Dear Dr. Charmaine Lamiel 

Have a nice day

The intensity of the X-ray diffraction line depends on the elemental composition of the sample and its preparation conditions.

During the preparation process, the crystallites may take a certain direction for their growing. Samples may have a preferred direction (preferred orientation) for their growth.

For thin films for example, the crystallinity of sample depends on the following: 

1- the substrate surface if it is hot or cold, is it crystalline or amorphous,

2- the deposition rate,

3- the deposition time,

4- temperature treatments. 

Really there are many factors control the formation of the sample and consequently the diffraction lines will affected.

For more details, please go to my RG page. There are some articles may help your study.

Good luck

Thank you very much for your response. I tried to do that too but I still have the same XRD peaks. 

In my opinion, the heigth of JCPDS planes may be not simple intensity but also the area of the peaks. 

Dear Dr. Charmaine Lamiel 

Have a nice day

The intensity of the X-ray diffraction line depends on the elemental composition of the sample and its preparation conditions.

During the preparation process, the crystallites may take a certain direction for their growing. Samples may have a preferred direction (preferred orientation) for their growth.

For thin films for example, the crystallinity of sample depends on the following: 

1- the substrate surface if it is hot or cold, is it crystalline or amorphous,

2- the deposition rate,

3- the deposition time,

4- temperature treatments. 

Really there are many factors control the formation of the sample and consequently the diffraction lines will affected.

For more details, please go to my RG page. There are some articles may help your study.

Good luck

Please read these articles:

http://www.scirp.org/journal/PaperInformation.aspx?paperID=23195

http://www.sciencedirect.com/science/article/pii/S0272884215011815

I think they can help u to get your answer.

From what I can see your XRD data matches very well with the standard. The relative intensity of the peaks might be due to the preferred crystalline orientation. But I would doubt that, since your SEM image shows a very random orientation. You can choose certain diffraction peak and do a rocking scan or even a pole figure scan in order to measure the orientation distributions.

It is due to the preferred orientation along . There are amorphous hump present (18-28 2theta). Interesting nano material. Try to increase the scan time.

It would be better if u attach the raw data.

I agree with Mr Manawan. Itis due to preferred orientation. In some experimental conditions the orientation may be different from the other and may lead to some different structures like rods, cones etc. i strongly suggest to take FESEM of your sample. if there is any specific structure then you can correlate.

 take FESEM/TEM at higher resolution. 

One method to confirm that this was a result of "preferred Orientation" using the conventional linear diffractogram, is to repeat the observations after turning the sample surface w.r.t. its surface normal.

https://www.flickr.com/photos/85210325@N04/7944785794/in/album-72157632728981912/

First, you sample might be grown with some directions favored. Then, anisotropic growth might be the reason. That is to say you have more periodicity than do other directions.

The second one is that other facets have weaker signal because of weaker x-ray interference between the these layers.   

Your peaks look broad. Is this an powder x-ray of nanoparticles? It might happen that when you drop your sample to the substrate glass slide, it dried up ending with nanoparticle with certain orientation. This is also possible. 

Hi Charmaine,

The specimen might be textured. The Xray intensities calculated using structure factor is valid only for powder samples. In powder specimen each plane has an equal probability of being oriented in any direction. While in a textured specimen, different specimen orientation will give you different intensities. 

Thank you very much for your responses.. This is an XRD of a powder sample. It is in amorphous state because of the conditions I made during the synthesis.

Dear colleague Charmaine Lamiel ,

Have a nice time day.

Herein fined two links contain much information about the increasing or decreasing XRD pattern intensity.

