The FWHM is that of the peak profile, without dividing by 2 as Marcella pointed out. Petr Brazda also mentioned some important aspects concerning the instrumental broadening. Another point to mention is that the factor (0.9) is evaluated under the assumption of a Gaussian line shape. Practically, the line shape is neither a pure Gaussian nor a pure Cauchy, and more sophisticated analysis is required. In practice, some people use a Voigt line shape, with the understanding that the Gaussian line is due to strain and the Lorentzian is due to particle size effect. The first step would then be to make a measurement on a standard sample to determine the instrumental effects. The patterns of the standard is fitted with a Voigt line shape, and the corrected integral breadths (Peak area/maximum intensity) of the Gaussin and Lorentian lines corresponding to the sample are evaluated following standard rules: The Lorentian widths due to particle size and instrumental broadening add up linearly, while for the Gaussian lines, the square of the observed breadth is given by the sum of the squares of the insrumental Gaussian and the strain Gaussian  (Langford work for example). This type of analysis is sometime prohibited by the fact that the signal quality should be very high, the background function is accurately determined, and the line shape fits very well at the tail. These effects could introduce high uncertainty in the evaluated particle size. The Stokes correction method relying on Fourier analysis is a more reliable method for the deconvolution of the particle size effect and the instrumental broadening. This method, however, requires much more work than other conventional methods; programming, however, could reduce the complexity of the analysis significantly.

The important point I wanted to make is that the Scherrer constant in the Scherrer's equation is evaluated on the assumption of a Gaussian line shape, however, the particle size effect is usually modeled with a Loretzian line shape. The use of the wrong line shape could introduce ~ 30-50% error in the evaluated particle size.     

The FWHM that you find with the fitting program is in degrees. So you need only to transform it in radians without any other division. In the case of your example the correct value is the first you reported.

Yes you have to divide only 2Theta by 2. I think your crystallite size is too big, it can not claim nano particle, right?

Thank you very much Marcella Bini for your answer. I know and am 100% sure the first option is correct but after I accidentally found this formula online that tells me to divide FWHM, I went a little in blurry. That's why I said to ask the question here, because someone may have theoretical deduction of Scherrer's equation.

Suphakan Kijamnajsuk, you have right, the angle must be divided and inserted in Scherrer eq. in theta not in 2theta, but my question is regarding FWHM because of the equation found on the internet.

Hello, I agree with the answers above. However, I think that there are other important things, which should be mentioned concerning your particle size values. The most important thing is the separation of the instrumental broadening (broadening of the diffraction maxima due to the experimental conditions). With your FWHM value it is very important because at about 40° 2theta it can be about 0.05°, which would shift your apparent crystallite size to a completely different values. I mention this because it seems you are really beginning with this. Another thing is the scherrer constant. It it there because of two things - 1) recalculating your FWHM value to integral breadth (so the Scherrer equation in fact calculates with the integral breadths but FWHM is easier to extract from the data) and 2) recalculating length of the coherently diffracting domain to its (apparent) diameter. This second thing implies that you have the knowledge of the particle shape and you can find some details in for example the article by Langford and Wilson in J. Appl. Cryst. (1978) 11, 102. The last thing I would like to mention is the precision of your size value. The fitting program usually gives you a taste of how precise is your 0.0819. You should then combine the standard deviation from the observed FWHM and that of the determination from the instrumental broadening and deduce the deviation of your size value based on these things.

The FWHM is that of the peak profile, without dividing by 2 as Marcella pointed out. Petr Brazda also mentioned some important aspects concerning the instrumental broadening. Another point to mention is that the factor (0.9) is evaluated under the assumption of a Gaussian line shape. Practically, the line shape is neither a pure Gaussian nor a pure Cauchy, and more sophisticated analysis is required. In practice, some people use a Voigt line shape, with the understanding that the Gaussian line is due to strain and the Lorentzian is due to particle size effect. The first step would then be to make a measurement on a standard sample to determine the instrumental effects. The patterns of the standard is fitted with a Voigt line shape, and the corrected integral breadths (Peak area/maximum intensity) of the Gaussin and Lorentian lines corresponding to the sample are evaluated following standard rules: The Lorentian widths due to particle size and instrumental broadening add up linearly, while for the Gaussian lines, the square of the observed breadth is given by the sum of the squares of the insrumental Gaussian and the strain Gaussian  (Langford work for example). This type of analysis is sometime prohibited by the fact that the signal quality should be very high, the background function is accurately determined, and the line shape fits very well at the tail. These effects could introduce high uncertainty in the evaluated particle size. The Stokes correction method relying on Fourier analysis is a more reliable method for the deconvolution of the particle size effect and the instrumental broadening. This method, however, requires much more work than other conventional methods; programming, however, could reduce the complexity of the analysis significantly.

