I have the following data from XRD of my nanoparticles. kindly tell me how to calculate size/dia of these particles:

No. Pos. [°2Th.] FWHM [°2Th.] d-spacing [Å]

1. 6.0974, 0.3149, 14.49535,

2. 40.9764, 0.576, 2.20077,

Question

# Please tell me how to calculate Crystal Size using XRD Data?

## All Answers (36)

- The equation to calculate the particle size is D=(0.9 Lambda)/(Beta*Cos Teta) that is known as Scherrers equation. where, lambda --- wavelength of x-rays

beta-FWHM of diffraction peak and theta--- angle corresponding to the peak.

hope this can help - use the Scherrer equation: (d = (K x lambda)/(fwhm x cos theta)

lambda is the X-ray wavelength in angstroms

K is a constant, usually set at 0.9.

FWHM is the full width of the Bragg peak at half maximum, in radians

theta is the Bragg peak, in radians. Just remember this isn't 2 x theta, but *theta*

http://prola.aps.org/abstract/PR/v56/i10/p978_1 - Thank you very much..please see the attached pic and kindly tell me is it Ok.?
- hello you must using theusing the Scherrer’s formula:
- You can't just use the Scherrer equation for the crystallite size calculation with these data. You have to substact the instrumental line broadening using a diffractogram from a well crystalline sample of Si or BaF2 for example.

Another point is that Scherrer not always gives you a real crystallite size value, this because the FWHM is affected not only by the particle size as by the structural strain. It must be pointed out also that the shape constant (k = 0.89 ~0.9) is used for spherical particles, so that you need to be sure that your crystallits are spherical. - @Ehsan Sadeghi Yes, I am using Scherrer’s formula:

@Mathias Strauss, Its confirmed that crystals are spherical - Please use Sh error's formula to do calculations.
- The Scherrer Equation is a quick method to obtain a particle size but as Mathias noted Reitveld refinement will give you the most accurate assessment since all of the experimental variables including the instrument effects are considered.

Even with refinement it is always worthwhile to characterize your sample with SEM and TEM to get a broader sense of the structure. For example, polycrystalline structures will have crystalline sizes that are smaller by sometimes several factors than the particles themselves. - Well to calculate the particle size using scherrer formula is a little tricky, since the peak broadening can be due to crystallite size (which gives a lorentzian contribution) or strain (Gaussian contribution). It will only be straight forward if the contribution to the broadening is due to one of the two factors, otherwise you should use Thomson-Cox-Hastings pseudo-Voigt function which is the sum of Lorentzian and Gaussian to perform your calculations.
- Crystallite size can be calculated using the Scherror's formula that one can get very easily from ant text book.
- All - please don't keep giving the same answers over and over. Mathias Strauss has already answered the question - i.e., these data cannot be used to determine size using the Scherrer formula (or by Rietveld refinement for that matter). This is not only because the peak broadening is a convolution of size and microstrain, but because the present data does not consider the instrumental broadening. So any size estimate is almost certainly wrong.

Of course some of the answers here do provide very useful advice for consideration once the instrumental broadening is known. - Hasna:

1) lambda is the wavelength of the radiation used to collect the data. For example 1.54Å for Cu K-alpha or 1.79Å for Co K-alpha. So it is independent of the sample chemistry.

2a) the answer might be: "both" - the so-called Scherrer constant is a function of the crystallite geometry so 0.94 and 0.89 are both potentially "correct" values of this constant. You can look this up in a variety of references.

2b) the useful answer might also be: "it probably doesn't matter" - because except in unusual circumstances the error in the derived crystallite size is likely much larger than the ~4% difference between the two constants. - For the X-ray diffraction analysis you need first correct for the instrumental line broadening and background line (background is represent amorphous content). Then you have diffraction intensity for crystal also you need to fit your data into a symmetrical Gaussian distribution as example. The crystallite dimensions is two one is called stack diameter (La) and other is stack height, (Lc) can be calculated from the full width at half maximum (FWHM) and Bragg’s peaks using Bragg’s and Deby-Sherrier equations. But you need to know which peak represent the stack high and stack diameter. You can use Trace software program from Diffraction Technology PTG LTD, Australia. This program refines the intensity of each peak (as a separate variable), smoothes the peak shape, as well as subtracting the background line and eliminating the K2-peak from the diffraction intensity
- Mr. Asim

You cannot calculate crystal size using this data. After you do correction of the above mentioned. To calculate stack high you need to know reflection peaks at (002), (004) (006) if it possible (001) and (003) and the Stack diameter reflection peaks at (111), (110) (100) (121) (220) and so on. Then directly use Sherrie equation: for stack high K factor =0.9 and for stack diameter K factor=1.84. - use the Scherror's formula
- The best option is SAXS data analysis for particle size from Guinier region of the Intensity pattern and XRD CAN NEVER A SOLUTION FOR NANO PARTICLE SIZE ESTIMATION.
- Use schemer equation and also inspect it by taking fesem. It is essential enough to validate the grain crystal size.
- yes Harrindra it will be ok in one way but still better if you can go for TEM & SAXS.
- Harindra,

It is ok but better to go for SAXS study. - According Professor Davos Balzar (towards the end of the XX century) there are 3 or more methods to determine size-strain analyzing the peak broadening of a material. There 3 more known: Scherrer equation, simplified integral breadth methods (Williamson-Hall, Langford,..) and Fourier methods. Others: double Voigt methods, Kojdecki algoritm, Leoni-Scardi algoritm,...). Overlapping reflections is a problem. Should get reflections, if possible, in the pattern diffraction with high intensity (>10000-20000 cps) obtained with speed of small step scan.
- What are the advantages of SAXS over WAXS for "particle size" (diffracting domain size) determination? Is the "strain" effect non-existent in SAXS?
- If you have X-ray diffraction data (* extension in raw), Just downloade

