# Scattering by an atom

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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
[1] International Tables For Crystallography Volume C, 3rd ed., E. Prince, Ed., Wiley, 2004.
[Bibtex]
@book{Prince2004,
doi = {10.1107/97809553602060000103},
edition = {3rd},
editor = {Prince, E},
publisher = {Wiley},
title = {{International Tables For Crystallography Volume C}},
year = {2004}
}