Dear Morteza,

First of, I would like to postulate that there are three separate issues which we are dealing with here.  Please see list below as well as abridged version of their origins:

Here is an unabridged version for those who's interested: 

The first on the list: a broad peak at ~ 14.5° is so called escape peak.  Please see '02 Lőrinczi ref. for more details.  It comes from the sample Φ-orientation, and can be mitigated if the sample is turned such that the peak is minimized, just like you had inadvertently discovered yourself :).  It is only a problem when the material of your interest has a peak in this region (i.e. 2D transition

metal dichalcogenides (TMDs) materials - I happen to have a pleasure working with).  If the peak of the interest is not there, I suggest starting your scan from 17 degrees on, so you do not have to worry about it. 

And here is the synopsis of the article: 

•  Lőrinczi presents an explanation regarding the nature of the escape peak, stating that it comes from the secondary florescence produced inside the counter which.  And futher disscimiator cannot distibguish between the signal from the sample and from the counter.
•  Lőrinczi presents two solutions to mitigate the broad peak at ~ 14.5°.
• The first method is to rotate the sample around it's axis.  He suggests that the optimal sample rotation is also film dependent, see Figs. 1 and 2. Lőrinczi suggests conducting a series of scans, finding the lowest substrate contribution and proceeding with the scans using these settings to mitigate the effect.
• The second method is to use lower high-voltage X-ray tube.  

I would like to note that not all silicon wafers I have worked with produced this phenomenon.  

The third on the list is a sharp peak at about 2Θ = 33°, which is a forbidden Si (200) reflection from the Si (100) wafer.  One can calculate easily a precise (200)'s position using the lattice plane spacing formula for cubic geometry and d-spacing for the (400) peak:

1/d2=(h2+k2+l2)/a2, and knowing that d(400)=1.3575 in my case.    

d(200)=(4/2)*1.3575=2.715 ~33o

d(100)=(4/1)*(1.3575)=5.43 ~16o - with this calculation, we had just ruled ruled out that the first peak on the list is a Si (100) forbidden reflection.   

Please see '15 Zaumseil ref. attached regarding the nature of the peak, and here is the synopsis:

• The forbidden Si (200) peak often appears on the XRD patterns of the Si (100) substrates due to the multiple diffraction (Umweganregung);
• It may have variable intensity and exhibit shoulders and sub-peaks at about ±2.5° 2Θ.  The peak(s) generally occur at 2Θ = 31° to 35°;
• Due to the particular Φ sample orientation, peak's intensity will change. For the particular Φ orientations outside the multiple diffraction (Umweganregung) conditions, the peaks will vanish completely. 
• Additional variance in ω orientation will create other artifacts (i.e. sub-peaks and both right and left shoulders). 
• For Si (111) substrates, forbidden Si (222) reflection always appears at about 2Θ = 58.9°.

I would like to note that not all silicon wafers I have worked with produced (200) forbidden reflection phenomenon.  

Please let me know if you have nay further questions. 

I do not know your exact diffraction geometry/instrumental setting. The peak shape is strainge. Can you exclude an instrumental artifact? Have you checked a naked substrate? It could be diffraction from white radiation and your Si substrate.

In fact the peak shape is very strange. If these peaks appear on different kinds of substrates they do not pertain to this material. The thin film deposition is performed in argon atmosphere? Because it seems that the peak positions could be compatible with some kind of vanadium oxide.

Dear Morteza;  Did you estimated  the mean d-spacing for  those two humps by taking their maximums as the quasi-peak positions using Bragg's law. Even though the thin vanadium coating might be amorphous initially due time it may recrystallize using structure as a template. But according my naive estimation the ration of the d-spacing between the first hump of V and  the first sharp peak of  W  is about  2.46, where  dW(111) =1.73A.  that  means dV  may be in the range of  4.26A    and which corresponds to  aV(111) = 7.24 A , and  that can't be associated with any known crystal structure of Vanadium (BCC). 

