Braggs law and x-ray crystallography

In summary, the equation tells you how to calculate the diffraction pattern from x-rays beamed at a crystal. You need to know the angle between the incident beam and the series of planes within the crystall which result in diffraction. You also need to know the distance between each layer of atoms in the crystal. Then you can use the bragg condition to calculate the diffraction pattern.
  • #1
mycotheology
89
0
I'm trying to figure out how Braggs law is useful for analysing crystal structures. So here's the equation:
nλ = 2dsinθ
where λ is the wavelength of incident radiation, d is the distance between each layer of atoms in the crystal and θ is the angle of incidence. So to keep things simple let's say I beam some x-rays at the crystal at a 90 degree angle of incidence so we can ignore the sinθ part of the equation. How does this equation tell me anything useful? Let's say I beam any old wavelength through the crystal onto a screen. There'll be a diffraction pattern. What then? How do I tell when I have found a wavelength that is an integer multiple of the distance between the atomic layers?
 
Physics news on Phys.org
  • #2
mycotheology said:
I'm trying to figure out how Braggs law is useful for analysing crystal structures. So here's the equation:
nλ = 2dsinθ
where λ is the wavelength of incident radiation, d is the distance between each layer of atoms in the crystal and θ is the angle of incidence. So to keep things simple let's say I beam some x-rays at the crystal at a 90 degree angle of incidence so we can ignore the sinθ part of the equation. How does this equation tell me anything useful? Let's say I beam any old wavelength through the crystal onto a screen. There'll be a diffraction pattern. What then? How do I tell when I have found a wavelength that is an integer multiple of the distance between the atomic layers?

θ is not the angle between the beam and the crystal but the angle between the beam and the series of planes within the crystall which result in diffraction. It is better to define θ as half the angle between the incident and diffracted ray. If θ, is 90 degrees it means the diffracted beam is 180 degrees away from the incident beam. Which means you do not have any diffraction spot for that specific series of planes at 90 degrees from your ray. However remember that the bragg condition applies to the diffraction produced by one series of planes not the whole crystal, therefore at anyone time, there are many different series of planes in your crystal at different incident angles from your ray some of which will obey the bragg condition and result in diffraction spots. That is why a diffraction pattern from a crystal usually contains many spots.

See: http://en.wikipedia.org/wiki/File:X-ray_diffraction_pattern_3clpro.jpg

Each one corresponds to a different series of planes which satisfied the bragg condition and from the position of the spot you can calculate what the incident angle was for the planes which resulted in the spot.

EDIT:
Also the point of x-ray crystallography is not "to find a wavelength that is an integer multiple of the distance between atomic layers". The point is to use the bragg condition to study the contents of the crystal. Using a different wavelength just changes the angles but mostly gives you the same diffraction pattern, only more/less spread-out.
 
  • #3
These planes are planes of atoms though aren't they. Do some of the light beams get reflected while others get diffracted? In all the explanations of Braggs rule itself I've read, they explain how it was originally used to describe the fact that x-rays beamed at a crystal will be reflected with a scattering angle equal to the incident angle. Thats not the same thing as beaming x-rays "through" the crystal and recording the diffraction pattern on a screen located on the other side of the crystal is it?

EDIT: Ah, I get it now. I was playing with this flash app:
http://www-outreach.phy.cam.ac.uk/camphy/xraydiffraction/xraydiffraction_exp.htm
and I see that I had the wrong idea of what d is. I thought d was the distance between each atom of a plane but its actually the distance between each plane. Now I can see how d will determine whether the waves interact constructively or destructively.
 
Last edited:
  • #4
Crystal diffraction spectroscopy has been used both as a reflection-type and as a transmission-type spectrometer to characterize x-rays, and gamma-ray beams up to ~ 1 MeV. See

http://nvlpubs.nist.gov/nistpubs/jres/105/1/j51des.pdf

Diffraction crystals have even been bent to focus beams from diffuse sources (Cauchois geometry, see Fig. 5) and from point sources (DuMond geometry, see Fig. 6). Crystal spectrometers have been used as a primary wavelength measuring standard, due to the ability to measure crystal planes and bending angles very accurately. The resolving power (resolution) is very high, and the efficiency very low.
 
  • #5
Crystal diffraction only works with x-rays (or neutrons or electrons) and not visible light because for a very large lambda you cannot find a Bragg angle to satisfy Bragg's law.

At each atom, a small part of the incident beam get absorbed and another small part gets scattered, more or less equally in all directions. The scattering is coherent, i.e. the beams retain their phase and can interfere. Bragg's law says that constructive interference happens when a certain relation between the wavelength, lattice spacing and angle of incidence (relative to the lattice planes) is satisfied.

If you do a more careful analysis you find that the scattered intensity depends on how the lattice planes are filled with atoms, and that there are many many sets of planes at different angles.

X-ray crystallography is to measure a large number of angles and intensities, either in single crystals or powders. From that information you can then reconstruct the position of the atoms in the crystal.

Amazingly this even works for protein crystals with thousands and thousands of atoms per unit cell.
 

What is Braggs law and how does it relate to x-ray crystallography?

Braggs law is a fundamental principle in x-ray crystallography that explains the relationship between the wavelength of x-rays and the spacing of atoms within a crystal lattice. It states that when x-rays are diffracted by a crystal, the angle of incidence is equal to the angle of reflection, and the path length difference between the diffracted rays is a multiple of the x-ray wavelength.

Why is x-ray crystallography an important tool in scientific research?

X-ray crystallography is a powerful technique for determining the structure of molecules and materials at the atomic level. It has applications in fields such as chemistry, biology, materials science, and pharmaceuticals, allowing scientists to study the shape, size, and arrangement of atoms within a crystal lattice.

What are the main steps involved in conducting an x-ray crystallography experiment?

The first step is to obtain a pure crystal of the substance being studied. This crystal is then placed in a beam of x-rays, and the diffraction pattern is recorded and analyzed. From this pattern, the positions of the atoms within the crystal can be determined using Braggs law and mathematical algorithms.

What are the limitations of x-ray crystallography?

X-ray crystallography is limited by the size and quality of the crystal being studied. Small or imperfect crystals may not produce clear or usable diffraction patterns. Additionally, x-ray crystallography cannot provide information about the dynamic behavior of molecules, as it captures a static snapshot of the crystal lattice.

What are some alternative techniques to x-ray crystallography for studying crystal structures?

Some alternative techniques include neutron diffraction, electron diffraction, and nuclear magnetic resonance (NMR) spectroscopy. Each of these techniques has its own advantages and limitations, and may be more suitable for certain types of samples or research questions.

Similar threads

Replies
54
Views
5K
Replies
3
Views
3K
  • Atomic and Condensed Matter
Replies
5
Views
1K
  • Atomic and Condensed Matter
Replies
1
Views
1K
Replies
2
Views
2K
  • Atomic and Condensed Matter
Replies
3
Views
2K
  • Quantum Physics
Replies
1
Views
2K
  • Atomic and Condensed Matter
Replies
7
Views
2K
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top