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X-Rays
Skills to develop
- Explain X-rays
- Interpret the symbols used in Bragg equation
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:
- 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 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.
Confidence Building Questions
©cchieh@uwaterloo.ca