Rules by which diffraction from certain indices cancels out and disappears because of the arrangement of atoms. In BCC, only planes with h + k + l even diffract; in FCC, only planes whose h, k and l are all even or all odd.
Japanese name: 消滅則
Extinction rules appears in 2 practice questions across 1 topic, including Materials characterization (SEM, TEM, XRD, XPS).
X-rays scattered by each atom in the unit cell overlap with phase differences that depend on the atoms' positions. Their sum is the structure factor F(hkl) = Σ fⱼ exp{2πi(hxⱼ + kyⱼ + lzⱼ)}, and the diffracted intensity is proportional to |F|². In BCC, the 2 atoms at the origin and the body center (1/2, 1/2, 1/2) give F = f{1 + (−1)^(h+k+l)}, which is 0 when h + k + l is odd. In FCC, the 4 atoms give F = 4f when h, k and l are all even or all odd, and 0 when they are mixed. The extinction rules describe how intensity becomes 0 through cancellation within the unit cell even when Bragg's condition is satisfied.
In cubic crystals sin²θ is proportional to h² + k² + l², so the lattice type can be found from the ratios of the peaks' sin²θ. Simple cubic gives 1, 2, 3, 4, 5, 6, 8… with 7 missing; BCC gives an evenly spaced 2, 4, 6, 8, 10… (ratios 1:2:3:4…); and FCC gives 3, 4, 8, 11, 12…, a sequence that comes in pairs. In diamond-structure silicon, planes with h + k + l = 4n + 2 also vanish in addition to the FCC condition, so (200) does not appear and the peaks run (111), (220), (311), (400).
The extinction rules assume that identical atoms sit at the positions that cancel. When different atoms occupy the body center and the corners, as in CsCl-type NiAl, reflections such as (100) that vanish in BCC appear as weak peaks proportional to the difference in scattering factors (superlattice reflections). The same happens on ordering of Cu₃Au, and the superlattice intensity serves as a measure of the degree of order. In the NaCl type, the intensity of planes with h, k and l all odd is set by the difference between the two ions' scattering factors; in KCl, where K⁺ and Cl⁻ have equal numbers of electrons, (111) nearly vanishes and the pattern looks simple cubic.
In real measurements, reflections that should be absent (forbidden reflections) are sometimes seen. In single crystals, multiple diffraction, where X-rays diffracted once are diffracted again by another set of planes, can produce sharp peaks at forbidden positions such as silicon (200). Silicon (222) is known to appear very weakly because the distribution of covalent bonding electrons is not spherically symmetric. In crystals with screw axes or glide planes, there are also extinction rules under which only particular rows of indices vanish, and these are used to determine the space group. It is important not to judge only by the presence or absence of a peak but to check the measurement conditions as well.
Answer: C. (111)
In FCC, diffraction appears only when h, k and l are all even or all odd; when even and odd are mixed, the scattering from the four atoms in the unit cell cancels and the intensity is zero (the extinction rules). Peaks therefore appear in the order (111), (200), (220), (311)…, and (111), with the smallest h² + k² + l², has the largest interplanar spacing and the lowest angle. (100) and (110) both mix even and odd indices and so do not appear in FCC. In BCC only planes with h + k + l even appear, so the first peak is (110), and the sequence of peaks distinguishes FCC from BCC.
Answer: C. About 0.362 nm
θ = 21.65° and sinθ ≈ 0.3689, so Bragg's law gives d = λ/(2 sinθ) ≈ 0.2088 nm. In FCC the lowest-angle peak is (111), so a = d × √(1² + 1² + 1²) = 0.2088 × √3 ≈ 0.362 nm, matching the lattice parameter of copper (about 0.3615 nm). About 0.209 nm is the interplanar spacing d itself, without converting to the lattice parameter. About 0.295 nm treats the first peak as (110) and multiplies by √2, the mistake of applying the BCC extinction rules. Correct indexing of the peaks is a prerequisite for determining the lattice parameter.
In HCP, planes with l odd and h + 2k a multiple of 3 vanish. For example, (0001) does not appear, and diffraction from the basal plane is observed as (0002).
The planes exist, but scattering from the atoms in the unit cell cancels and the intensity becomes 0. For example, the reflection from the BCC (100) planes cancels because it is opposite in phase to the reflection from the planes of body-center atoms lying halfway between them.
Yes. The conditions for absence come from the same structure factor. However, electrons scatter strongly, and forbidden reflections appear easily through multiple scattering, so care is needed when indexing.
In FCC, diffraction appears only when h, k and l are all even or all odd; when even and odd are mixed, the scattering from the four atoms in the unit cell cancels and the intensity is zero (the extinction rules).
In powder XRD of a metal with the FCC structure, what are t…About 0.295 nm treats the first peak as (110) and multiplies by √2, the mistake of applying the BCC extinction rules.
In powder XRD of copper (FCC) with Cu Kα radiation (wavelen…XRD (X-ray diffraction) Bragg's law EDS (energy-dispersive X-ray spectroscopy) Scherrer equation SEM (scanning electron microscope) SIMS (secondary ion mass spectrometry) XPS (X-ray photoelectron spectroscopy)