Lecture - Chapter 3

# Lecture - Chapter 3 - Chapter 3 The Structure of...

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Unformatted text preview: Chapter 3 The Structure of Crystalline Solids Why study the structures of crystalline solids? The properties of materials are directly related to their structures. Example: Aircraft skin is made of Al but not of Mg both being light metals. The crystals structure of Al and Mg are totally different, as a result Al is ductile and Mg is very brittle. Learning Objectives Describe difference between crystalline and non-crystalline structures. Define a unit cell of different crystal structure. Derive the relationship between the unit cell edge-length and atomic or ionic radius. Calculate densities of materials. Show crystallographic directions. Show crystallographic planes (Miller Indices). Crystal Structures Atoms in metals and ceramics are arranged in a regular manner, reference Figure 3.2 (c), just like tennis balls packed in a box. This regular arrangement is repeated many times so that there exists a long range order, which is called crystalline structure. (c) An aggregate of many BCC unit cells. Crystal Structures Unit cell - The smallest volume which is repeated in three dimensions is defined as the unit cell. Cubic means that all three sides are equal. For the simple cubic structure there is an atom at each of the eight corners of the cube. For example, Figures 3.40, a hard sphere model of simple cubic structure is shown. Metallic Crystal Structures Metals typically have one of three different crystal structures: Face-centered cubic Body-centered cubic Hexagonal close-packed Table 3.1 Gives the atomic radius and crystal structure for some commonly known metals. Face Centered Cubic Structure (FCC) Figure 3.1 (a) shows the hard sphere model of face centered cubic unit cell. Atoms touch each other along the center of the face or the face diagonal. Figure 3.1 (b) & (c) also show other views of the FCC structure for visualization purposes. FACE CENTERED CUBIC STRUCTURE (fcc) fd a a Fig 3.1(a) A hard sphere unit cell (b) a reduced sphere unit cell. Fig 3.1(c) An aggregate of many atoms. FCC Structure cont. Where: a = the length of the side of the unit cell R = radius of the atom fd = face diagonal. fd2 = a2+ a2 = 2 a2 fd = 2 a = 4R a = 2R2 (FCC) FCC Structure cont. No. of atoms per unit cell (FCC) = 8 corner atoms / 8 unit cell + 6 face centered atoms/ 2 unit cells. = 4 atoms / unit cell Atomic Packing Factor Atomic packing factor (APF) gives an indication of how efficiently the atoms are packed in a unit cell. APF= Volume of atoms in a unit cell Total unit cell volume Atomic Packing Factor cont. For an FCC unit cell (4 atoms per unit cell) APF= 4*(4/3 R3 ) / a3 4/3 R3 = Volume of an atom a3 = Volume of the unit cell But since a = 22 R then, APF (FCC) = 16/3 R3 / (22 R) 3 APF (FCC) = .74 Body centered cubic crystal structure (BCC) In the body centered cubic structure, in addition to a corner atoms there is an entire atom contained at the center of the cubic unit cell. Reference Figure (c). BCC structure cont. Relation between a and R Body diagonal = bd = 4R Therefore: bd2 = a2 + (2 a) 2 = 3 a2 bd = 3 a 4R = 3 a a = 4R (BCC) 3 BODY CENTERED CUBIC STRUCTURE (BCC) Fig 3.2 (a) A hard sphere unit cell. (b) A reduced sphere unit cell. BODY CENTERED CUBIC (BCC) STRUCTURE No. of atoms per unit cell = (8 corner atoms / 8 unit cells) +1 center atom = 2 atoms/unit cell Group Discussion Problem Show that the atomic packing factor for BCC (body centered cubic) is 0.68 APF = vol. of the atoms in the unit cell unit cell volume Unit cell volume = a3 = ( 4R )3 3 HEXAGONAL CLOSE PACKED CRYSTAL STRUCTURE (HCP) c Fig 3.3 (a) a reduced-sphere unit cell Fig 3.3(b) An aggregate of many HCP unit cells. Hexagonal Closed Packed Structure Some metals like Zn, Cd, Mg. etc. have a crystal structure with a hexagonal unit cell. Base of the unit cell is a regular hexagon (having 6 sides) with side a. c is the height of the unit cell. Relation between c and a is c = 1.633 a. Hexagonal Closed Packed Structure cont. No. of atoms / unit cell = ( 12 corner atoms / 6 unit cells) + (2 base centered atoms /2 unit cell ) + 3 atoms at height c/2 within the unit cell = 6 atoms/unit cell Hexagonal Closed Packed Structure Relation between `a' (edge of the unit cell) and `R' (radius of the atom) is a = 2 R, since the atoms touch each other along all six edges of the hexagonal base. Note: The area of a hexagon is: 6 * 2 * * R * 3 R = 63 R2 Therefore the volume of the HCP unit cell is: Area of the base * height (c) = 63 R2 * 1.633 * 2R = 33.941 R3 Density of Metals Consider the unit cell of a