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X-Rays

Skills to develop

X-rays

Like light, X-rays are electromagnetic radiation with very short wavelengths. Thus, X-rays photons have high energy, and they penetrate opaque material, but absorbed by materials containing heavy elements.

X-ray Diffactions

When light passes through a series of equal-spaced pinholes, it gives rise to a pattern due to wave interference, and such a phenomenon is known as diffraction. X-rays have wavelengths comparable to the inter atomic distances of crystals, and the interference patterns are developed by crystals when a beam of X-rays passes a crystal or a sample of crystal powder. The phenomena are known as diffraction of X-rays by crystals. More theory is given in Introduction to X-ray Diffraction.

X-ray diffraction, discovered by von Laue in 1912, is a well establised technique for material analysis. This link is the home page of Lambda Research, which provide various services using X-ray diffractions. For example:

In 1913, the father and son team of W.H. Bragg and W.L. Bragg gave the equation for the interpretation of of X-ray diffraction, and this is known as the Bragg equation. 2 d sin q = n l where d is the distance between crystallographic planes, and q is half the angle of diffraction, n is an integer, and l is the wavelength of the X-ray. A set of planes gives several diffraction beams, each is known as the nth order.

An animated illustration of Bragg equation shows the graphical relationship of the variables in the equation. The units used for wavelength and distances are &176; and 1 &176; = 100 pm.

Example 1

The X-ray wavelength from a copper X-ray is 154.2 pm. If the inter-planar distance from NaCl is 286 pm, what is the angle q?

Solution

sin q = l / (2 d)
    = 154 / 2*282
    = 0.273
q = 15.8°

Example 2

The X-ray of unknown wavelength is used. If the inter-planar distance from NaCl is 286 pm, and the angle q is found to be 7.23°?

Solution

l = 2 d sinl
    = 2*282*sin(7.23°)
    = 71 pm

Example 3

The X-ray of wavelength 71 pm is used. If the inter-planar distance from KI is 353 pm, what is the angle q for the second order diffracted beam?

Solution
The calculation is shown below:

sin q = l / (2 d)
    = 71 / 2*353
    = 0.100
q = 5.8°
These examples illustrate some example of the applications of X-rays diffraction for the study of solids.

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