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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. οƒ˜ The three indices are not separated by commas and are enclosed in open brackets. οƒ˜ If any of the indices is negative, a bar is placed in top of that index οƒ˜ 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€² , 𝛽 β€² π‘Žπ‘›π‘‘ 𝛾 β€² with π‘₯, 𝑦 π‘Žπ‘›π‘‘ 𝑧 axes 𝑑 𝑑 respectively. Then cos 𝛼 β€² = 𝑂𝐴 = π‘Ž β„β„Ž 𝑑 𝑑 cos 𝛽 β€² = 𝑂𝐡 = π‘Ž β„π‘˜ 𝑑 = π‘‘β„Ž π‘Ž π‘‘π‘˜ π‘Ž 𝑑 𝑑𝑙 ⁄𝑙 π‘Ž cos 𝛾 β€² = 𝑂𝐢 = π‘Ž = = Now π‘π‘œπ‘  2 𝛼 β€² + π‘π‘œπ‘  2 𝛽 β€² + π‘π‘œπ‘  2 𝛾 β€² = 1 π‘‘β„Ž 2 π‘‘π‘˜ 2 𝑑𝑙 2 Hence ( π‘Ž ) + ( π‘Ž ) + ( π‘Ž ) = 1 (or) 𝑑 = 𝑂𝑁 = βˆšβ„Ž2 π‘Ž +π‘˜ 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 2π‘Ž 2π‘Ž β„Ž , π‘˜ π‘Žπ‘›π‘‘ 2π‘Ž 𝑙 Therefore 𝑂𝑀 = 2𝑑 = βˆšβ„Ž2 2π‘Ž +π‘˜ 2 +𝑙2 Hence the spacing between the adjacent planes= OM-ON=NM i.e., 5. 𝑑 = βˆšβ„Ž2 π‘Ž +π‘˜ 2 +𝑙2 Derive Bragg’ s Law (3M) Consider a set of parallel planes of a crystal separated by a distance β€˜d’ apart. Suppose a beam of monochromatic X-rays incident at an angle β€˜ΞΈβ€™ on these planes. The beam PA is reflected from atom β€˜A’ in plane-1 whereas beam RB is reflected from atom β€˜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 sinΞΈ + d sinΞΈ= 2d sinΞΈ 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 Ξ» i.e. 2d sinΞΈ=nΞ» 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 4πœƒ = 𝑅 ( πœ‹ ), a list of ΞΈ values can be obtained. Where β€˜R’ is the radius of the camera. Since the wavelength Ξ» is known, substituting these ΞΈ 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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Dr. Shaik kareem Ahmed 2018 2. Line defects Line defects are called dislocations. These are one-dimensional imperfections in the geometrical sense. The two basic types of dislocations are a. edge dislocation and b. screw dislocation. Edge dislocation: An edge location is created in the crystal by introducing an extra half plane. The atoms below the edge of the extra plane are squeezed together and are in a state of compression. Screw dislocation: Screw dislocation results from a displacement of the atoms in one part of a crystal relative to the rest of the crystal, forming a spiral ramp around the dislocation line. . 3. Surface defects The defects that occur along the surface of the crystal is said to be surface defects. The surface defects are two dimensional defects. The different types of surface defects are Grain boundaries: Grain boundaries are those surface imperfections which Separate crystals of different orientations in a polycrystalline aggregate. Tilt boundaries: Tilt boundary is another type of surface imperfection and it may be regarded as an array of edge dislocations. It is also a class of low angle boundaries. Twin boundaries: Surface imperfections which separate two orientations that are mirror images of one another are called twin boundaries. Twin boundaries occur in pairs, such that the orientaion change introduced by one boundary is restored by the other as shown in figure. Stacking faults: A stacking fault is a surface defect that results from the stacking of one atomic plane out of sequence on another, while the lattice on either side of the fault is perfect. The stacking fault is a discrepancy in the packing sequence of the layers of atoms. 4. Volume Defects Volume defects are 3-dimensional defects. These include pores, cracks, foreign inclusions and other phases. These defects are normally introduced during processing and fabrication steps. All these defects are capable of acting as stress raisers, and thus deleterious to parent metal’s mechanical behavior. However, in some cases foreign particles are added purposefully to strengthen the parent material. 5

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