metal crystal structure. Density = Mass of the unit cell / volume of the unit cell ( No.of atoms /unit cell)*(Atomic weight in grams) = (Volume of unit cell in cm3)*(Avogadro's number) Avogadro's number = 6.02 x 1023 atoms / mole Group Discussion Problem Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its density with the experimental value found inside the front cover. Group Discussion Problem Titanium has an HCP crystal structure, an atomic radius of 0.1445 nm, and an atomic weight of 47.90 g/mol. Compute and compare its density with the experimental value found inside the front cover. Polymorphism and Allotropy Polymorphism some metals, as well as non-metals may have more than one crystal structure. When this occurs in elemental solids it is termed "Allotropy." Generally for polymorphs and allotrops, different structures are possible for the same element but only when exposed to extreme temperatures and/or pressures. Polymorphism and Allotropy cont. Ex. Under normal ambient conditions Carbon exists as Graphite, but under extreme temperatures and pressures can form a diamond. Also, pure iron changes from a BCC crystal structure at room temperature to an FCC crystal structure above 1674F. Crystal Systems The crystal systems are defined by the unit cell geometry which is defined by 6 parameters, 3 edge lengths a, b and c and 3 angles , and (Table 3.6). Triclinic crystal structure is list symmetric. None of the 3 sides are equal, none of the 3 angles are equal or equal to 90. Monoclinic : a # b # c = = 90 # and so on. Fig 3.19. A unit cell with x, y and z axes, showing axial length (a, b and c) and inter axial angles (, and ). Table 3.6 (a) Table 3.6(a) contd.. Crystallographic directions and planes Crystallographic directions A crystallographic direction is defined as a line starting from the origin and passing through a point within a unit cell. Crystallographic Directions Example: The line shown in this figure is starting from the origin and passing through 1 0 point. What is the crystallographic direction? Z 0,0,0 . ,1,0 X Y [1 2 0] Clear all fractions by multiplying across by 2, 1 2 0 The direction is shown in the third bracket [1 2 0]. Crystallographic Directions Example: Show [1 2 3] direction within the unit cell Z [1 2 3] 1/3,2/3,1 Y 0,0,0 X [1 2 3] direction passes through (divide by the largest no.) 1/3 2/3 1 point within a unit cell. Draw a line starting from the origin and passing through the 1/3 2/3 1 point in the unit cell. Crystallographic planes Crystallographic planes are shown in miller indices hkl. Crystallographic Planes Example: Indicate MI (Miller Indices) of the following crystallographic plane) Z (0 1 0) Y X Consider a unit cell and origin. Intercepts of a plane with X, Y and Z are 1 . Reciprocals are 0 1 0. Miller Indices are (0 1 0) shown by the first bracket. Crystallographic Planes Example: Indicate MI (Miller Indices) of the following crystallographic plane) Z (6 3 0) 2/3 Y 1/3 X Intercepts of the second plane are Reciprocals are - 3 3/2 0 Clear all fractions- 6 3 0 MI are - 6 3 0 1/3 2/3 Crystallographic Directions Examples: Z (4 2 0) 1/2 1/4 Y X Show 4 2 0 plane within a unit cell Reciprocals are - Intercepts are - Show and intercepts along the x and y axes and equivalents within the unit cell and connect them. Group Discussion Problem Draw a cubic unit cell, and within that cell a [121] direction and a (210) plane. Hint: Draw the unit cell and show the point with respect to the unit cell. Go along negative direction of z axis one time to show the 12-1 point. Connect that point to the origin. Draw the plane in a separate unit cell Atomic Arrangement The arrangement of atoms in a plane depends on the crystals structure. For example Figure 3.24 shows the arrangement of atoms in (110) plane in FCC structure. Atomic Arrangement cont. Atoms are touching each other along the face diagonal. Figure 3.25 shows the arrangement of atoms in the (110) plane in BCC structure. In this case the atoms are touching each other along the body diagonal. Linear Density LD = # of whole atoms centered on a direction vector Length of direction vector Ex. Reference the FCC unit cell shown in Fig 3.24, in the direction [1 -1 0] or [-1 1 0] there are two halves of an atom and one whole atom lying on the planar direction vector, which equals two atoms. The length of the vector is a2, where a = 2R2, so that the length of the direction vector is 4R. LD then is equal to 2 or LD = 1 4R 2R Planar Density PD = # of whole atoms centered on a unit cell plane Area of unit cell plane Ex. Again, referencing the FCC unit cell shown in Fig 3.24, for the (110) plane, there are two halves of an atom and