Bragg's Law, Miller Indices and X-ray Diffraction of Crystals

X-ray Diffraction: Why Crystals Act as Three-Dimensional Gratings

To investigate the internal architecture of crystalline materials, radiation with an appropriate wavelength is required. In optical spectroscopy, diffraction (विवर्तन) occurs noticeably only when the wavelength of incident radiation is comparable in magnitude to the spacing between diffracting obstacles. In crystals, repeating atoms, ions, or molecules are separated by interatomic distances on the order of 0.1 to 0.4 nanometers (1 to 4 Angstroms).

Visible light possesses wavelengths between 400 and 700 nm, which are roughly three orders of magnitude larger than interatomic distances. Consequently, visible light passes through a crystal without producing diffraction effects from individual atomic planes. Monochromatic X-rays, however, possess wavelengths typically falling between 0.05 and 0.25 nm (such as copper Kα radiation at approximately 0.154 nm or 1.54 Å). Because the X-ray wavelength closely matches the interplanar spacing of crystalline lattices, crystals serve as natural three-dimensional diffraction gratings.

When an incident X-ray beam strikes an electron cloud within an atom, the oscillating electromagnetic field forces the electrons to oscillate at the same frequency. These oscillating charges re-emit electromagnetic radiation elastically in all directions. In an amorphous material, these scattered waves interfere destructively in almost all directions due to random atomic positioning. In an ordered crystal, however, constructive interference (रचनात्मक व्यतिकरण) occurs along specific geometric directions, giving rise to distinct diffraction maxima known as Bragg reflections.

Derivation of the Bragg Condition

In 1913, W. H. Bragg and W. L. Bragg simplified the complex three-dimensional scattering problem by conceptualizing diffraction as the specular reflection of X-rays from sets of parallel, equidistant crystallographic planes. Although true optical reflection occurs only at a surface, X-ray "reflection" involves partial penetration through millions of successive parallel lattice planes.

Consider two parallel lattice planes belonging to the same crystallographic family, separated by an interplanar spacing d (अंतर-समतलीय दूरी). Let a collimated, parallel beam of monochromatic X-rays of wavelength λ strike these planes at a glancing angle θ:

1. Ray 1 strikes an atom located at point A on the uppermost plane and is scattered at an angle θ relative to the plane.
2. Ray 2 strikes an atom located at point B on the adjacent lower plane, positioned directly beneath point A along the normal to the planes, such that the distance AB = d. Ray 2 is also scattered at angle θ.
3. To determine whether the scattered rays emerge in phase, we evaluate the difference in the path lengths traversed by Ray 1 and Ray 2.
4. Drop a perpendicular from point A onto the incident path of Ray 2, meeting Ray 2 at point C. Similarly, drop a perpendicular from point A onto the scattered path of Ray 2, meeting it at point D.
5. Up to the wavefront line AC, both rays travel identical distances. Beyond the wavefront line AD, both scattered rays travel identical distances.
6. The extra path length traversed by Ray 2 inside the crystal is the segment CB + BD.

From the right-angled triangle CAB, the angle ∠CAB equals θ. Therefore:

CB = AB · sin θ = d sin θ

From the right-angled triangle DAB, the angle ∠DAB also equals θ. Therefore:

BD = AB · sin θ = d sin θ

The total path difference Δ between the two waves is:

Δ = CB + BD = d sin θ + d sin θ = 2d sin θ

For constructive interference to occur and produce a sharp, observable diffraction peak, this total path difference must equal an integral number of whole wavelengths:

nλ = 2d sin θ

Here, n represents the order of reflection (n = 1, 2, 3...). An order of n = 1 denotes a first-order reflection where the path difference equals exactly one wavelength; n = 2 indicates a path difference of two wavelengths, and so forth.

In modern crystallographic practice, it is conventional to absorb the order of reflection n into the interplanar spacing. An n-th order reflection from planes with spacing d is mathematically equivalent to a first-order reflection (n = 1) from a fictitious set of parallel planes having spacing d_hkl = d / n. Under this convention, Bragg's law is written simply as:

λ = 2d_hkl sin θ

Miller Indices: Defining Planes from Lattice Intercepts

A crystal lattice consists of an infinite array of points that can be sliced by sets of equidistant, parallel planes. To identify and quantify these planes uniquely, crystallographers use a system of notation known as Miller indices, denoted by the triplet (h k l).