The important point I wanted to make is that the Scherrer constant in the Scherrer's equation is evaluated on the assumption of a Gaussian line shape, however, the particle size effect is usually modeled with a Loretzian line shape. The use of the wrong line shape could introduce ~ 30-50% error in the evaluated particle size.     

I agree with Sami Mahmood.

I agree with above comments. In addition to them one can emphasize that Debye-Scherrer calculation is an "estimation" of the particles' size, (actually grain sizes'). It may not result a definite value. TEM results are usually used to prove size calculations.

Hi everyone,

The simplest way to calculate the correlation length from the FWHM of a peak is:

1) use wave vector q instead of 2theta. That means q = 4*pi*sin(theta)/lambda.

2) plot intensity vs q

3) fir the peak with a gaussina, lorentzian or voigt function

4) get the FWHM of the peak from the fitting

5) correlation length = 2*pi/FWHM

and forget about the Scherrer equation...

A practical approach Antoni ;)

Thanks.

Hi again,

2theta axis is a "tradition" used in XRD coming from crystallography.

q axis is much easier to handle since it gives you a direct information about the d spacing; d = 2*pi/q. This comes from the scattering "tradition" from SAXS.

At the end both techniques are the same. My more than 16 years of experience tells me that q wave vector is the best for dealing with X-ray (SAXS, MAXS, WAXS, XRD...) data.

Welcome!

Thank you very much for your answers, Prof. Sami Mahmood I totally agree with you. After these discussions I will change the lorentzian function to fit the peak. In addition, I want to use SEM measurements to verify the results obtained by Scherrer formula.

Thank you Antoni Sanchez-Ferrer, I did not know this method but i will try it soon.

Dear Daniel,

Please, have a look to my papers where you will see the use of I proposed you for the correlation length evaluation of domains...

If you have any further question, please conctact me.

Glad to assist you.

SALUT!

Tony

Be careful Daniel when comparing sizes evaluated by different techniques. If your single particle is polycrystalline then SEM reveals the physical particle size while XRD reveals the structural coherence length (crystallite length perpendicular to the reflecting plane), which is in this case smaller than the size obtained by SEM measurements. Also, particle agglomeration may introduce difficulty in measuring particle size by SEM.   

Dear Tony,

Thank you for all the information shared with me. I played around with the last diffractogram acquired and I tried your model. It is extremely simple and fast. I'll start reading your articles and if I have questions I will contact you.

Salut,

Dani

Cu placere. La revedere!

SALUT! NOROC!

Tony

Hello. I think that we are approaching to the real point, thus the answer to the question "What do we get when we we will use the profile analysis of the diffraction maxima?". This is important question and the answer is not easy. We may begin with the simplest case - a sample consisting of monodisperse spheres. In this case, the length of the crystallite along the scattering vector (vector normal to the diffracting "planes") is the same for all the hkl diffractions (and all the particles). However, imagine a circle (our crystallite's crosssection) and a vector (the scattering vector). The thickness of the circle is not the same all across the circle (it goes for diameter in the centre to zero at the edges). Therefore, if you integrate the lengths over the whole volume of the sphere you get that the average length is only 3/4 of the diameter of the sphere. Therefore, one part of the Scherrer constant for spherical particles is 4/3 (you convert the lenght to diameter). The other part for the value 0.9, which is used is ((pi/ln(2))^0.5)/2 (here is the point in Prof. Mahmood's comment - Scherrer worked on the neutron diffraction, where the profiles are mainly Gaussin, for Lorentzian is the constant for recalculation of the FWHM to integral breadth pi/2 - can you see the difference?) and another part (due to the definition; lee Langford and Wilson's article) is (pi/6)^1/3. Antoni Sanchez-Ferrer mentioned another technique for extracting the length - the so called Williamson-Hall plot. The drawback of this procedure (and also the procedure of taking only an arbitrary maximum) is that when your crystallites are not spherical, your plot will not be monotoneous. A nice example would be spherical Au particles with 111 twins. These defects will (depending on the concentration) decrease your crystal length and in the W-H plot you will get for 111, 222,.. maxima shorter lengths, because you crystals are thin discs, eventhough they are stacked to form a sphere. This was the simple case. In reality, you have distribution of diameters of the particles/crystals of more or less non-uniform morphology. So the real samples are very complicated and you have to think about your diffraction pattern as a sum of the signal from all your particles/crystals. Here you have to realize that the diffraction is volume technique - so the most prominent signal will come from the biggest particles/crystals in your sample. This is a difference from TEM/SEM where you get a histogram (you count the particles (not crystals usually!) according to their diameter). Well, so back to the question - "What do we get when we we will use the profile analysis of the diffraction maxima?" - we can get many things. When we will evaluate only one diffraction maximum and the crystals in the sample will not be spherical and there will be microstrain (and we will evaluate the maximum at higher 2theta angles), we will have a polydisperse particles and we will use wrong Scherrer constant we can get completely wrong results. Though the problem is complex, do not worry. Look for example in Robert L. Snyder, Jaroslav Fiala, and Hans J. Bunge Defect and Microstructure Analysis by Diffraction. The bast thing you can do is thinking about what your values are derived from and what is the error your values have.