I wish this is useful and help you.

yours

A. EL-Denglawey

https://www.researchgate.net/post/What_would_be_the_reason_for_an_increase_in_XRD_peak_intensity_of_pure_metal_oxide_Co3O4_compared_to_mixed_metal_oxide_NiCo2O4

https://www.researchgate.net/post/What_are_the_the_basic_reasons_for_the_increase_or_decrease_in_intensity_of_a_particular_peak_with_doping_level_of_a_dopant_in_XRD_pattern3

Powder diffraction peaks should ideally be sharp. But however sharp a peak may be it will still have a distinctive shape. If the peak is very sharp then it simply means that we must scan with small steps (typically 6 to 10 steps as discussed earlier) through the region of 2θ within which the peak resides in order to have sufficient sampling to define its shape. The next figure shows the 211 and 886 peaks of yttrium oxide using a laboratory powder diffractometer to scan slowly through the two peaks, counting for 10 seconds every 0.01° 2θ for the 211 and 20 seconds every 0.01° 2θ for the 886. The experimental procedures for examining peak shapes are obviously a little different: here we have spent 45 minutes collecting just two peaks, whereas this would normally be sufficient to collect nearly the whole pattern. In point of fact you will notice the high angle peak is not as clear and smooth as the low angle peak. This is because it is much less intense (a peak height 37 times smaller) yet has only been scanned for twice as long. In order for the profile data to be statistically equivalent we would have to have collected for far longer, resulting in a collection time of many hours for this peak alone.

Low Angle Peak (hkl = 2 1 1)

High Angle Peak (hkl = 8 8 6)

Careful Inspection of the scan reveals the followings:

1- The (111) reflections peak is not clear and the identification is in doubt.

2- There is a broad hump between 15-28 degree. Similar to the amorphous behavior

3- There is another shoulder at the right side of the (311) peak, which can not be ignored.

4- The reflections (222) smeared out.

5- The reflections (400) is quite high.

6- The reflections (422) disappeared.

All these behaviors are characteristics of a multi-phase nano-paricles.

Dear Charmain, would u mind to attach ur raw data?

As already mentioned, your material simply does not has a sufficient crystal size to form sharp peaks. The plate-like shape is probably the reason for the slightly sharper 004 (please don't use parenthesis since Laue indices don't use them in order to distinguish lattice planes (hkl) from their interferences hkl ) although I would expect a more drastic effect but I am not sure what the 3µm scale means, the entire scale or the distance between two adjacent ticks only? Are these the raw data or did you already subtract background radiation? Altogether the intensity does not seem that high. Surprising for such material, except it is only a very small amount...

The intensity description of the lines drawn below indicates the entire intensity of this intereference, i.e., not the hight. This matches quite well when the peaks have comparable width but you can see that 004 is more narrow so that the intensity below the peak is definitely smaller than the  311. Since there is certainly texture expected (because of the plate-like shape) this is a further uncertainty. On the other hand, I guess it is a cubic phase (spinel?) so that I do not expect that the plate plane is described by {311}. It seems to me more reasonable that this is probably the {001} and the crystallization conditions did not preferred a formation of isometric grains. How did you prepare the sample for  XRD? Did you mill it additionally? The strong broadening expresses a very "bad" crystallinity which I wouldn't expect for this phase if I see the SEM picture. In other words, the SEM picture does not match (to me) at the moment to the XRD results. 

Gert! Just to get it right, please clarify "(please don't use parenthesis since Laue indices don't use them in order to distinguish lattice planes (hkl) from their interferences hkl )"

What do you mean for us non-conformists :-)

"Since there is certainly texture expected (because of the plate-like shape) this is a further uncertainty."

In my opinion, this uncertainty will be drastically ameleorated and the resulting analyses enhanced significantly if the entire 2D diffraction signal is observed instead of a convoluted (smudged) conventional equatorial scan linear powder diffractogram, yes? The use of WPPM (Scardi et al) would be helpful as well in this regard.

Dear Ravi, Miller indices (hkl) and {hkl} define single or all symmetry-equivalent lattice planes. For (hkl) and {hkl} h, k, and l have to be prime to each other. You can also say that this direction in reciprocal space only indicates direction and distance to the first reciprocal lattice point along this direction. In contrast, Laue indices are used to describe the interferences of lattice planes. In International Tables for Crystallography the given reflection conditions are deirected to Laue and not to Miller indices. Laue indices can be seen as reciprocal lattice points which are preferably used in TEM to index spot patterns (SAED), e.g. 000 (as origin) and 010, 020, 030 etc along the same direction.

