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\begin{document}

\begin{titlepage}
\newcommand{\HRule}{\rule{\linewidth}{0.5mm}}
\center
\textsc{\LARGE Universität Tübingen}\\[1.5cm] 
\textsc{\Large F-Praktikum}\\[0.5cm] 
\textsc{\large RBE}\\[0.5cm] 
\HRule \\[0.4cm]
{\huge \bfseries X-Ray Diffraction On Crystals}\\[0.4cm] 
Version 1
\HRule \\[1.0cm]
\begin{minipage}{0.4\textwidth}
    \begin{flushleft} \large
        \emph{Von:}\\
        Group 22\newline
        \underline{Fabian Kraus}\newline
        Simon Fischer
    \end{flushleft}
\end{minipage}
~
\begin{minipage}{0.4\textwidth}
    \begin{flushright} \large
        \emph{Supervisor:}\\
        Ivan Zaluzhnyy
        \end{flushright}
\end{minipage}\\[2cm]
{\large Date: 17.06.2021}\\
{\large Version 1: \today}\\

\end{titlepage}


\newpage
\tableofcontents
\newpage


\section{Introduction}
X-rays are most well known for their usage in medicine, where they can make the inner structure of bodies visible. In physics, we can similarly use these short-wavelength rays to make the inner structure of condensed matter visible.

In this experimant we probe the lattice structure of Copper, Tungsten and NaCl using an X-Ray-Tube and photographic film, called the Debye-Scherrer method.