You have to return in time to the first explanations:

The Scherrer Formula for X-Ray Particle Size Determination

Phys. Rev. 56, 978 – Published 15 November 1939

A. L. Patterson

The XRD size analysis can be divided in two basic steps:

1.       Extraction of the peak profile due to size broadening.

Here instrument and eventually strain and stacking faults broadening have to be “subtracted”. Since the measured diffraction peak is a result of a convolution of different functions, the most accurate approach would be a convolution based approach to profile fitting (see TOPAS program by Alan Coelho). Basically, the problem is to get rid of additional contributions to XRD peak profile other than size effect. This can be done reasonably well with existing software, and this procedure is reproducible in different instruments.

2.       Modeling of the peak profile.

This step already involves the introduction of a morphological model (size distribution and shape of crystallites) of the system of crystallites. Unfortunately, the information hidden inside the peak profile is, in general, not enough to find an unambiguous (unique) morphological solution. There are purely mathematical reasons behind.  The physically meaningful information that we really measure (can extract), volume weighted or area weighted thicknesses, (I will come latter on to these sizes) is the result of the "integration" or averaging of what we want to know, crystallite morphology description. In order to obtain the latter, one needs to solve, basically, an integral equation which belongs to so-called ill-posed problem. One of the characteristics of this type of problem is a non-uniqueness of the solution. In other words, different morphological solutions (size distributions and shapes) can equally well describe the experimentally measured profiles. That is why, it is always a good idea to check the morphological model presented in XRD analysis by microscopy.

Obviously, if the real situation is close enough to some simple model (like reasonably well defined shape and reasonably narrow size dispersion) one can get pretty nice XRD size analysis. The example already discussed is monodispersed spherical crystallites. Volume weighted or area weighed measurable sizes are just connected to the diameter of the crystallite by a coefficients 4/3 or 3/2, respectively.

If the real morphology of crystallites is more complex, the morphological model has to follow this complexity and, for example, one needs to provide (assume) a more complex model for the shape (cubic, cylindrical, tetragonal, oval, parallelepiped and so on) or/and one needs to provide a model for the size distribution (normal, lognormal, gauss etc.) .

Complex shape, other than spherical, immediately means anisotropy in size broadening. If one measures different peaks (different directions of scattering vector), the size broadening will be different (the broadening as a function of scattering vector modulus). This situation can be smeared of by so-called crystallographic multiplicity effect.

 This crystallographic factor arises from the fact that the crystal lattice of the structure may have, in general, an arbitrary orientation with respect to the shape of a crystallite. The contribution to the XRD powder peak profile of the particularly oriented atomic planes of the same family is, in general, will be different, even for one single crystallite. More simply, the crystallographic planes belonging to the same family are oriented differently in a crystallite and the average thicknesses of the crystallite along the normal to these planes are, in general, different.

In many cases, however, this crystallographic factor is averaged off, because of the eventual arbitrariness of the orientation of the crystal lattice with respect to the crystallite shape in a system of many crystallites. As a result, not really spherical crystallites observed by microscopy produce isotropic broadening of XRD peaks. However, if this is not the case, and crystal lattice of crystallites has a defined orientation with respect to the shape, XRD becomes a tool not only to see the size anisotropy, but also to judge about a relationship between crystal lattice and the crystal shape.

To be continued …(more about sizes, shapes and size distribution modeling and, also, practical approach to XRD size analysis).

Dear Daniel,

you have to use the Omega scale, not the two theta scale. Then the second answer is correct. The value that you obtain is approximately the average size along the scattering vector, usually the normal to the sample surface. Minor corrections arise from the assumption of the peak shape and on the assumed size distribution of particles.

Not necessarily you need to have crystallites to observe a peak broadening. The formula works also for thin films. Is a consequence of the fact that you obsreve the square of the fourier transform of the crystal density.

