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Dr. Shaik kareem Ahmed 2018 Crystallography Unit-I 1. Write a short note on crystal systems (3M) On the basis of values of lattice parameters, crystals are classified into seven categories S.No Crystal System 1 2 Cubic Tetragonal Rhombic (Orthorhombic) 3 Lattice Constants a=b=c a=b≠c Crystallography Angles α =β= γ=90o α =β= γ=90o a≠b≠c α =β= γ=90o 4 Monoclinic a≠b≠c α =β= γ≠90o 5 Triclinic a≠b≠c α≠β≠ γ≠90o Examples Po, Ag, Au, Cu,Pb, TiO2, SnO2, KH2PO4 KNO3, K2SO4 Gypsum, K2MgSO4.5H2O K2Cr2O7, CuSO4.5H2O Rhombohedral a=b=c α =β= γ≠90o As, Sb, Bi (Trigonal) 7 Hexagonal a=b≠c α =β=90o, γ=120 o Quartz, Zn, Cd The Unit cells of all the crystal systems are shown below 6 2. Explain Bravais lattice (3M) According to Bravais, there are only fourteen independent ways of arranging points in three-dimensional space such that each arrangement confirms to the definitions of a space lattice. Thus there are fourteen (14) possible types of space lattices among the seven crystal systems and are called Bravais Lattices. S.NO CRYSTAL SYSTEM Bravais Lattices 1 Cubic 3 2 Tetragonal 2 3 Rhombic 4 Name of Bravais Lattice Simple, Body centered and face centered Simple, Body centered Simple, Body Centered, face centered Base Centered 4 Monoclinic 2 Simple Base Centered 5 6 Triclinic Rhombohedral 1 1 Simple Simple 7 Hexagonal 1 Simple 1

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Dr. Shaik kareem Ahmed 2018 3. Write as short notes on crystal planes and miller Indices (3M) Crystallographic planes are specified by 3 indices (h k l) called Miller indices.  The three indices are not separated by commas and are enclosed in open brackets.  If any of the indices is negative, a bar is placed in top of that index  If plane is parallel to that axis, intercept is infinity Procedure for determining miller indices (h k l): 1. Record the intercept values in order x,y,z. 2. Take reciprocals of the intercept values. 3. Convert the reciprocalsinto the smallest integers by taking L.C.M of denominators if necessary. Example: Plane ABC has intercepts of 2 units along X-axis, 3 units along Y-axis and 2 units along Z-axis.find the miller indices for a plane shown in figure. Solution: Intercepts are (2 3 2) 111 Reciprocal of the three intercepts are( 2 3 2) Convert reciprocals into small integers by multiplying with LCM of denominators i.e., with 6. Then we get miller indices of the plane ABC is (3 2 3). 4. 4. Derive interplanar spacing in crystals (2M) Let (h k l) be the miller indices of the plane ABC. Let ON=d be a normal to the plane passing through the origin O. Let this ON make angles 𝛼 ′ , 𝛽 ′ 𝑎𝑛𝑑 𝛾 ′ with 𝑥, 𝑦 𝑎𝑛𝑑 𝑧 axes 𝑑 𝑑 respectively. Then cos 𝛼 ′ = 𝑂𝐴 = 𝑎 ⁄ℎ 𝑑 𝑑 cos 𝛽 ′ = 𝑂𝐵 = 𝑎 ⁄𝑘 𝑑 = 𝑑ℎ 𝑎 𝑑𝑘 𝑎 𝑑 𝑑𝑙 ⁄𝑙 𝑎 cos 𝛾 ′ = 𝑂𝐶 = 𝑎 = = Now 𝑐𝑜𝑠 2 𝛼 ′ + 𝑐𝑜𝑠 2 𝛽 ′ + 𝑐𝑜𝑠 2 𝛾 ′ = 1 𝑑ℎ 2 𝑑𝑘 2 𝑑𝑙 2 Hence ( 𝑎 ) + ( 𝑎 ) + ( 𝑎 ) = 1 (or) 𝑑 = 𝑂𝑁 = √ℎ2 𝑎 +𝑘 2 +𝑙2 2

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Dr. Shaik kareem Ahmed 2018 Let the next plane A’B’C’ parallel to ABC be at a distance OM from the origin. Then its intercepts are 2𝑎 2𝑎 ℎ , 𝑘 𝑎𝑛𝑑 2𝑎 𝑙 Therefore 𝑂𝑀 = 2𝑑 = √ℎ2 2𝑎 +𝑘 2 +𝑙2 Hence the spacing between the adjacent planes= OM-ON=NM i.e., 5. 𝑑 = √ℎ2 𝑎 +𝑘 2 +𝑙2 Derive Bragg’ s Law (3M) Consider a set of parallel planes of a crystal separated by a distance ‘d’ apart. Suppose a beam of monochromatic X-rays incident at an angle ‘θ’ on these planes. The beam PA is reflected from atom ‘A’ in plane-1 whereas beam RB is reflected from atom ‘B’ in plane-2 as shown in figure. These two reflected rays will be in phase or out of phase with each other depending on their path difference. This path difference can be found by drawing perpendiculars AM and AN. It is obvious that second ray travels an extra distance = MB+BN Hence the path difference the two reflected beams =MB+BN=d sinθ + d sinθ= 2d sinθ Bragg’s law states that the two reflected beams will be in-phase to each other, if this path difference equals to integral multiple of λ i.e. 2d sinθ=nλ Where n=1,2,3 for the first order, second order and third order maxima respectively. 6. Explain powder diffraction method for determination of lattice constant (5M) 3

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Dr. Shaik kareem Ahmed 2018 (Debye-Scherrer method) The experimental arrangement of powder method is shown in figure. The powdered specimen is kept inside a small capillary tube present at the centre of the camera. A photographic film strip placed along the drum of the camera. A finepencil beam X-rays is made to fall on the powdered sample. The powder diffracts the x-rays in accordance with Bragg’s law to produce cones of diffracted beams. These cones intersect the photographic film. Due to narrow width of the film only pair of arcs of the circle are recorded on the film. The non-diffracted rays leave the camera via the exit port. The film is removed from the camera, processed and flattened. It shows the diffraction lines and the holes for the incident and transmitted beams. The distance between two 𝑆 180 successive arcs S is measured and using the relation 4𝜃 = 𝑅 ( 𝜋 ), a list of θ values can be obtained. Where ‘R’ is the radius of the camera. Since the wavelength λ is known, substituting these θ values in Bragg’s formula, a list of inter-planar spacingd can be calculated. From the ratio of interplanar spacing, the type of the lattice can be identified as well as lattice constant a can be calculated. 7. Write a short note on Classification of defects (5M) 1. Point Defects (a) Vacancies: Vacancies also called Shottky defects Schottky defect: Ion vacancies are called Shottky defects. A shottky defect is the combination of one cation vacancy and one anion vacancy. A pair of one cation and one anion missing from an ionic crystal(a schottky defect) is shown in figure. These defects are normally generated in equal number of anion and cation vacancies hence electrical neutrality is maintained in the crystal. (b) Interstitialcies: Interstitial defect also called Frenkel defect Frenkel defect:In the case of ionic crystal, an ion displaces from the lattice into an interstitial site is called a Frenkel defect. A Frenkel defect is the combination of one cation vacancy and one cation interstial defect. 4

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