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In the description of scattering by an atom, we need to take into account the interference of x-rays radiated from different positions in the atom since an electron spreads around the atomic nucleus in quantum-mechanical picture even when the number of electrons is one.

Figure 1 shows the configuration of the scattering process by an atom, whose atomic number is \(Z\). \(\vec{k}\) and \(\vec{k’}\) are the wave number vectors, which lengths are the same, i.e., \(|\vec{k}| = |\vec{k’}| =k = 2\pi/\lambda\), \(\vec{r}\) is a position where we evaluate the interference of x-ray, \(\rho(\vec{r})\) is an electron density at \(\vec{r}\), and \(\diff V\) is a volume element at \(\vec{r}\). The phase difference, \(\Delta \phi(\vec{r})\) between x-rays radiated at the origin and position \(\vec{r}\) is described to be

\begin{align}

\label{Eq:phase_difference_in_an_atom}

\tag{1}

\Delta \phi(\vec{r}) = \left( \vec{k} – \vec{k}’ \right) \cdot \vec{r} = \vec{Q} \cdot \vec{r},

\end{align}

where

\begin{align}

\label{Eq:dif_Q_vector}

\tag{2}

\vec{Q} = \vec{k} – \vec{k}’.

\end{align}

This interference occurs in the whole atom; therefore, the scattering amplitude is as follows:

\begin{align}

\label{Eq:integral_of_phases}

\tag{3}

-r_{0}\int \rho(\vec{r}) \e^{i \Delta \phi (\vec{r})} \diff V =

-r_{0}\int \rho(\vec{r}) \e^{i \vec{Q}\cdot \vec{r}} \diff V

\equiv -r_{0}f^{0}(\vec{Q}),

\end{align}

where the integration is carried out in the whole atom and \(f^{0}(\vec{Q})\) is known as the “*atomic form factor*“. From the definition, it is definitely expected that \(f^{0}(\vec{Q} = 0)\) is identical to the atomic number, i.e., the number of electrons, \(Z\). Thus, \(f^{0}(\vec{Q})\) value is scattering power described with a unit of the number of electrons called “*electron unit (eu)*, which is frequently used to discuss the absorption amplitude in the DAFS method as well as the conventional diffraction technique.

When we assume the spherical electron density, i.e., \(\rho(\vec{r}) \to \rho(r)\), the atomic form factor is described and evaluated in a simpler form because the electric density is just a function of distance \(r\). With the absolute value1 of \(|\vec{Q}| = 4\pi \sin \theta / \lambda\), the atomic form factor can be reduced2 into

\begin{align}

\label{Eq:integration_f_zero}

\tag{4}

f^{0}(Q) = \int_{0}^{\infty} 4\pi r^{2} \rho(r) \frac{\sin Qr}{Qr} \diff r.

\end{align}

Thus, if one knows the electron density, \(\rho(r)\), by some theoretical calculations, the atomic form factors of each element and ions can be evaluated and used for the structure analysis by x-ray diffraction technique. As seen in Eq. \eqref{Eq:integration_f_zero}, the atomic factor is a function of only \(Q\); therefore, the values determined from the electron density calculated by the quantum mechanical approaches such as Hartree-Fock or Fermi-Thomas-Dirac are available as a function of \((\sin \theta/ \lambda)\) in *International tables for crystallography; vol. C* [1] as in the following form:

\begin{align}

\label{Eq:model_atomic_form_factor}

\tag{5}

f^{0}\left(\frac{\sin \theta}{\lambda}\right) = \sum_{i =1}^{4 \mathrm{\ or\ } 5} a_{i} \exp \left\{ -b_{i} \left( \frac{\sin \theta}{\lambda} \right)^{2} \right\} + c,

\end{align}

where \(a_{i}\), \(b_{i}\) and \(c\) values of each element and ion are given in the book.

###### For further study…(This article was written based on the following book)

- J. Als-Nielsen and D. McMorrow, Elements of Modern X-ray Physics, 2nd ed., John Wiley & Sons, 2011.

[Bibtex]`@book{Als-Nielsen2011, author = {Als-Nielsen, J and McMorrow, D}, edition = {2nd}, publisher = {John Wiley \& Sons}, title = {{Elements of Modern X-ray Physics}}, year = {2011} }`

###### References

[Bibtex]

```
@book{Prince2004,
doi = {10.1107/97809553602060000103},
edition = {3rd},
editor = {Prince, E},
publisher = {Wiley},
title = {{International Tables For Crystallography Volume C}},
year = {2004}
}
```