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How to analyse XRD data?

  1. Jun 2, 2006 #1
    Hi friends,

    What is the best way to analyze the XRD data for phase identification and particle size calculation? so far I am using debye-scherrer's formula. Is there any advanced formula?

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  3. Jun 2, 2006 #2


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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 ?
  4. Jun 8, 2006 #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 debye-scherrer formula

    d = 0.9*lambda/delta*cos(theta) ,

    lambda --- wavelength of x-rays
    delta-FWHM of diffraction peak
    theta--- angle corresponding to the peak
  5. Jun 8, 2006 #4


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    The one drawback of the above simple method is that it works only if stress-related and instrument-related 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.
  6. Oct 18, 2006 #5

    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

  7. Oct 18, 2006 #6


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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 stress-related broadening?
    4. Is the powder analyzed directly after milling, or is it annealed before it is measured?
  8. Oct 19, 2006 #7
    I am using Iron and nickel powder. The X axis (2- Theta) range for both of them is (40-100 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
  9. Oct 19, 2006 #8


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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 2-column ascii (a .txt file will work) format and attach that too, it will be useful.
  10. Oct 19, 2006 #9
    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

    Attached Files:

  11. Oct 21, 2006 #10


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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 instrument-related 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 Cu-K_alpha1. Is that right?
    3. What is the model of diffractometer this was measured with?
  12. Oct 22, 2006 #11


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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 particle-size 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 3-sigma 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 half-maximum 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 10-hrs 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(1-sin^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 stress-related 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 intrument-related 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.
    Last edited: Oct 22, 2006
  13. Oct 23, 2006 #12

    Thank you very much for being so patient and replying me.The diffractometer which i am using is Philips 1830, and i am using Cu-K_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
  14. Feb 20, 2007 #13
    particle size calculation

    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??
  15. Mar 6, 2007 #14
    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.
  16. May 14, 2007 #15
    XRD of microsized TiO2

    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

  17. Aug 20, 2007 #16
    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.
  18. Nov 21, 2007 #17
    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??


  19. Jan 25, 2008 #18
    How to estimate particle size....

    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?
  20. Mar 11, 2008 #19
    How do I use the Debye Scherrer Equation

    Hi! I plan to determine the size of my deposits using XRD and the debye scherrer equation. Could anyone tell me how this is done? Thanks!

    - Buknoy
  21. May 23, 2008 #20
    @ kpullab
    Thank You very much for the .xls file really appreciate it.
  22. Nov 25, 2008 #21
    also I need to know the value of K in Scherrer formula when the system is tetragonal chalcopyrite semiconductor.plaese send the answer on my e-mail
  23. Dec 9, 2008 #22
    I have XRD measurements data of (004),(111) and (115). How to calculate the FWHM value and dislocation density of my samples by Williamson-Hall plot. Please guide me how to calculate.
  24. Apr 13, 2009 #23
    I am trying to use the information from your post #11 to use with my data. In following the derivation in your post, you quote the final formula as A = 0.9 Lamda/d.

    I'm wondering if you are implyin that A = Beta/cos(theta) as that would match your derivation, or maybe you are showing a simplification here.

    I have particles on a surface that are between 200 - 700 nm, but do not know how they may recombine under the surface. I originally was using the
    A = 0.9 Lamda*cos(theta) /d
    formula that I have seen others use, but am now concerned since my particles are clearly greater than 1000 Angstroms.... can you please advise?
  25. Apr 27, 2009 #24
    Re: how to analyse particle size from XRD data by scherrer formula with data

    how to determine particle size by scherrer formula from xrd data?
  26. May 6, 2009 #25
    Dear gokul,
    we have got a program named xpert plus, which gives us fwhm value for the selected peak. is the program feasible. can we get a program which could directly find the particle size.
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