The position and orientation of any crystal plane are initially defined by its intercepts along the three crystallographic axes x, y, and z, expressed in units of the unit cell edges a, b, and c. These direct intercepts are historically known as Weiss indices. However, direct intercepts present mathematical inconveniences—planes parallel to an axis intersect it at infinity (∞), which cannot be manipulated conveniently in algebraic formulas.

Miller indices resolve this by taking the reciprocals of the fractional intercepts. The standard procedure to determine the Miller indices (h k l) of any crystal plane consists of four reproducible steps:

• Step 1: Determine the intercepts. Find the points where the plane intersects the crystallographic axes, expressed as multiples of the unit cell parameters: p · a, q · b, and r · c.
• Step 2: Express intercepts as fractional coordinates. Drop the unit cell parameters to leave the fractional intercepts (p, q, r). If a plane is parallel to an axis, its intercept is taken as infinity (∞).
• Step 3: Invert the values. Take the reciprocal of each intercept: (1/p, 1/q, 1/r). The reciprocal of infinity is zero (1/∞ = 0), indicating that a zero index corresponds to a plane parallel to that particular crystallographic axis.
• Step 4: Clear fractions. If fractions remain, multiply all three reciprocals by their lowest common denominator (LCD) to obtain the smallest set of coprime integers h, k, and l.

If a plane intercepts an axis on the negative side of the origin, the negative sign is indicated by placing a bar over the corresponding index (e.g., h̄, k̄, or l̄, pronounced "h-bar"). Parentheses (h k l) denote a specific plane or a set of parallel planes; curly brackets {h k l} denote a complete family of symmetry-equivalent planes; square brackets [u v w] denote a crystallographic direction vector.

Interplanar Spacing (d_hkl) in Cubic and Other Crystal Systems

The perpendicular distance separating adjacent parallel planes in a family (h k l) is known as the interplanar spacing, d_hkl. This distance depends directly on the unit cell dimensions and the Miller indices.

For an orthogonal coordinate system where all interaxial angles are 90° (α = β = γ = 90°), the general expression relating interplanar spacing to unit cell dimensions is:

1 / (d_hkl)² = (h / a)² + (k / b)² + (l / c)²

In a cubic crystal system, all three edge lengths are equal (a = b = c). The formula simplifies directly to:

1 / (d_hkl)² = (h² + k² + l²) / a²

Taking the reciprocal and square root gives the standard cubic interplanar spacing equation:

d_hkl = a / √(h² + k² + l²)

By substituting this expression for d_hkl into Bragg's law (for n = 1), we establish the direct bridge between diffraction angle θ and the cubic lattice parameter a:

sin θ = λ / (2d_hkl) = [λ / (2a)] · √(h² + k² + l²)

Squaring both sides yields:

sin² θ = [λ² / (4a²)] · (h² + k² + l²)

Because the term [λ² / (4a²)] is a constant for a given crystal and a fixed monochromatic X-ray wavelength, sin² θ is directly proportional to (h² + k² + l²). This proportionality is the foundational mathematical tool used to index experimental diffraction peaks.

For comparison, in a tetragonal system (a = b ≠ c), the expression is:

1 / (d_hkl)² = (h² + k²) / a² + l² / c²

And for an orthorhombic system (a ≠ b ≠ c):

1 / (d_hkl)² = h² / a² + k² / b² + l² / c²

The Origin of Systematic Absences and Selection Rules

Bragg's law predicts the geometric angles at which constructive interference can occur based entirely on the size and shape of the unit cell. However, Bragg's law alone does not account for the placement of atoms inside the unit cell. In non-primitive lattices, such as body-centered cubic (bcc) and face-centered cubic (fcc), additional atoms positioned at cell centers or face centers scatter X-rays that can interfere destructively with the rays scattered by corner atoms.

This destructive phase cancellation causes the intensity of certain theoretical reflections to drop to exactly zero. Such missing diffraction peaks are called systematic absences or extinctions.

The total amplitude scattered by all atoms in a unit cell in the direction of the reflection (h k l) is governed by the structure factor, F_hkl:

F_hkl = Σ_j f_j · exp[2πi(h·u_j + k·v_j + l·w_j)]

Here, f_j is the atomic scattering factor of the j-th atom, and (u_j, v_j, w_j) are its fractional coordinates within the unit cell. The intensity of an X-ray reflection is proportional to the square of the absolute structure factor: I_hkl ∝ |F_hkl|². If F_hkl = 0, the reflection is systematically absent.

