I have deposited SiC thin film on n-type Silicon substrate (100). I have recorded the XRD diffractogram in two different conditions. In the first case, I have performed detector scan and observed two peaks, centred around 2theta = 35.5° and 56.6°. I have assigned the lower one to SiC and a higher one to Si substrate. In the second case, firstly I have performed a phi scan corresponding to 2theta= 35.5° and then did detector scan, for this case I have observed only one peak in the diffractogram centred around 2theta=35.5°. The XRD diffractogram has been attached for your reference. Kindly explain the physics. How the diffraction is taking place in both the cases?
In your 1st scan, the 2nd peak observed at ~56.5 deg (2theta) corresponds to (311)Si of Si(100) substrate. The (311) and (100) planes in Si are oriented at 25.39 degrees. Also, the Bragg angle theta (and not 2*theta) for (311) is 28.28. Thus, if you perform the detector-scan at a glancing angle of 2-3 degrees or so (~ 28.28-25.39 ), the x-rays will incident on (311) plane at nearly the correct Bragg angle (with the FWHM of ( 311) peak), and hence the x-rays diffracted from (311) plane will be detected by detector during the scan, PROVIDED, the ''normal' to the (311) planes lies in the plane of diffraction (containing the incident, diffracted x-rays and film-normal). This is case in your first scan.
By doing a small phi-rotation (or in-plane rotation), you essentially move the (311) plane's normal out of the diffraction plane, which means X-rays are still being diffracted from (311) planes but they are not in the plane where detector moves during the scan). The result is that you do not observe this diffraction, as is the case in your 2nd scan.
Note that had the Si substrate being perfectly polycrystalline powder pellet the two patterns (with regards to the substrate peaks) will be similar irrespective of the phi-rotation.
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