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Interplanar Spacing for Cubic Crystals

Calculates the spacing between parallel lattice planes in a cubic crystal from the lattice parameter and Miller indices.

Materials ScienceCrystallographyResearch

Interplanar Spacing Calculatordhkl = a / √(h² + k² + l²)

dhkl = a / √( h² + k² + l² )
dhkl = interplanar spacing (Å)  ·  a = lattice parameter (Å)  ·  h, k, l = Miller indices
⟹ Solved, a, h, k, l
Å
Å
Solve for:
Presets:
d
d: a: h: k: l:
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Interplanar Spacing Gauge
Small (< 1.0 Å) Medium (1.0–3.0 Å) Large (3.0–5.0 Å) Very Large (> 5.0 Å)
dhkl = a / √(h² + k² + l²)  ·  Units: Å for d and a, Miller indices are integers

Interpretation

d_hkl = a / √(h² + k² + l²). Distance between parallel planes in a cubic lattice. Used in XRD to identify crystal structure and lattice parameter. Essential for crystallography.

d_hkl = a / sqrt(h^2 + k^2 + l^2)
Interplanar Spacing for Cubic Crystals

Variables

SymbolQuantityUnit
d_hklInterplanar spacingangstrom
aCubic lattice parameterangstrom
h, k, lMiller indices of the crystal plane

What it means

In a cubic crystal system, the interplanar spacing d_hkl for the plane with Miller indices (hkl) is given by d_hkl = a / √(h² + k² + l²), where a is the lattice constant. This formula is derived from the geometry of the cubic unit cell. It is used in X‑ray diffraction (Bragg’s law) to determine the positions of diffraction peaks, enabling identification of crystal structure and accurate lattice parameters. By measuring the angles of diffraction peaks, one can index the pattern and determine the crystal system. This is fundamental in materials characterisation, mineralogy, and solid‑state chemistry. Understanding this spacing is crucial for interpreting diffraction patterns and for analysing polycrystalline materials.

Worked example

Interplanar Spacing – Two Examples

Real‑World
Scenario: A cubic crystal with lattice parameter a = 3.615 Å. The crystallographer calculates the interplanar spacing for the (111) plane to interpret an XRD pattern and identify the crystal structure.
ParameterValue
a3.615 Å
(h,k,l)(1,1,1)
1d₁₁₁ = 3.615/√(1+1+1) = 3.615/√3 = 3.615/1.732 = 2.087 Å
Result 2.087 Å ✓ d‑spacing
Scenario: A silicon crystal has a = 5.43 Å. The semiconductor engineer calculates the interplanar spacing for the (220) plane to verify the crystal orientation in a wafer.
ParameterValue
a5.43 Å
(h,k,l)(2,2,0)
1d₂₂₀ = 5.43/√(4+4+0) = 5.43/√8 = 5.43/2.828 = 1.92 Å
Result 1.92 Å ✓ Smaller spacing
Materials insight: Interplanar spacing depends on the lattice parameter and Miller indices. It is critical for X‑ray diffraction analysis of crystal structures.

Common mistakes

  • Interplanar spacing for cubic crystals: d_hkl = a / √(h²+k²+l²).
  • Lattice parameter a: The length of the cubic unit cell edge – in m or Å.
  • Miller indices h, k, l: Integers – do not reduce fractions incorrectly.
  • Assumes orthogonal axes: Valid only for cubic systems – for other crystal systems, use the more general formula.
  • Forbidden reflections: Some combinations may have zero intensity (e.g., FCC has systematic absences).

Applications

The interplanar spacing for cubic crystals, d_hkl = a / √(h² + k² + l²), gives the distance between parallel planes with Miller indices (hkl). It is used in X‑ray diffraction to identify peaks, to index patterns, and to determine crystal structure. Materials scientists rely on this to analyse polycrystalline and single‑crystal materials, to measure lattice parameters, and to study phase transformations. By calculating d_hkl, they can assign reflections in diffraction patterns, enabling phase identification and microstructural characterisation. This formula is a workhorse of crystallographic analysis.

  • Indexing of X‑ray and electron diffraction patterns
  • Phase identification in metals, ceramics, and minerals
  • Lattice parameter determination from diffraction data
  • Study of epitaxial relationships and texture
  • Residual stress measurement via peak shift

Frequently Asked Questions

Q01What is the formula for interplanar spacing in cubic crystals?
A01

The interplanar spacing d_hkl for a cubic crystal with lattice parameter a is d_hkl = a / √(h² + k² + l²), where (hkl) are the Miller indices of the plane.

Q02What is the common mistake when using this formula?
A02

Using the cubic‑only formula for a non‑cubic crystal system (tetragonal, orthorhombic, etc.), where a different, more complex spacing equation is required.

Q03What is the significance of the denominator in the formula?
A03

The denominator √(h² + k² + l²) is the length of the reciprocal lattice vector. The interplanar spacing is inversely proportional to this length.

Q04How do you determine the lattice parameter a from X‑ray diffraction?
A04

Using Bragg's law: nλ = 2d sinθ. For a known peak (hkl), d = a / √(h²+k²+l²). Measure θ, then solve for a.

Q05What are the d‑spacings for common cubic structures?
A05

For a simple cubic (a=1): d_100 = 1; d_110 = 1/√2 ≈ 0.707; d_111 = 1/√3 ≈ 0.577. These ratios are characteristic of the structure.

Q06How does the interplanar spacing change with the order of diffraction?
A06

The formula gives the spacing for the actual lattice planes. For a given plane, higher order diffraction (n>1) corresponds to the same d but different path difference.

Q07What are the limitations of the formula?
A07

  • Only for cubic systems.
  • Assumes perfect crystals.
  • Does not account for thermal expansion (a changes with temperature).

Q08How do you index a diffraction pattern from a cubic crystal?
A08

Measure the d‑spacings of the peaks. For a cubic crystal, the ratio of d² values (or sin²θ) corresponds to reciprocal of (h²+k²+l²). Compare with allowed reflections for the structure (FCC, BCC, etc.).

Q09What is the relation between interplanar spacing and the reciprocal lattice?
A09

The interplanar spacing is the reciprocal of the magnitude of the reciprocal lattice vector G_hkl: d = 2π / |G| (in some conventions).

Q10How do you calculate the interplanar spacing for a plane with negative indices?
A10

The formula uses the squares of the indices, so the sign does not matter. For example, d_1̄10 = d_110.