2. Crystal Structure

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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(2<strong>2.</strong>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
Page 2: crystal systemonly 7 different type
Page 5 and 6: BCCFCCcoordination number = 8total
Page 7 and 8: the closest packingHCPa b a b a …
Page 9 and 10: directions are crystallographically
Page 11 and 12: ex. draw the following crystallogra
Page 13 and 14: hexagonal structureMiller-Bravais i
Page 15 and 16: volume, planar, linear densityvolum
Page 17 and 18: no. of atoms = ½ + 1 + ½ = 2 atom
Page 19: X-ray diffractionreflected wave pat
Page 23: BCC {110) and {200}sin 2 θ A h 2 A

2. Crystal Structure

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