1. Primitive Cubic Lattice (sc)

A primitive cubic unit cell contains one lattice point per unit cell, situated at fractional coordinates (0, 0, 0). Evaluating the structure factor for a single atom at the origin gives:

F_hkl = f · exp[2πi(0)] = f ≠ 0

Because the structure factor never vanishes, all combinations of (h k l) are allowed. The observable values of (h² + k² + l²) include:

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 16...

Notice that 7 and 15 are missing. This is not a systematic extinction caused by centering; rather, it is a mathematical property of integers: no sum of three squared integers can ever equal 7 (since the only squares available are 0, 1, 4, and their combinations cannot yield 7) or 15.

2. Body-Centered Cubic Lattice (bcc)

A body-centered cubic unit cell contains two identical atoms at fractional positions (0, 0, 0) and (1/2, 1/2, 1/2). The structure factor becomes:

F_hkl = f · [exp(0) + exp(2πi(h/2 + k/2 + l/2))] = f · [1 + exp(πi(h + k + l))]

Using Euler's identity, exp(πi · m) = (-1)^m where m is an integer. Thus:

• If (h + k + l) is an even integer (सम): exp(πi(h + k + l)) = +1, giving F_hkl = 2f, and |F_hkl|² = 4f². The reflection is allowed.
• If (h + k + l) is an odd integer (विषम): exp(πi(h + k + l)) = -1, giving F_hkl = f(1 - 1) = 0. The scattered wave from the body-centered atom is exactly 180° out of phase with the corner atoms, resulting in complete destructive interference.

Therefore, the selection rule for bcc is: (h + k + l) must be even. The allowed values of (h² + k² + l²) are:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20...

If we divide this entire sequence by its first term (2), the ratio of sin² θ values yields the characteristic series:

1 : 2 : 3 : 4 : 5 : 6 : 7 : 8...

3. Face-Centered Cubic Lattice (fcc)

An fcc unit cell contains four identical atoms at positions (0, 0, 0), (1/2, 1/2, 0), (1/2, 0, 1/2), and (0, 1/2, 1/2). Evaluating the structure factor:

F_hkl = f · [1 + exp(πi(h + k)) + exp(πi(h + l)) + exp(πi(k + l))]

• If h, k, l are all even or all odd (unmixed indices): the sum of any two indices is always even. Thus, each exponential term evaluates to +1, yielding F_hkl = 4f. The reflection is allowed.
• If h, k, l are mixed (some even, some odd): two of the index pairs sum to an odd number and one sums to an even number (e.g., for 100, h+k=1, h+l=1, k+l=0). This yields F_hkl = f · [1 - 1 - 1 + 1] = 0. The reflection is systematically absent.

Therefore, the selection rule for fcc is: h, k, l must be unmixed (all even or all odd). Zero is treated as an even integer. The allowed values of (h² + k² + l²) are:

3, 4, 8, 11, 12, 16, 19, 20, 24, 27...

Normalizing this sequence against the first peak gives the distinctive fcc ratio:

3 : 4 : 8 : 11 : 12 : 16 : 19 : 20...

For more detailed experimental analysis of diffractometer traces and peak shapes, consult our detailed article on powder XRD patterns and crystal structure determination. Additionally, real crystals often deviate from idealized lattices due to vacancies or interstitial ions, as explored in our guide on crystal defects and non-stoichiometry.

Solved Problems in Crystal Diffraction and Miller Indices

Problem 1: Determining Miller Indices from Crystallographic Intercepts

Question: A crystal plane intersects the crystallographic axes at intercepts 2a, 3b, and is parallel to the c-axis. A second plane makes intercepts at a/2, -2b, and 3c/2. Determine the Miller indices for both planes.

Solution:

For Plane 1:

• Intercepts in terms of unit cell lengths: 2, 3, ∞ (parallel to c means intercept is at infinity).
• Reciprocals of intercepts: 1/2, 1/3, 1/∞ = 1/2, 1/3, 0.
• Multiply by the lowest common multiple (LCM = 6): 6 × (1/2) = 3; 6 × (1/3) = 2; 6 × 0 = 0.
• The Miller indices are (3 2 0).

For Plane 2:

• Intercepts in terms of unit cell lengths: 1/2, -2, 3/2.
• Reciprocals of intercepts: 1/(1/2) = 2; 1/(-2) = -1/2; 1/(3/2) = 2/3.
• Multiply by the lowest common multiple (LCM = 6): 6 × 2 = 12; 6 × (-1/2) = -3; 6 × (2/3) = 4.
• The negative sign is placed as a bar over the index. The Miller indices are (12 3̄ 4).