Yes, Ravi, if everyone would have a 2D detector, many things would be easier. But they are either too expensive or have resolution or adjustment limitations (you are the expert to judge this better than me), i.e., some other challanges. Nevertheless, the statistic should and can be improved also here by using a standard 0D or 1D detector.

Certainly, the SEM image does not allow to make any reliable conclusion but the visible plates suggest that the crystals are of good quality and don't really fit to the visible broadening. Therefore my question: Is the shown XRD pattern the result of the crystals presented in the SEM image or not?

Hello, please see attached for the xrdml file.. Thank you..

I have already removed the background from the figure I have posted earlier.. The reason for having a not highly crystalline peaks is that I purposely synthesized this material amorphously. Just that the intensity near 45 degree seems not to match the standard. The ratio of my precursor was at 1:2 (Ni:Co) so I was expecting it will match the standard. However, I was also thinking if not all of my precursors reacted, It might happened that Ni/Co acted as dopant. But Ni (111) and Co(111) shares similar position in 2 theta,, Or that, since I had a flake structure, the high intensity shown in (400) might have been the reason for this.

We need to know about crystal structures because structure, to a large extent, determines

properties. X-ray diffraction (XRD) is one of a number of experimental tools that are used to

identify the structures of crystalline solids.

The XRD patterns, the product of an XRD experiment, are somewhat like fingerprints in that

they are unique to the material that is being examined. The information in an XRD pattern is a

direct result of two things:

(1) The size and shape of the unit cells, which determine the relative positions of the

diffraction peaks;

(2) Atomic positions within the unit cell, which determine the relative intensities of the

diffraction peaks (remember the structure factor?).

Taking these things into account, we can calculate the size and shape of a unit cell from the

positions of the XRD peaks and we can determine the positions of the atoms in the unit cell from

the intensities of the diffraction peaks.

Full identification of crystal structures is a multi-step process that consists of:

(1) Calculation of the size and shape of the unit cell from the XRD peak positions;

(2) Computation of the number of atoms/unit cell from the size and shape of the cell,

chemical composition, and measured density;

(3) Determination of atom positions from the relative intensities of the XRD peaks

We will only concern ourselves with step (1), calculation of the size and shape of the unit cell

from XRD peak positions. We loosely refer to this as “indexing.”

I have already removed the background from the figure I have posted earlier. The reason for having a not highly crystalline peaks is that I purposely synthesized this material amorphously.  Dont remove the background because that diffuse scattering contain the information of short range order (amourphous).

Just that the intensity near 45 degree seems not to match the standard. As explained before, it's caused by preferred orientation, SEM image may confirmed the distorted plate-like shape that cause it. It also has anisotropic shape broadening that cause (00l) to be sharper.

The ratio of my precursor was at 1:2 (Ni:Co) so I was expecting it will match the standard. This is due to geometry not chemical. If chemically unbalance, u may expect more than one phase. Maybe u should use elemetal analysis.

However, I was also thinking if not all of my precursors reacted, It might happened that Ni/Co acted as dopant. But Ni (111) and Co(111) shares similar position in 2 theta. XRD cannot distinguish this effect because the form factor of Co and Ni is very close (one e- difference). Co-Ni share 50:50 at 16k, but even all Ni replace by Co or in the opposite term, it wont give significant effect.

Or that, since I had a flake structure, the high intensity shown in (400) might have been the reason for this. Yes it is

I suggest to anneal the sample or re-take the data by increase the scan time, look for instrument with energy discrimination detector to suppress the fluorercence.

Or the best way is to combine refinement with Neutron due to the form factor of Co : Ni has a contras of 2.49 : 10.3. Or combine X-ray X-ray by changing the wavelength close to the absorbtion edge of Co.

I had quick look of your data and it seems that you may have more complex spinel structure: NixCo1-xO4 with x=0.35 approximately like in this paper (http://pubs.rsc.org/En/content/articlelanding/2013/ta/c3ta12713c#!divAbstract), but i am not 'spinel master' so it can be overinterpretation;)

In attachment some plots from my Rietveld fitting ,

Good luck,

Thank you very much for your responses. I will repeat the XRD first at a faster scan rate.. I appreaciate all of your efforts..