By various substrates I mean p-Si, n-Si and p++-Si  with different coated metals as electrode for the vanadium. And I can see those humps in either oxygen content and non-oxygen content vacuum of sputtering (Ms. Bini)

Mr. Leinweber, I didn't checked the naked substrate yet and as I noticed now my all substrate were Si based, so it could be the "diffraction from white radiation and Si" as you said. Will you please explain more about this phenomenon?  Also I just checked other's XRD patterns from the same machine for other metals (Zn,Ce) on Si substrate and found those humps and the peak of 33 degree in those patterns again.

Thanks for the answers.

Dear Morteza,

These two humps are originated whether from your sample or from an instrumental artifact. 

My first remark is that these two humps are too broad for a diffraction pattern from metals although you have a  small sized crystallite/or internal stress.

The second remark is even if they are from a crystalline structure, the interlayer distance calculated from Bragg's law would be  for the first hump (2-theta=14°) is around 6-7 A° (depending on the X-ray wavelength that you use), which is very large for metallic Vanadium as said by Prof. Ogurtani.     

If you have these two humps even you use different substrates (with and without SiO2/Silicon!), it is possible that they have an instrumental origin. I just suggest you to check your XRD system by using a well known metal (Cu, steel, Ni, ...) as sample and observe if these two humps always exist. 

Another explanation would be, the presence of other chemical species, like oxides as proposed by Prof. Bini. 

What is the origin of the pattern at 2-theta=33°? metallic Vanadium? 

Dr. Atli, as for the other substrates, sapphire and glass coated materials do not show such humps in their XRD pattern and I just noticed that the peak on 33° plus the two humps can be found in other metals (Zn,Ce) XRD patterns coated on Si-based substrates (p,p++,n-Si,SiO2). I think this error is due to interaction of XRD system with Si-based substrate as Prof.Leineweber mentioned but I don't know what is happening and how. 

Dear Morteza,  Please go through the XRPD literature published in Google  related to  the amorphous substances including the melts, and  please compare those  patterns with   yours, which you   have gotten them with your own hands !!!!! I think you were  very luck by getting beautiful sharp  substantial peak which can only be obtained under the drastic quenching in thin films.  BEST REGARDS

Notice, I'm intentionally being provocative for the dramatic effect :-) No offense meant. But repent and change your ways to avoid all this ambivalence at the higher echelons of XRD R&D with one of the most precise century aged techniques, X-ray Diffractometery :-) Check out the attached YouTube video playlist!

Don't keep us all guessing! Please share all the details. The feedback from the "expert" RG membership will be a lot better :-)

Rip Van Winkle!

https://www.flickr.com/photos/85210325@N04/10659886546/in/set-72157645018820696

https://www.flickr.com/photos/85210325@N04/10221065324/in/set-72157645018820696

https://www.flickr.com/photos/85210325@N04/11343736043/in/set-72157645018820696

https://www.youtube.com/watch?v=QRcj_t1WgVs&list=PL7032E2DAF1F3941F&index=2

Depending on you primary and secondary optics you do not perform your diffraction experiments with monochromatic radiation. For simple powders, you do not see the effects, becase the contribution to a powder pattern is often negligible as compared to the background. But an extremely powerfully diffractig substrate like a single crystal can reveal these effects.  So if at least a part of these effect is also revealed by other Si samples it is very probable that it is Si(400) by some white radiation or some dynamic Si(200)/Si(100) (I cannot explain much more). Things to check:

-What is the orientation of the crystal substrate?

-can you tilt you specimen slightly in omega or chi (depending on your diffractometer!) to get rid of these peaks?