WinFit X-ray diffraction program and then use it to calculate crystal size - Ravim

Yes there is strain effect in SAXS too but needs quite attention towards the SAXS data. - Qualitative information about average crystallite (attention with the answer of Mathias Strauss) can be obtained using Scherrer's equation:

D = 0.89 lambda / beta cos theta

where D = the crystallite size (usually in nm), lambda = the wavelength of X-ray diffraction (in nm), beta = the full width at half-maximum, and theta = the Bragg diffraction angle. - José! "beta = the full width at half-maximum".

For a Gaussian Bragg profile, the integral breadth, β, is related to the FWHM peak width, H, by β = 0.5 H (π / log e 2)^1/2. Please note that in modern diffractometers with advanced incident beam conditioners, the Bragg profile is seldom Gaussian. Besides, the sample Nano structure convolutes to change the profile shape. In addition, one needs to know the "back-ground" signal precisely prior and post the Bragg profile "shoulders".

The Sherrer formula is a simple relationship that needs tweaking based on other experimental parameters. The availability of the raw diffractogram to review would eliminate a huge handicap for further analyses and suggestions. - Thanks Ravi:

Yes, use integral breadth (= >beta = the full width at half-maximum => FWHM) is sometimes better in the Scherrer formula.

Regards - Dillip! Please post some references for SAXS strain analysis.
- Ravi,

OK I am doing very soon. - The FWHM and Bragg Peak position are sufficient for use of the Scherrer formula. But the values you get need to be understood in terms of some "known standard" for identical experimental conditions. That is the most precise method for using XRD successfully!

From our experimental experiences and observations the FWHM of powders or single crystals is related to the average dislocation density or "average cell structure/substructure size" (coherently diffracting domain size in reference to a particular crystallographic plane) in the crystal/crystals. From literature analyses and our own micrographic observations with substrates/epi's of materials like GaAs, GaSb, LaSrCuO4, LaSrAlO4, AlGaAs/GaAs, InAs/InAsSb, InGaAs, LiF, LaB6, Quartz, Si, AL2024, AL7075, ZnSe https://www.youtube.com/watch?v=dFCQS8oUyT0&list=PL7032E2DAF1F3941F it appears that defects such as twins or stacking faults affect the region below the FWHM. We see some interesting features, for example, when we map the FW12.5%M (@12.5% Maxima, bottom of the shoulder) instead of the FWHM (@50%) for each pixel. https://www.flickr.com/photos/85210325@N04/10647827636/

These observations do undoubtedly need supplementary substantiation with AFM, SEM & TEM, etc..

We're highly interested in correlating Nano structure with XRD reciprocal space parameters and eventually the materials' quality, performance & yield for manufacturing. Please post any suggestions, references or comments that may help us in this quest.

BraggXRDMicroscopy@gmail.com

We can interface with all existing XRD equipment and give precise/quantitative "sight to the blind" in your XRD Lab! Here's what we are up to at 40um spatial resolution, 25mm diameter X-ray input window size, 8-16Bit (HDRI-High Dynamic Range Imaging is possible) with up to 10^5 gain: - Hi Ravi Ananth:

A relatively recent publication explains the use of the Scherrer equation that I have tried to indicate which you have properly corrected.

Regards. - Thanks José! Appreciate the PDF version as I have very little access to literature other than Google.

In this case the particles appear to be homogeneously near spherical. It is also important to note that not only is the "average" particle size info present in the XRD signal but so is the size distribution. However, with the conventional diffractometery this information is smudged due to either the sample rotation or the large sample area integration (focusing geometry) inherent to the conventional "spatially blind" method. It is really a miracle that through mathematical machination we are able to extract meaningful information from this super convoluted smudge of a XRD signal. The Scherrer formula is an effective empirical approach validated over decades but does require appropriate assumptions in Nano structure. - Hi Ravi, more information:

Depending on the type of chemical synthesis crystallite morphology can change (spherical, prismatic, ...). These crystallographic programs based on mathematical algorithms Kojdeck and Leonii it may show and visualized by electron microscopy techniques. An example, the attached article.

Regards. - Using XRD data, You can calculate particle size with the help of Scherrer formula and using the value of FWHM, peak position and wavelength of x-ray used.
- How to find the percentage of crystallinity of semi-crystalline materials ?

## Popular Answers

Mathias Strauss· University of CampinasAnother point is that Scherrer not always gives you a real crystallite size value, this because the FWHM is affected not only by the particle size as by the structural strain. It must be pointed out also that the shape constant (k = 0.89 ~0.9) is used for spherical particles, so that you need to be sure that your crystallits are spherical.

Edward Payzant· Oak Ridge National Laboratory1) lambda is the wavelength of the radiation used to collect the data. For example 1.54Å for Cu K-alpha or 1.79Å for Co K-alpha. So it is independent of the sample chemistry.

2a) the answer might be: "both" - the so-called Scherrer constant is a function of the crystallite geometry so 0.94 and 0.89 are both potentially "correct" values of this constant. You can look this up in a variety of references.

2b) the useful answer might also be: "it probably doesn't matter" - because except in unusual circumstances the error in the derived crystallite size is likely much larger than the ~4% difference between the two constants.