Here the MAUD refinement with the aid of harmonic PO and POPA rules for the anisotropic broadening. Still not perfect due to the nature of the sample.

As discussed, the pattern has significant fluorescence effect. And because of the detector opening is ~3.4 2theta, the scan time become ~65 second per data, but the more fluorescence photons enter the detector.

Change to Co-tube if u have it, or change the 1D to 0D detector-mode, use smaller antiscatter and receiving slit. And to prevent Preferred Orientation, mix the sample with amorphous material such as wax, vaseline etc.

Dear Dr. Charmaine 

While taking the diffraction pattern, plz do the followings steps:

1. After power on (KV, mA), leave for sometime (half an hour) for stable X-ray intensity.

2. Don't Scan your sample with very high speed, set at  1deg/min or less.

3. "d" value is the  indication of new phase not the intensity. Sometimes, variation of d may be due to strain in the crystallites. Roughly, the crystallites and their condition, size are responsible for the intensity variation.

Best of luck!

The preferred orientation of crystal can influence the height of peak in certain plane. If the crystal preferred orientation is in one direction normally the height of peak will be high, but if the arrangement of crystal is random orientation, the height of peak will be low.

Thanks Gert! I'll assimilate that soon.

BTW one doesn't need a fancy expensive 2D detector. Cheap extinct dental film would do. :-)

Back in 1979-1984 we used film despite having a 0D and 1D (Braun) for basic alignment. Folks using 0D and 1D detectors have really no way of making sure that the equatorial plane is parallel to the incident beam direction. Am I correct? I just found that out recently in a HRXRD system which had precisely this challenge. The only way to detect this is by using a 2D detector or a piece of dental film to image the cross-section of the incident beam at various distances from the beam exit slit. 

Just curious, how does one make sure that the diffractometer equatorial plane is parallel to the exit beam direction presently using a 0D or 1D detector? How does one then determine the orientation of the 1D detector w.r.t. diffractometer plane? Typically the 1D detector is placed either perpendicular or parallel to the diffractometer plane. This alignment was not easy or trivial back in those days and is a critical starting point for all further XRD analyses, right?

Charmaine! Is the xrdml file diffraction intensity data?

Any chance you could post an Excel version?

Do you have additional scans after rotating the sample about the surface normal to positively identify the existence of the ubiquitous "preferred orientation"?

Hello Dr. Ravi,

Here is the .xls file..

Thank you everyone..

Dear Dr Lamiel,

If the sample has pronounced crystallographic preferred orientation, some of XRD intensities increases the other intensities decreases [Klug and Alexander (1974); Cullity, B. D. (1978)]. The preferred orientation can be reduced by preparing the samples (with 50% by weight silica gel) and mounting the specimens by side drifting. Additionally, the intensities can be corrected due to the preferred orientation effects by using either March-Dollase model or generalized spherical harmonics description, for details - see https://scripts.iucr.org/cgi-bin/paper?cg5013

Cullity, B. D. (1978). Elements of X-ray Diffraction, (Addition-Weseley Publishing Company, Inc. London, 1978).

Klug, H.P. and Alexander, L.E. (1974) . X-Ray Diffraction Procedures for Polycrystalline and Amorphous Materials. 2nd Edition, John Wiley and Sons, New York.

Simply, it reflects a preferential orientation of some facets on the expense of others

Dear Charmaine Lamiel,

As others mentioned, this is related to preferentially exposed facets, chemical composition, and/or vacancies. The paper below combines experimental and computational XRD studies to show what you are looking for. This paper is open access - enjoy.

Article Identification of preferentially exposed crystal facets by X...

Best Regards,

Alexandre A. S. Gonçalves

At that particular plane, the crystallites will be more when compared to other plane

Intensity is an arbitrary unit, varies with instrument, source detector etc.and conditions during measurement as well. Please ensure you take readings from the same instrument if you want a comparative study. Or else go for relative intensity. Regarding higher intensity of a peak as compared to other, it defines the order of crystallinity in a certain plane of orientation.