"Depending on you primary and secondary optics you do not perform your diffraction experiments with monochromatic radiation." Quite to the contrary, we always perform XRD observations with monochromatic radiation (e.g., I'd think a Ni filter for Cu target at the least) to minimize the distortion of the reciprocal space due to multiple wavelengths. Managing this without the synchrotron is a challenge we have overcome through ultrasensitive, cost effective 2D real time XRD imaging systems . The reason why most prefer to keep all the wavelengths in XRD observations is due to poor signal from samples and poor sensitivity of detectors, especially to spatial variations in intensity. Folks in XRD Topography routinely use "white radiation" to smudge the reciprocal space in order to get better image contrast. This is a fallacy. The best way to improve contrast in reciprocal space is by actually probing the reciprocal space with a single or a "delta function" monochromatized beam and the rocking curve method (standard Theta-2Theta would do:-). This becomes even more critical in powder diffractograms with closely spaced peaks. The wavelength dispersion of the incident beam will determine the reciprocal space resolution ultimately.

As Andreas points out check for the sensitivity of the diffractogram to Phi, Chi, Omega, Psi(?). Make sure your diffractogram is unchanged with x,y & z (sample surface displacement) translations also.

@ Mr Anath

"I'd think a Ni filter for Cu target at the least..."

This is not strictly monochromatic! Check a strongly diffractic sample (e.g. a single crystal) with such a configuration and you will see a lot of effects.... Not only around the ordinary Bragg reflections....

@Ananth @Leineweber

The orientation of the Si crystal is 100, and the X-ray tube has a wavelength of 1.5418740 (Cu-Ka). For the other technical details I need to have some inquiry. 

¿Where are the V paks on the diffraction pattern? Moreover the peak at around 33o coincides with the strongest peak of V2O3. Could the broader peaks coming from some vanadium oxide?

Andreas! That is right. Near delta functions may only be obtained through synchrotron. In normal lab sources

OK Mr.Ananth, I'll try to get diffractogram of a naked Si substrate placed in different direction on the holder of XRD system. 

Dr.Gallardo the peak of 33° exist either in oxidized and metal Vanadium XRD pattern and most probably belongs to Si substrate along with two other humps. 

Seems like this is GIXRD pattern.

IMHO, there's an induced strain and stress to SiO2 perpendicular to the plane cause by Si and W. But need to confirm by pattern from naked Si/SiO2.

As the simulation below,  by changing the lattice constant we can reconstruct the diffraction pattern.

It can be seen, there's strong preferred orientation at 001 to match the Si substrate, as well as W-beta. The second peak intensity is about ~1/3 from the first, quite similar to the actual pattern.

I am not sure, but if this would be a Si-reflection using white radiation as brought into discussion shouldn't intensity appear everywhere and not as hump since the reflecting lattice plane doesn't change and theta is simply using the white spectra to fullfill Bragg's law? In so far I would expect an imaging of the intensity distribution of the X-radiation with 2theta inclusing sharp peaks for Kalpha and perhaps Kbeta but no hump at all.

Such artifacts are typical if not exactly aligned substrates are measured and only the tails of peaks are detected (of characteristiv radiation). However I don't believe that theses tails will be that broad. Dynamical effects can be also excluded since there is no "forbidden" peak at 15° (using characteristic Cu Kalpha).

Monochromators typically reflect only parts of the spectrum ....

There is nowhere written that a monochromator was used, isn't it?

you should check for slit size and alignment (what is your geometry?) and also run a measurement on empty sample holder

I tend to agree with Gert about each reflection having the convoluted components of the incident beam wavelength dispersion and not just a selected few reflections, as is being hypothesized.

However, it would be helpful if Morteza could fill us in on the details of the XRD optics and experiment as requested earlier for better feedback.

"can you tilt you specimen slightly in omega or chi (depending on your diffractometer!) to get rid of these peaks?" I don't think so, not by just "slightly" tilting. The subject peaks are distinct and broad with excellent relative SNR (signal to noise ratio). The root cause of this ambiguity is the use of the archaic "spatially blind" 0D scintillation detector in this case, as by the majority of the XRD community still (a century after the Braggs). As I suggested earlier, a few seconds exposure with dental film (an extinct and unknown art today) would tell us a whole lot more immidiately. Unfortunately, besides not using the correct 2D detectors the ubiquitous use of ancient technology (0D) in high precision XRD measurements is a huge handicap for data interpretation. Absence of the diffractogram from the substrate without the film is not helpful either :-)

The fact that Morteza has not yet resorted to the simple "dental film" and that no one else in this erudite audience is suggesting it implies:

https://www.researchgate.net/post/What_is_the_best_way_to_interpret_REAL_TIME_3D_X_Y_Omega_2Theta_reciprocal_space_relative_intensity_data_from_HRXRD_rocking_curve_analyses

OK I am going to run XRD system today and I will put the results here by tonight. 

