT 4 LECTURE 234 ILLUSTRATION PLANES IN A CRYSTAL Plane ABC has intercepts of 2 units along X-axis, 3 units along Y-axis and 2 units along Z-axis.

35 PH 0101 UNIT 4 LECTURE 235 DETERMINATION OF ‘MILLER INDICES’ Step 1:The intercepts are 2,3 and 2 on the three axes. Step 2:The reciprocals are 1/2, 1/3 and 1/2. Step 3:The least common denominator is ‘6’. Multiplying each reciprocal by lcd, we get, 3,2 and 3. Step 4:Hence Miller indices for the plane ABC is (3 2 3) ILLUSTRATION

36 PH 0101 UNIT 4 LECTURE 236 IMPORTANT FEATURES OF MILLER INDICES For the cubic crystal especially, the important features of Miller indices are, A plane which is parallel to any one of the co-ordinate axes has an intercept of infinity (  ). Therefore the Miller index for that axis is zero; i.e. for an intercept at infinity, the corresponding index is zero. MILLER INDICES

37 PH 0101 UNIT 4 LECTURE 237 EXAMPLE ( 1 0 0 ) plane Plane parallel to Y and Z axes

38 PH 0101 UNIT 4 LECTURE 238 EXAMPLE In the above plane, the intercept along X axis is 1 unit. The plane is parallel to Y and Z axes. So, the intercepts along Y and Z axes are ‘  ’. Now the intercepts are 1,  and . The reciprocals of the intercepts are = 1/1, 1/  and 1/ . Therefore the Miller indices for the above plane is (1 0 0).

39 PH 0101 UNIT 4 LECTURE 239 MILLER INDICES IMPORTANT FEATURES OF MILLER INDICES A plane passing through the origin is defined in terms of a parallel plane having non zero intercepts. All equally spaced parallel planes have same ‘Miller indices’ i.e. The Miller indices do not only define a particular plane but also a set of parallel planes. Thus the planes whose intercepts are 1, 1,1; 2,2,2; -3,-3,-3 etc., are all represented by the same set of Miller indices.

40 PH 0101 UNIT 4 LECTURE 240 MILLER INDICES IMPORTANT FEATURES OF MILLER INDICES It is only the ratio of the indices which is important in this notation. The (6 2 2) planes are the same as (3 1 1) planes. If a plane cuts an axis on the negative side of the origin, corresponding index is negative. It is represented by a bar, like (1 0 0). i.e. Miller indices (1 0 0) indicates that the plane has an intercept in the –ve X –axis.

41 PH 0101 UNIT 4 LECTURE 241 MILLER INDICES OF SOME IMPORTANT PLANES

42 Spacing between planes in a cubic crystal where d hkl = inter-planar spacing between planes with Miller indices h,k,and l. a = lattice constant (edge of the cube) h, k, l = Miller indices of cubic planes being considered.

43 PH 0101 UNIT 4 LECTURE 243 PROBLEMS Worked Example: Calculate the miller indices for the plane with intercepts 2a, - 3b and 4c the along the crystallographic axes. The intercepts are 2, - 3 and 4 Step 1:The intercepts are 2, -3 and 4 along the 3 axes Step 2: The reciprocals are Step 3: The least common denominator is 12. Multiplying each reciprocal by lcd, we get 6 -4 and 3 Step 4: Hence the Miller indices for the plane is

44 PH 0101 UNIT 4 LECTURE 244 Worked Example The lattice constant for a unit cell of aluminum is 4.031Å Calculate the interplanar space of (2 1 1) plane. a = 4.031 Å (h k l) = (2 1 1) Interplanar spacing  d = 1.6456 Å PROBLEMS

45 PH 0101 UNIT 4 LECTURE 245 PROBLEMS Worked Example: Find the perpendicular distance between the two planes indicated by the Miller indices (1 2 1) and (2 1 2) in a unit cell of a cubic lattice with a lattice constant parameter ‘a’. We know the perpendicular distance between the origin and the plane is (1 2 1) and the perpendicular distance between the origin and the plane (2 1 2),

46 PH 0101 UNIT 4 LECTURE 246 PROBLEMS The perpendicular distance between the planes (1 2 1) and (2 1 2) are, d = d 1 – d 2 = (or)d = 0.0749 a.


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