My name is Alexander FufaeV and here I write about:

Bragg's Law: How X-rays Are Reflected by the Crystal Lattice

Important Formula

Formula: Bragg's Law

$$m \, \class{violet}{\lambda} ~=~ 2\class{blue}{d} \, \sin(\class{gray}{\theta})$$ $$m ~=~ \frac{ 2\class{blue}{d} \, \sin(\theta) }{ \class{violet}{\lambda} }$$ $$\class{violet}{\lambda} ~=~ \frac{2\class{blue}{d}}{m} \, \sin(\class{gray}{\theta})$$ $$\class{blue}{d} ~=~ \frac{ m \, \class{violet}{\lambda} }{ 2 \sin(\class{gray}{\theta}) }$$ $$\class{gray}{\theta} ~=~ \arcsin\left(\frac{ m \, \class{violet}{\lambda} }{ 2\class{blue}{d} }\right)$$

Rearrange formula

What do the formula symbols mean?

Diffraction order

$$ m $$ Unit $$ - $$

An integer number ($m$ = 0,1,2, ...) indicating a multiple of the wavelength $ m \, \class{violet}{\lambda} $ at which constructive interference occurs.

Wavelength

$$ \lambda $$ Unit $$ \mathrm{m} $$

The wavelength characterizes the energy of the monochromatic light used to irradiate the crystal under investigation at an angle. X-rays are used in the study of crystals so that the wavelength is short enough to be within the range of the lattice constant. Only in this case diffraction phenomena can be observed.

Lattice constant

$$ \class{blue}{d} $$ Unit $$ \mathrm{m} $$

Distance between two adjacent lattice planes of the crystal.

Angle

$$ \class{gray}{\theta} $$ Unit $$ \mathrm{rad} $$

The angle between the incident X-ray beam and the grating plane.

Bragg reflection on two lattice planes

The incident X-ray wave (depicted as a beam in Illustration 1) with the wavelength $ \class{violet}{\lambda} $ is reflected at two adjacent lattice planes (points represent lattice atoms). The glancing angle $ \theta $ is the angle at which an interference maximum occurs in the interference pattern.

The X-ray wave, which is reflected at the deeper lattice plane, has to travel a greater distance. This path difference corresponds twice $ \class{blue}{d}\,\sin(\theta)$. However, the path difference must also correspond to a multiple of the wavelength so that constructive interference of the two waves can take place: $ m \, \class{violet}{\lambda} $

The Bragg's Law states the condition under which constructive interference $ m \, \class{violet}{\lambda} $ can occur on the lattice.

$$ \begin{align} m \, \class{violet}{\lambda} ~=~ 2\class{blue}{d} \, \sin(\class{gray}{\theta}) \end{align} $$

The term $ \sin(\class{gray}{\theta}) $ can have a maximum value of 1.

Why is visible light unsuitable for diffraction experiments on crystals?

For example, consider the first diffraction order: $ m = 1 $. Then the Bragg condition becomes:

$$ \begin{align} 2\class{blue}{d} ~=~ \lambda_{\text{max}} \end{align} $$

It follows from this that the wavelength can be at most twice the lattice constant $ \class{blue}{d} $ so that something can be observed at all in the diffraction experiment. However, since the lattice constant is very small (for silicon, for example, in the order of $10^{-12} \, \mathrm{m} $), the visible light (wavelength of the order of $ 10^{-9} \, \mathrm{m} $) has far too long wavelength to fulfill the condition 2:

$$ \begin{align} 2 \cdot 10^{-12} \, \mathrm{m} ~<~ 10^{-9} \, \mathrm{m} \end{align} $$

To fulfill Eq. 2, you need short-wave light, otherwise you cannot examine crystal structures such as those of silicon.

Which type of radiation is used to determine the structure of a crystal?

X-rays - good for examining "thick" crystals. Electron radiation - e.g. for examining small samples. Neutron radiation - e.g. for localizing hydrogen atoms in a crystal lattice.