\end{abstract}
\section{Results and Analysis}
-\begin{table}[h]
+\subsection{Collected Data}
+\begin{table}[H]
\caption{Rotational absorbances for the fundamental transition.}
\label{tbl:fundamental-results}
\centering
\end{tabular}
\end{table}
-\begin{table}[h]
+\begin{table}[H]
\caption{Rotational absorbances for the overtone transition.}
\label{tbl:overtone-results}
\centering
\cline{2-6}
& 7 & 5786.65 & 5493.84 & 292.81 & 310.49 \\
\cline{2-6}
- & 8 & 5464.63 & & & \\
+ & 8 & & 5464.63 & & \\
\hline
\end{tabular}
\end{table}
+\subsection{Determination of Rotational Constants}
The values of $\tilde{B_0}$, $\tilde{B_1}$ and $\tilde{B_2}$ were calculated
accounting for the centrifugal distortion of the molecules by using equations
\ref{eq:comb-diff-upper} and \ref{eq:comb-diff-lower}.
\ref{tbl:centrifugal-distortion-const} and hence the rotational constants,
$\tilde{B_{\nu}}$ shown in table \ref{tbl:rot-const-bond-lengths}.
-The bond lengths were calculated from the $\tilde{B_{\nu}}$ values using equation \ref{eq:bond-length}.
+The bond lengths were then determined from the $\tilde{B_{\nu}}$ values using equation \ref{eq:bond-length}.
\begin{equation}\label{eq:bond-length}
r_{\nu} = \sqrt{4 \pi c \hbar \mu \tilde{B_{\nu}}}
\hline
& $\nu$ & $\tilde{B_{\nu}}$ / \si{\per\centi\metre} & $r_{\nu}$ / \si{\pico\meter} \\
\hline
- \multirow{3}{*}{\ce{^{35}Cl}} & 0 & & \\
+ \multirow{3}{*}{\ce{H^{35}Cl}} & 0 & & \\
\cline{2-4}
& 1 & & \\
\cline{2-4}
& 2 & & \\
\hline
- \multirow{3}{*}{\ce{^{37}Cl}} & 0 & & \\
+ \multirow{3}{*}{\ce{H^{37}Cl}} & 0 & & \\
\cline{2-4}
& 1 & & \\
\cline{2-4}
\subsection{Determination of Vibrational Constants}
The values for the harmonic constant $\tilde{\nu_e}$ and the dimensionless anharmonicity constant $x_e$ in equation
-\ref{eq:vib-energy} (which gives the expected vibrational energy levels if the rotational spectrum is
-ignored).
-
-\begin{equation}\label{eq:vib-energy}
- \tilde{E_{\nu}} = \tilde{\nu_e} \left( \nu + \frac{1}{2} \right) -
- \tilde{\nu_e} x_e \left( \nu + \frac{1}{2} \right)^2
-\end{equation}
-
-From equation \ref{eq:vib-energy} we can obtain the energy related to the pure vibrational transition
-$\tilde{E}(\nu_f \leftarrow 0)$ (the $\nu = 0$ to $\nu = \nu_f$ transition) as equation \ref{eq:delta-e-coeff}.
+\ref{eq:vib-energy} was determined using equations \ref{eq:nu-e} and \ref{eq:xe}. The derivation of these equations is
+included within section \ref{sec:vib-const-deriv} of the supplementary information.
\begin{align}
- \tilde{E}\left( \nu_f \leftarrow 0 \right) &= \tilde{E}_{\nu_f} - \tilde{E}_0 \nonumber \\
- &= \tilde{\nu_e}\left( \nu_f + \frac{1}{2} - \frac{1}{2} \right) - \tilde{\nu_e} x_e \left(
- \frac{1}{4} - \left( \nu_f + \frac{1}{2} \right)^2 \right) \nonumber \\
- \label{eq:delta-e-coeff}
- &= \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
+ \label{eq:nu-e}
+ \tilde{\nu_e} &= R(0) - 3 \tilde{B_1} + \tilde{B_2} \\
+ \label{eq:xe}
+ x_e &= \frac{1}{2} \frac{\tilde{B_2} - \tilde{B_1}}{R(0) - 3 \tilde{B_1} + \tilde{B_2}}
\end{align}
-The energy related to the R branch transitions can be determined to yield
-equation \ref{eq:r-trans-energy} where $J$ is the rotational state adopted in
-the lower vibrational state ($\nu = 0$).
