
#1
Jun206, 05:23 AM

P: 3

Hi friends,
What is the best way to analyze the XRD data for phase identification and particle size calculation? so far I am using debyescherrer's formula. Is there any advanced formula? thankyou 



#2
Jun206, 10:07 AM

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Can you please provide more complete details?
What system are you studying?What phases do you expect? How many phases (down to what trace levels) do you wish to identify and to what accuracy? What instrumentation (just a DS camera, or maybe a diffractometer) are you using/do you have at your disposal? What software is avaliable at your XRD lab? Do you have the PDF database on a computer? Do you also have any search/match software? Do you have a Rietveld Analysis (the only way I know to determine particle size) program ? 



#3
Jun806, 04:18 AM

P: 3

Thankyou gokul for your willingness....!!
I use PANalytical XRay Diffractometer. I am comfortable with phase identification. But I wanted to know a way to calculate particle size. Yes we have PDF database and Rietveld Analysis program. So far I used to determine particle size using debyescherrer formula d = 0.9*lambda/delta*cos(theta) , where lambda  wavelength of xrays deltaFWHM of diffraction peak theta angle corresponding to the peak 



#4
Jun806, 09:03 AM

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how to analyse XRD data?
The one drawback of the above simple method is that it works only if stressrelated and instrumentrelated broadening are negligible in comparison to particle size effects. This condition is often met with particle sizes that are in the 10  100 nm range. At um particle sizes, you will have to be careful to separate broadening effects from particle size (~ [itex]1/cos \theta[/itex]), strains fields (~ [itex]tan \theta[/itex]) and all instrumental effects (~ some constant additive number). By plotting FWHM vs angle, and fitting the peak width vs Bragg angle curve to parameters A,B and C in :
[tex]FWHM^2 = \left( \frac{A}{cos \theta} \right) ^2 + (Btan \theta)^2 + C^2 [/tex] From the best fit, you can extract A and calculate the particle size from the DS formula above. PANalytical itself makes pretty good particle size analysis software (it's part of the software package they try to sell with their diffractometers) and it may just be that your XRD lab already has this software. 



#5
Oct1806, 03:01 PM

P: 4

Hi,
I am using the Scherrer's formula for calculating the grain size of powders, I was not getting proper results. Is there any way that i can know what is going wrong in my calculations. Thank you priya 



#6
Oct1806, 06:51 PM

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We might be able to help if you can give us more details:
1. What is the form of your data (x, y, ranges)? 2. How exactly did you do the calculation? 3. Did you ignore stressrelated broadening? 4. Is the powder analyzed directly after milling, or is it annealed before it is measured? 



#7
Oct1906, 01:02 PM

P: 4

I am using Iron and nickel powder. The X axis (2 Theta) range for both of them is (40100 degrees). The Y axis is the counts(which varies). I actually did the Xrd of my as received powder which is stress free. I did a (2 Theta) Scan of the powder. I can send you one of my sample calculation so that you have a look at it. The problem is my as recieved powder should give me a higher value but i am getting the grain size about 15 nm which i dont think is possible. I have checked the Xrd of both as received as well as 10 hrs ball milling. I have not annealed my powder.
Thnak you krishna priya 



#8
Oct1906, 04:37 PM

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Yes, if you include the calculation as an attachment in your next post (click on "Go Advanced" below the Reply box, and scroll down to "Manage Attachments") I can take a look at it. Also, if you can include the raw data in 2column ascii (a .txt file will work) format and attach that too, it will be useful.




#9
Oct1906, 07:25 PM

P: 4

Hai Gokul,
I am sending you a word file as i cannot attach an Excel file. So as soon as you receive my attachment you have to change the extension to"Grain size.xls".I calculated the grain size of both as received and 10 hrs of ball milled powder. I hope you can understand my calculation. If not please let me know. I have also enclosed charts in them, you can have a look at them too. Thank you krishna priya 



#10
Oct2106, 10:21 PM

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Just glanced at it now. My initial estimate agrees with yours ~ 30nm for the "as received" powder. I'll need to do a more careful calculation when I find a little more time  there may be a large instrumentrelated broadening that is making the particle size look small.
A few questions: 1. What is the expected range of the particle size? Have you tried resolving particles under an optical microscope? 2. This isn't stated explicitly, but it appears you are using CuK_alpha1. Is that right? 3. What is the model of diffractometer this was measured with? 