\section{Theory}

\subsection{Lattice Scattering}

\subsubsection{Interference in terms of scattering vector}
In order to obtain a quantitative expression for the intensity in a scattering scenarion, let us look at \autoref{rbe:fig:scattering} below. It shows an incident, plane wave with wave vector $k_0$ that is scattered by two neighboring atoms which result in an outgoing wave vector $k_1$. 
\begin{figure}[H]
    \centering
    \includegraphics[width=100mm]{05_RBE/interference_sketch.png}
    \caption{Geometry of the system under consideration. | \textit{Source:} Manual.}
    \label{rbe:fig:scattering}
\end{figure}
For elastic scattering, there is no momentum transfer to the atoms. Therefore, the modulus of the wave vector is conserved before and after the scattering event:
\begin{equation}
    \mid\Vec{k}_0\mid\,=\,\mid\Vec{k}_1\mid\,=\,\frac{1}{\lambda},
\end{equation}
where we use the crystallographic notation, which implies us omitting the factor $2\pi$ in the expression above. Instead, it reappears in later exponents, such that the amplitude $U$ of the wave is still $U\,\propto\,e^{2\pi\Vec{k}\Vec{r}-\omega t}$.\\
Now, for us to obtain the phase shift between the two waves caused by the scattering event, we can write down the optical path (wave vector times distance) that the waves take. A closer look at \autoref{rbe:fig:scattering} reveals the realtions $\Vec{d}\cdot\Vec{k}_1\,=\,kd\cos\alpha$ and $\Vec{d}\cdot\Vec{k}_0\,=\,kd\cos\beta$, just for geometrical reasoning. With that we define the \textit{scattering wave vector} $\Vec{q}\,\equiv\,\Vec{k}_1\,-\,\Vec{k}_0$ and we get by combining those three equations
\begin{equation}
    2\pi kd\left(\cos\alpha-\cos\beta\right).
\end{equation}
For constructive interference, this phase shift should be equal to
\begin{equation}
    2\pi kd\left(\cos\alpha-\cos\beta\right)\,=\,2\pi n,
\end{equation}
where $n\in\mathbb{Z}$. For constructive interference, we require $\Vec{d}\cdot\Vec{q}=n$, which leads us, under consideration of \autoref{rbe:fig:scattering} to
\begin{equation}
    \sin(\alpha)=\frac{\lambda |\Vec{q}|}{2},
    \label{rbe_bragg_1}
\end{equation}
with $\Vec{q}=\Vec{k_1}-\Vec{k_0}$.\\
Moving to the case of $N$ atoms, i.e. more than two, we can generalize this by summing over all atoms inside a unit cell $\Vec{r}_j$ and also by summing over all unit cells $\Vec{R}_N$. This yields the expression for the scattering amplitude as
\begin{equation}
    U_{tot}\,=\,U_0\sum_{\Vec{r}_j}f_j(\Vec{q})e^{2\pi i\Vec{q}\cdot\Vec{r}_j}\sum_{\Vec{R}_n}e^{2\pi i\Vec{q}\cdot\Vec{R}_n}.
    \label{rbe:eq_ampl1}
\end{equation}
The first sum is called the \textit{structure factor}, whereas $f_j(\Vec{q})$ is the \textit{atomic form factor} and takes different kind of atoms into account.\\
The periodicity of the lattice allows us to define the \textit{lattice vector}
\begin{equation}
    \Vec{R}_n\,=\,\nu_1\Vec{a}_1\,+\,\nu_2\Vec{a}_2\,+\,\nu_3\Vec{a}_3,
\end{equation}
which we can exploit to evaluate the second sum on \eqref{rbe:eq_ampl1} by the geometric series:
\begin{align}
    \sum_{\nu\,=\,0}^{N}\left(e^{2\pi i q a}\right)^{\nu}\,=&\,\frac{\left(e^{2\pi i q a}\right)^{N+1}-1}{e^{2\pi i q a} -1}\,=\,\frac{\left(e^{\pi i q a}\right)^{N+1}}{e^{\pi i q a}}\frac{e^{\pi i q a(N+1)}\,-\,e^{-\pi i q a(N+1)}}{e^{\pi i q a}\,-\,e^{-\pi i q a}}\\\
    =&\,\frac{\left(e^{\pi i q a}\right)^{N+1}}{e^{\pi i q a}}\frac{\sin\left(\pi qa(N+1)\right)}{\sin\left(\pi qa\right)} 
\end{align}
Finally, we obtain for the intensity, that is, the amplitude squared:
\begin{equation}
   I\,\propto\left\vert\frac{\sin\left(\pi qa(N+1)\right)}{\sin\left(\pi qa\right)}\right\vert^2.
\end{equation}
For a large number of unit cells $N$, the intensity behaves like delta functions with sharp peaks at $q\,=\,0,\pm\frac{1}{a},\pm\frac{2}{a},\dots$. But since every relaistic crystal has a finite length, those perfect delta distributions will wash out and not peak to infinity, but will rather have a finite height. In the case above, the vector $\Vec{q}$ corresponded to what we call a \textit{reciprocal} vector, so it is worth briefly introducing the reciprocal space. It is a spae of great importance, which is why we will also need it in the following chapters. Actually, the scattered pattern which will be visible corresponds directly to points in the reciprocal space.\\
A recirprocal vector w.r.t. a vector $\Vec{a}_i$ is defined as
\begin{equation}
   \Vec{a}_i\cdot\Vec{a}_j^*\,=\,\delta_{ij},
\end{equation}
where $\Vec{a}_j^*$ is the reciprocal lattice vector. For different types of lattices this reciprocal lattice vector can be arbitrarily complicated. For simplicity, we choose a simple cubic lattice, i.e.
\begin{equation}
   a\cdot\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}\, ,\,\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,\,\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}\,\rightarrow \frac{1}{a}\left\{\begin{pmatrix}1\\0\\0\end{pmatrix}\, ,\,\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,\,\begin{pmatrix}0\\0\\1\end{pmatrix}\right\}
\end{equation}
and then add more atoms with a multidimensional basis, that is, we prefer to extend the atom basis instead of making the reciprocal lattice vectors more complicated. In general, we see that for a sc crystal lattice, the reciprocal lattice vector has the form
\begin{equation}
    \Vec{q}=\frac{1}{a}\left( h \Vec{a}_1^*+ k \Vec{a}_2^* + l \Vec{a}_3^* \right),
\end{equation}
where $h$, $k$, $l$ $\in \mathbb{N}$. By plugging it into \eqref{rbe_bragg_1}, we arrive at the important condition for constructive interference
\begin{equation}
    a = \frac{\lambda}{2 \sin\Theta} \sqrt{h^2+k^2+l^2}.
\end{equation}