Let us consider the seemingly simple question. What measurand does one choose for the broadening of the XRD powder peak? Full width at half the maximum of the peak is just one of, in principle, many possible choices.  To my opinion, the answer should be based on the mathematical model that relates the dimensional parameter of a crystallite with the experimentally observed characteristic of the diffraction peak. It is quite important for the model to provide a physically meaningful definition of the size characteristic. As was already mentioned in this forum, Langford and Wilson in their paper (1978) considered in details this question. Two sizes with clear physical meanings and mathematical definitions can be obtained from XRD powder peak: one is called in the literature volume weighted thickness (Lv), the other is area weighted thickness (La) of the crystallite.

Definition of Lv:

Stokes and Wilson (1942) deduced the equation, very similar to the Scherrer´s one, that relates the Lv with the peak broadening measured as the integrated peak intensity (peak area) divided by the peak intensity at the maximum. This broadening, called integral breadth has very simple geometrical meaning: it is a base of an isosceles triangle having the same area as the measured peak and the height equal to the maximum of the peak intensity. The meaning of Lv is the volume average thickness of the crystallite measured perpendicular to the hkl crystallographic plane (or along the direction of the scattering vector). The averaging to obtain the Lv is done in the following manner: one divides the crystallite in infinitely small volumes, takes the length of the line from one border to the other border of the crystallite along the scattering vector passing through a particular volume element and multiplies this length by the elementary volume. Then, one makes a sum of all such a products over all elementary volumes. Mathematically, this is the integral over the volume of the crystallite of the length of the infinitely small strips of crystallite taken along the scattering vector.  For practical calculations, the equivalent definition of Lv is more useful: instead of a volume integration of the length one does the line integration of the volume. The latter volume is the volume common to the crystallite and its imaginary copy (ghost) displaced some distance apart along the scattering vector. The integral of this volume is taken along the line of the scattering vector from minus to plus infinity. Obviously, the integration outside the limits of the displacements where there is no overlap between crystallite and its ghost is unnecessary.

 About the units of broadening:

The common between different peak broadening definitions is that it is some angular distance (which is 2theta) between two points somehow related to the diffraction peak intensity.  It is more convenient, in many cases, to use directly the modulus of a scattering vector instead of the 2theta angle as the unit to describe the X-ray scattering magnitude. The space of a scattering vector (s-space) is the famous reciprocal space. Sometimes, the factor 2*pi is introduced in the reciprocal space and so-called q-space is more used in some applications.

If one expresses the Stokes and Wilson equation for the Lv in the units of a scattering vector, it simplifies to Lv=1/B, were B is an integral breadth expressed in the units of a scattering vector. The conversion from s-space peak broadening to 2theta-space broadening is done by a factor cos(theta)/lambda:

 B(s)=B(2theta)* cos(theta)/lambda

Isotropic peak broadening means a constancy of B(s):  B(s) = constant. This constancy of size broadening as a function of scattering vector has the important implication for the separation of different contribution to powder XRD peak broadening. For example, the peak broadening due to microstrains is not constant (linear with s in a first approximation). These different dependences allows one to separate the strain and size broadenings, and it is a basis of the Williamson-Hall method of strain-size analysis.

To be continued …

Dear Alexi

please give me a hint on the size estimation of quantum dots since their peak is too broad and how we can calculate the radial breath and strain induced effects separately,

thank you

Dear Masoud

There is a couple of points that should be clarified before you can use X-ray diffraction method for the size-strain analysis. I presume that the quantum dots are nanocrystals (deposited on a substrate). If the structure of the dots is amorphous, the atomic scattering is not coherent and you cannot use this approach. Second point, in order to separate strain and size contributions to the peak broadening you will need to measure at least two diffraction peaks. There is no way to do this using just a single diffraction peak. I also did not understand what do you mean by radial breath?

Anyway, if the quantum dots are crystalline and you can measure few not overlapped diffraction peaks from dots, try the following:

1.       Determine the area of each diffraction peak and the respective intensities at maximums.

2.       Calculate the integral breadth for each peak: B(2theta)=A/I, where A is the peak area and I is the peak intensity at the maximum.

3.       Convert B(2theta) into B(d*):  B(d*)=B(2theta) *pi/180 *Cos(theta)/λ  , where d* is an inverse of d-spacing for a given reflection (reciprocal space variable),  λ is the wave length of X-rays and theta is the Bragg angle (the half of the scattering angle)

4.       Plot B as a function of d* for the measured peaks.

Ideally, the points on your plot should be close to a straight line. If you have more than two points on your plot you can make a fit by a straight line B(d*)=a+b d*

The meaning of coefficients a and b follows from the following equivalent equation:

B(d*)= 1/L+ e d*, where L is an estimate for the average thickness of crystallites and e is an estimate for the strain. The exact physical meaning of L and e will depend on the assumptions about size model and strain model.

