We did XRD of p-type piranha cleaned pristine (100) silicon wafer in two different modes. One is in Bragg Brentano (theta-2theta) geometry and other is in grazing angle. While doing grazing angle omega was kept in 0.5 degree. Scan rate of the XRD in both the cases were 2 deg/min. It is quite evident that (100) silicon should have a peak at 69.25 deg which corresponds to (400). I am able to get the peak in theta-2theta geometry. However, there is a strong peak appearing at 51.25 for the same silicon when XRD was done in grazing angle mode. This peak (51.25) appears both the cases when silicon does or does not have any film on top. I have attached one image for your reference. The sample size is 12.5mm x12.5mm square. The incident slit was 10 mm. Could anyone let me know what could be the reason?
Dear Ankur,
when using grazing incidence XRD you are very surface sensitive.
So you may see the tiny oxide layer and one of the SiO2 peaks.
When googling for SiO2 XRD pattern some time XRD pattern show up having a peak around 51 deg.
For example when looking at the first link and press the 'Read More' button or at the second link Fig. 4 curve a and b.
The fact that only one of the 'SiO2' peak shows up may be due to preferred orientation due to your single crystal wafer.
http://www.axt.com.au/x-ray-diffraction-for-materials-science/
Article One-Pot Synthesis of Core-Shell Cu@SiO2 Nanospheres and Thei...
As I interprete your statement correctly, you have a (001) oriented SINGLE crystal and you want measure it on a powder diffractometer using two modes: a) the typical Bragg-Brentano geometry, i.e. in theta-2theta mode, and b) at gracing incident: i.e. 0.5°-2theta.
If this would be correct, both modes are useless since they are both developed for powders and not for single crystals. In fact, I am surprised that you got any signal since single crystal interferences have a very, very narrow peak ...:-)
The condition for diffraction is, that for a certain 2theta position the normal of the reflected (interferring) lattice plane is EXACTLY between and therefore in the plane formed by the primary direction and the sample-detector direction (often called diffracted beam which is actually not correct since there are an infinite number of beams bearing intensity. You only select one (or a bit more by the selection of the apertures) by positioning your counter).
Why do you get intensity in case of a powder but not for a single crystal? The answer is that you have to align a single crystal damn good to fulfill this conditions. Si is so perfect that you practically always measure somewhere in the tail of the 2-dimensional intensity distribution. In case of a powder, a few crystals are always in this condition. It also means, that for another interference totally different crystals generate the signal in your diffractogram. Only for the different interference orders of a (hkl) the identical crystals contribute to the signal (in first approximation since you have to consider that the illuminated area changes as well as the interaction depth).
I assume that you accidentally acquired intensity caused by Umweganregung, i.e. a multiple reflection which sometimes appears. One indication is that your 004 should be a) MUCH more intense and narrow than you found. This also is an indication that you measured in the tail, i.e. (001) as normal vector was not exactly between and in the plane of primary and reflected beam.
If your Si crystal is masde for microselectronic applications I would not expect a surface influence, but i am not an expert so that Gerhard might be correct. Since you also underlined that all effects occur also without a coating (anyway which one) I don't want speculate which reflection you see could come from another phase.
I would like to add (cf G Martens answer) that piranha cleaning is also used to make thin SiO2 passivation layer in image sensor applications. So i'll recommend to add a HF based cleaning step to remove that oxyde.
Good luck.
Thanks Ali.
Do you know the typical thicknesses of SiO2 by that method?
Hi all
Thanks for your reply. I found some different answer which most likely true. At 51.4 degree, (113) forbidden peak of silicon appears at certain phi angle. If you rotate the substrate by just 5 to 10 deg it disappears. Have a look into the attached data.
Ankur Goswami - Thank you for sharing the interesting result! Although the post is a year old, I would like to share this paper by P. Zaumseil in the Journal of App. Cryst, 2015. I thought it may be of use to you.
Article High-resolution characterization of the forbidden Si 200 and...
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