\subsubsection{BCC Structure Factor}
A body centered cubic (bcc) unit cell can be described with a two atom basis, where the two atoms are at
\begin{equation}
   \Vec{r}_j\,\in\,\left\{a\begin{pmatrix}0\\0\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}1\\1\\1\end{pmatrix}\right\},
\end{equation}
while we choose a sc lattice basis and therefore also a reciprocal space that can be described in sc. This comes from the fact that the reciprocal lattice of a sc lattice is again of the form of a sc lattice. For the scattering vector $\Vec{q}$, which is a recirpocal vector, this means that it holds:
\begin{equation}
   \Vec{q}\,=\,\frac{1}{a}\begin{pmatrix}h\\k\\l\end{pmatrix},
\end{equation}
with $h,k,l\,\in\,\mathbb{N}$. Hence, this results in a structure factor of
\begin{equation}
   F_{bcc}^{sc}\,=\,f\cdot\left(1\,+\,e^{i\pi(h+k+l)}\right)\,=\,f\cdot\left(1\,+\,(-1)^{h+k+l}\right).
\end{equation}
The lower index stands for the lattice we describe, the upper index stands for the crystal lattice which requires a corresponding choice of the atom basis, just as we did before. From the opinion of the author it is important to always note the upper index because otherwise this information is lost, although it is crucial to determine the amplitude / intensity from \eqref{rbe:eq_ampl1}.\\
In words the result for the structure factor tells us: we only get a peak when the sum of the $h,k,l$ indices is an even number and the peaks for the odd case are vanishing completely.


\subsubsection{FCC structure factor}
We can treat the case of a fcc cell analogously by chosing the atom basis to be
\begin{equation}
   \Vec{r}_j\,\in\,\left\{a\begin{pmatrix}0\\0\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}1\\1\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}1\\0\\1\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}0\\1\\1\end{pmatrix}\right\}.
\end{equation}
With that we end up with the structure factor
\begin{multline}
   F_{fcc}\,=\,f\cdot\left(1\,+\,(-1)^{h+k}\,+\,(-1)^{h+l}\,+\,(-1)^{k+l}\right)
   =\,
   \begin{cases}
   4f &\text{if h,k,l all even/odd}\\ 
   0 &\text{else.}
   \end{cases}
\end{multline}


\subsubsection{Structure Factor of a Mixed Unit Cell}
Up to now, we treated cells consisiting of only one type of atoms. To generalize, and like most things, complicate it further, we can choose a unit cells with two different kinds of atoms at positions $\Vec{r}_\alpha ,\Vec{r}_\beta$. As we said earlier, the atom factors $f_\alpha , f_\beta$ take this change into account. For one kind of atoms, denoted with the index $\alpha$, we have
\begin{equation}
   \Vec{r}_\alpha\,\in\,\left\{
   a\begin{pmatrix}0\\0\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}1\\1\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}1\\0\\1\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}0\\1\\1\end{pmatrix}
   \right\}
\end{equation}
and for the other kind, denoted with $\beta$:
\begin{equation}
   \Vec{r}_\beta\,\in\,\left\{
   \frac{a}{2}\begin{pmatrix}1\\0\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}0\\1\\0\end{pmatrix}\, ,\,\frac{a}{2}\begin{pmatrix}0\\0\\1\end{pmatrix}\,
   ,\,\frac{a}{2}\begin{pmatrix}1\\1\\1\end{pmatrix}
   \right\}.
\end{equation}
With that we end up with the structure factor
\begin{align}
   F_{fcc(\alpha,\beta)}^{sc}\,=&\,
   f_\alpha\left(1\,+\,e^{i\pi(h+k)}\,+\,e^{i\pi(h+l)}\,+\,e^{i\pi(k+l)}\right)
   +\,f_\beta\left(e^{i\pi(h+k+l)}\,+\,e^{i\pi h}\,+\,e^{i\pi k}\,+\,e^{i\pi l}\right)\\\
   =&\left(f_\alpha\,+\,f_\beta e^{i\pi(h+k+l)}\right)\cdot
   \left(1\,+\,e^{i\pi(h+k)}\,+\,e^{i\pi(h+l)}\,+\,e^{i\pi(k+l)}\right)
\end{align}
where we used that $e^{2\pi n}\,=\,1$ for all $n\,\in\,\mathbb{N}$. Again, we can express this in a case disticntion:
\begin{equation}
   F_{fcc(\alpha,\beta)}^{sc}\,=\,
   \begin{cases}
          4\left(f_\alpha\,+\,f_\beta\right) &\text{if h,k,l all even} \\
          4\left(f_\alpha\,-\,f_\beta\right) &\text{if h,k,l all odd} \\
          0 &\text{else}
   \end{cases}
\end{equation}
Thus, we see a similar pattern as in the case of the fcc lattice with a single atom, but just with varying intensity. In fact, the case of h,k,l akk odd now reveals a new visible peak, which, surprisingly, was not visible before in the fcc single atom basis case. We also observe that if $f_\alpha=f_\beta$, i.e. we return to the case of a single atom basis, we obtain the same structure factor as before.


