2. Crystal Structure

crystal systemonly 7 different types of unit cells14 standard (Bravais) unit cells coulddescribe all possible lattice networkscrystalsystemaxial lengths &interaxial angles space latticecubic a = b = c simple cubicα = β = γ = 90 o body-centered cubicface-centered cubictetragonal a = b≠ c simple tetragonalα = β = γ = 90 o body-centered tetragonalorthorhombic a ≠ b ≠ c simple orthorhombicα = β = γ = 90 o body-centered orthorhombicbase-centered orthorhombicface-centered orthorhombicrhombohedral a = b = c simple rhombohedralα = β = γ ≠ 90 ohexagonal a = b ≠ c simple hexagonalα = β = 90 oγ = 120 omonoclinic a ≠ b ≠ c simple monoclinicα = γ = 90 o ≠ β base-centered monoclinictriclinic a ≠ b ≠ c simple triclinicα ≠ β ≠ γ ≠ 90 o2

metallic crystal structures90% elemental metals crystallize into threecrystal structures:․ body-centered cubic (BCC)a (nm) R (nm)Cr 0.289 0.125Fe 0.287 0.124Mo 0.315 0.136K 0.533 0.231Na 0.429 0.186Ta 0.330 0.143W 0.316 0.137V 0.304 0.132․face-centered cubic (FCC)a (nm)R (nm)Al 0.405 0.143Cu 0.3615 0.128Au 0.408 0.144Pb 0.495 0.175Ni 0.352 0.125Pt 0.393 0.139Ag 0.409 0.144․hexagonal close-packed (HCP)a c (nm) R (nm)Al 0.2973 0.5618 0.143Zn 0.2665 0.4947 0.133Mg 0.3209 0.5209 0.160Co 0.2507 0.4069 0.125Zr 0.3231 0.5148 0.160Ti 0.2950 0.4683 0.147Be 0.2286 0.3584 0.1134

BCCFCCcoordination number = 8total 2 atoms per unit cell4 R√3 a = 4 R a = —— √3atomic packing factor (APF)volume of atoms in unit cellAPF = ——————————volume of unit cellAPF = 0.685

HCPcoordination number = 12total 4 atoms per unit cell4 R√2 a = 4 R a = ——√2APF = 0.74the closest packing possible of sphericalatoms cubic closest-packedcoordination number = 12total 2 atoms per unit cellAPF = 0.74the closest packing possible of sphericalatomsc/a ratio for ideal HCP structure is 1.6336

the closest packingHCPa b a b a …..FCCa b c a b c ….7

atom positions in cubic unit cellBCC unit cellcoordinates of eight corners:(0, 0, 0) (1, 0, 0) (0, 1, 0) (0, 0, 1)(1, 1, 0) (1, 0, 1) (0, 1, 1) (1, 1, 1)coordinate of the center: (½, ½, ½)direction index – the vector components of thedirection resolved along each of thecoordinate axes and reduced to the smallestintegersall parallel direction vectors have the samedirection indices8

directions are crystallographically equivalent ifthe atom spacing along each direction is thesameex. cubic edge directions:[100] [010] [001] [010] [001] [100]≣ equivalent directions are called indices of afamily or formex. draw the following directions:(a) [112](b) [110](c) [321]9

Miller indices for crystallographic planesMiller notation system (hkl)Miller index – the reciprocals of the fractionalintercepts that the plane makes with the x, y,and z axes of the three nonparallel edges ofthe cubic unit cellprocedure for determining Miller index:(1) choose a plane not pass through (0, 0, 0)(2) determine the intercepts of the plane withx, y, and z axes(3) form the reciprocals of these intercepts(4) find the smallest set of whole numbersthat are in the same ratio as the interceptsex.10

ex. draw the following crystallographic planesin cubic unit cell:(a) (101) (b) (110)(c) (221)planes of a family or form {hkl}ex. (100), (010), (001) are a family{100}an important relationship for cubic system,the direction indices of a direction perpendicularto a crystal plane are the same as the Millerindices of that plane11

interplanar spacing between two closest parallelplanes with the same Miller indices isdesignated d hkl (h, k, l are the Miller indices)ad hkl = ——————√h 2 + k 2 + l 2a: = lattice constantex. determine the Miller indices of the planesshown as follow:(a)(b)(5120) (646)12

hexagonal structureMiller-Bravais indices – HCP crystal planeindices (hkil) h + k + i = 0three basal axes a 1 , a 2 , a 3 and c axisbasal planes (0001)prism planes(ABCD) (1010)(ABEF) (1100)(CDGH) (0110)13

direction indices in HCP unit cellfour indices [uvtw] u + v + t =0u = ⅓(2u – v) v = ⅓(2u – v) t = -(u + v)directions a 1 , a 2 , a 3+a 3 direction incorporating c axisdirections on the upper basal planes14

volume, planar, linear densityvolume densitymass/unit cellρ v = ————————volume/unit cellex. Cu has FCC structure, atomic radius of0.1278 nm, atomic mass of 63.54 g/molcalculate the density of Cu in Mg/m 3 .FCC structure √2 a = 4 Ra = 2 √2 R = 2 √2 (1.278 × 10 -10 )= 3.61 × 10 -10 mV = (3.61 × 10 -10 m) 3 = 4.70 × 10 -29 m 34 Cu per unit cellm = 4 × 63.54 × 1.66 × 10 -30 Mg = 4.22 × 10 -28 Mgρ v = 4.22 × 10 -28 Mg / 4.70 × 10 -29 m 3= 8.98 Mg/m 3 (exp. = 8.96 Mg/m 3 )planar atomic densityequiv. no. of atoms whose centersare intersected by selected areaρ p = ————————————selected areaex. calculate planar atomic density ρ p on (110)plane of the α-Fe in BCC lattice inatoms/mm 2 . (lattice constant a = 0.287 nm)15

