Equivalence of the Laue equations and the Bragg equation

The Laue equations are equal to the Bragg equation if we denote the angle between the wave vectors s and s₀ by 2θ (see diagram below), which makes the vector s – s₀ normal to the lattice plane.

As the scattering of X-rays by an atom is assumed to be elastic (no loss in momentum), the magnitudes of the wave vectors are the same. Thus, s – s₀ becomes the base of an isosceles triangle with equal sides of Is₀I and IsI. We have:

$\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |=\left | \textbf{\textit{s}}\right |sin\theta +\left | \textbf{\textit{s}}_0\right |sin\theta$

Since $\left | \textbf{\textit{s}}\right |=\left | \textbf{\textit{s}}_0\right |=\frac{1}{\lambda }$ (see previous article)

$\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |=\frac{2sin\theta }{\lambda }\; \; \; \; \; \; \; (23)$

To find an expression for d_nh,nk,nl (see above diagram), we determine the scalar projection of a/h (the vector linking two plane-intercept points on the a-axis) on the vector s – s₀ to give:

$d_{nh,nk,nl}=\frac{\textbf{\textit{a}}}{h}\cdot \frac{\textbf{\textit{s}}-\textbf{\textit{s}}_0}{\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |}$

Dividing both sides of eq20 by h, we have $\frac{\textbf{\textit{a}}}{h}\cdot ( \textbf{\textit{s}}-\textbf{\textit{s}}_0) =1$ . So,

$d_{nh,nk,nl}=\frac{1}{\left | \textbf{\textit{s}}-\textbf{\textit{s}}_0\right |}\; \; \; \; \; \; \; (24)$

Substitute eq23 in eq24, we have the Bragg equation:

$2d_{nh,nk,nl}\: sin\theta =\lambda$

Since the Laue equations are equal to the Bragg equation if we denote the angle between the wave vectors s and s₀ by 2θ, the Bragg equation is a specific form of the Laue equations with the condition of 2θ imposed.

Equivalence of the Laue equations and the Bragg equation

next article: reciprocal lattice

Previous article: Laue equations

Content page of X-ray crystallography

Content page of advanced chemistry

Main content page