Q1(b). Spacing of the (123) planes of an orthorhombic cell

1. Miller indices (123): the plane nearest the origin cuts the axes at a/1, b/2 and c/3. All planes of the family are parallel and equally spaced; the perpendicular distance between two successive planes is d₁₂₃, the plane through O being the first.
2. Spacing formula for axes at right angles (the normal to the plane makes angles whose direction cosines are hd/a, kd/b, ld/c, and their squares add to 1): 1dₕₖₗ²=h²a²+k²b²+l²c²
3. Substitution: 1d²=1²(0.5)²+2²(1.0)²+3²(1.5)²=4+4+4=12 Å⁻²
4. Result: d₁₂₃=(1)/(√(12)) Å, d₁₂₃=0.289 Å=28.9 pm
5. Check: each of the three intercepts equals 0.5 Å here, so the (123) plane cuts the three edges at the same distance from O; d is smaller than every intercept, as it must be for the perpendicular.
6. Higher orders: planes written (246) have half this spacing (0.144 Å), but they are the second-order reflection from the same (123) planes (2d₁₂₃θ=2λ), not a new set of lattice planes. The spacing between successive 123-planes of the lattice is 0.289 Å.

Application. with Bragg's law, λ=2dθ, the measured angles of an X-ray powder pattern give the set of dₕₖₗ; fitting them to the orthorhombic formula gives a, b and c (indexing the pattern).