SSRAO

Explanation:
For BCC: 2 atoms per unit cell. Atoms touch along body diagonal: 4r = a√3, so r = a√3/4. Volume of atoms = 2×(4/3)π(a√3/4)³ = π√3a³/8. APF = π√3/8 ≈ 0.6802 ≈ 0.68.

Explanation:
For FCC: 4 atoms per unit cell. Atoms touch along face diagonal: 4r = a√2, so r = a√2/4. APF = 4×(4/3)π(a√2/4)³/a³ = π√2/6 ≈ 0.7405 ≈ 0.74.

Explanation:
For SC: 1 atom per unit cell. Atoms touch along edge: 2r = a, so r = a/2. APF = (4/3)π(a/2)³/a³ = π/6 ≈ 0.5236 ≈ 0.52.

Explanation:
BCC: 8 corner atoms × (1/8) = 1, plus 1 body-center atom = 1. Total = 2 atoms per unit cell. Examples: α-Fe, Cr, W, Mo, V.

Explanation:
FCC: 8 corner atoms × (1/8) = 1, plus 6 face-center atoms × (1/2) = 3. Total = 4 atoms per unit cell. Examples: Al, Cu, Ni, Ag, Au, γ-Fe.

Explanation:
In FCC, each atom has 12 nearest neighbors: 4 in the same layer, 4 above, 4 below. CN = 12. FCC and HCP both have CN = 12 and APF = 0.74.

Explanation:
In BCC, each atom is surrounded by 8 nearest neighbors at the corners of the cube (body-center to corners). CN = 8. SC = 6, FCC = HCP = 12.

Explanation:
Parallel to x and y axes means intercepts at ∞, ∞, 1. Reciprocals: 0, 0, 1. Miller indices = (0 0 1). This plane is perpendicular to the z-axis.

Explanation:
APF values: SC = 0.52, BCC = 0.68, FCC = 0.74, Diamond = 0.34. FCC (and HCP) have the highest APF of 0.74.

Explanation:
Diamond cubic = FCC lattice (4 atoms) + 4 atoms inside at (1/4,1/4,1/4) type positions. Total = 8 atoms per unit cell. Examples: Si, Ge, diamond (C).

Explanation:
Diamond cubic: 8 atoms per unit cell. Atoms touch along tetrahedral bond: 8r = a√3, r = a√3/8. APF = 8×(4/3)π(a√3/8)³/a³ = π√3/16 ≈ 0.3401 ≈ 0.34. Lowest APF due to directional covalent bonds.

Explanation:
For ideal HCP: c/a = √(8/3) = 1.6330. Real metals deviate: Zn (1.856), Cd (1.886), Mg (1.624), Ti (1.587). Metals with c/a > 1.633 are elongated along c-axis.

Explanation:
HCP: 12 corner atoms × (1/6) = 2, 2 face-center atoms × (1/2) = 1, 3 interior atoms × 1 = 3. Total = 6 atoms per unit cell. Examples: Mg, Zn, Ti, Co, Cd, Be.

Explanation:
HCP: each atom has 6 neighbors in same layer + 3 above + 3 below = 12. HCP and FCC both have CN = 12 and APF = 0.74. Difference: FCC stacking ABCABC, HCP stacking ABABAB.

Explanation:
For cubic crystal: d_hkl = a / √(h² + k² + l²). Examples: d₁₀₀ = a, d₁₁₀ = a/√2, d₁₁₁ = a/√3. Used in Bragg's law: 2d sinθ = nλ.

Explanation:
Bragg's law: nλ = 2d sinθ. n = diffraction order, λ = X-ray wavelength, d = interplanar spacing, θ = glancing angle. Path difference = 2d sinθ; constructive interference when path difference = nλ.

Explanation:
In FCC, atoms touch along face diagonal [110]: 4r = a√2. Close-packed directions are <110> family (12 directions). Close-packed planes are {111}. In BCC, close-packed direction is [111] (body diagonal).

Explanation:
In BCC, atoms touch along body diagonal [111]: 4r = a√3. Close-packed directions are <111> family (8 directions). Primary slip system in BCC: {110}<111>.

Explanation:
FCC: Octahedral voids = 12 edge-centers × (1/4) + 1 body-center = 4 per unit cell. Atoms per unit cell = 4. Ratio = 4/4 = 1 octahedral void per atom. Tetrahedral voids = 8 per unit cell → 2 per atom.

Explanation:
FCC: 8 tetrahedral voids per unit cell (one in each octant at positions like a/4,a/4,a/4). With 4 atoms per unit cell: 8/4 = 2 tetrahedral voids per atom. Tetrahedral void radius ratio: r/R = 0.225.

Explanation:
Octahedral void (CN=6): r + R = R√2, so r/R = √2 - 1 = 0.414. Summary: Triangular (CN=3): 0.155, Tetrahedral (CN=4): 0.225, Octahedral (CN=6): 0.414, Cubic (CN=8): 0.732.

Explanation:
Tetrahedral void (CN=4): r/R = √(3/2) - 1 = 0.2247 ≈ 0.225. Only small atoms can fit in tetrahedral voids. Octahedral void: r/R = 0.414. Cubic void: r/R = 0.732.

Explanation:
Copper (Cu) has FCC structure at room temperature. FCC metals: Al, Cu, Ni, Ag, Au, Pb, Pt. BCC metals: α-Fe, Cr, W, Mo, V, Nb, Ta. HCP metals: Mg, Zn, Ti, Co, Cd, Be.

Explanation:
In cubic systems only, the direction perpendicular to plane (hkl) has the same indices [hkl]. This is because cubic basis vectors are orthogonal and equal. Direction perpendicular to (112) is [112]. This property does not apply to non-cubic systems.