\subsubsection{Influence of Temperature - Debye-Waller-Factor}
Only at absolute zero temperature (ideal case $T=0\,\SI{}{K}$) the atoms remain at their place with respect to time. On the other hand, if we increase the temperature, the atoms start to "wiggle" around their initial positions. We call their initial positions $\Vec{r}_j^{(0)}$ and their deviation from that $\Vec{u}_j(t)$, which obviously depends on time. Thus, we can describe the position of the $j$'th atom as
\begin{equation}
   \Vec{r}_j(t)\,=\,\Vec{r}_j^{(0)}\,+\,\Vec{u}_j(t).
\end{equation}
Now, since the position is time dependent, it stands to reason to average the structure factor over time, i.e.
\begin{equation}
   \langle F\rangle\,=\,\sum_j f_j\left\langle e^{2\pi i\Vec{q}\cdot\Vec{r}_j(t)}\right\rangle\,=\,\sum_j f_j e^{2\pi i \Vec{q}\cdot\Vec{r}_j^{(0)}}\left\langle e^{2\pi i\Vec{q}\cdot\Vec{u}_j(t)}\right\rangle.
\end{equation}
We assume that $|\Vec{u}_j(t)|<<\Vec{r}_j^{(0)}$, that is, we assume the atoms to wiggle over distances much shorter than the interatomic distances. With that assumption, we can expand latter expression and obtain
\begin{equation}
   \left\langle e^{2\pi i\Vec{q}\cdot\Vec{u}_j(t)}\right\rangle\,\approx\,1\,+\,2\pi i \langle\Vec{q}\cdot\Vec{u}_j(t)\rangle\,-\,\frac{4\pi^2}{2}\langle(\Vec{q}\cdot\Vec{u}_j(t))^2\rangle.
\end{equation}
The first term is constant, the second one vanishes as there is no preferred direction of thermal fluctuation and it averages out. Assuming also that the thermal motion is on average the same for all $j$ atoms, we find that
\begin{equation}
   \langle(\Vec{q}\cdot\Vec{u}_j(t))^2\rangle\,=\,q^2\langle u^2\rangle\langle\cos^2\theta\rangle.
\end{equation}
The last part can be calculated explicitely as
\begin{equation}
   \frac{1}{4\pi}\int_0^{2\pi}\mathrm{d}\varphi\int_0^\pi\mathrm{d}\theta\sin\theta\cos^2\theta\,=\,\frac{1}{2}\int_{-1}^1u^2\mathrm{d}u\,=\,\frac{1}{3},
\end{equation}
where we substituted $\cos\theta\,=\,u$. Combining those results, we end up with an intensity of
\begin{equation}
   I(T)\,=\,I(T\rightarrow 0)e^{-\frac{2\pi}{3}q^2\langle u^2\rangle},
\end{equation}
where the exponential factor is called the \textit{Debye-Waller factor}. Importantly, the peaks will not "wash out" or smear out as one would expect, but rather the intensity drops with the peaks still being as sharp as before.


