Questions about PDF Analysis

    Some of the questions we have received from customers are listed below.

    Q1. For lab instrument-based PDF analysis, should I monochromatize the $\mathrm{K\alpha}$ radiation to $\mathrm{K\alpha_{1}}$ radiation?

    A. No, it is not necessary to monochromatize. The reason is that the real space resolution $\Delta r$ is calculated to be only 0.001 Å or less between the monochromatized $\mathrm{K\alpha_{1}}$ radiation and the general $\mathrm{K\alpha_{1}}$ radiation data. Furthermore, monochromatizing to $\mathrm{K\alpha_{1}}$ results in 1/10 of the intensity before monochromatization; therefore, measurement time takes 10 times longer.

     

    Table. $\Delta\ r$ at each wavelength

    Characteristic X-ray $\lambda\ (Å)$ $Q\ (Å^{-1})$ $\Delta r$ (Å)
    $\mathrm{Ag\ K\alpha}$ 0.560880 22.0644 0.1428
    $\mathrm{Ag\ K\alpha_{1}}$ 0.559422 22.1219 0.1420
    $\mathrm{Ag\ K\alpha_{2}}$ 0563813 21.9496 0.1431

     

    Q2. Can I perform PDF analysis of thin film samples?

    A. It is currently difficult with a lab machine because Mo and Ag wavelengths have deeper penetration depth, so information from substrates other than thin films will be introduced.

    Q3. Can I perform PDF analysis of compounds?

    A. Yes; however, the weight ratio or composition ratio may make it difficult to see the desired interatomic distance. Also, when fitting with PDFgui, the weight ratio can be obtained from the scale factor.

    Q4. How long does it take to analyze a PDF?

    A. A test measurement takes from a few minutes to 10 minutes. Full-scale measurement often takes about 4 hours. Full-scale modeling analysis may take up to a day. The measurement time will increase or decrease depending on the analysis and quality desired.

    Q5. Are detectors equipped with CdTe sensors suitable for PDF analysis?

    A. CdTe detectors certainly have good absorption at high-energy wavelengths, but their detection efficiency is not 100%. In the case of low-energy X-rays such as $\mathrm{Cu\ K}\alpha$, a detector equipped with CdTe sensors will hav lower detection efficiency, and another detector is required. Furthermore, commercially available detectors equipped with CdTe sensors require greater maintenance compared to Si sensors, because of a water-cooling system for device temperature control and a dry air circulation system for keeping dry condition.

    Q6. Can the reflection method be used for analysis?

    A. The reflection method can also be used for analysis, but it is a little difficult to estimate the amount of BG components such as air scattering in the measured total scattering data. Therefore, we generally recommend the transmission method.

    Q7. Is it possible to calculate the partial correlations from the observed data of a multi-element system?

    A. Yes, the partial correlations can be calculated by common modeling tools.

    Q8. What are the advantages and disadvantages of EXAFS (Extended X-ray Absorption Fine Structure) and PDF?

    A. The advantage of XAFS is its element selectivity.
    Therefore, it is possible to directly observe relatively short-range information about the first and second neighbors of correlations that include the element of interest.
    The disadvantages of EXAFS are the unreliability of data on the long-range side and the need for phase correction to obtain absolute distances. On the other hand, the advantage of PDF is that it can capture all information from short to long distances, allowing structural analysis.
    The disadvantage of PDF is that correlations are observed for all atoms in the system, especially since most real materials are composed of multiple elements and it is impossible to extract specific correlations only from the measured values.

     

    Table. Comparizon between PDF and EXAFS

      PDF EXAFS
    Element selectivity No Yes
    Distance information Obtained without correction Phase correction is required
    Obtained distance information Weighted correlation for all elements Correlation for specific elements

     

    Q9. Can I use SmartLab Studio $\mathrm{I\hspace{-.01em}I}$?

    A. Yes, it is possible. The data can be analyzed by providing wavelength information and horizontal polarization ratio to the measured data.

    Q10. In the case of a mixture of amorphous and crystalline, is it possible to extract structural information from each?

    A. Since diffraction from amorphous is much weaker in intensity than from crystalline materials, information from amorphous materials is likely to be buried by information from crystals, making extraction difficult.

    Q11. Can we observe Li-Li correlations in Li₂S?
    A. As it turns out, Li-Li correlations do not appear in total scattering data or PDF analysis results.