Explanation:
ρ = nM/(Nₐ × a³). For FCC: n = 4. ρ = 4M/(Nₐa³). For BCC: ρ = 2M/(Nₐa³). For SC: ρ = M/(Nₐa³). Used to calculate theoretical density and compare with experimental values.

Explanation:
FCC has stacking sequence ABCABC... of close-packed {111} planes. Each layer is offset from the previous, repeating every 3 layers. HCP has ABABAB... Both have APF = 0.74 and CN = 12.

Explanation:
HCP has stacking sequence ABABAB... of close-packed basal planes. The 3rd layer (A) is directly above the 1st layer. FCC has ABCABC... Both structures have APF = 0.74 and CN = 12.

Explanation:
In FCC, shortest lattice vector (Burgers vector) is a/2<110>. Magnitude = a/√2. For BCC: b = a/2<111>, magnitude = a√3/2. FCC dislocations can dissociate: a/2<110> → a/6<211> + a/6<12̄1> (Shockley partials).

Explanation:
FCC slip: {111} planes (4) × <110> directions (3 per plane) = 12 slip systems. BCC: up to 48. HCP basal: 3. FCC has 12 slip systems making it more ductile than HCP.

Explanation:
BCC {110}<111>: {110} family has 6 planes × 2 <111> directions each = 12 slip systems. BCC can also slip on {112} (12) and {123} (24), giving up to 48 total. Primary slip system is {110}<111>.

Explanation:
HCP basal slip: {0001} (1 plane) × <11̄20> (3 directions) = 3 slip systems. This limits ductility of HCP metals at room temperature. FCC has 12 slip systems. HCP can also have prismatic (3) and pyramidal slip at elevated temperatures.

Explanation:
cosθ = (h₁h₂+k₁k₂+l₁l₂)/[√(h₁²+k₁²+l₁²)×√(h₂²+k₂²+l₂²)]. For [100] and [110]: cosθ = 1/[1×√2] = 1/√2. θ = 45°.

Explanation:
cosθ = (1+1+0)/[√3×√2] = 2/√6 = 0.8165. θ = cos⁻¹(0.8165) = 35.26°. This angle is important for understanding the relationship between close-packed planes and directions in FCC metals.

Explanation:
d₁₁₀ = a/√(1²+1²+0²) = 0.287/√2 = 0.287/1.414 = 0.2029 ≈ 0.203 nm. The (110) plane is the most densely packed plane in BCC iron and the primary slip plane.

Explanation:
FCC: 4r = a√2, so a = 4r/√2 = 2√2 × r = 2×1.414×0.128 = 0.3620 nm. Actual lattice parameter of Cu = 0.3615 nm. For BCC: a = 4r/√3.

Explanation:
BCC: 4r = a√3, so a = 4r/√3 = 4×0.124/1.732 = 0.496/1.732 = 0.2864 ≈ 0.286 nm. Actual lattice parameter of α-Fe = 0.2866 nm.

Explanation:
ρ = 4M/(Nₐa³). a = 3.615×10⁻⁸ cm. a³ = 47.27×10⁻²⁴ cm³. ρ = (4×63.5)/(6.023×10²³×47.27×10⁻²⁴) = 254/28.46 = 8.93 ≈ 8.94 g/cm³. Experimental density of Cu = 8.96 g/cm³.

Explanation:
Face Centered Tetragonal (FCT) is NOT a Bravais lattice because it reduces to Body Centered Tetragonal (BCT) with a smaller unit cell. There are 14 Bravais lattices: Cubic (3), Tetragonal (2), Orthorhombic (4), Hexagonal (1), Trigonal (1), Monoclinic (2), Triclinic (1).

Explanation:
14 Bravais lattices in 3D across 7 crystal systems: Triclinic (1), Monoclinic (2), Orthorhombic (4), Tetragonal (2), Cubic (3), Hexagonal (1), Rhombohedral (1). Total = 14. There are 32 point groups and 230 space groups.

Explanation:
7 crystal systems: (1) Triclinic: a≠b≠c, α≠β≠γ; (2) Monoclinic: a≠b≠c, α=γ=90°≠β; (3) Orthorhombic: a≠b≠c, α=β=γ=90°; (4) Tetragonal: a=b≠c, α=β=γ=90°; (5) Cubic: a=b=c, α=β=γ=90°; (6) Hexagonal: a=b≠c, γ=120°; (7) Rhombohedral: a=b=c, α=β=γ≠90°.

Explanation:
Void space = 1 - APF = 1 - 0.74 = 0.26 = 26%. FCC has APF = 0.74. Compare: SC void = 48%, BCC void = 32%, Diamond void = 66%. FCC and HCP have minimum void space (26%).

Explanation:
Void space = 1 - APF = 1 - 0.68 = 0.32 = 32%. BCC has APF = 0.68. Despite larger void fraction, BCC octahedral voids are smaller than FCC octahedral voids, limiting interstitial solubility.

Explanation:
BCC: atoms at (0,0,0) and (1/2,1/2,1/2). F = f[1 + e^(iπ(h+k+l))]. When h+k+l = odd: F = 0 (systematic absence). When h+k+l = even: F = 2f (allowed). Selection rule: h+k+l must be even for BCC diffraction.

Explanation:
FCC: atoms at (0,0,0), (1/2,1/2,0), (1/2,0,1/2), (0,1/2,1/2). Systematic absence when h,k,l are mixed. Allowed: h,k,l all odd (111, 311) or all even (200, 220, 222). Absent: 100, 110, 210, 211.

Explanation:
BCC allowed reflections: h+k+l = even. Smallest h²+k²+l²: (110) h+k+l=2 ✓ → first peak. (100) h+k+l=1 → absent. (111) h+k+l=3 → absent. (200) h+k+l=2 ✓ → second peak.