#11
Oct2206, 10:33 AM

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After spending a little more time with the data, I'm beginnig to think the scan was too fast to be able to do a reasonable particlesize calculation. The error bars on the data may be too large to do a meaningful analysis of the various contributions to the peak broadening (see post#4, above).
In any case, here's how I would do it from the given data: 1. For each of the 5 peaks, do a Gaussian fit to the data (limit it to about 3sigma from the mean) and determine the width([itex]\beta[/itex]) and the peak value ([itex]\theta[/itex]) from the Gaussian (not from the raw data), as well as the errors in these values [itex]\delta\beta~,~~ \delta \theta[/itex]. For the baseline subtraction in halfmaximum determination, I would use two sets of baseline data about 3 degrees on either side of each peak and use a mean of both sets (this is to minimize any angular dependence in the background). 2. Calculate the 5 values of [itex]\beta ^2 cos ^2\theta [/itex] and plot these points (along with their error bars) against [itex]sin^2 \theta [/itex], for each of the two data sets (as received, and 10hrs milled). 3. If you are lucky, and the error bars are small enough, you will be able to fit a straight line to both sets and extract a slope and intercept for each one. 4. What these 4 numbers represent: (from post #4) [tex]\beta^2 = \left( \frac{A}{cos \theta} \right) ^2 + (Btan \theta)^2 + C^2 [/tex] [tex]\implies \beta^2 cos^2 \theta = A^2 + (Bsin \theta)^2 + (Ccos \theta )^2 [/tex] [tex] = A^2 + (Bsin \theta)^2 + C^2(1sin^2 \theta ) [/tex] [tex] = A^2 + C^2 + sin^2 \theta(B^2  C^2) [/tex] From the linear fit, you have [itex]slope=B^2  C^2~, ~~intercept=A^2 + C^2 [/itex]. (i) For the as received powder, B (the stressrelated broadening coefficient) should be small and hence it might be reasonable to neglect it. Note that, this would mean the slope for this line should be negative. If you do not have a negative slope, you can not neglect the stress in the as received sample; this makes the calculation more tricky. (ii) The broadening coefficient from instrumental factors should be sample independent, so C is the same number for both samples. If the intercepts for both samples are close to each other, the broadening is probably dominated by intrumentrelated factors. That leaves you with 4 unknowns : A(as rec'd), A'(milled), B'(milled) and C(instr). From the two data sets you have 4 equations for these 4 unknowns (2 slopes and 2 intercepts). 5. Finally, using [itex]A=0.9 \lambda / d [/itex], you can determine the approximate particle size, d. 



#12
Oct2306, 01:01 PM

P: 4

Hai,
Thank you very much for being so patient and replying me.The diffractometer which i am using is Philips 1830, and i am using CuK_alpha1. I have done a detailed analysis of about 45 min for my powder. Anyways i will try your suggested method. Can you do me a small favour can u post me a sample calculation so that it might be helpful forme to proceed. I have read in some books about the method which you posted (Suryanarayana). I also tried a Gaussian fit to my curves but i still end up with the same value. So, i will be very thankful if you will send me any sample claculation. Thank you krishna priya 



#13
Feb2007, 10:28 PM

P: 2

To calculate particle size, I want to use the Scherrer equation. My problem is that i don't know what are the value of constant k that should be used. In the note that I read the value of k varies between 0.89 to 1.39 depending on crystallite size and shape.. For a small cubic crystal of uniform size normally used k=0.94. Now i'm calculating alumina so what value of k should be used??




#14
Mar607, 10:08 AM

P: 1

Dear frinds,
I have a problem in structure analysis. The PXRD (Philips Diffractometer) data are with me. Can any one plz help me in the analysis? I can provide you *.uxd files, with other details. I have used Mn2O3 and VO2 to form the compound. Regards, Girish 



#15
May1407, 11:04 AM

P: 1

Hello to everybody,
I performed some XRD patterns of two sample of titanium dioxide, one with micron particles and the other one with nano particles. While with the nanosized one I didn't have any problems to apply the Scherrer equation to evaluate the particle size, with the microsized I couldn't. I know that Scherrer equation is valid in a range 10nm  100nm and my microsized sample should have bigger particles, between 100nm and 200nm. Could anyone help me with this matter? Thanks a lot in advance Shaka 



#16
Aug2007, 03:02 PM

P: 1

Hi guys and gals, I have a question. As the axis is in 2θ degrees, do we have to half the FWHM values as we do for the cosθ.




#17
Nov2107, 11:22 PM

P: 1

Hai,
I want to calculate the particle size, by Scherrer equation. My problem is that i don't know what are the value of constant k that should be used. In the note that I read the value of k varies between 0.89 to 1.39 depending on crystallite size and shape.. For a small cubic crystal of uniform size normally used k=0.94. Now i'm calculating nano Zro2 so what value of k should be used?? Thankyou anbu 



#18
Jan2508, 02:28 AM

P: 2

Hi friends,
I'm studying dusty plasmas. I made in situ FTIR measurements...is it possible to estimate particle radius and density if absorption and diffusion (diffusion means scattering) are present on infrared spectra? How? 


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