# X ray diffraction ```X-ray diffraction
The measurement of X-ray wavelengths
proved to be very difficult because they
are so short. In 1912, however, von
Laue used a crystal in an attempt to
diffract the X-rays after failing to do this
with ordinary optical-style diffraction
gratings, and was successful.
X ray beam


A
B
C
His work was followed up by Friedrich
and Knipping, and good X-ray diffraction
Crystal atoms
patterns were produced. To give
d
diffraction the obstacles must be only a
few wavelengths apart and so the atoms
in a crystal lattice were ideal for X-ray diffraction since their separation is about 10-10 m (0.1
nm). Sir William and his son Sir Lawrence Bragg used a crystal as a reflection diffraction
grating.
}
In the diagram a beam of X-rays of wavelength  is incident at a glancing angle  on a
crystal where the atomic planes are separated by a distance d.
The path difference between the waves reflected at the top plane and those reflected at the
second plane is ABC = 2dsin.
Constructive interference occurs when this path difference is equal to a whole number of
wavelengths. Therefore:
X ray diffraction:
2dsin = m
This equation is useful in determining the structure of a crystal, because if the X-ray
wavelength is known the atomic spacing in the crystal can be found.
If the inter-atomic spacing can be found from X-ray diffraction, then the Avogadro constant
(NA)can be calculated. The electron charge can then be found, from F = NAe.
Alternatively X-ray diffraction can provide a means of calculating X-ray wavelengths if the
inter-atomic spacing is known.
Example
A beam of X-rays of wavelength 0.3 nm is incident on a crystal, and gives a first-order
maximum when the glancing angle is 9.0 degrees. Find the atomic spacing.
d = /2sin = 0.3/2x0.156 = 0.96 nm
```