Go for it Morteza! That's the only way to find out! XRD is the best studied NDE tool to characterize the in situ Nano structure of your composite down to FEMTOMETERS in reciprocal space!

BTW what is the end use of this type of material. Please post some literature references for all of us to get more familiar with what to expect in terms of materials Nano structure.

If you are still in the XRD Jurassic Park like the majority of folks dabbling with XRD at the highest echelon of R&D then fall back to tools that stalwarts like Laue, Bragg, Ewald used, photographic film! You can easily digitize the dental film using a trusted modern scanner. Use MS Paint or Origin to analyze the image. You'll be amazed what each frame would tell you. Just a simple Debye-Scherrer image in 2D can enlighten you more than the hour long 1D conventional diffractogram with 0D  point counters in terms of preferred orientation or epitaxy and to some extent "particle" size or CDD (coherently diffracting domain) size. Besides composition!

I continue to encourage the use of dental film in all XRD labs that cannot afford to get a 2D real time XRD imaging system! 

https://www.flickr.com/photos/85210325@N04/7988760159/in/set-72157632728981912

I have taken those two pictures from Google, which were ontained and published by © 2013 Liquidmetal® Technologies, Inc.

The first XRD belongs to the crystalline metal-alloy, and the second XRD obtained from the same metal-alloy, which was in the form of liquid/metal having amorphous structure but still keeping the short range order.

If one wouldn't get lazy to go through the data base of Google,  one could find  enormous amount of comparative cases to reveal the XRPD spectral structural  transitions  when a  substance goes through  from the one crystalline form towards the final amorphous shape sequentially through the meta-stable states,  depending upon the quenching rates as well as the size.  

BEST REGARDS

Here is the XRD patterns of n-Si (100) as a naked substrate for it's first positioned state and rotated about the surface normal states and also a XRD result for an empty sample holder (the inset).

The configurations of the XRD system are in following images (slit size and filter type of X-Ray).

You are most likely looking at the substrate topograph. Use the dental film in front of your detector as I suggested and you'll see. I'll show you examples with the NIST 2000 SRM. The only thing out of place is the large diffracting range in Omega and 2-Theta most likely due to the divergence in the low angle geometry. I'll have to make some computations to gain a higher certitude. I personally don't think it has to do with the "amorphous" phase but would remain open-minded :-)

What is the approximate footprint of the beam on the sample for the geometry pertaining to that (hkl) reflection?

Are you using a conventional point counter or 1D or 2D detector to acquire the linear diffractogram?

BTW index all your reflections correctly!

Peak at ~33 is belong to Si(100). It only appear at first orientation, while rotate 45 and 90 it disappeared.

I still thinking that those 2 broad hump may from ultrathin SiO2 layer that thermally grown as the mask. Maybe u cud try to etch it and take the XRD pattern again. Attachment is one of the method.

No one could  explain such a broad hump coming from a crystalline substance regardless its micro structure unless one goes to the  sub nano-metric randomly distributed aggregates. What should we call  then that aggregate?  AMORPHOUS having short range order !!!!  Isn't it right?  or the mosaic structure? 

I still don't get the peaks of W and the very narrow peak. Did you use Cu radiation (as I assumed until now)? Then you W peaks do not match at all, or which kind of phase this is? 

Tarik! If that signal in question was from the "AMORPHOUS" constituent then why should it display preferred orientation (periodic maxima when rotated about "Phi")? Why is its relative intensity so high?