Explanation:
FCC allowed: h,k,l all odd or all even. Smallest h²+k²+l²: (111) all odd ✓, h²+k²+l²=3 → first peak. (100) mixed → absent. (110) mixed → absent. (200) all even ✓ → second peak.

Explanation:
HCP has same packing efficiency as FCC = 74% (APF = 0.74). Both are close-packed with CN = 12. For ideal HCP: 6 atoms per unit cell, APF = π/(3√2) = π√2/6 ≈ 0.7405.

Explanation:
d₂₀₀ = a/√(4+0+0) = 0.3615/2 = 0.18075 ≈ 0.1808 nm. The (200) reflection is the second peak in FCC X-ray diffraction. d₂₀₀ = d₁₀₀/2 = a/2.

Explanation:
d₁₁₁ = a/√3, d₁₁₀ = a/√2, d₁₀₀ = a. Ratio = (1/√3) : (1/√2) : 1. Equivalently √2 : √3 : √6 (multiply by √6). The (100) plane has the largest spacing in cubic crystals.

Explanation:
{111} planes in FCC are close-packed (highest planar density ≈ 2.31/a²). Atoms arranged in hexagonal pattern. {110} is most densely packed in BCC. {0001} is close-packed in HCP. Close-packed planes are preferred slip planes.

Explanation:
{110} planes in BCC have the highest planar density = 1/(√2 a²). Primary slip system in BCC is {110}<111>. In FCC, {111} is the close-packed plane. In HCP, {0001} is the close-packed plane.

Explanation:
Iron allotropic transformations: α-Fe (BCC) → γ-Fe (FCC) at 912°C. γ-Fe (FCC) → δ-Fe (BCC) at 1394°C. δ-Fe → Liquid at 1538°C. The α→γ transformation is critical for steel heat treatment.

Explanation:
Max C solubility: γ-Fe (FCC) = 2.14 wt% at 1147°C; α-Fe (BCC) = 0.022 wt% at 727°C; δ-Fe (BCC) = 0.09 wt% at 1493°C. FCC has larger octahedral voids, allowing more carbon dissolution.

Explanation:
Weiss zone law: hu + kv + lw = 0. Example: Does [110] lie in (001)? 0×1+0×1+1×0 = 0 ✓. Used to find zone axis and determine which directions lie in a given plane.

Explanation:
Zone axis = cross product of plane normals: u = k₁l₂-k₂l₁, v = l₁h₂-l₂h₁, w = h₁k₂-h₂k₁. Example: Zone axis of (100) and (010): u=0, v=0, w=1 → [001].

Explanation:
SC: 8 corner atoms × (1/8) = 1 atom per unit cell. Each corner atom is shared by 8 unit cells. Only polonium (Po) has SC structure at room temperature. SC has lowest CN (6), lowest APF (0.52).

Explanation:
Intercepts: a, 2a, ∞. In units of a: 1, 2, ∞. Reciprocals: 1, 1/2, 0. Clear fractions (×2): 2, 1, 0. Miller indices = (2 1 0).

Explanation:
{110} family: (110), (101), (011), (1̄10), (1̄01), (01̄1) = 6 distinct planes. {100} has 3 planes, {111} has 4 planes, {112} has 12 planes, {123} has 24 planes.

Explanation:
<111> family: [111], [11̄1̄], [1̄11̄], [1̄1̄1], [1̄1̄1̄], [1̄11], [11̄1], [111̄] = 8 directions (4 body diagonals × 2 directions each). <100> has 6 directions, <110> has 12 directions.

Explanation:
In FCC, atoms touch along [110] (face diagonal = a√2). Adjacent atoms along [110]: distance = a√2/2 = a/√2. This equals 2r since 4r = a√2 → 2r = a/√2. This is the Burgers vector magnitude for FCC: |b| = a/√2.

Explanation:
In BCC, atoms touch along [111] (body diagonal = a√3). Corner at (0,0,0) to body-center at (a/2,a/2,a/2): distance = a√3/2. This is the repeat distance = 2r since 4r = a√3 → 2r = a√3/2. Burgers vector magnitude for BCC: |b| = a√3/2.

Explanation:
NaCl (rock salt) = two interpenetrating FCC lattices (Na⁺ and Cl⁻) offset by a/2 along [100]. Each ion surrounded by 6 opposite ions (CN=6, octahedral). Radius ratio r_Na/r_Cl = 0.536. 4 formula units per unit cell. Examples: NaCl, KCl, MgO, FeO.

Explanation:
CsCl: Cs⁺ at body center, Cl⁻ at corners. Each Cs⁺ surrounded by 8 Cl⁻ → CN = 8. NOT a BCC structure (BCC has same atoms). 1 formula unit per unit cell. Radius ratio r_Cs/r_Cl = 0.96 > 0.732, allowing CN=8. Examples: CsCl, CsBr, CsI.

Explanation:
Zinc blende: S²⁻ in FCC positions, Zn²⁺ in half the tetrahedral voids. Each Zn²⁺ surrounded by 4 S²⁻ (CN=4, tetrahedral). Radius ratio r_Zn/r_S = 0.40 (range 0.225-0.414). 4 formula units per unit cell. Examples: ZnS, GaAs, InP, SiC.

Explanation:
Fluorite (CaF₂): Ca²⁺ in FCC (4 per unit cell), F⁻ in ALL 8 tetrahedral voids. Ca:F = 4:8 = 1:2 ✓. Each Ca²⁺ has CN=8, each F⁻ has CN=4. Examples: CaF₂, UO₂, ThO₂. Anti-fluorite (Li₂O): anions in FCC, cations in all tetrahedral voids.

Explanation:
Perovskite (ABO₃): A cation (large, e.g., Ca, Ba) at corners (1 per unit cell), B cation (small, e.g., Ti) at body center (1), O²⁻ at face centers (3). B has CN=6 (octahedral), A has CN=12. Examples: BaTiO₃, SrTiO₃, PbTiO₃.