Problem 2: Interplanar Spacing and Bragg Angle Calculation

Question: A metallic element crystallizes in a face-centered cubic structure with a lattice parameter of a = 4.00 Å. A monochromatic X-ray beam with wavelength λ = 1.54 Å is used. Calculate the interplanar spacing for the (220) planes and determine the glancing angle θ for first-order reflection.

Solution:

First, calculate the interplanar spacing d₂₂₀ using the cubic formula:

d₂₂₀ = a / √(h² + k² + l²) = 4.00 Å / √(2² + 2² + 0²) = 4.00 / √8 = 4.00 / 2.8284 ≈ 1.414 Å

Next, apply Bragg's law for first-order reflection (n = 1):

λ = 2 · d₂₂₀ · sin θ

sin θ = λ / (2 · d₂₂₀) = 1.54 Å / (2 × 1.414 Å) = 1.54 / 2.828 ≈ 0.5446

Taking the inverse sine:

θ = arcsin(0.5446) ≈ 33.0°

Therefore, the interplanar spacing is 1.414 Å and the glancing angle is 33.0° (corresponding to a diffractometer 2θ angle of 66.0°).

Problem 3: Indexing Powder XRD Data and Lattice Identification

Question: In a powder diffraction experiment on a cubic metal using X-rays of wavelength λ = 1.54 Å, the first four reflection peaks appear at diffractometer angles (2θ) of 38.4°, 44.6°, 65.0°, and 78.1°. Identify the Bravais lattice type (sc, bcc, or fcc), index each peak, and calculate the unit cell edge length a.

Solution:

Step 1: Convert 2θ angles to θ, and compute sin θ and sin² θ for each reflection:

• Peak 1: 2θ = 38.4° → θ = 19.2° → sin(19.2°) = 0.3289 → sin² θ = 0.1082
• Peak 2: 2θ = 44.6° → θ = 22.3° → sin(22.3°) = 0.3795 → sin² θ = 0.1440
• Peak 3: 2θ = 65.0° → θ = 32.5° → sin(32.5°) = 0.5373 → sin² θ = 0.2887
• Peak 4: 2θ = 78.1° → θ = 39.05° → sin(39.05°) = 0.6300 → sin² θ = 0.3969

Step 2: Test the ratio of the sin² θ values to identify the lattice selection rule:

• Divide each value by a common factor to test candidate ratios.
• Dividing by 0.0360 gives:
  0.1082 / 0.0360 ≈ 3.00
  0.1440 / 0.0360 ≈ 4.00
  0.2887 / 0.0360 ≈ 8.02
  0.3969 / 0.0360 ≈ 11.02

The sequence of (h² + k² + l²) values is 3, 4, 8, 11. Comparing this with our selection rules table, this sequence matches the Face-Centered Cubic (fcc) lattice, where indices must be all odd or all even:

• Peak 1: h² + k² + l² = 3 → (111)
• Peak 2: h² + k² + l² = 4 → (200)
• Peak 3: h² + k² + l² = 8 → (220)
• Peak 4: h² + k² + l² = 11 → (311)

Step 3: Calculate the lattice parameter a using Peak 1:

sin² θ = [λ² / (4a²)] · (h² + k² + l²)

a² = [λ² · (h² + k² + l²)] / [4 · sin² θ] = [(1.54 Å)² × 3] / [4 × 0.1082] = [2.3716 × 3] / 0.4328 = 7.1148 / 0.4328 ≈ 16.44 Ų

a = √(16.44 Ų) ≈ 4.05 Å

The crystal lattice is face-centered cubic (fcc) with an edge length of 4.05 Å (consistent with metallic aluminum).

Problem 4: Finding the Lowest Angle Reflection in a BCC Crystal

Question: Metallic tungsten crystallizes in a body-centered cubic lattice with unit cell edge length a = 3.16 Å. Determine the lowest glancing angle θ at which a diffraction peak can be observed when using radiation of wavelength λ = 1.54 Å.

Solution:

The lowest glancing angle θ corresponds to the smallest non-zero value of sin θ, which requires the smallest value of (h² + k² + l²) allowed by the lattice selection rule.

For a body-centered cubic lattice, reflections must satisfy h + k + l = even. The smallest positive integer combination satisfying this rule is (110), for which:

h² + k² + l² = 1² + 1² + 0² = 2

Now evaluate the interplanar spacing for (110):

d₁₁₀ = a / √2 = 3.16 Å / 1.4142 ≈ 2.234 Å

Using Bragg's equation (n = 1):

sin θ = λ / (2 · d₁₁₀) = 1.54 Å / (2 × 2.234 Å) = 1.54 / 4.468 ≈ 0.3447

θ = arcsin(0.3447) ≈ 20.16°

The lowest glancing angle for tungsten under these conditions is 20.16°.