    TT. E. Faber and J. M. Ziman reported the idea of classifying total correlations calculated by Debye's scattering formula by correlation (partial correlation)₁)。
    The structure factors and PDFs expressed in this approach are called Faber-Ziman type structure factors and PDFs.
    Here, Faber-Ziman type structure factors are explained using Li₂S as an example.

    \begin{equation} S(Q) = \frac{c_{\mathrm{Li}}f_{\mathrm{Li}}c_{\mathrm{Li}}f_{\mathrm{Li}}}{\langle f \rangle^{2}}S_{\mathrm{LiLi}}(Q) + \frac{c_{\mathrm{Li}}f_{\mathrm{Li}}c_{\mathrm{S}}f_{\mathrm{S}}}{\langle f \rangle^{2}}S_{\mathrm{LiS}}(Q) + \frac{c_{\mathrm{S}}f_{\mathrm{S}}c_{\mathrm{Li}}f_{\mathrm{Li}}}{\langle f \rangle^{2}}S_{\mathrm{SLi}}(Q) + \frac{c_{\mathrm{S}}f_{\mathrm{S}}c_{\mathrm{S}}f_{\mathrm{S}}}{\langle f \rangle^{2}}S_{\mathrm{SS}}(Q) \nonumber \end{equation} \begin{equation} \langle f \rangle = \sum_{i}^{n}c_if_i \nonumber \end{equation}

    $S(Q)$ is the structure factor of $\mathrm{Li_{2}S}$, and $c_{i}$ and $f_{i}$ are the concentration of $i$ atoms and the atomic scattering factor, respectively.
    Furthermore, Li-O and O-Li in the equation are equivalent (indistinguishable) and can be combined into one.

    \begin{equation} S(Q) = \frac{c_{\mathrm{Li}}f_{\mathrm{Li}}c_{\mathrm{Li}}f_{\mathrm{Li}}}{\langle f \rangle^{2}}S_{\mathrm{LiLi}}(Q) + \frac{2c_{\mathrm{Li}}f_{\mathrm{Li}}c_{\mathrm{S}}f_{\mathrm{S}}}{\langle f \rangle^{2}}S_{\mathrm{LiS}}(Q) + \frac{c_{\mathrm{S}}f_{\mathrm{S}}c_{\mathrm{S}}f_{\mathrm{S}}}{\langle f \rangle^{2}}S_{\mathrm{SS}}(Q) \nonumber \end{equation}

    From this equation, we see that the correlation observed in the total scattering intensity is the sum of partial correlations with the atomic concentration and atomic scattering factor as coefficients. Since the structure factor $S(Q)$ and the PDF $G(r)$ can be interconverted by the Fourier transform, the Faber-Ziman type structure factor concept can also be applied to $G(r)$. However, for $G(r)$, the atomic scattering factor $f$ is the atomic number $Z$.

    \begin{equation} G(r) = \frac{c_{\mathrm{Li}}Z_{\mathrm{Li}}c_{\mathrm{Li}}Z_{\mathrm{Li}}}{\langle Z \rangle^{2}}G_{\mathrm{LiLi}}(r) + \frac{2c_{\mathrm{Li}}Z_{\mathrm{Li}}c_{\mathrm{S}}Z_{\mathrm{S}}}{\langle Z \rangle^{2}}G_{\mathrm{LiS}}(r) + \frac{c_{\mathrm{S}}Z_{\mathrm{S}}c_{\mathrm{S}}Z_{\mathrm{S}}}{\langle Z \rangle^{2}}G_{\mathrm{SS}}(r) \nonumber \end{equation} \begin{equation} \langle Z \rangle = \sum_{i}^{n}c_iZ_i \nonumber \end{equation}

    The table summarizes the coefficients of the Faber-Ziman type structure factor for the partial correlation of $\mathrm{Li_{2}S}$.
    Since Li-Li only accounts for about 7% of the total, we can see that it is likely to be difficult to observe peaks of only Li-Li correlations in the actual $G(r)$.

     

    Table Faber-Ziman coeficient of Li₂S in $G(r)$

    Correlation Li-Li Li-S S-S
    Coefficient % 7.44 39.67 52.89

     

    Finally, the Faber-Ziman type structure factor and the general form of the PDF are included.

    In the case of $S(Q)$

    \begin{equation} S(Q)=\sum_{i, j\ge i}(2-\delta_{ij})\frac{c_{i}f_{i}c_{j}f_{j}}{\langle f \rangle} S_{ij}(Q) \nonumber \end{equation}

    In canse of $G(r)$

    \begin{equation} G(r)=\sum_{i, j\ge i}(2-\delta_{ij})\frac{c_{i}Z_{i}c_{j}Z_{j}}{\langle Z \rangle} G_{ij}(r) \nonumber \end{equation}

    $\delta_{ij}$ is Kronecker's delta. Thus the correlation for homologous atoms is$(c_{i}f_{i}c_{j}f_{j})/\langle f \rangle$ and between dissimilar atoms is $(2c_{i}f_{i}c_{j}f_{j})/\langle f \rangle$.

     

    References:

      [1] T. E. Faber and J. M. Ziman, Phil. Mag., 11(1965), 153-173. (https://doi.org/10.1080/14786436508211931)