I want to compare my XRD data (generated form Cr anode) with an existing data(but from Cu anode).
Normally in XRD analysis, we use 2theta as the x-axis to see the peak positions. However, the Q index is the independent from the wavelength. The Q can be calculated by 4*Pi*sin(theta)/lambda, where theta is the Bragg angle and the lambda is the wavelength. Q index, in solid state physics, shows the momentum transfer in reciprocal space. When do the calculation, for example, if you have a intensity I0 at 2theta=10 degrees, you just calculate the Q by 4*pi*sin(5 degrees)/lambda and use it as the new Abscissa of I0. After you do this for both Cr and Cu, you can just plot them to see the difference.
Another way is just use the Bragg's formula, 2dsin(theta)=lambda, and calculate the d values as the abscissa. This tells exactly the interplanar distances in real space.
Neel Ghosh What is the purpose of the comparison? Cr and Cu will give different diffratograms in XRD, so I have difficulty understanding the purpose of the comparison.
Normally in XRD analysis, we use 2theta as the x-axis to see the peak positions. However, the Q index is the independent from the wavelength. The Q can be calculated by 4*Pi*sin(theta)/lambda, where theta is the Bragg angle and the lambda is the wavelength. Q index, in solid state physics, shows the momentum transfer in reciprocal space. When do the calculation, for example, if you have a intensity I0 at 2theta=10 degrees, you just calculate the Q by 4*pi*sin(5 degrees)/lambda and use it as the new Abscissa of I0. After you do this for both Cr and Cu, you can just plot them to see the difference.
Another way is just use the Bragg's formula, 2dsin(theta)=lambda, and calculate the d values as the abscissa. This tells exactly the interplanar distances in real space.
Yizhou Lu's answer explains everything. In most of the cases, XRD analysis software displays the diffraction patterns using the scheme: intensity vs. doubled Bragg's angle (2*theta). I would recommend the second way proposed by Yizhou Lu . If you want to compare the patterns from two different tubes, you must calculate the interplanar spacings, anyway. Of course, if you have 2*theta angles from one tube with step of, say, 0.05 deg., after conversion to the other tube the step will be different.
I am not sure that i understand ur q?
Anyway, u can plot all data on the same layer
Jarosław Ferenc I have made the assumption that the poster knows about Bragg angles and the purpose of XRD for lattice contents and the format that the results are presented in. Plus the software these days does all of this for you. If I’m wrong then the poster had no idea why they carried out XRD..... Comparing 2 different metals via XRD isn’t particularly illuminating. They’re different. Period.
Alan F Rawle , having some experience in reading the RG forums, I assume that sometimes perfect strangers to the subjects ask questions. I reckon this is the case, therefore I posted my response as it stands. IMO, you simply misinterpreted the question: Neel Ghosh asked about the patterns of similar set of samples obtained with different tubes. And I agree, the software used nowadays gives incredible possibilities, but for those less familiar with XRD analysis the simple ways of reaching the final results are more enlighting.
Jarosław Ferenc , yes, I am new to the subject and also I thank you and Yizhou Lu for the answer.
Alan F Rawle you misinterpreted my question. I am talking about comparing xrd data from two different tubes with different metal anodes( namely Cr and Cu).
Neel Ghosh OK, understood. You have 2 different X-ray wavelength sources (with copper K alpha being one). I assume you have limited or no experience with XRD.
Neel Ghosh So, when you compare your data what will be the markers to show that the measurements are similar or diffextent?
Alan F Rawle I am converting the thetas(from the Cr anode xray to the Cu anode) using the formula mentioned(by equating the Q values) and then checking the theta values for the two xrd data.
Neel Ghosh You are not understanding my question. How do you define 'matching'? I'm trying to find out your criteria for defining that the results are the same or different - the correct route is via math/statistics and I'm trying to understand your approach.
Neel Ghosh , if you wish to compare the results from two different tubes (or even two different diffractometers), you must have the very same sample measured in both diffractometers (or even the same one, if only the tubes were replaced). Different position of tubes (adjustable, but yet never perfectly repeatable) will yield different position of peaks. Even the slits play their role, since they will give larger or smaller scatter of rays, which results in different broadening of peaks. If you do not have the same sample compared in both pieces of equipment, you will not be able to correct (unify) your results, therefore your results will be fine for illustration only, but not for comparison of any parameter of the structure.
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