Braggs law: lattice constant

  • Thread starter vebbie
  • Start date
  • #1
4
0

Main Question or Discussion Point

need help working out this problem. if you have a simple cubic lattice characterised using x-ray diffraction with a wavelegth of 1.6[tex]\dot{A}[/tex]. The main peak in the scan is (222) and the angle is 32 degrees. By using braggs law to find the lattice constant do you just sub in the values and solve for d? (n[tex]\lambda[/tex]=dSin[tex]\theta[/tex]. If so what value do you use for "n".
 

Answers and Replies

  • #2
Dr Transport
Science Advisor
Gold Member
2,306
428
It is the diffraction order, use 1 for the first order diffraction lobe.
 
  • #3
4
0
but does the peak have any use for the calculation. suppos it was the 422 peak
 
  • #4
418
3
d is the spacing of the (222) planes. If you were looking at the (422) reflection, then d would be the spacing of the (422) planes. You might have to do some geometry to get the lattice spacing from the data.
 
  • #5
135
0
"but does the peak have any use for the calculation. suppos it was the 422 peak"

The index n is a consequence of the fact that reflections occur when (k - k') = G. Braggs Law diffraction condition is a simplification of the Laue Condition for diffraction: (k - k') = G. The higher order peaks are forward scattering instead of backwards scattering so you won't be able to see them in a standard diffractometer. You would nee to generate a Laue pattern to see these peaks. To answer your question you don't need these peaks.

modey3
 

Related Threads for: Braggs law: lattice constant

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
5K
Top