Distance between planes in crystals

1. Apr 27, 2006

Stephan Hoyer

I'm working on a lab report on powder X-ray diffraction off of some relatively straight-forward crystals (Si, NaCl, CsCl) for an introductary course on modern physics.

I thought it would be useful to include a partial derivation of the formula relating the distance between parallel planes, d, the length of a cell edge, a, and the miller indices (hkl) for a cubic lattice:

$$d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}}$$

I would be happy (and it would be sufficient for my purposes) to do a basic derivation of the spacing between lines in a hypothetical two dimensional square lattice. I've thought a lot about this problem, however, and what I thought would be a clear geometrical fact is turning out to be not so obvious.

Does anyone have any hints or links to a derivation? I got several texts on X-ray diffraction from my college's library, including a text, "Interpretation of x-ray powder diffraction patterns" but none of them include a clear derivation. What I've found online seems to be generally cursory, as well. I've drawn out a two dimensional square lattice and sample parallel lines going through it and I can see that the equation holds, but I'd like a simple proof, from first principles if possible.

2. Apr 27, 2006

Andrew Mason

I am not sure if this helps you but have a look at this site

AM

3. Apr 28, 2006

Gokul43201

Staff Emeritus
I couldn't find it there with a quick look. So, anyway, it's short enough that I can write it down in a few lines.

Consider two adjacent planes, one of which goes through the origin. The second plane makes intercepts a/h, b/k, c/l (by definition of the Miller Indices). Let the point on this plane that's nearest the origin (O) be P. Then OP is the required d-spacing.

Let the line OP make angles A, B and C with each of the three axes. From trig, we have cos2(A)+cos2(B)+cos2(C)=1
But cos(A) = OP/OX = d/(a/h) = dh/a

Similarly, plug in for cos(B) and cos(C) and you will get the required result.

4. Apr 29, 2006

Stephan Hoyer

Thanks for you help. It looks like the general proof isn't actually so tedius after all, so I guess I'll include that instead.

5. Dec 31, 2007

mconn86

Nice visual explanation here.
Slides 23+24

6. Nov 14, 2011

Fullbase

Really Thanks. Thank you so much!