\subsection{X-Rays}\label{rbe:ch_x_rays}
In the experiment we will be using an x-ray tube. The setup can be seen in \autoref{fig1}.
\begin{figure}[H]
    \centering
    \includegraphics[width=140mm]{05_RBE/x_ray_tube.png}
    \caption{Setup of the X-ray tube | \textit{Source:} \url{https://sites.google.com/site/frcrphysicsnotes/production-of-x-rays}}
    \label{fig1}
\end{figure}
As one can see in the figure, a current traveling through the cathode causes electrons to be emitted via thermic emission. A focusing cup, as the name suggests, focuses the electron beam towards the anode. An applied voltage $U$ leads then to a kinetic energy of the electrons of
\begin{equation}
    E_{kin}=eU.
\end{equation}
When the electrons hit the target, they slow down emitting a \textit{continuous spectrum} due to Bremsstrahlung. On top of that, they hit other electrons wihtin deeper shells of the atoms, which then emit a \textit{discrete} spectrum as electrons from higher energy levels fill the hole. However, the majority of the energy of the incoming electrons is converted to heat. Therefore, one needs the water to conduct the heat away from the tube and to cool it. The X-ray beam finally leaves the tube through a window made of beryllium. The reason for the choice of beryllium is simple: it is simply (almost) non-reactive. It can then be observed or measured by the experimenter.
    
    
\subsubsection{Spectrum}
As previously mentioned, the electrons in the tube have two contributions to the spectrum: One is continous, which comes from Bremsstrahlung, i.e. electrons being decelerated when hitting the target. On the other hand, the impact of the electrons can also remove other electrons in lower shells of the target atoms, which then fill up those holes with higher lying electrons under the emission of X-rays. Both contributions can be seen in \autoref{fig2} below.

\begin{figure}[H]
    \centering
    \includegraphics[width=100mm]{05_RBE/x_ray_spectrum.png}
    \caption{Intensity over wavelength diagram of a x-ray spectrum. Both contributions, namely Bremsstrahlung (green, dashed line) and the characteristic radiation from the target (green, solid lide), can be seen adding up to the total intensity (red) | \textit{Source:} \url{https://lp.uni-goettingen.de/get/text/4385}}
    \label{fig2}
\end{figure}
The sharp peaks in the characteristic spectrum can be understood by looking at \autoref{fig3}. There, we can see different lines in the X-ray production.
\begin{figure}[H]
    \centering
    \includegraphics[width=80mm]{05_RBE/copper_energy_levels.png}
    \caption{Energy levels of copper | \textit{Source:} \url{http://xrayweb.chem.ou.edu/notes/xray.html}}
    \label{fig3}
\end{figure}
Finally, it can be observed that the spectrum cuts off at a critical wavelength $\lambda_G$, as seen in \autoref{fig2} above. The reason for that is the maximum energy a photon can get, which is the total kinetic energy of an incident electron. Because the wavelength scales inversely with frequency we obtain a minimum wavelength for a maximum energy and get
\begin{equation}
    \lambda_{min} \,=\,\frac{c}{\nu_{max}}\,=\,\frac{hc}{h\nu_{max}}\,=\,\frac{hc}{E_{max}}\,=\,\frac{hc}{eU}.
\end{equation}



