This is the XRD map obtained from 2 dimensional XRD tool with crystal rotating. I know the lattice parameters, however I don't know the single crystal orientation. I want to know the crystal orientation from this graph. We don't have access to Laue diffraction tool right now, that's why we are trying with 2D-XRD.
You could, e.g., calculate all theoretical deflection reflexes and compare it with your data. The theoretical values can be obtained from the Laue condition, where the reciprocal lattice vector G is equal to the scattering vector q. q is defined by the 2theta angle, G by the reciprocal unit vectors and, thus, hkl and the lattice parameters.
However, a 2D map of 2theta and omega would be maybe more appropriate for your problem.
Dear Martin, we also do have 2D map of 2theta and omega. Please find the graph in the attachment. I really didn't get how to calculate the theoretical reflexes? Could you please give a simple illustration, if you don't mind?
Does each spot on the 2D graph represent one set of planes or all spots represent same set of planes?
In chapter 2, there are some basics of x-ray diffraction theory and the respective formulas in the case of a wurtzite crystal. Anyway, you can easily apply that for any crystal structure.
From my experience in 2D XRD (we have a GADDS system with microdiffraction and an Eurelian cradle), I must say that is not easy neither to identify the hkl indexes of the measured spots nor to identify the nature of the single crystal. There is not enough space in this forum to teach this. The easiest way is to take an small piece of your single crystal and molten it to powder and analyze (without phi scan). Prior to determine the orientation of your single crystal, you must know what your single crystal is.
Roughly, the orientation of the single crystal is as follows:
Once you know what is your single crystal is, then you must perform several Omega scans on the single crystal (in Scan mode) to cover a wide enough 2theta range, without phi rotation nor any oscillation. In the equatorial plane of the resulting frames you will have the with this analysis,the spots of the hkl planes paralels to the analyzed surface. Once you know the hkl indexes of such planes, you start to know the orientation of the single crystal. Depending on the symmetry of the single crystal, you must measure several frames changing the 2theta, phi and chi angles in order to determine completely the orientation of the single crystal. All the measuremets take time to planify them and understand what are you doing. It is very helpful to have some software with an stereographic projection (for instance CarInev) to planify the measurements and index the spots that appear in the frames.
I don't have any paper that explains the procedure.
The spots in the equatorial plane of the frame are those that are located at gamma=-90º (in my diffractometer system). By pressing F9, you can read the position of the mouse in 2theta and gamma angle. If one spot is in the equatorial plane of the frame, and the frame has been measured at chi=90º (or 0º depending on your set up), it means that the corresponding plane is paralel to the sample surface. In other words, the uvw vector of the corresponding plane is normal to the sample surface.
By knowing the hkl indexes of the analyzed sample surface, you can start to locate the axes orientation of your single crystal. For instance, if your sample is cubic and the analyzed surface is 100, then you must search 110 (with the corresponding 2theta angle) by measuring several frames in the phi angle (from 0 to 90º, step size 1º and SCAN mode) and chi=45 fixed. You can improve the phi angle by decreasing the phi step size. The 110 spot should appear in any frame of this serie but just in the center (at the expected 2theta angle and gamma=90º). Once you have determined the correct phi angle for the 110, you can check it with an Omega scan with the previous angles. If everything is OK, you should detect the reflections 110, 220, 330, etc.
Once again, this a rough description of the procedure. As you can imagine, the procedure for a monoclinic structure is more tedious.