Find the interplanar spacing and diffraction angle 2θ from the lattice parameter a and Miller indices (hkl), and check with the extinction rules for each crystal structure whether a peak appears. The defaults are BCC iron (a = 0.2866 nm) and Cu Kα radiation (λ = 0.15406 nm).
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Results
Interplanar spacing d
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Diffraction angle 2θ
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Extinction rule
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Peaks from the low-angle side (up to 6)
(hkl) and 2θ
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Calculated for first-order diffraction (n = 1). Planes for which λ/2d exceeds 1 do not diffract at that wavelength. Peak intensities (structure factor magnitude, multiplicity and temperature factor) are not calculated.
See it as a graph
Other inputs stay at their current values while the chosen input varies from 1/4 to 4 times its current value. The red dot is the current value.
Bragg's law and cubic interplanar spacing
When X-rays of wavelength λ strike a crystal, strong diffraction occurs at angles θ where reflections from atomic planes spaced d apart reinforce each other. This condition is Bragg's law, nλ = 2d·sinθ. In a cubic crystal, the spacing of planes with Miller indices (hkl) is d = a/√(h² + k² + l²). For the (110) planes of the default BCC iron (a = 0.2866 nm), d = 0.2866/√2 ≈ 0.2027 nm, and with Cu Kα radiation (λ = 0.15406 nm), sinθ = 0.15406/(2 × 0.2027) ≈ 0.380, giving 2θ ≈ 44.68°. XRD instruments output patterns with 2θ on the horizontal axis, so calculations are usually compared in 2θ as well.
Extinction rules: why expected peaks are missing
When the unit cell contains several atoms, the waves scattered from them can cancel, making the diffracted intensity from certain planes zero. These are the extinction rules, which can be derived by calculating the structure factor F = Σ f·exp(2πi(hx + ky + lz)). In body-centered cubic, only planes with h + k + l even appear: (100) is absent, and the sequence runs (110), (200), (211), (220)…. In face-centered cubic, only planes with h, k and l all even or all odd appear: (111), (200), (220), (311)…. In the diamond structure (Si, Ge and so on), planes with all-even indices must in addition have h + k + l divisible by 4, so (200) is absent and the sequence is (111), (220), (311), (400)….
Identifying the crystal structure from the peak sequence
In cubic crystals sin²θ is proportional to h² + k² + l², so taking the ratios of sin²θ for the peaks in order from low angle identifies the structure. For BCC the ratios run 2 : 4 : 6 : 8 … (1 : 2 : 3 : 4 …), and for FCC 3 : 4 : 8 : 11 : 12 …. SC gives 1 : 2 : 3 : 4 : 5 : 6 : 8 …, characterized by the missing 7. If the first two peaks are in the ratio 3 : 4, it is likely FCC; if 1 : 2, BCC or SC (SC if 7 is missing from the sequence). Switch the structure in this calculator to see how the list of peaks changes.
When calculation and measurement disagree
Measured peak positions can deviate from calculation by around 0.1° because of specimen height displacement, zero-point error, changes in lattice parameter due to alloying elements, residual stress and so on. To determine lattice parameters accurately, several high-angle peaks are used and extrapolated. Cu Kα radiation also consists of Kα1 (0.15406 nm) and Kα2 (0.15444 nm), and at high angles peaks appear split in two. Peak widths broaden with crystallite size and strain, and the Scherrer equation D = Kλ/(β·cosθ) gives an estimate of crystallite size.
From Bragg's law 2d sinθ = nλ (n = 1), sinθ = 0.15406/(2 × 0.3135) ≈ 0.2457, so θ ≈ 14.2° and the diffraction angle is 2θ ≈ 28.4°. XRD instruments and data normally plot 2θ, the angle between the incident and diffracted X-rays, on the horizontal axis. About 14.2° is the mistake of answering θ itself, and about 58.9° drops the 2 in Bragg's equation and uses sinθ = λ/d. The peak of silicon powder near 28.4° is strong and sharp, so it is also widely used as a standard for calibrating diffraction angles.
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.
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.
FAQ
What is n in Bragg's law?
It is the order of diffraction. Diffraction from (110) planes with n = 2 appears at the same angle as first-order diffraction from (220) planes, which have half the spacing. In XRD, therefore, n is usually taken as 1 and the reflection is labeled by planes with doubled indices, such as (220).
Why does BCC show no (100) peak?
The body-center atoms form planes of the same density exactly halfway between the (100) planes formed by the corner atoms, and the waves scattered from them are half a wavelength out of step and cancel. All planes with h + k + l odd vanish for the same reason.
What is the wavelength of Cu Kα radiation?
Kα1 is 0.15406 nm and Kα2 is 0.15444 nm, with an intensity ratio of about 2 : 1. When they are not separated, the weighted average of about 0.15418 nm is sometimes used. Co Kα (about 0.1789 nm) and Mo Kα (about 0.0709 nm) are also commonly used.