Please have a look at the Bragg law:
lambda=2*d*sin(theta)
This equation has to be valid for both wave lengths (Cu and Fe -K-alpha).
So you have:
lambdaCu = 2*d*sin(thetaCu) and lambdaFe = 2*d*sin(thetaFe)
Taking the ratio you will have:
lambdaCu/lambdaFe = sin(thetaCu)/sin(thetaFe)
Solving for thetaFe you have to go via:
sin(thetaFe) = sin(thetaCu)*lambdaFe/lambdaCu
That's it.....
Immanuel Paulraj What is your purpose??? Convert is not the same as an experiment.
Maykel Manawan because our XRD target is Fe. All the literature reported in Cu target. So I would like to convert Fe to Cu
Immanuel Paulraj There's an option in search-match software where you can select the source before indexing. Fe-source is quite rare unless your sample is fluorescence to Cu-source.
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