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My textbook, Solid-State Physics, Fluidics, and Analytical Techniques in Micro- and Nanotechnology, by Madou, says the following:

X-ray analysis reveals the symmetries of crystals (lattice type), distances between atomic planes (lattice parameter), the positions of atoms in crystals, the types of atoms from the intensities of diffracted x-rays, and the degree of crystallinity (ordering).

I found this quite fascinating. Do we really have technology with the resolution to distinguish between layers of single atoms of materials? In other words, we can tell when one layer of single atoms ends and the next layer of single atoms begins, and analyse them as such?

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    $\begingroup$ Yes, it does exist, but the meaning of the quotation was not that. $\endgroup$ – Greg Jan 18 at 4:03
  • $\begingroup$ @Greg oh? My apologies. Can you please explain what the author meant? $\endgroup$ – The Pointer Jan 18 at 4:04
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    $\begingroup$ If I wanted to nitpick, I would say that this short text piece already contains two errors. Is the rest of the book written as sloppily? ;) $\endgroup$ – Karl Jan 18 at 14:55
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    $\begingroup$ Well, you cannot identify the elementar composition of a sample by xray diffraction. That is utterly ridiculous. XPS is a spectroscopic technique, hence the "S", not diffraction. And strictly speaking, XRD works on the distribution of electron density in your sample, not on "atoms". But that is indeed nitpicking. ;) $\endgroup$ – Karl Jan 18 at 15:11
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    $\begingroup$ @Karl, the point about electron density is more significant than just nitpicking imo; as I understand it, this is exactly why XRD can not identify exactly what atom it is looking at. Although I am happy to be corrected or nitpicked at if I am wrong. :) $\endgroup$ – orthocresol Jan 18 at 20:45
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X-rays do not allow you to look at the material just like an microscope. You get the spacing information from the diffraction pattern of only crystalline materials and a lot of mathematics. On the other hand, atomic force microscopy may actually be able to tell you where one atomic layer starts and where it ends on a surface. For example, imagine a gold surface, onto which you deposit a layer of molecules (say on local spot by some means). AFM would allow you to create an image (again by indirect means, as it is not an optical technique) of that molecular layer. See google images generated by AFM and you will see how fancy they are.

To the OP: A good way to studying is to rely on multiple references of the subject. Do not take everything literally. The authors are human beings after all with limited knowledge and of course they are not the experts of $every$ topic covered in the textbook. When I was a student, I used to spend a lot of time trying to decipher what the author was saying in a particular sentence. With maturity you learn that the better way (read: more scholarly way) is to look at the same topic in another book and see what the other author is saying.

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  • $\begingroup$ Thanks for the answer. Hmm, those atomic force microscopes look almost identical to optical microscopes, no? $\endgroup$ – The Pointer Jan 18 at 4:40
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    $\begingroup$ Sort of. Optical microscopes have objective lens and a telescope (eye piece), there is nothing like that. I have never used AFM but seen the images. People have imaged single molecules on surfaces. $\endgroup$ – M. Farooq Jan 18 at 4:45
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    $\begingroup$ Yes, that was my point. X-rays are an indirect way of observing distances and positions. I am not sure if you can study single layers of atoms by X-rays. TEM is good for extremely thin layers. $\endgroup$ – M. Farooq Jan 18 at 6:00
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    $\begingroup$ @ThePointer An AFM is (very basically) scanning a very sharp probe tip across a surface in 2 dimensions and measuring the surface hight at each point. This is nothing like how an optical microscope works and doesnt provide the same sort of information that an optical microscope does. $\endgroup$ – Matt Jan 18 at 13:46
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    $\begingroup$ @Karl, I thought the textbook writer wrote that. The proper word would be refraction at the lenses. Diffraction requires interference. I think you just mistranslated some German terminology but meant refraction. $\endgroup$ – M. Farooq Jan 22 at 4:46

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