Hello Sir Runguang Li

Yes, for the determination of the crystallite size and lattice strain we need to know the value of full width at half maxima. However, the observed broadening of the peaks is due to crystallite size, lattice strain and instrumental factor. Therefore, we need to remove the instrumental factor to determine the crystallite size and lattice strain with the least error.

To remove the instrumental factor, we need to know the peak broadening for the standard SiO2, CeO2 or LaB6. For the removal of the instrumental factor in the broadening of the XRD peaks we used the relation:

B(resultant)^2=B(observed)^2-(instrumental)^2, where B represents the full width at half maxima.

And honestly, I don't have any idea about the peaks that have smaller peaks than the standard sample.

I am referring to this paper(S K Mishra et al 2015 IOP Conf. Ser.: Mater. Sci. Eng. 75 012001) that I have read about this topic.

Thank You.

Hi Runguang Li,

First, please ignore Mikhail Kendin's response as it is likely a bot stringing a bunch of phrases that often occur together regardless of meaning. Either that or the poster is not sufficiently skilled in English to coherently communicate (potentially valid) thoughts. The idea that the instrument contribution to broadening varies with the sample is nonsense by the definition of the instrument contribution! Unfortunately, we've seen a sudden increase is such garbage here.

Second, what Haobam Samananda Singh says is valid, and I will try to clarify what he says. The formula

Bsample2 = Bobserved2 - Binstrumental2

[square of observed breadth is sum of squares of instrumental and sample breadths]

is correct, but B usually represents breadth, which is the peak integral divided by the peak height (the width a rectangular peak would have with the same area and height). This formula is appropriate when fit to Gaussian peak profiles.

On the other hand, if peaks are better described as Lorentzian, the formula is

Wsample = Wobserved - Winstrumental

[observed width is sum of instrumental and sample widths]

where W is the full width at half maximum (FWHM). This formula is appropriate when fit to Lorentzian = Cauchy peak profiles.

Note that real profiles are often between Gaussian and Lorentzian and Voight or pseudo-Voight profile is used to fit real profiles. In this case, the formula is

Wsamplea = Wobserveda - Winstrumentala,

where a is a number between 1 and 2 that depends on how much like Gaussian (a=2) vs how much like Lorentzian (a=1) the profiles are. Unfortunately, the variation between a=1 and =2 is very nonlinear, so there is not a simple relationship to use for a as a function of the pseudo-Voight mixing parameters (which is the fraction of Gaussian or fraction Lorentzian contribution to the peak profile).

So I suggest the following

Now to your question about observed widths equal or less than your standard peak widths.

• Do not use - or be very cautious using; see point below - CeO2 as a peak width reference. There is no NIST SRM CeO2. If you have material certified for that use by another national standards body, I would think that is ok, but don't assume there is negligible sample broadening otherwise.
• Do not assume line position standards or other material with certified lattice parameters or other certified reference characteristics have negligible sample broadening.
• Do use NIST SRM 660c (or 660b, etc. predecessors) as a peak width reference. NIST certifies it as a "Line Position and Line Shape Standard for Powder Diffraction" (italics added to point out the indication that it is appropriate for peak instrumental broadening determination.
• In my opinion, if you have compared your own material (such as CeO2, or other) to a genuine line shape SRM or line width SRM and you find the peak widths to be identical (at same angles; see point below) to those of the SRM in the identical instrument that you will use for your experiments with the identical configuration you will use for your experimental data, then fine, go ahead and use that as if it was an SRM.
• When you compare peak angle widths or subtract instrumental broadening, be aware that the instrumental broadening will vary with angle, so you need to fit your observed instrumental broadening (of the standard material) to a function of angle and use that function to calculate the instrumental broadening at the peak angles of your samples.

If you are careful of all of the above, there are two possibilities (that I can think of) if your sample gives narrower (or equal width) peaks than your standard:

• Your sample may have crystallites that are too big (or equivalently, beam is too small), resulting in poor particle statistics, which can cause some of your peaks to look like single crystal peaks and thus be very narrow. In this case your results will not be valid when analyzed by methods appropriate for powder diffraction because the basic assumption of powder diffraction that there are many crystallites in the sample is not fulfilled.
• The other possibility is that your sample has a negligible contribution to the observed peak width and simple measurement uncertainty can make the observed width sometimes narrower than the peak widths of the standard. In this case, you can conclude that you have negligible microstrain and crystallite sizes too big to measure by this technique. You can estimate a lower limit of the crystallite size (size is ≥ this estimate) and an upper limit of the microstrain (microstrain is ≤ this estimate) by running various calculations with sample broadening greater than that observed to estimate the sensitivity of your setup.

Hope this helps. Feel free to ask further questions.

In a DM Haobam Singh asked "I want to ask you sir that the peak positions of our sample and that of the standard sample are different. If the peak positions are nearly equal than i can remove the instrumental factor.

So, i am confusing due to the difference in the peak positions. Will you help me in solving this problem?"

I'll answer here for the benefit of all.

Fit the observed peak widths of your standard to a function. Then use the value of that function at the peak angles of your sample to subtract instrumental broadening. This has the additional advantage of averaging the standard widths.

One common equation for widths as a function of 2θ, implemented in all Rietveld whole pattern fitting programs is the Caglioti formula:

wcaglioti = (U tan2θ + V tanθ + W)1/2,

where θ is just (2θ)/2 and U, V, W are fitted.

I believe this was derived for the expected peak width variation with angle for neutron scattering, in which peaks are gaussian, so it may not fit your data perfectly.

I find that adding a term that contributes additional broadening at low angles gives excellent fits (to my standards on my diffractometer with rather large divergence slits and a large linear detector - maybe good for you, maybe not, I haven't checked)

wcaglioti+empirical = (U tan2θ + V tanθ + W + Z/sin2θ)1/2

You can use any function for w(2θ) that nicely fits your reference peak width data. Don't worry about whether there is a sound theoretical basis for it. You will just use it to average out scatter in your measured standard peak widths and to interpolate between data points of the instrumental broadening curve. (Do not use it to extrapolate!)

Thanks w.s.l. Boyer for your valuable answer and Runguang Li for starting a discussion about this topic.

Thanks w.s.l. Boyer for the sufficient discussion. My steel sample is rolled-annealed with an average grain size of 5 um and the sample thickness is 1.2 mm. I just noticed that my LaB6 rings are quite mosaic and seem to have relatively large crystallites in the powders, which may be one reason for the observed broadened LaB6 peaks.

Since the parameter I want from the peak broadening analysis is actually only the microstrain, I think maybe I can still perform the W-H analysis with my LaB6 FWHMs (assuming it contains some degree of size-broadening but no strain-broadening). The sample microstrain is only determined from a final slope in W-H, which is independent of the size-broadening of LaB6 and will therefore not be detrimental. Of course, the grain size deduced is not reliable.