1 atom (center) + ¼ atom (corner) × 4 = 2 atomsarea = a ×√2 a = √2 a 2 = √2 (2.87 × 10 -7 ) 2= 1.164 × 10 -13 mm 22 atomsρ p = ———————— = 1.72 × 10 13 atoms/mm 21.164 × 10 -13 mm 2linear atomic densityno. of atoms diam. intersected by selectedlength of line in direction of interestρ l = ———————————————selected length of lineex. calculate linear atomic density ρ l in [110]direction in Cu crystal lattice in atoms/mm.(Cu is FCC and lattice constant a = 0.361 nm)16

no. of atoms = ½ + 1 + ½ = 2 atomslength = √2 a = √2 (3.61 × 10 -7 )= 5.104 × 10 -7 mm2 atomsρ l = ——————— = 3.92 × 10 6 atoms/mm5.104 × 10 -7 mmpolymorphism or allotropyelement or compound exists in more than onecrystalline form under different conditions oftemperature and pressureex.metalcrystal structureat room temperatureat othertemperature 1Ca FCC BCC (> 447 o C)Co HCP FCC (> 427 o C)Hf HCP BCC (> 1742 o C)Fe BCC (α) FCC (912-1394 o C) (γ)BCC (> 1394 o C) (δ)Li BCC HCP (< -193 o C)Na BCC HCP (< -233 o C)Tl HCP BCC (> 234 o C)Ti HCP BCC (> 883 o C)Y HCP BCC (> 1481 o C)Zr HCP BCC (> 872 o C) 117

crystal structure analysisX-ray sourcesx-rays used for diffraction are radiations withwavelengths 0.05 ~ 0.25 nma voltage of 35 kV is applied between cathode(W filament) and anode (Mo target)x-ray spectrum 0.2 ~ 1.4 nmwavelength of K α line 0.07 nm18

X-ray diffractionreflected wave patterns of beam are not in phase,no reinforced beam will be produceddestructive interference occursreflected wave patterns of beam are in phase,reinforcement of the beam or constructiveinterference occursnλ = MP + PN n = 1, 2, 3…..19nλ = 2 d hkl sinθ

ex. BCC Fe placed in an x-ray diffractometerusing x-ray with λ = 0.1541 nm. diffractionfrom {110} planes was obtained at 2θ =44.704 o . calculate lattice constant a.λ = 2 d 110 sinθλ 0.1541 nmd 110 = ——— = —————— = 0.2026 nm2sinθ 2sin(22.352 o )a = d hkl √h 2 + k 2 + l 2 = 0.2026√2 = 0.287 nmX-ray diffraction analysis of crystal structurespowder diffraction methoddiffractometer20

diffraction pattern for cubic unit cellad hkl = ——————√h 2 + k 2 + l 2and λ = 2d sinθ2 a sinθλ = ———————√h 2 + k 2 + l 2rules for determining the diffracting {hkl}planes in cubic crystalsreflection present reflection absentBCC (h + k + l) = even (h + k + l) = oddFCC (h, k, l) all odd orall even(h, k, l) not all oddor all evenex. Diffraction pattern for W sample by theuse of a diffractometer with Cu radiationW : BCC structure21

2 a sinθλ = ——————√h 2 + k 2 + l 2λ 2 (h 2 + k 2 + l 2 )sin 2 θ = ———————4a 2sin 2 θ A h 2 A + k 2 A + l 2 A——— = ———————sin 2 θ Β h 2 B + k 2 B + l 2 BMiller indices of the diffracting planes for BCCand FCC{hkl} Σ[h 2 + k 2 + l 2 ] FCC BCC{100} 1 ….. …..{110} 2 ….. 110{111} 3 111 …..{200} 4 200 200{210} 5 ….. …..{211} 6 ….. 211first two sets of diffraction planesFCC {111) and {200}sin 2 θ A h 2 A + k 2 A + l 2 A——— = ——————— = 0.75sin 2 θ Β h 2 B + k 2 B + l 2 B22

BCC {110) and {200}sin 2 θ A h 2 A + k 2 A + l 2 A——— = ——————— = 0. 5sin 2 θ Β h 2 B + k 2 B + l 2 Bex. an element that has either BCC or FCCstructure shows diffraction peaks atfollowing 2θ angles: 40, 58, 73, 86.8, 100.4and 114.7. wavelength of x-ray λ = 0.154a. BCC or FCC?b. determine the lattice constant a.c. identify the element.(a) 2θ θ sinθ sin 2 θ40 20 0.3420 0.117058 29 0.4848 0.235073 36.5 0.5948 0.353886.8 43.4 0.6871 0.4721100.4 50.2 0.7683 0.5903first and second angles0.1170/0.3420 = 0.5 BCC structure(b) λ √h 2 + k 2 + l 2 0.154 √2a = — —————— = ——— ———2 sinθ 2 0.342= 0.318 nm(c) W23