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We use Scherrer's equation to calculate crystalline size of a particle:

$$d = \frac{Bλ}{β\cos{θ}}$$

But which theta should we take from the XRD pattern for this calculation?

For example, the following plot shows the diffraction pattern of nano (above) and bulk (below) silicon and it has two diffraction peaks at $2θ \approx 69^\circ$ and $2θ \approx 76^\circ$. Which theta should we take for calculating crystalline size of nano silicon and why?

Diffractograms of nano and bulk silicon

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  • $\begingroup$ for the polycrystalline materials the instrumental error can be eliminated by subtracting the peak width of polycrystalline materials by single crystals peak broadening in same machine. $\endgroup$ – user1256 Feb 18 '13 at 7:54
  • $\begingroup$ You should take the theta for the peak for which you measured the peak width. For an ideal case, the peak widths should be Bragg angle (i.e. resolution)-dependent. Or you could refine beta for the entire data set (Michael Green's answer) keeping in mind that this might be overkill given the approximations and caveats (answer by F'x). $\endgroup$ – Karsten Theis Jul 19 at 21:01
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Scherrer’s formula is a simple equation for the simple case of a single peak broadened only due to crystalline particle size. Application of the formula to other, more complex cases, gives only an estimate. Quoting Wikipedia:

It is important to realize that the Scherrer formula provides a lower bound on the particle size. The reason for this is that a variety of factors can contribute to the width of a diffraction peak; besides crystallite size, the most important of these are usually inhomogeneous strain and instrumental effects. If all of these other contributions to the peak width were zero, then the peak width would be determined solely by the crystallite size and the Scherrer formula would apply. If the other contributions to the width are non-zero, then the crystallite size can be larger than that predicted by the Scherrer formula, with the "extra" peak width coming from the other factors.

Scherrer formula gives an estimate, an order of magnitude. Here, calculate the particle sizes for the two peaks, and they should be close. The differences are covered by the large uncertainty of your resulting value.

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You can use a simple mathematical manipulation to adjust the scherrer equation to take data from an XRD plot directly...

$$d = \frac{B \lambda}{\beta \cdot cos( \frac{2\theta}{2}))}$$

...where $2\theta$ is the bragg angle.

For the lower XRD pattern set, you'll need to deconvolute one of the peaks. If you know how to use python, the easiest way this can be done is that you can use a least-squares regression analysis to fit a set of pseudo-vogit functions to the XRD data and then use one of the functions to determine the FWHM and $2\theta$ value, where the pseudo-vogit is...

$$PV(x,A,\mu,\sigma,\alpha) = \frac{(1-\alpha)A}{\frac{\sigma}{2\cdot ln(2)}\cdot\sqrt{2\pi}}\cdot e^{\frac{-1}{2}\cdot \frac{(x-\mu)^2}{(\frac{\sigma}{2 \cdot ln(2)})^2}}+\frac{\alpha A}{\pi}\cdot[\frac{\sigma}{(x-\mu)^2 + \sigma^2}]$$

...and the minimizing condition is...

$$PV_{sum} = \sum_{i=1}^{n_{peaks}} PV_i(x_i,A_i,\mu_i,\sigma_i,\alpha_i) $$

...finding the condition which...

$$(data_i - PV_{sum}(data_i))^2 \rightarrow 0 $$

which is something a computer can do pretty easily. This'll output the set of parameters which fit $PV_{sum}$ to the entire data set - just use one of the PV conditions to calculate FWHM and $2\theta$ to determine the particle size.

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