-
-\begin{equation}\label{eq:r-trans-energy}
- R(J) = \Delta \tilde{E}(\nu_f \leftarrow 0) + (\tilde{B_{\nu_f}} + \tilde{B_0}) (J + 1) + (\tilde{B_{\nu_f}} -
- \tilde{B_0}) ( J + 1)^2
-\end{equation}
-
-Setting $J = 0$ hence gives equation \ref{eq:r-trans-energy-j0}.
-
-\begin{equation}\label{eq:r-trans-energy-j0}
- R(J) - 2 \tilde{B_{\nu_f}} = \Delta \tilde{E}(\nu_f \leftarrow 0)
-\end{equation}
-
-Subsituting equation \ref{eq:delta-e-coeff} into equation \ref{eq:r-trans-energy-j0} yields equation
-\ref{eq:r-trans-energy-sub}.
-
-\begin{equation}\label{eq:r-trans-energy-sub}
- R(0) - 2 \tilde{B_{\nu_f}} = \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
-\end{equation}
-
-Now from equation \ref{eq:r-trans-energy-sub} a system of linear equations can be
-obtained by setting $\nu_f = 1$ and $\nu_f = 2$ (equations xx and xx) for which we can solve to give equations xx and xx.
-
-
+The uncertainties in these values was hence estimated using equations \ref{eq:nu-e-uncert} and
+\ref{eq:xe-uncert} where the uncertainty in $R(0)$ assumed to be negligible
+since it is determined by reading the wavenumber directly from the spectrum.
+\begin{align}
+ \label{eq:nu-e-uncert}
+ \alpha_{\tilde{\nu_e}} &= \sqrt{ \left( 3 \alpha_{\tilde{B_1}} \right)^2 + \left( \alpha_{\tilde{B_1}}
+ \right)^2} \\
+ \label{eq:xe-uncert}
+ \alpha_{x_e} &= x_e \sqrt{ \frac{ \left( \alpha_{\tilde{B_1}} \right)^2 + \left(
+ \alpha_{\tilde{B_2}} \right)^2 }{ \left( \tilde{B_2} - \tilde{B_1} \right)^2} + \left( \frac{\alpha_{\tilde{\nu_e}}}{\tilde{\nu_e}} \right)^2}
+\end{align}
+\subsection{Determination of Bond Force Constants}
+The bond force constants, $k$, shown in table \ref{tbl:force-constants} were determined using equation
+\ref{eq:force-constant} where $\mu$ is the reduced mass of the molecule and the error in $k$, $\alpha_k$, was
+determined using the calculus approximation (equation \ref{eq:force-constant-err}) where the error in $\mu$
+was assumed negligible compared to that in $\nu_e$.
+\begin{align}
+ \label{eq:force-constant}
+ k &= 4 \pi^2 c^2 \mu \tilde{\nu_e}^2 \\
+ \label{eq:force-constant-err}
+ \alpha_k &= 8 \pi^2 c^2 \mu \tilde{\nu_e} \alpha_{\tilde{\nu_e}}
+\end{align}
-for which setting $J = 0$ and $\nu_f = 1$ and subsequently $\nu_f = 2$ gives the
-system of linear equations (equations xx and xx) for which we can solve to give equations xx and xx.
+\begin{table}[H]
+ \caption{Force Constants.}
+ \label{tbl:force-constants}
+ \centering
+ \begin{tabular}{|c|c|c|}
+ \hline
+ & \ce{H^{35}Cl} & \ce{H^{37}Cl} \\
+ \hline
+ k / \si{\newton\per\centi\metre} & & \\
+ \hline
+ \end{tabular}
-%TODO: Total energy equation, determination of \nu_e x_e, calculation of values, determination of k with error
-%propagation and table.