four quarters of an atom lying in the (110) unit cell plane, which equals again two whole atoms. The area of the plane is a2 * a, where a = 2R2, so that the area of the plane is 8R22. LD then is equal to 2 or PD = 1 8R22 4R22 CLOSE-PACKED CRYSTAL STRUCTURES A A A A A A A CLOSE PACKED CRYSTAL STRUCTURE METALS Close Packing of Atoms (Fig. 3.13) Let the centers of all the atoms in one close packed plane be labeled A. In the next plane there could be three atoms in the empty spaces, labeled B. In the third plane, the atoms could be either in the A or C position . CLOSE PACKED CRYSTAL STRUCTURE METALS cont. If the atoms are in the A position in the third plane, then the sequence of close packing is A B A B ---------- which is the hexagonal closed pack crystal structure. On the other hand , if the atoms are in the C position in the third plane the sequence is A B C A B C ----------- which is the FCC structure (1 1 1) plane. Anisotropic Isotropic materials have the same properties along any direction. But crystalline materials will have different properties along different directions since the atomic arrangement is different along different directions. Table 3.7 shows that the modulus of elasticity of iron is highest along the [1 1 1] direction and lowest along the [1 0 0] direction. Polycrystalline materials In general, most crystalline solids are made up of many crystals, rather than just one single crystal. But there are exceptions: In nature, gems and minerals are often found as giant crystals which are indicative of the microscopic crystal structure which they possess. Polycrystalline materials cont. In electronics, silicon is grown as a single crystal to achieve optimal electrical properties. In industry, parts which can suffer from creep and high temperature corrosion can be made from single crystals to increase their useful life. Polycrystalline materials cont. Polycrystalline materials are comprised of many small grains which are single crystals with perfect atomic arrangement but crystallographic planes are oriented at varying and often random angles with respect to each other. These boundaries between individual grains are called grain boundaries. Fig 3.33 (a) Small crystallite nuclei. (b) Growth of the crystallites; obstruction of some grains that are adjacent to one another is also shown. (c) Upon completion of solidification, grains having irregular shapes have formed. (d) The grain structures as it would appear under the microscope; Dark lines are the grain boundaries. 3.16 XRay Diffraction A materials crystallographic structure can be discovered through x-ray diffraction. Bragg's Law states that only waves which correspond to certain crystallographic planes will be reflected due to constructive interference. Other wavelengths will destructively interfere with one another, thereby canceling themselves out, and giving an image which can then be used to infer the materials crystallographic structure. Bragg's Law n d sin 2 = n = order of the reflection (positive integer) = wavelength of the x-rays d = distance between crystallographic planes = angle of incidence of x-ray Interplanar Spacing (dhkl) The spacing between crystallographic planes of miller indices (hkl) is dhkl, where: dhkl = a _ h2 + k2 + l2 Note: This equation is applicable to simple cubic structures, with atoms only at the corner positions, however in real crystals the additional atoms add scatter. Interplanar Spacing (dhkl) cont. As a result, in FCC crystals h, k, & l must all be either odd or even in order for diffraction to occur, and in BCC crystals h+k+l must equal an even number in order for diffraction to occur. Amorphous / NonCrystalline Solids Glasses and some polymers may or may not have a crystalline structure. Fig 3.38 (a) shows crystalline structure or regular arrangement of atoms where in each ring of SiO2 there are six silicon and six Oxygen atoms. Amorphous / NonCrystalline Solids cont. Fig 3.38(b) shows non crystalline silicon dioxide where in one of the ring 5 silicon and 5 oxygen atoms and in another ring 7 silicon and 7 oxygen atoms or irregular arrangement of atoms. Amorphous or Non Crystalline solid. Fig 3.38 (a) crystalline silicon dioxide and (b) noncrystalline silicon dioxide Oxygen atom Silicon atom 5 oxygen and 5 silicon atoms 6 oxygen and 6 silicon atoms 7 oxygen and 7 silicon atoms Homework HW #2 - 3.10, 3.19, 3.33, 3.36, 3.41(a,b,c,d), 3.48 - Due 2/7/05 ...
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## This note was uploaded on 05/03/2010 for the course ME 250-750 taught by Professor Signer during the Summer '10 term at Wichita State.

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