\subsubsection{Filtering}
When electromagnetic radiation, where X-rays fall into this category, pass through a medium that is not vacuum, the intensity decays. The exponential decay law is also known as the \textit{Beer–Lambert law}. Assuming a beam traveling in X-direction, we can write down the differential equation
\begin{equation}
    \frac{\mathrm{d}I}{\mathrm{d}x}\,=\,-\mu I_0,
\end{equation}
where $I_0$ is the initial intensity, that is, the intensity before the beam enters the medium. This can be integrated to
\begin{equation}
    I(x)\,=\,I(0)e^{-\mu x},
\end{equation}
where $\mu$ defines a typical length scale on which the intensity decays. More precisely, we call
\begin{equation}
    d_{\frac{1}{2}}\,\equiv\,\frac{\ln(2)}{\mu}
\end{equation}
as the location where the intensity reaches exactly half of its initial value $I_0$, i.e. $I(d_{\frac{1}{2}})=I_0$.\\
The parameter $\mu$ contains several contributions from interaction of light with matter, namely the classical scattering ($\sigma_k$), photo effect ($\tau$), compton effect ($\sigma_C$) and pair creation ($\kappa$):
\begin{equation}
    \mu\,=\,\sigma_k\,+\,\tau\,+\,\sigma_C\,+\,\kappa.
\end{equation}
Each of these coefficients has a different dependence on the photon energy.

\newpage

%\section{Experimental Setup}
%The experimental setup is extensively discussed in section \ref{rbe:ch_x_rays}.

%\newpage




\newpage
\section{Results and Analysis}
To evaluate the data taken, we mesaure the distance between two opposite maxima
of the same order. The angle can be determined by the relation
\begin{equation}
    1\° \leftrightarrow 1 \, \SI{}{mm}.
\end{equation}
Furthermore, we have the given wavelength of $\lambda=0.154178\,\SI{}{nm}$ and the relation introduced in the theory section
\begin{equation}
    a = \frac{\lambda}{2 \sin\Theta} \sqrt{h^2+k^2+l^2}.
\end{equation}
Dividing two maxima therefore leads to:
\begin{equation}
    \frac{\sin^2\Theta}{\sin^2\Theta'} = \frac{h^2+k^2+l^2}{h'^2+k'^2+l'^2}.
    \label{rbe_sin_by_sin}
\end{equation}
With all those ingredients, we can determine the lattice constant for each h,k,l that do not vanish. Afterwards we take the mean value and give the statistical deviation from it.

\subsection{Tungsten}
For Tungsten, the result is summarized in the table below. For all following calculations, we use for $\Theta'$ the first angle of the table, i.e. all values in \eqref{rbe_sin_by_sin} are given w.r.t. the first angle. The distances are given in the appendix and are visible on the images themselves. For every probe we used the first eight lines as they are the best visible ones.

\begin{table}[H]
    \centering
    \begin{tabular}{c|c|c|c|c|c}
        hkl & $\Theta\pm \Delta \Theta\,$ [$\SI{}{MHz}$] & $\frac{\sin^2\Theta}{\sin^2\Theta'}$ & $a\pm \Delta a\,$ [$\SI{}{\mathring{A}}$] \\ \hline
        
        $110$ & $20.0 \pm 0.25$ & - & $3.188\pm0.039$\\
        \hline
        $200$ & $29.25\pm 0.25$ & $1.777$ & $3.155\pm0.25$\\
        \hline
        $211$ & $40.2 \pm 0.25$ & $3.108$ & $2.922\pm0.016$\\
        \hline
        $220$ & $46.7 \pm 0.25$ & $3.943$ & $2.996\pm0.0.013$\\
        \hline
        $310$ & $54.0 \pm 0.25$ & $4.873$ & $3.013\pm0.010$\\
        \hline
        $222$ & $61.75 \pm 0.25$ & $5.777$ & $3.032\pm0.008$\\
        \hline
        $321$ & $66.0 \pm 0.25$ & $6.213$ & $3.157\pm0.007$\\
        \hline
        $400$ & $77.0 \pm 0.25$ & $7.068$ & $3.156\pm0.004$\\
    \end{tabular}
    \caption{Results for tungsten.}
\end{table}
The errors for $a$ have been determined with Gaussian error propagation.\\
We can now determine the median of those values by
\begin{equation}
    <a> = \frac{1}{N}\sum_{i=0}^N a_i,
\end{equation}
where $N$ takes values from 1 to 8 (eight measurements) and we obtain
\begin{equation}
    <a>\pm \sigma_{<a>} = \left(3.093 \pm 0.04\right)\,\SI{}{\mathring{A}},
\end{equation}
whereas $\sigma_{<a>}$ is the standard deviation calculated by
\begin{equation}
    \sigma_{<a>} = \frac{\sigma_a}{\sqrt{N}} = \frac{1}{\sqrt{N}}\sqrt{\frac{1}{N-1} \sum_{i=0}^N\left( a_i - <a> \right)^2 }.
\end{equation}
Comparing the above result to the literature [\ref{RBE_Lit1}], we see that our value is close to the one given as
\begin{equation}
    a_{literature}=3.16475 \mathring{A},
\end{equation}
but not quite in the error range with an upper boundary of $3.133 \mathring{A}$. Still, the value determined by our mesaurements comes close to this result. It should be noted, that reading off the length scales by hand can produce significant errors for $\Theta$, which is then transferrred on the caltucation of the lattice constant $a$.\\
In order to minimize the systematic error we extrapolate once again linearly.

