XRays
 Page ID
 35135
Skills to Develop
 Explain Xrays.
 Interpret the symbols used in the Bragg equation.
Like light, Xrays are electromagnetic radiation with very short wavelengths. Thus, Xray photons have high energy, and they penetrate opaque material, but are absorbed by materials containing heavy elements.
Xray Diffraction
When light passes through a series of equalspaced pinholes, it gives rise to a pattern due to wave interference, and such a phenomenon is known as diffraction. Xrays have wavelengths comparable to the interatomic distances of crystals, and the interference patterns are developed by crystals when a beam of Xrays passes a crystal or a sample of crystal powder. The phenomena are known as diffraction of Xrays by crystals. More theory is given in Introduction to Xray Diffraction.
Xray diffraction, discovered by von Laue in 1912, is a well established technique for material analysis. This link is the home page of Lambda Research, which provide various services using Xray diffractions. For example:
 Residual Stress Measurement
 Qualitative Phase Analysis
 Quantitative Phase Analysis
 Precise Lattice Parameter Determination
In 1913, the father and son team of W.H. Bragg and W.L. Bragg gave the equation for the interpretation of Xray diffraction, and this is known as the Bragg equation.
\(2\, d\, \sin \theta = n\, \lambda\)
where d is the distance between crystallographic planes, \(\theta\) is half the angle of diffraction, n is an integer, and \(\lambda\) is the wavelength of the Xray. A set of planes gives several diffraction beams; each is known as the nth order.
Example 1
The Xray wavelength from a copper Xray is 154.2 pm. If the interplanar distance from \(\ce{NaCl}\) is 286 pm, what is the angle \(\theta\)?
Solution
\(\begin{align}
\sin \theta &= \dfrac{\lambda}{2 d}\\
&= \dfrac{154}{2\times282}\\
&= 0.273
\end{align}\)
\(\theta = 15.8^\circ\)
Example 2
An Xray of unknown wavelength is used. If the interplanar distance from \(\ce{NaCl}\) is 286 pm, and the angle \(\theta\) is found to be 7.23°, what is \(\lambda\)?
Solution
\(\begin{align}
\lambda &= 2\, d\, \sin\theta\\
&= 2\times282\times\sin(7.23^\circ)\\
&= \mathrm{71\: pm}
\end{align}\)
Example 3
The Xray of wavelength 71 pm is used. If the interplanar distance from \(\ce{KI}\) is 353 pm, what is the angle \(\theta\) for the second order diffracted beam?
Solution
The calculation is shown below:
\(\begin{align}
\sin \theta &= \dfrac{\lambda}{2 d}\\
&= \dfrac{71}{2\times353}\\
&= 0.100\\
\theta &= 5.8^\circ
\end{align}\)
These examples illustrate some example of the applications of Xray diffraction for the study of solids.
Questions
 If the wavelength is 300 pm and the interplanar distance d is 300 pm, what is the angle \(\theta\) in the Bragg equation, for n = 1?
Hint: 30 degrees

\(\ce{NaCl}\) and \(\ce{KI}\) have the same structure type, but the unit cells' edge lengths a are 564 and 706 pm respectively. The second order diffracted beam is observable, (n = 2). For such a diffraction, which compound should have a higher angle?
Hint: \(\ce{NaCl}\)
Discussion 
The larger the interplanar distance d, the smaller the angle.
Contributors
Chung (Peter) Chieh (Chemistry, University of Waterloo)