Please note that I assumed that the instrument contribution to peak broadening can be neglected.

Regarding the question, XRD features provided FWHM for each 2 theta and in the calculation 2 theta is divided by 2 to get theta value for Cos(theta) but along with this B value is not divided by 2. Should one divide B value by 2? if not,what will be the explanation behind it?

Please respond.Thank you very much in advance

Answer is no, we do not divide the width by 2. This is because the quoted peak width in Sherrer formula is based on evaluating the intensity at Delta (2 theta). Further, the integral breadth in the method of Stokes and Wilson is based on the ratio of the integrated intensity of the peak (which is spanned by 2 thata) to the maximum peak intensity, which is effectively evaluating the width of an "equivalent" rectangle with the same peak height and area. The width of that rectangle is again correlated with Delta (2 theta). The constant appearing in Sherrer formula is the proportionality constant between the FWHM and the integral breadth, and is dependent on the line shape of the peak profile (Gaussian, Lorentzian, ...). 

Hi,

First to your question: you can use lambda=(2/Q)sin(theta). Mind some1 adds 2pi to scattering vector... But are you sure the platelet boundaries will not destroy coherence between the scattered photons from different crystals? It is not like in a twin! For reflections, which are not affected by twinnig you can use Scherrer equation to deduce "apparent" size (mind all previous comments). But for particles, where the distance between them violates the periodicity (which is the a priory case) of the layers this will not in my opinion work.

β is the line broadening at half the maximum intensity (FWHM), after subtracting the instrumental line broadening, in radians. This quantity is also sometimes denoted as Δ(2θ);

θ is the Bragg angle (in degrees).

A good example for you can be: A AM CHEM SOC, 2005, 127,17808-17813

Hello Yasine

Your SAXS measurements are already in the units of the length (better to say in the units of the inverse length, A-1). Normally, SAXS measurements are presented in so-called q-space which is related to reciprocal space units by coefficient 2*pi.  So, the Scherrer formula in these q units is:

B=2*pi * k/L,

where B is the broadening of the peak you measure in q-space, k is the Scherrer constant and L is a dimension of crystallites (in A). Take a look at my previous posts that consider in details the conversion procedures between different units of peak broadening.

The basic problem that you may have when calculate the size of the stacks of clay platelets by using the above equation is an underestimation of the clay particles dimension along the stacking direction (I suppose that you will use the respective diffraction peak to calculate the thickness of stacks, in this case k=1) as compared with the estimations that you may have from microscopy images. Even though the stacks of clay platelets appear in micrographs as regular, it does not necessary mean that they are regular at atomic scale. X-ray scattering is much sensitive to atomic scale imperfections than TEM or even HRTEM. Microstrains may be present and this will produce an additional broadening in the diffraction peak.

Few other eventual effects will also contribute to a distortion of your size measurements by using the Scherrer equation. For example, the intercalation of clay particles is not a perfect (you may have a fraction of clay only partially intercalated or completely exfoliated) or the size of intercalated particles is dispersed. All these will affect the calculated result. In general, a comprehensive morphological characterization of clays from X-ray scattering is a challenging task, but as a rough estimation, the Scherrer equation should work.

I see a lot of contributions to this discussion. I'll read in further detail to better learn the conventional use of the Scherrer Equation & the W-H Technique to quantify Nanostructural periodicities and strain fields.

Observation 1:  β Radians (NOT DEGREE), is known as integrated breadth which also is the line broadening denoted in the Scherrer Equation. β is NOT EQUAL to the full width at half the maximum intensity (FWHM). These two parameters are related to each other based on the shape of the Bragg profile. Gaussian, Voigt, Lorentzian, Pearson VII etc.

Observation 2: The presence of "preferred orientation" or non isotropic particle shapes will introduce some variations that need to be included in the assumed model.

The value beta in sherrer equation is in radians. If the scale is in 2Theta it must be halved. The value 0.9 comes from the fact the the peak is not a Gaussian nor a Lorentian. The peak shape is that horiginating form Laue equation. Instrumental broadening can in fact affect the value of the FWHM.

I thought the 0.9 factor had to do with the "particle shape" factor :-)

In any case, the "instrumental shape" is the key. Practically, this "instrumental profile" is characteristic of the beam conditioning apparatus and the other optics used. The shape may be Lorentzian or Gaussian or other depending on the optics. One may also use Bruker LEPTOS  to numerically compute the "strain free state" Bragg profile aka "instrumental profile".