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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 *n*th 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

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