Explanation:
Schmid's law: τ_resolved = σ × cosφ × cosλ = σ × m. φ = angle between tensile axis and slip plane normal, λ = angle between tensile axis and slip direction. Maximum m = 0.5 when φ = λ = 45°.

Explanation:
m = cosφ × cosλ. Maximum when φ = λ = 45°: m_max = cos45° × cos45° = (1/√2)² = 1/2 = 0.5. This occurs when slip plane normal and slip direction are both at 45° to the tensile axis.

Explanation:
Tetragonal: a = b ≠ c, α = β = γ = 90°. Square prism unit cell. Examples: In (white tin), TiO₂ (rutile), SnO₂, martensite (BCT). Has 2 Bravais lattices: Simple Tetragonal and Body-Centered Tetragonal.

Explanation:
Orthorhombic: a ≠ b ≠ c, α = β = γ = 90°. Rectangular parallelepiped with all different edge lengths. Examples: α-sulfur, α-uranium, cementite (Fe₃C), BaSO₄. Has 4 Bravais lattices.

Explanation:
d₁₁₁ = a/√(1+1+1) = 0.405/√3 = 0.405/1.732 = 0.2338 ≈ 0.234 nm. The (111) plane is the close-packed plane in FCC Al. Actual a(Al) = 0.4049 nm.

Explanation:
SC: each atom has 6 nearest neighbors: 2 along ±x, 2 along ±y, 2 along ±z, all at distance a. CN: SC=6, BCC=8, FCC=HCP=12, Diamond=4.

Explanation:
Diamond cubic: CN = 4 (tetrahedral bonding). Each C atom forms 4 covalent bonds (sp³ hybridization). Bond angle = 109.47°. Low CN and directional bonds → low APF (0.34) and extreme hardness. Si and Ge also have diamond cubic structure.

Explanation:
Debye-Scherrer (powder diffraction): monochromatic X-rays on polycrystalline powder. Randomly oriented crystallites present all diffraction planes simultaneously. Diffracted beams form Debye-Scherrer cones. Used for lattice parameter determination, phase identification, grain size (Scherrer equation: D = Kλ/(β cosθ)).

Explanation:
Scherrer equation: D = Kλ/(β cosθ). D = crystallite size, K ≈ 0.9 (shape factor), λ = wavelength, β = peak FWHM (radians), θ = Bragg angle. Smaller crystallites → broader peaks. Valid for D < 100 nm.

Explanation:
Laue method: White (polychromatic) X-rays on a stationary single crystal. Each plane selects the appropriate wavelength to satisfy Bragg's law. Produces Laue spots used to determine crystal orientation and symmetry. Used for crystal orientation determination in industry.

Explanation:
FCC: atoms touch along face diagonal. 4r = a√2, so r = a√2/4 = a/(2√2) = 0.3536a. BCC: 4r = a√3, r = a√3/4 = 0.4330a. SC: 2r = a, r = a/2. Diamond: 8r = a√3, r = a√3/8.

Explanation:
BCC: atoms touch along body diagonal. 4r = a√3, so r = a√3/4 = 0.4330a. FCC: r = a√2/4 = 0.3536a. SC: r = a/2. BCC atoms are larger relative to lattice parameter than FCC atoms.

Explanation:
FCC (100): 4 corners × (1/4) = 1, 1 face-center × 1 = 1. Total = 2 atoms. Area = a². PD = 2/a². Ranking for FCC: PD(111) = 4/(√3a²) > PD(100) = 2/a² > PD(110) = √2/a².

Explanation:
FCC (110): rectangle a × a√2. Atoms: 4 corners × (1/4) = 1, 2 edge-centers × (1/2) = 1. Total = 2. Area = √2 a². PD = 2/(√2 a²) = √2/a² ≈ 1.414/a².

Explanation:
BCC (110): rectangle a × a√2. Atoms: 4 corners × (1/4) = 1, body-center at (a/2,a/2,a/2) lies on (110) since 1/2+1/2=1 ✓, shared between 2 cells → 1/2. Total = 1.5? Standard result: 1 atom in area √2a². PD = 1/(√2 a²).

Explanation:
BCC: shortest lattice vector = a/2<111>. Magnitude = a√3/2. For FCC: b = a/2<110>, magnitude = a/√2. Smaller |b| → lower dislocation energy (∝ b²). BCC dislocations can dissociate into a/8<111> + a/4<110> (non-planar).

Explanation:
Dislocation energy per unit length: E ≈ Gb²/2 (approximate). More precisely: E = (Gb²/4π) × ln(R/r₀). Since E ∝ b², dislocations prefer shortest Burgers vector. Frank's rule: b₁² > b₂² + b₃² → dissociation is energetically favorable.

Explanation:
Angle between planes = angle between normals. cosθ = (1+1+0)/[√3×√2] = 2/√6 = 0.8165. θ = cos⁻¹(0.8165) = 35.26°. Important for understanding slip plane-direction relationships in FCC metals.

Explanation:
cosθ = (1×1+0×1+0×1)/[√1×√3] = 1/√3 = 0.5774. θ = cos⁻¹(0.5774) = 54.74°. This is the angle between cube face and octahedral plane in FCC. Complementary angle = 90°-54.74° = 35.26°.

Explanation:
BCT: 8 corners × (1/8) = 1, plus 1 body-center = 1. Total = 2 atoms per unit cell. Same count as BCC but with a ≠ c. Martensite (BCT) in steel: c/a = 1 + 0.046 × (wt% C).

Explanation:
Martensite has BCT structure. Forms by diffusionless transformation from FCC austenite. Carbon atoms trapped in octahedral sites along c-axis distort BCC → BCT. c/a = 1 + 0.046 × (wt% C). Tetragonal distortion causes high hardness and brittleness.

