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I'm solving problems related to X-ray diffraction, and I get these results which apparently are correct. My question may be dumb, but I just can't see how the interplanar distance might be shorter than the atoms radius.

What's up with this?

I think it might make sense that its shorter than the cells parameter, but it's not immediately obvious to me, so if someone could prove it to me, i'd be quite glad.

Edit:

For example:

The atomic radius of chromium is 124pm. THe distance between (310) planes is .0912nm. See problem 6

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Does this help?

X-Ray diffraction from a cubic crystal

The distance between planes will be less than the distance between atoms in many cases because the atoms don't have to stack directly on top of one another in each plane.

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First of all, diffraction is best thought of in reciprocal space rather than from planes of atoms, but let me answer your question directly.

You may not understand the meaning of a reflection index. If one imagines a simple cubic lattice made up of atoms of diameter a, then the lattice dimension of the cube is also a. The spacing between (100) planes is also a. The spacing between (200) planes is a/2, as these planes have double the density of (100) planes. As we keep increasing h,k or l the planes get closer.

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  • $\begingroup$ Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question. $\endgroup$
    – Philipp
    Commented Oct 23, 2014 at 22:39
  • $\begingroup$ Now, that the formalities are out of the way... :) Good answer, +1 for that. However, I fear it won't help the OP very much. He seemed to look for a more visual sort of answer. If you could add any pictures that illustrate your point, I think it would greatly help to make the answer more intuitively understandable. $\endgroup$
    – Philipp
    Commented Oct 23, 2014 at 22:44

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