\begin{figure}[H]
    \centering
    \includegraphics[width=120mm]{05_RBE/Tungsten_plot.png}
    \caption{Extrapolation for Tungsten.}
    \label{rbe_tungsten_plot}
\end{figure}
The offset of the linear fit yields
\begin{equation}
    a_{extraplo}=\left(3.072 \pm 0.02\right)\,\SI{}{\mathring{A}},
\end{equation}
which again lies close to the value found previously.



\subsection{Copper}
For copper, we proceed in the same way and obtain the following results, which are presented in the table below.

\begin{table}[H]
    \centering
    \begin{tabular}{c|c|c|c|c|c}
        hkl & $\Theta\pm \Delta \Theta\,$ [$\SI{}{MHz}$] & $\frac{\sin^2\Theta}{\sin^2\Theta'}$ & $a\pm \Delta a\,$ [$\SI{}{\mathring{A}}$] \\ \hline
        
        $111$ & $21.5 \pm 0.25$ & - & $3.643\pm0.041$\\
        \hline
        $200$ & $25.25\pm 0.25$ & $1.354$ & $3.614\pm0.33$\\
        \hline
        $220$ & $37.2 \pm 0.25$ & $2.721$ & $3.610\pm0.021$\\
        \hline
        $311$ & $45.0 \pm 0.25$ & $3.722$ & $3.616\pm0.0.016$\\
        \hline
        $222$ & $47.75 \pm 0.25$ & $4.079$ & $3.608\pm0.015$\\
        \hline
        $400$ & $56.125 \pm 0.25$ & $5.131$ & $3.714\pm0.011$\\
        \hline
        $331$ & $64.5 \pm 0.25$ & $6.065$ & $3.722\pm0.008$\\
        \hline
        $420$ & $72 \pm 0.25$ & $6.734$ & $3.625\pm0.006$\\
    \end{tabular}
    \caption{Results for copper.}
\end{table}
As above, we determined the median for the lattice constant $a$ of copper and deviation from it as
\begin{equation}
    <a>\pm \sigma_{<a>} = \left(3.603 \pm 0.02\right)\,\SI{}{\mathring{A}}.
\end{equation}
Compared to the literature [\ref{RBE_Lit2}], we see that our value is rather close to the one given as
\begin{equation}
    a_{literature}=3.627 \mathring{A},
\end{equation}
but just not lying inside the error intervall by a difference of $0.004\mathring{A}$, which turns out to be a good result. Reasons for this deviation are firstly the uncertainty of reading the values off the scale and secondly, the fact that some lines were not very sharply defined, which results in a larger deviation from the expected value.\\
Here we also extrapolate linearly.

\begin{figure}[H]
    \centering
    \includegraphics[width=140mm]{05_RBE/Copper_plot.png}
    \caption{Extrapolation for Copper.}
    \label{rbe_tungsten_plot}
\end{figure}
The offset of the linear fit yields
\begin{equation}
    a_{extraplo}=\left(3.682 \pm 0.08\right)\,\SI{}{\mathring{A}},
\end{equation}
which lies within the error interval of the literature and therefore yields a good result.