Explanation:
Austenite (γ-iron) has FCC structure, stable 912°C–1394°C. Can dissolve up to 2.14 wt% C (at 1147°C) due to larger octahedral voids. Rapid cooling → martensite (BCT); slow cooling → pearlite (ferrite + cementite).

Explanation:
Ferrite (α-iron) has BCC structure, stable below 912°C. Dissolves only 0.022 wt% C at 727°C. Soft, ductile, ferromagnetic below 770°C (Curie temperature). 12 slip systems on {110}<111>.

Explanation:
Cementite (Fe₃C) has orthorhombic crystal structure (space group Pnma). Lattice parameters: a=0.452 nm, b=0.509 nm, c=0.674 nm. Very hard (HV≈1000) and brittle. Hardest phase in steel. Forms lamellae with ferrite in pearlite.

Explanation:
α-Ti has HCP structure at room temperature (a=0.295 nm, c=0.468 nm, c/a=1.587). Transforms to BCC (β-Ti) above 882°C (β-transus). This allotropic transformation is exploited in Ti alloys: α (HCP), β (BCC), α+β (Ti-6Al-4V).

Explanation:
Mg has HCP structure: a=0.321 nm, c=0.521 nm, c/a=1.624 (close to ideal 1.633). Density = 1.74 g/cm³ (lightest structural metal). Limited ductility at RT due to 3 basal slip systems. Ductility improves above 200°C.

Explanation:
Zn has HCP structure: a=0.266 nm, c=0.495 nm, c/a=1.856 (significantly > ideal 1.633). High c/a ratio makes Zn anisotropic. Used for galvanizing steel (corrosion protection). Zn is brittle at room temperature due to high c/a ratio.

Explanation:
FCC octahedral voids: 12 edge-centers × (1/4) = 3, plus 1 body-center = 1. Total = 4 per unit cell. With 4 atoms: 1 octahedral void per atom. In NaCl, Na⁺ occupies all octahedral voids in Cl⁻ FCC lattice.

Explanation:
FCC: 8 tetrahedral voids per unit cell (one in each octant at positions like a/4,a/4,a/4). With 4 atoms: 8/4 = 2 tetrahedral voids per atom. In zinc blende, Zn²⁺ occupies half (4) of these tetrahedral voids.

Explanation:
Reciprocal lattice vector G_hkl = h·a* + k·b* + l·c* is perpendicular to real-space plane (hkl). |G_hkl| = 1/d_hkl. The reciprocal lattice is the Fourier transform of the real lattice. Bragg's law in reciprocal space: diffraction when reciprocal lattice point lies on Ewald sphere.

Explanation:
d₁₁₀ = 0.287/√2 = 0.2029 nm. Bragg's law: sinθ = λ/(2d) = 0.154/(2×0.2029) = 0.154/0.4058 = 0.3794. θ = sin⁻¹(0.3794) = 22.3° (Bragg angle). The diffraction angle 2θ = 44.6°.

Explanation:
FCC (111): equilateral triangle with side a/√2 × √2 = a√2... Atoms: 3 corners × (1/6) = 1/2, 3 edge-centers × (1/2) = 3/2. Total = 2. Area = (√3/4)(a√2)² = √3a²/2. PD = 2/(√3a²/2) = 4/(√3a²) ≈ 2.31/a². Highest planar density in FCC.

Explanation:
There are 230 space groups in 3D, combining 14 Bravais lattices with 32 point groups and translational symmetry elements (screw axes, glide planes). 7 crystal systems → 14 Bravais lattices → 32 point groups → 230 space groups. All crystal structures belong to one of these 230 space groups.

Explanation:
There are 32 crystallographic point groups in 3D, also called crystal classes. They describe the symmetry of the external form of crystals. 7 crystal systems → 32 point groups → 230 space groups. Point groups include rotation axes, mirror planes, inversion centers, and improper rotation axes.

Explanation:
Tungsten (W) has BCC crystal structure with a = 0.316 nm. W has the highest melting point of all metals (3422°C) and highest tensile strength at high temperatures. BCC metals: W, Mo, Cr, V, Nb, Ta, α-Fe. W is used in filaments, cutting tools, and high-temperature applications.

Explanation:
Aluminum (Al) has FCC crystal structure with a = 0.4049 nm, density = 2.70 g/cm³. FCC Al has 12 slip systems ({111}<110>), giving good ductility. Al is widely used in aerospace and automotive industries due to its low density and good corrosion resistance.

Explanation:
Nickel (Ni) has FCC crystal structure with a = 0.352 nm. Ni is ferromagnetic below its Curie temperature (358°C). FCC Ni has good ductility and corrosion resistance. Used in superalloys (Inconel, Hastelloy) for high-temperature applications.

Explanation:
Cobalt (Co) has HCP structure at room temperature (α-Co), transforming to FCC (β-Co) above 417°C. HCP Co: a=0.251 nm, c=0.407 nm, c/a=1.623. Co is ferromagnetic and used in hard magnets, cutting tools (WC-Co), and superalloys.

Explanation:
Beryllium (Be) has HCP structure: a=0.229 nm, c=0.358 nm, c/a=1.568 (less than ideal 1.633). Be has the highest stiffness-to-weight ratio of any metal. Used in aerospace, nuclear reactors (neutron moderator/reflector), and X-ray windows.

Explanation:
Lead (Pb) has FCC crystal structure with a = 0.495 nm, density = 11.35 g/cm³ (one of the densest common metals). Pb has a very low melting point (327°C) for an FCC metal. Used in batteries, radiation shielding, and solders.

Explanation:
Silver (Ag) has FCC crystal structure with a = 0.409 nm. Ag has the highest electrical and thermal conductivity of all metals. FCC structure gives good ductility. Used in electrical contacts, jewelry, photography, and antimicrobial applications.

