Braggs law and x-ray crystallography

Click For Summary

Discussion Overview

The discussion revolves around the application of Bragg's law in analyzing crystal structures through x-ray crystallography. Participants explore the implications of the law, the nature of diffraction patterns, and the relationship between wavelength, atomic layer spacing, and angles of incidence.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions how Bragg's law can provide useful information when using x-rays at a 90-degree angle of incidence, suggesting that the sinθ component can be ignored.
  • Another participant clarifies that θ should be defined as half the angle between the incident and diffracted rays, indicating that a 90-degree angle does not yield a diffraction spot for that series of planes.
  • It is noted that the Bragg condition applies to individual series of planes within the crystal, leading to multiple diffraction spots in a pattern.
  • One participant expresses confusion regarding the distinction between reflection and diffraction, questioning whether x-rays are beamed through the crystal or reflected off it.
  • A later reply acknowledges a misunderstanding about the definition of d, realizing it refers to the distance between planes rather than between individual atoms.
  • Another participant discusses the use of crystal diffraction spectroscopy for characterizing x-rays and gamma-ray beams, highlighting its applications in measuring wavelengths accurately.
  • It is mentioned that Bragg's law is specific to x-rays, neutrons, or electrons, as visible light does not satisfy the conditions due to larger wavelengths.
  • Participants discuss the coherent scattering of x-rays at atomic planes and how this leads to constructive interference based on the arrangement of atoms.
  • One participant notes that x-ray crystallography can reconstruct atomic positions in crystals, including complex structures like protein crystals.

Areas of Agreement / Disagreement

Participants express various viewpoints on the interpretation of Bragg's law and its application. There is no consensus on the initial understanding of the law's implications, and multiple competing views regarding the nature of diffraction and the role of angles remain present.

Contextual Notes

Some participants have differing interpretations of the definitions and applications of Bragg's law, particularly concerning the angles and distances involved. There are also unresolved questions about the relationship between the diffraction pattern and the underlying crystal structure.

mycotheology
Messages
86
Reaction score
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
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.
 
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:
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.
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 54 ·
2
Replies
54
Views
11K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K