The asymmetry of the rocking curve means that lots of the grains are not perpendicular (with small deviations as in the first case) to the film surface, but more in a "postive" direction. This happens for instance, when you deposit a thin film under a different angle than 90° relatively to the substrate surface. with best regards, uk

The shape you show is indeed a bit weird mainly because it's non-symmetrical. It appears to me as  a combination of a crystalline and an amorphous peak, although I haven't heard of amorphous Pt.  Alternatively, you may have a second amorphous peak of some oxide.

The asymmetry of the rocking curve means that lots of the grains are not perpendicular (with small deviations as in the first case) to the film surface, but more in a "postive" direction. This happens for instance, when you deposit a thin film under a different angle than 90° relatively to the substrate surface. with best regards, uk

As it clearly indicates from the XRD rocking curve that the Pt is not fully crystallized, it seems some amorphous phase still present. You may also try to increase the annealing temperature and annealing time to improve the crystalline quality of Pt. Once you improve the crystalline quality of Pt, you will get a symmetric Pt rocking curve.

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Regards

The asymmetry of the rocking curve does not say much about crystallinity, but as Hans-Ulrich wrote, more grains are oriented not parallel to the surface in the "positive" direction. 

The asymmetry of the rocking curve is not a very unusual phenomenon. If you take a single crystal with FWHM close to the resolution limit, the shape of the rocking curve will be unpredictable as, by rotating a crystal around.some axis, you change the illuminated area and the proportion of domains contributing to reflection.

Now about your system, Christophe:

I believe the 100 nm thick Pt films deposited at high temperature and at room temperature and then annealed differ in the stress state.

Actually I have found a paper (see the link below) for (Pt/Ti/SiO2/Si system) that says the difference can be dramatic (tensile in plane in one case and compressive in-plane in another).

I am not certain whether these considerations are relevant to your observations.

The difference in the stress state may cause changes in orientation if the stress is beyond the elastic limit., but this should lead to just broadening of the rocking curve (not an asymmetry).

Can you measure the rocking curves of other constituents (Si and SiO2)?

http://link.springer.com/article/10.1007%2Fs11664-998-0171-7

I agree with Mark O Rikel:

"The asymmetry of the rocking curve does not say much about crystallinity, but as Hans-Ulrich wrote, more grains are oriented not parallel to the surface in the 'positive' direction."

I suspect those talking about amorphous peaks or not fully crystalline are forgetting that it is a rocking curve.  If a scan in 2theta looked like that, it might be an overlap of sharp Bragg peak (from larger 1000 nm crystals), broad Bragg peak (from nanocrystals) and/or amorphous peak.  But it is not a 2theta scan, it is a non-crystalline portions just won't show up (or will just be background).

If you can get some additional data, I suggest double-checking that the acceptance angle covers the whole peak, even at the maximum tilts in both directions.

Also, try repeating the rocking curve with the sample rotated 90 and 180 degrees from original setting.  If the amorphous or partially crystalline suggestions are true, the curve will stay the same no matter the orientation (I don't expect this).  If the sample has grains not oriented parallel to surface, the tail (asymmetry direction) will follow the sample orientation (I think this is likely).  The latter might be expected because the lower temperature deposition allows less atomic rearrangement to align with the substrate after the atoms impact the surface.  Thus, any small deviation from atoms arriving perpendicular to the surface will have an effect on the sample microstructure at low temperature, but not at high temperature.  Does that make sense to you?

Thank you all for your answers. I have now to think about it.

Regards

I have noticed a noticeable shift in the peak positions of the two samples. Such shift could arise from lattice strain, and could be responsible for the observed asymmetry in the room-temperature deposited film. This seems to support the viewpoint of Mark Rikel regarding the stress in the sample. Also, the relative orientation of the film surface with respect to the substrate and zero beam direction would obviously result in the observed asymmetry as Ulrich Krebs suggested.