Explanation:
Gold (Au) has FCC crystal structure with a = 0.408 nm. Au is extremely ductile (can be beaten into sheets 0.1 μm thick) due to FCC structure with 12 slip systems. Au is chemically inert and used in electronics, jewelry, and dentistry.

Explanation:
Chromium (Cr) has BCC crystal structure with a = 0.288 nm. Cr is hard and brittle at room temperature due to BCC structure and limited slip systems. Used in stainless steel (>10.5% Cr for corrosion resistance), hard chrome plating, and tool steels.

Explanation:
Molybdenum (Mo) has BCC crystal structure with a = 0.315 nm, melting point = 2623°C. Mo is used in high-temperature alloys, electrical contacts, and as a lubricant (MoS₂). BCC Mo has high strength at elevated temperatures.

Explanation:
d₂₂₀ = a/√(4+4+0) = 0.405/√8 = 0.405/2.828 = 0.1432 ≈ 0.143 nm. The (220) reflection is allowed in FCC (all even). This is the 3rd peak in FCC Al diffraction pattern. d₂₂₀ = d₁₁₀/2 = a/(2√2).

Explanation:
d₃₁₁ = a/√(9+1+1) = 0.3615/√11 = 0.3615/3.317 = 0.1090 ≈ 0.109 nm. The (311) reflection is allowed in FCC (all odd). This is the 4th peak in FCC Cu diffraction pattern.

Explanation:
ρ = 2M/(Nₐa³). a = 2.87×10⁻⁸ cm. a³ = 23.65×10⁻²⁴ cm³. ρ = (2×55.85)/(6.023×10²³×23.65×10⁻²⁴) = 111.7/14.24 = 7.84 ≈ 7.87 g/cm³. Experimental density of Fe = 7.87 g/cm³.

Explanation:
ρ = 4M/(Nₐa³). a = 4.05×10⁻⁸ cm. a³ = 66.43×10⁻²⁴ cm³. ρ = (4×26.98)/(6.023×10²³×66.43×10⁻²⁴) = 107.92/40.01 = 2.698 ≈ 2.70 g/cm³. Experimental density of Al = 2.70 g/cm³.

Explanation:
FCC [100] direction: length = a. Atoms: 2 corner atoms × (1/2) = 1 atom. Linear density = 1/a. Compare: [110] LD = √2/a (close-packed direction, higher density). [111] LD = 0 in FCC (no atom at body center). The [110] direction has the highest linear density in FCC.

Explanation:
BCC [111] direction (close-packed): length = a√3. Atoms: 2 corner atoms × (1/2) + 1 body-center atom = 1 + 1 = 2 atoms. Linear density = 2/(a√3) = 2/(√3 a). This is the highest linear density in BCC, confirming [111] as the close-packed direction.

Explanation:
BCC [110] direction: length = a√2 (face diagonal). Atoms: 2 corner atoms × (1/2) = 1 atom (body center does NOT lie on [110] face diagonal). Linear density = 1/(a√2) = 1/(√2 a). Compare: BCC [111] LD = 2/(√3 a) > BCC [110] LD = 1/(√2 a), confirming [111] is close-packed.

Explanation:
cosθ = (1×0+0×1+0×1)/[√1×√2] = 0/√2 = 0. θ = 90°. Wait: (1×0+0×1+0×1) = 0. θ = 90°. But option D is 90°. Let me recalculate: [100]·[011] = 1×0+0×1+0×1 = 0. cosθ = 0/[1×√2] = 0. θ = 90°. Answer is D=90°. Correcting: correct answer is D.

Explanation:
cosθ = (1×0+1×0+0×1)/[√2×√1] = 0/√2 = 0. θ = cos⁻¹(0) = 90°. The [110] direction lies in the (001) plane (Weiss zone law: 0×1+0×1+1×0=0 ✓), so [110] is perpendicular to [001].

Explanation:
cosθ = (1×0+0×1+0×1)/[√1×√2] = 0/√2 = 0. θ = cos⁻¹(0) = 90°. The [100] and [011] directions are perpendicular. Verification using Weiss zone law: [011] lies in (100) plane since 1×0+0×1+0×1=0 ✓.

Explanation:
Face-centered orthorhombic: 8 corners × (1/8) = 1, plus 6 face-centers × (1/2) = 3. Total = 4 atoms per unit cell. Same count as FCC but with a ≠ b ≠ c. Examples: α-sulfur (orthorhombic), some intermetallic compounds.

Explanation:
Silicon (Si) has diamond cubic structure with a = 0.543 nm. Each Si atom forms 4 covalent bonds (sp³ hybridization, CN=4). APF = 0.34. Si is a semiconductor with band gap = 1.12 eV. Used in microelectronics, solar cells, and MEMS.

Explanation:
Germanium (Ge) has diamond cubic structure with a = 0.566 nm. Like Si, each Ge atom forms 4 covalent bonds (sp³). Ge is a semiconductor with band gap = 0.67 eV. Used in transistors, infrared optics, and fiber optic systems.

Explanation:
GaAs has zinc blende structure: As³⁻ in FCC positions, Ga³⁺ in half the tetrahedral voids. CN = 4 for both ions. GaAs is a direct band gap semiconductor (Eg = 1.42 eV) used in LEDs, laser diodes, solar cells, and high-frequency transistors.

Explanation:
MgO has NaCl (rock salt) structure: two interpenetrating FCC lattices of Mg²⁺ and O²⁻. CN = 6 for both ions. Radius ratio r_Mg/r_O = 0.072/0.140 = 0.514 (range 0.414-0.732 for octahedral). MgO is a refractory ceramic with melting point 2852°C.