Maykel! An "ultra-thin SiO2 layer" couldn't have such a high relative intensity either without GIXRD geometry, could it?  :-)

This does not have a complicated solution but a simple obvious straight-forward explanation, in my opinion.

Chances are that, if one is sputtering metals dissimilar to the substrate then, it is most likely going to be poly-crystalline or some sort  of a "low order" thin film with some amorphous content. The probability of some degree of "preferred orientation" is extremely high as well. 

Creating "amorphous" thin films simply by sputtering dissimilar metals seems to be a "tall order". I would think that in order to get "amorphous" constituent one would have to rapidly cool to solid state from vapor or liquid phase. Is that a possibility with "RF magnetron sputtering" as Tarik postulates? If that is the case then it would be prudent to estimate the crystalline/amorphous content ratio as well.

A simple dental film exposure would yield some immediate clues. Of course 2D real time XRD imaging is the correct  & ideal method for studying this issue.

 Ravi, what is the peak at 33°?

Good question Gert! Maybe Morteza could look it up using the diffractometer software. I see it waxing and waning with "Phi" rotation of the substrate. It would be good to see the same for the substrate without the sputtered film on it. Maybe, flip the sample over and examine the substrate on the back side if possible.

We are all engaging in lots of futile pontification in the absence of the simple good old dental film, aren't we? Finding a local old time dentist could end all the speculation instantaneously :-)

OFF TOPIC! Just to add some color to a drab technical discussion, here is a political joke I can get away with in the US without being prosecuted like the poor young chick Turkish model. That's like jail time for wrong RG answers. His uninsultable immagesty has even threatened the present pope with Sirhan Sirhan clones :-)

Look what I found! "Sultan Erdogan's baby photo"! Could be Pope Jean Paul's too? Or for that matter Hindu Cow Protector PM Modi's or Gandhi's but not Obama's! Equal opportunity basher! LOL! There is no insult in this joke unless one wants to perceive it so or if one's self-image is puny! This kid probably grew up to be Arnold Schwarzenegger unlike me!

I'll judge your sense of humor based on the feedback :-)

MR.Mananwan, I cleaned the Si sample with HF acid before acquiring the XRD patterns. I also have some patterns for oxidized Si which shows those humps and peaks again (following image). So I dont think they belong to ultrathin SiO2 layer.

The thickness of the Si substrates is about 750µm. 

You need to identify a few experimental parameters to get a better feel. See diagram attached.

Also remember that W & V film will attenuate the substrate signal. Both are very high attenuating constituents with large atomic numbers. It would be important to understand the approximate thickness of the film (50nm) and the related attenuation of the substrate reflections.

"I also have some patterns for oxidized Si which shows those humps and peaks." Do you see them in regular Si substrates without film?

I also notice you mention the Si substrate but show Si/SiO2 in figures. Please explain.

@Ravi, I'm not sure about it, I'll try do the same experiment, but without etch the wafer. Yet the peaks intensity is not much reduced after coated with Tungsten and Vanadium (30-50nm) compared to Si (100) peak in the first image, mean something on the top of the wafer.

@Morteza, is it the omega angle >1 ?

Because GIXRD scan at constant omega arc, the best is to get the reciprocal map; as mentioned by Andreas about white radiation streak, maybe a thin layers, d-space variation (strain/stress), mosaic spread or dislocation will appear.

What u can do is:

- Scan with the same 2theta range as the first position but changing the omega for each scan, maybe 0.5 or smaller. Stack the data and use Isolines view to observe.

- GIRD with omega of 0.1, assumed packing of Vanadium is 0.8 so the penetration depth is ~37 nm. If u wanna get rid those 2 broad humps.

- In-plane diffraction for the waver

- X-ray reflectivity

Please also do a Gonio scan.

@Gertz, it's the peak of Si (100), because Si wafer is single crystal it doesn't appear after rotate 45 and 90. Here the articles contain those (100) peak.