\subsection{Sodium Chloride}
The results for Sodium Chloride can be found below.

\begin{table}[H]
    \centering
    \begin{tabular}{c|c|c|c|c|c}
        hkl & $\Theta\pm \Delta \Theta\,$ $\SI{}{MHz}$ & $\frac{\sin^2\Theta}{\sin^2\Theta'}$ & $a\pm \Delta a\,$ $\SI{}{\mathring{A}}$ \\ \hline
        
        $200$ & $16 \pm 0.25$ & - & $5.594\pm0.086$\\
        \hline
        $220$ & $22.5\pm 0.25$ & $1.777$ & $5.698\pm0.060$\\
        \hline
        $222$ & $28.25 \pm 0.25$ & $3.108$ & $5.642\pm0.045$\\
        \hline
        $400$ & $33.15 \pm 0.25$ & $3.943$ & $5.634\pm0.0.038$\\
        \hline
        $420$ & $38.1 \pm 0.25$ & $4.873$ & $5.587\pm0.032$\\
        \hline
        $422$ & $42.00 \pm 0.25$ & $5.777$ & $5.644\pm0.289$\\
        \hline
        $440$ & $50.5 \pm 0.25$ & $6.213$ & $5.651\pm0.020$\\
        \hline
        $442$ & $54.2 \pm 0.25$ & $7.058$ & $5.703\pm0.184$\\
    \end{tabular}
    \caption{Results for Sodium Chloride.}
\end{table}
Again, we determined the median for the lattice constant $a$ of Sodium Chloride to be
\begin{equation}
    <a>\pm \sigma_{<a>} = \left(5.643 \pm 0.015\right)\,\SI{}{\mathring{A}}.
\end{equation}
Compared to the literature [\ref{RBE_Lit3}], we see that our value is very close to the one given as
\begin{equation}
    a_{literature}=5.673 \mathring{A},
\end{equation}
as it almost coincides with the result. Again, it is not lying inside the error interval, but it comes close the border of the error interval's upper limit.
Again, we extrapolate linearly.

\begin{figure}[H]
    \centering
    \includegraphics[width=140mm]{05_RBE/NaCl_plot.png}
    \caption{Extrapolation for Sodium Chloride.}
    \label{rbe_tungsten_plot}
\end{figure}
The offset of the linear fit yields
\begin{equation}
    a_{extraplo}=\left(5.691 \pm 0.05\right)\,\SI{}{\mathring{A}},
\end{equation}
which lies within the error interval of the literature.


\newpage

\section{Conclusion}
The results for Sodium Chloride as well as the ones for Copper match very well with the results that can be found in the literature. Only the results for Tungsten obviously do not match closely to the ones given by literature. The possible mistakes were already discussed above, whereas we expect the largest error to arise from reading off the values in \autoref{RBE_W} where the millimeter scale is hardly visible. Apart from Tunsgten, all results turn out to be as expected.




\newpage

\section{Appendix}

\begin{figure}[H]
    \centering
    \includegraphics[width=180mm]{05_RBE/RBE_NaCl.png}
    \caption{Image of NaCl.}
    \label{RBE_C}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=180mm]{05_RBE/RBE_Copper.jpg}
    \caption{Image of copper.}
    \label{RBE_NaCl}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=180mm]{05_RBE/RBE_W.png}
    \caption{Image of tungsten.}
    \label{RBE_W}
\end{figure}


\newpage

\section{Literaure}

\begin{enumerate}
    \item \label{RBE_Lit1} Jette, E.; Foote, F., Journal of Chemical Physics (1935) 3, (*) Article ID * (p. 605-616)
    
    \item \label{RBE_Lit2} Holgersson, S., Annalen der Physik (Berlin, Germany) (1926) 79, (*) Article ID * (p. 35-54)
    
    \item \label{RBE_Lit3} Llamas Jansa, I.; Aliouane, N.; Deledda, S.; Fonnelop, J.E.; Frommen, C.; Humphries, T.;
Lieutenant, K.; Sartori, S.; Sorby, M.H.; Hauback, B.C., Journal of Alloys and Compounds
(2012) 530, (*) Article ID * (p. 186-192)
    
\end{enumerate}





\end{document}