Explanation:
TiO₂ (rutile) has tetragonal crystal structure: a = b = 0.459 nm, c = 0.296 nm, c/a = 0.644. Ti⁴⁺ at corners and body center, O²⁻ at specific positions. Each Ti⁴⁺ has CN=6 (octahedral), each O²⁻ has CN=3 (trigonal planar). Used as white pigment, photocatalyst, and in solar cells.

Explanation:
α-Al₂O₃ (corundum) has rhombohedral (trigonal) crystal structure. O²⁻ in HCP arrangement, Al³⁺ in 2/3 of octahedral voids. CN(Al) = 6, CN(O) = 4. Melting point = 2054°C. Used as abrasive (emery, corundum), refractory, and substrate for electronics. Ruby and sapphire are Al₂O₃ with trace impurities.

Explanation:
The Ewald sphere (radius = 1/λ) in reciprocal space shows diffraction conditions: diffraction occurs when a reciprocal lattice point lies on the Ewald sphere surface. This is equivalent to Bragg's law. The construction visualizes which reflections are accessible for a given wavelength and crystal orientation.

Explanation:
BCC: 4r = a√3, so a = 4r/√3 = 4×0.132/1.732 = 0.528/1.732 = 0.3049 ≈ 0.305 nm. Actual lattice parameter of V = 0.303 nm. V is used in vanadium steel alloys and as a catalyst.

Explanation:
FCC: 4r = a√2, so a = 4r/√2 = 2√2 × r = 2×1.414×0.125 = 0.3536 ≈ 0.354 nm. Actual lattice parameter of Ni = 0.352 nm. Good agreement confirms hard-sphere model.

Explanation:
NaCl: Na⁺ at edge centers (12×1/4=3) + body center (1) = 4 Na⁺. Cl⁻ at corners (8×1/8=1) + face centers (6×1/2=3) = 4 Cl⁻. Total = 4 NaCl formula units per unit cell. Alternatively: two interpenetrating FCC lattices, each with 4 atoms → 4 formula units.

Explanation:
CsCl: Cs⁺ at body center (1 atom), Cl⁻ at corners (8×1/8=1 atom). Total = 1 CsCl formula unit per unit cell. This is a simple cubic lattice with a 2-atom basis, NOT a BCC structure (BCC has same atoms at corners and body center).

Explanation:
ZnS zinc blende: S²⁻ in FCC positions (4 per unit cell), Zn²⁺ in 4 of 8 tetrahedral voids (4 per unit cell). Total = 4 ZnS formula units per unit cell. The 4 Zn²⁺ are at positions (1/4,1/4,1/4), (3/4,3/4,1/4), (3/4,1/4,3/4), (1/4,3/4,3/4).

Explanation:
The Curie temperature of α-Fe (BCC ferrite) is 770°C. Below 770°C, Fe is ferromagnetic. Above 770°C (but below 912°C), Fe is paramagnetic but still BCC. At 912°C, Fe transforms from BCC (α) to FCC (γ). The Curie temperature is NOT a crystal structure change, only a magnetic transition.

Explanation:
δ-iron has BCC crystal structure, stable from 1394°C to 1538°C (melting point). Iron allotropes: α-Fe (BCC, <912°C), γ-Fe (FCC, 912-1394°C), δ-Fe (BCC, 1394-1538°C). δ-Fe has a larger lattice parameter (a=0.293 nm) than α-Fe (a=0.287 nm) due to higher temperature.

Explanation:
Max C solubility in α-Fe (BCC ferrite) = 0.022 wt% at 727°C (eutectoid temperature). Very low because BCC octahedral voids are smaller (r_void/r_Fe ≈ 0.154) than FCC octahedral voids (r_void/r_Fe ≈ 0.414). Carbon atoms cause significant lattice distortion in BCC.

Explanation:
Zn has c/a = 1.856, significantly greater than ideal (1.633). This means Zn is elongated along c-axis. Compare: Be (1.568), Ti (1.587), Mg (1.624), ideal (1.633), Co (1.623), Zn (1.856), Cd (1.886). Metals with c/a > 1.633 tend to be more anisotropic.

Explanation:
Mg has c/a = 1.624, close to ideal (1.633). This near-ideal ratio means Mg has nearly isotropic properties in the basal plane. Compare: Be (1.568), Ti (1.587), Mg (1.624), ideal (1.633), Zn (1.856). Mg's near-ideal c/a contributes to its relatively good ductility among HCP metals.

Explanation:
Platinum (Pt) has FCC crystal structure with a = 0.392 nm, density = 21.45 g/cm³ (one of the densest metals). Pt is chemically inert and used as a catalyst (catalytic converters, fuel cells), in jewelry, and in laboratory equipment.

Explanation:
Cadmium (Cd) has HCP structure with a=0.298 nm, c=0.562 nm, c/a=1.886 (highest among common HCP metals). Cd is used in Ni-Cd batteries, electroplating, and as a neutron absorber in nuclear reactors. High c/a makes Cd highly anisotropic.

Explanation:
Niobium (Nb) has BCC crystal structure with a=0.330 nm. Nb is a superconductor below 9.2 K and is used in superconducting magnets, superalloys (for jet engines), and as a steel microalloying element for grain refinement.

Explanation:
Tantalum (Ta) has BCC crystal structure with a=0.330 nm, melting point=2996°C. Ta is used in capacitors (tantalum capacitors), surgical implants (biocompatible), and high-temperature alloys. BCC Ta has good ductility despite high melting point.

Explanation:
α-Zr has HCP structure at room temperature (a=0.323 nm, c=0.515 nm, c/a=1.593). Transforms to BCC (β-Zr) above 863°C. Zr has very low neutron absorption cross-section, making it ideal for nuclear reactor fuel cladding. Zircaloy alloys are used in nuclear applications.

Explanation:
d₂₀₀ = a/√(4+0+0) = 0.287/2 = 0.1435 nm. The (200) reflection is allowed in BCC (h+k+l=2, even). This is the second peak in BCC iron diffraction pattern. d₂₀₀ = a/2.