Exam Perspective: Solid State Chemistry in IIT-JAM, CSIR-NET & GATE

Solid-state chemistry and crystallography form a foundational part of the physical chemistry syllabus in major national examinations across India, including IIT-JAM, CSIR-NET, GATE, and CUET-PG. Because the fundamental physics of diffraction relies on rigorous geometric relationships, questions in this area are typically numerical and objective.

To master this topic for national entrance examinations:

• Understand the derivation: Do not merely memorize nλ = 2d sin θ. Be prepared to identify how glancing angles, path differences, and refractive indices interact.
• Master selection rules: Be able to identify sc, bcc, and fcc lattices instantly from their sin² θ peak ratios or from lists of observed (h k l) reflections.
• Work with past papers: Solve actual previous exam problems directly from official sources to understand how questions are framed. Aspirants are advised to review official question papers and download our compiled previous year papers to evaluate question styles independently.
• Confirm syllabus details: Weightages and question counts vary across test sessions. Always verify current exam patterns and official guidelines directly from the relevant organizing institute's brochure.

For structured preparation, ABC Chemistry offers comprehensive solid state chemistry study material, targeted mock assessments, and personalized mentoring led by founder Mr. Aniket Bhardwaj, who brings over 20 years of dedicated chemistry teaching experience. We offer classroom-centre and online classes tailored to university students and research aspirants across India. To discuss admission details or academic planning, feel free to contact our academic team at 9212142427.

Frequently Asked Questions

Why is reflection from the (100) plane absent in a body-centered cubic (bcc) crystal?

In a bcc lattice, there is an atom at the center of the unit cell at coordinates (1/2, 1/2, 1/2) in addition to the corner atoms. When X-rays reflect from the (100) planes, the plane passing through the body-centered atoms lies exactly halfway between the adjacent (100) planes. Rays scattered by this intervening plane travel a path length that is exactly half a wavelength (λ/2) out of phase with rays scattered by the corner planes. This creates complete destructive interference, extinguishing the (100) peak. Mathematically, h + k + l = 1 + 0 + 0 = 1 (odd), yielding a structure factor of zero.

What is the difference between the Bragg glancing angle θ and the diffractometer angle 2θ?

The Bragg angle θ is the glancing angle measured between the incident X-ray beam and the parallel lattice planes. In standard powder diffractometers, the detector rotates through twice this angle relative to the transmitted beam direction to intercept the diffracted beam. Therefore, experimental diffractometer charts are plotted against 2θ on the horizontal axis. When performing numerical calculations with Bragg's law, you must always divide the reported 2θ value by two to obtain θ before taking the sine.

Why is there no peak corresponding to h² + k² + l² = 7 in simple cubic crystals?

The absence of a peak at h² + k² + l² = 7 is not due to destructive interference or lattice centering (systematic absence). In simple cubic lattices, all combinations of h, k, and l have non-zero structure factors. However, the number 7 cannot be formed by summing three squared integers (the only available squares are 0, 1, 4, 9...). The same mathematical restriction applies to 15, 23, 28, 31, and other integers defined by Legendre's three-square theorem.

What do Miller indices with a zero mean geometrically?

A Miller index of zero indicates that the crystallographic plane is parallel to that specific axis. Because the plane never intersects that axis, the theoretical intercept is at infinity (∞). Taking the reciprocal of infinity yields zero (1/∞ = 0). For example, the (110) plane intersects the x and y axes at unit distances but runs parallel to the z-axis.

Can Bragg's law be applied to amorphous solids or liquids?

Bragg's law strictly describes constructive interference from long-range, periodic, three-dimensional arrays of lattice planes. Amorphous solids (such as glasses) and liquids lack long-range periodic order, so they do not produce sharp Bragg peaks. Instead, X-ray scattering from amorphous materials yields broad, diffuse halos reflecting only short-range average nearest-neighbor distances.

How does the order of reflection n relate to Miller indices?

An n-th order reflection from a crystallographic plane with Miller indices (h k l) and spacing d has a path difference of nλ = 2d sin θ. In modern notation, this reflection is treated as a first-order reflection (n = 1) from an imaginary set of parallel planes having spacing d / n and indices (nh nk nl). For instance, the second-order reflection (n = 2) from the (100) plane corresponds mathematically to the first-order reflection from the (200) plane.