# Braggs law: lattice constant

1. May 3, 2009

### vebbie

need help working out this problem. if you have a simple cubic lattice characterised using x-ray diffraction with a wavelegth of 1.6$$\dot{A}$$. 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$$\lambda$$=dSin$$\theta$$. If so what value do you use for "n".

2. May 3, 2009

### Dr Transport

It is the diffraction order, use 1 for the first order diffraction lobe.

3. May 4, 2009

### vebbie

but does the peak have any use for the calculation. suppos it was the 422 peak

4. May 4, 2009

### nbo10

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. May 5, 2009

### Modey3

"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