Explanation:
d₂₁₁ = a/√(4+1+1) = 0.287/√6 = 0.287/2.449 = 0.1172 ≈ 0.117 nm. The (211) reflection is allowed in BCC (h+k+l=4, even). This is the third peak in BCC iron diffraction pattern.

Explanation:
Octahedral void radius ratio: r_oct/R = 0.414. Tetrahedral void radius ratio: r_tet/R = 0.225. Ratio of octahedral to tetrahedral void size = 0.414/0.225 ≈ 1.84. Octahedral voids are larger than tetrahedral voids. This is why larger interstitial atoms (like C in Fe) prefer octahedral sites.

Explanation:
In BCC, octahedral voids are at face centers and edge centers. The void at face center (1/2,1/2,0): surrounded by 4 atoms at distance a/2 in the plane and 2 atoms at distance a/2 perpendicular. Void radius = a/2 - r_Fe = 0.287/2 - 0.124 = 0.1435 - 0.124 = 0.0195 ≈ 0.019 nm. Wait: the BCC octahedral void is non-regular; the smaller dimension gives r_void = a/2 - r_Fe ≈ 0.020 nm. Carbon radius = 0.077 nm >> 0.020 nm, causing large distortion.

Explanation:
In FCC, octahedral void at body center: r_void = a/2 - r_Fe = 0.365/2 - 0.129 = 0.1825 - 0.129 = 0.0535 ≈ 0.053 nm. Alternatively: r_void = 0.414 × r_Fe = 0.414 × 0.129 = 0.053 nm. Carbon radius = 0.077 nm > 0.053 nm, but FCC can accommodate more C than BCC due to larger void size.

Explanation:
Cubic SiC (β-SiC or 3C-SiC) has zinc blende structure: C atoms in FCC positions, Si atoms in half the tetrahedral voids (or vice versa). CN = 4 for both Si and C. SiC is a wide band gap semiconductor (Eg = 2.36 eV for 3C-SiC) used in power electronics and high-temperature applications.

Explanation:
BaTiO₃ above Curie temperature (120°C) has cubic perovskite structure: Ba²⁺ at corners, Ti⁴⁺ at body center, O²⁻ at face centers. Below 120°C: tetragonal (ferroelectric). Below 5°C: orthorhombic. Below -90°C: rhombohedral. The cubic phase is paraelectric (centrosymmetric).

Explanation:
FeO (wustite) has NaCl (rock salt) structure: two interpenetrating FCC lattices of Fe²⁺ and O²⁻. CN = 6 for both ions. FeO is always non-stoichiometric (Fe₁₋ₓO, x = 0.05-0.17) due to Fe²⁺ vacancies compensated by Fe³⁺. Used as a precursor in iron production.

Explanation:
UO₂ has fluorite structure: U⁴⁺ in FCC positions, O²⁻ in all 8 tetrahedral voids. U:O = 4:8 = 1:2 ✓. CN(U) = 8, CN(O) = 4. UO₂ is the primary nuclear fuel. Its fluorite structure can accommodate excess oxygen (hyperstoichiometric UO₂₊ₓ) due to available interstitial sites.

Explanation:
Normal spinel (MgAl₂O₄): O²⁻ in FCC (32 per unit cell with 8 formula units). Mg²⁺ in 1/8 of tetrahedral voids (8 of 64 = 8 per unit cell). Al³⁺ in 1/2 of octahedral voids (16 of 32 = 16 per unit cell). Formula: 8 Mg + 16 Al + 32 O = 8 MgAl₂O₄. Inverse spinel: half Al³⁺ in tetrahedral, Mg²⁺ + half Al³⁺ in octahedral.

Explanation:
Graphite has hexagonal layered structure with ABABAB stacking of graphene layers. Within each layer: sp² hybridized C atoms in hexagonal arrangement with strong covalent bonds (C-C = 0.142 nm). Between layers: weak van der Waals forces (interlayer spacing = 0.335 nm). This anisotropy makes graphite a good lubricant and electrical conductor along layers.

Explanation:
C₆₀ (Buckminsterfullerene) molecules pack in FCC arrangement at room temperature with a = 1.417 nm. Each C₆₀ molecule is a truncated icosahedron with 60 carbon atoms. The FCC packing of C₆₀ molecules is stabilized by van der Waals interactions. Above 249 K, C₆₀ molecules rotate freely (orientationally disordered FCC).

Explanation:
NiAl has CsCl-type (B2) ordered structure: Ni at body center, Al at corners (or vice versa). This is an ordered BCC-based structure. NiAl has high melting point (1638°C), low density (5.86 g/cm³), and good oxidation resistance. Used in high-temperature coatings and as a structural intermetallic.

Explanation:
Ordered Cu₃Au has L1₂ structure: Au at corners, Cu at face centers of FCC unit cell. This is an ordered FCC-based structure. Above the order-disorder transition temperature (390°C), Cu and Au randomly occupy FCC sites (disordered FCC). Ordering increases hardness and decreases electrical conductivity.

Explanation:
Ordered Fe₃Al has DO₃ structure: an ordered BCC-based superstructure. The unit cell is 8 times the BCC unit cell. Fe₃Al has good oxidation and sulfidation resistance, low density, and moderate strength. Used in high-temperature structural applications. Disordering occurs above ~540°C (B2 structure) and ~820°C (disordered BCC).

Explanation:
γ-TiAl has L1₀ (ordered face-centered tetragonal) structure: alternating layers of Ti and Al atoms on {001} planes of an FCC-based structure. c/a ≈ 1.02 (slightly tetragonal). γ-TiAl has low density (3.9 g/cm³), high melting point (1460°C), and good high-temperature strength. Used in jet engine components.