This is the first time I've seen the forbidden Si 002 that strong :-). Are the dynamical effects that strong, or why 002 is visible although it should be forbidden?

BTW any sample with FWHM of 1.5 degrees, as in this case at about 14o 2-Theta, is bordering on "poly-crystalline" besides Gert's notion of "forbidden reflection". Please review the RG post on NIST 2000 SRM to get an idea of what typical Si FWHM should be for various (hkl)'s. So, having the virgin substrate diffractogram (without any film) would eliminate this ambiguity. I've attached some data from a ZnSe (224) reflection (ACT - Asymmetric Crystal Topography) which has a similar 1-1.5 degree FWHM (without the "humps"). You would never see such details with the conventional "spatially blind" 0D point counter. Do review. This data was collected on a Panalytical X'Pert/MRD type of instrumentation with a PIXcel 3D area detector by Panalytical in collaboration with the EU synchrotron facilities many years ago. Blissfully, we have the software to analyze such 3D (x, y, Omega) relative intensity data in great detail, they didn't!

Never-the-less, days of futile pontification by us so-called experts cannot substitute a quick simple exposure with dental film. It seems quite obvious that "spatially blind" 0D RSM's (in-plane and out-of-plane) are preferred over 2D dental film or that such archaic tools have become extinct in the modern pre-historic XRD labs :-) Which is it? I'm yet to see folks display the photographic image of the incident beam used, just for reference at least. I have never acquired a diffractogram using a 2D detector without recording the "incident beam" in its complete vivid details at the least! Most folks cannot even confirm that the diffractometer axis (rocking axis or the z-axis) is parallel to the "line source" or that it bisects the incident beam. This would be basic requirement for such observations, wouldn't it?

Now, if you do use a 0D detector then the correct source to use is a collimated 0D point source (micro focus), isn't it? Ever tried optimizing a reflection for rocking curve with a 0D source and a 0D detector? What criterion did you use? Maximum integrated intensity? :-)

Once we know the "beam foot print on the sample" and the "receiving" slit dimensions in front of the detector along with the detector active area, we will understand the area integration involved with these measurements. If the detector (20-25mm diameter active area typically) is wide open without any receiving slit then the large FWHM would make a bit more sense. Otherwise, it is an enigma!

At a 240 mm typical  diffractometer radius a 25 mm diameter active detector area would subtend nearly 6o in 2-Theta without receiving slits!

The original data for the ZnSe (224) rocking curve reciprocal space imaging is attached as a YouTube video.  If I was to dumb down the 55um pixel size PIXcel detector into a 0D scintillation detector, as most folks do for the RSM's, then the vivid topograph will reduce to a 0.8o "hump" (Figure 1) as well even with a "receiving slit" (Figure 2) :-)

Figure 1- https://www.flickr.com/photos/85210325@N04/14702658058/in/set-72157647811010651

Figure 2 - https://www.flickr.com/photos/85210325@N04/14889281565/in/set-72157647811010651

Figure 2 

https://www.youtube.com/watch?v=dFCQS8oUyT0&index=37&list=PL7032E2DAF1F3941F

https://www.flickr.com/photos/85210325@N04/14702658058/in/set-72157647811010651

https://www.flickr.com/photos/85210325@N04/15099411688/in/album-72157647811010651/

https://www.flickr.com/photos/85210325@N04/14889281565/in/album-72157647811010651/

Maykel, we are not using the GIXRD analysis. Our system works with conventional θ/2θ scanning method.

What is the beam size and slit dimensions?

At  6o Omega the foot print on the sample surface would be huge for the 13 degree reflection. It is almost like GIXRD :-)

Calculate the beam expansion/contraction due to the geometry! Most important compute the extinction depths for the various constituents on the composite sample surface. This will give you an idea of their effect on the relative intensity.

I though it's GIXRD because the broad peak has a similar character as Si-wafer with omega 0.5 degree and peak at ~33, but with additional peak from my substance. Unfortunately the data is only from 20-100 2theta.