+\end{table}
\section{Discussion}
Two methods able to be used to determine the value of $\tilde{B_0}$: using
\section{Supplementary Information}
+\subsection{Centrifugal Distortion Coefficients}
\begin{table}[h]
\caption{Centrifugal distortion coefficients.}
\label{tbl:centrifugal-distortion-const}
\hline
\end{tabular}
\end{table}
+
+\subsection{Derivation of Vibrational Constant Equations}\label{sec:vib-const-deriv}
+The energies associated with the discrete vibrational energy levels in a molecule are given by equation
+\ref{eq:vib-energy}.
+
+\begin{equation}\label{eq:vib-energy}
+ \tilde{E_{\nu}} = \tilde{\nu_e} \left( \nu + \frac{1}{2} \right) -
+ \tilde{\nu_e} x_e \left( \nu + \frac{1}{2} \right)^2
+\end{equation}
+
+From equation \ref{eq:vib-energy} we can obtain the energy related to the pure vibrational transition
+$\tilde{E}(\nu_f \leftarrow 0)$ (the $\nu = 0$ to $\nu = \nu_f$ transition) as equation \ref{eq:delta-e-coeff}.
+
+\begin{align}
+ \tilde{E}\left( \nu_f \leftarrow 0 \right) &= \tilde{E}_{\nu_f} - \tilde{E}_0 \nonumber \\
+ &= \tilde{\nu_e}\left( \nu_f + \frac{1}{2} - \frac{1}{2} \right) - \tilde{\nu_e} x_e \left(
+ \frac{1}{4} - \left( \nu_f + \frac{1}{2} \right)^2 \right) \nonumber \\
+ \label{eq:delta-e-coeff}
+ &= \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
+\end{align}
+
+The energy related to the R branch transitions can be determined to yield
+equation \ref{eq:r-trans-energy} where $J$ is the rotational state adopted in
+the lower vibrational state ($\nu = 0$).
+
+\begin{equation}\label{eq:r-trans-energy}
+ R(J) = \Delta \tilde{E}(\nu_f \leftarrow 0) + (\tilde{B_{\nu_f}} + \tilde{B_0}) (J + 1) + (\tilde{B_{\nu_f}} -
+ \tilde{B_0}) ( J + 1)^2
+\end{equation}
+
+Setting $J = 0$ hence gives equation \ref{eq:r-trans-energy-j0}.
+
+\begin{equation}\label{eq:r-trans-energy-j0}
+ R(J) - 2 \tilde{B_{\nu_f}} = \Delta \tilde{E}(\nu_f \leftarrow 0)
+\end{equation}
+
+Substituting equation \ref{eq:delta-e-coeff} into equation \ref{eq:r-trans-energy-j0} yields equation
+\ref{eq:r-trans-energy-sub}.
+
+\begin{equation}\label{eq:r-trans-energy-sub}
+ R(0) - 2 \tilde{B_{\nu_f}} = \tilde{\nu_e} \left( \nu_f - \left( \nu_f^2 + \nu_f \right) x_e \right)
+\end{equation}
+
+Now from equation \ref{eq:r-trans-energy-sub} a system of linear equations can be
+obtained by setting $\nu_f = 1$ and $\nu_f = 2$ (equations \ref{eq:coeff-1} and
+\ref{eq:coeff-2}).
+
+\begin{align}
+ \label{eq:coeff-1}
+ R(0) - 2 \tilde{B_1} &= \tilde{\nu_e} \left( 1 - 2 x_e \right) \\
+ \label{eq:coeff-2}
+ R(0) - 2 \tilde{B_2} &= \tilde{\nu_e} \left( 1 - 6 x_e \right)
+\end{align}
+
+Solving equations \ref{eq:coeff-1} and \ref{eq:coeff-2} for $\tilde{\nu_e}$ and
+$x_e$ then gives equations \ref{eq:nu-e} and \ref{eq:xe}.
+
\end{document}
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