That is so nice of you to say! I really appreciate it :)

Thank you, I'm glad it was helpful!

I'm glad it was helpful!

Glad to hear it!

Thanks!

Glad it was helpful!

I'm happy to hear that! And honestly, comments like yours mean more to me than I can say and encourage me to continue making videos, so thank you.

Thank you for the kind words! And I apologize for the slow response, I was very busy preparing for and then taking a vacation. Regarding your question, let me direct you to this paper from Rigaku (www.rigaku.com/newsletters/mabu/pdf/TF4_In-Plane.pdf). This does a better job explaining these concepts with pictures than I could do using only words here in the comment section.

Have you ever heard of indexing peaks? If you search KZfaq, I'm sure you can find some examples. Essentially, different types of unit cells have certain allowed and forbidden reflections (see here for some examples: groups.mrl.uiuc.edu/chiang/czoschke/diffraction-selection-rules.html) based off of symmetry/geometry. What this means is that peaks from certain planes will appear, peaks from certain planes won't appear in the pattern. Different types of symmetry have different equations to explain the relationship between the d-spacing and the lattice parameters. You can use the 2theta positions of the peaks to solve Bragg's law for the d-spacings of the peaks, and then you can use the d-spacings to solve for the lattice parameters. This might require some trial and error as you first have to assume a crystal structure and use that assumption to solve for the lattice parameters. If a single set of lattice parameters works in conjunction with your assumed crystal structure, then you are likely good. If a single set of lattice parameters doesn't work, then your assumption of the crystal structure is likely incorrect, and you need a new assumption. This is much easier to understand when shown rather than told, which is why I suggest looking up a video on indexing. The math isn't all that difficult, assuming that you don't have a very low-symmetry structure, it is just a bit time consuming to work everything out.

I hope it went well!

Thanks!

That would mean that the X-ray source and detector would occupy the same physical space. This is what limits us in how high in 2theta we can collect data.

I suppose "ignored" was the wrong term to use. In reality, we are just assuming it is 1. Essentially, (hkl) studied at a value of n other than 1 gives the same results as using n = 1 for the (n*h n*k n*l) plane. For that reason, we can just assume that n = 1 for each plane. Personally, I like to work things out with numbers to see how it works. If you are similar, let's imagine that we are studying TiO2. For the (101) plane, d = 3.51455 angstroms. Let's also assume that we are using Cu Kalpha radiation (lambda = 1.5406 angstroms). If we use n = 1 and enter the values listed above for the wavelength and d-spacing, Bragg's law looks like (1)*(1.5406) = 2*(3.51455)*sin(theta) Solving for theta and multiplying by 2 (since peak locations are reported in 2theta), we get 25.321 degrees. If we do the same thing but make n = 3, we get a 2theta value of 82.222 degrees. Let's now see what happens if we look at the (303) plane, which has a d-spacing of 1.17152 angstroms. If we choose n = 1, we get (1)*(1.5406) = 2*(1.17152)*sin(theta) Solving for theta and multiplying by 2, we get 82.222 degrees, which is the same value we obtained when looking at the (101) plane with n = 3. This shows that n=3 for (101) gives the same results as n=1 for (303). I hope that makes sense!

@@IAMMDiffractionFacility that’s incredible, thanks for your rapid response with a great walkthrough! ❤️

My apologies. I originally made this to be presented during a seminar. I then simply uploaded it to KZfaq, not really thinking about subtitles.

The simplified way of looking at a diffraction experiment and explaining Bragg's law makes it look like it is just reflection, but it isn't. What XRD makes use of is elastic scattering, in which the X-rays begin oscillating the electrons in atoms. Those oscillations then release X-rays of the same wavelength as the X-rays that cause them. These new X-rays are what leave the sample.

@@IAMMDiffractionFacility shouldn't a photon of x ray have enough energy to dislodge an electron from the atom. or do these crystals have very high binding energy to their electrons? And don't electrons only absorb a photon if its energy can allow the electron to do an energy level transition otherwise they just let the photon pass? Or is this a misconception i have

Very good questions. Let me preface my thoughts with the admission that it has been many years since I have really thought on these topics in such depth. Most of my work with X-rays and whatnot has been more practical (setting up experiments, analyzing data, etc.), so I've likely lost some of the knowledge I used to have on such topics. Sometimes, the X-ray does have enough energy to dislodge an inner-shell electron from the atom. Such a situation is how you get very undesirable fluorescence in your diffraction patterns when using certain X-ray sources to study samples with certain elements (e.g., Cu source strongly fluoresces Co and Fe). I can't explain why, but this problem seems to strongly occur for only two or three elements that depend upon the type of source you are using. Regarding absorption of a photon, it is my understanding that moving/removing an electron from its shell (see previous paragraph) and oscillating electrons are two options.

@@IAMMDiffractionFacility ok thank you for your help! I've tried looking for deeper answers to these questions but have struggled to find anything concrete online😅