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Bragg's Law and X-Ray Diffraction

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Physical Chemistry • Analytical Techniques

Bragg's Law and X-Ray Diffraction

The Geometry of Crystals, Path Difference, and the Glancing Angle Trap.

By chemca Team • Updated Sep 2026

How do we know the exact arrangement of atoms inside a solid crystal? We cannot see them with a standard microscope. Instead, we use X-ray crystallography. When X-rays strike a crystal, they reflect off the different planes of atoms. W.L. Bragg and W.H. Bragg derived a beautiful mathematical relationship that connects the wavelength of the X-rays, the angle of reflection, and the distance between the atomic planes.

1. The Master Equation

When parallel X-ray beams strike adjacent, parallel planes of atoms in a crystal, the beam hitting the deeper plane travels a slightly longer distance. For these reflected beams to emerge in phase and produce a bright spot (Constructive Interference), this path difference must be an integral multiple of the wavelength ($n\lambda$).

$n\lambda = 2d \sin \theta$
Decoding the Variables:
  • $n$: The order of diffraction (an integer: 1, 2, 3...). Usually, $n=1$ for first-order reflections.
  • $\lambda$: The wavelength of the incident X-rays.
  • $d$: The interplanar spacing (perpendicular distance between two adjacent parallel planes of atoms).
  • $\theta$: The Glancing Angle (the angle the incident beam makes with the crystal plane).
Geometry of Bragg's Law Plane 1 Plane 2 d A B C Path Difference = AB + BC AB = d sinΞΈ, BC = d sinΞΈ → Total = 2d sinΞΈ ΞΈ ΞΈ

2. The Grand Exam Trap: The Glancing Angle ($\theta$)

In standard optics (like Snell's Law or reflection), the angle of incidence is measured from the normal (the perpendicular line). Bragg's Law is different!

In Bragg's Law, $\theta$ is the Glancing Angle: the angle between the X-ray beam and the CRYSTAL PLANE itself.

If an exam question states: "X-rays strike a crystal such that the angle of incidence with the NORMAL is $60^\circ$," you MUST NOT use $60^\circ$ in the formula.

Correct calculation: $\theta = 90^\circ - 60^\circ = 30^\circ$. You must use $\sin(30^\circ)$.

3. Interplanar Spacing ($d_{hkl}$) and Miller Indices

A crystal lattice contains many different sets of parallel planes, cut at different angles. We identify these specific sets of planes using Miller Indices ($h, k, l$).

For a simple Cubic Crystal System (where edge lengths $a = b = c$), the distance $d$ between adjacent parallel planes with Miller indices $(h, k, l)$ is directly related to the edge length $a$ by a simple geometric formula:

$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$
Examples of $d_{hkl}$ in Cubic Lattices:
  • For $(1 0 0)$ planes:
    $d_{100} = \frac{a}{\sqrt{1^2 + 0^2 + 0^2}} = \frac{a}{1} = \mathbf{a}$
  • For $(1 1 0)$ planes:
    $d_{110} = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \mathbf{\frac{a}{\sqrt{2}}}$
  • For $(1 1 1)$ planes:
    $d_{111} = \frac{a}{\sqrt{1^2 + 1^2 + 1^2}} = \mathbf{\frac{a}{\sqrt{3}}}$

4. Maximum Order of Diffraction ($n_{max}$)

The sine function has a maximum value of 1 ($\sin \theta \leq 1$). Therefore, substituting this into Bragg's Law gives the theoretical limit for the maximum order of diffraction ($n$) that can be observed for a given crystal and X-ray wavelength.

$n\lambda = 2d \sin \theta$

Since $\sin \theta \leq 1 \implies n\lambda \leq 2d$

$\mathbf{n_{max} = \lfloor \frac{2d}{\lambda} \rfloor}$

The value of $n_{max}$ must be a whole number (integer floor).

Mastery Check: Bragg's Law

15 High-Yield Questions to test your JEE/NEET Preparation

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