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+%\chapter*{Spectrum of Sodium }
+%\addcontentsline{toc}{chapter}{Spectrum of Sodium}
+
+\documentclass{article}
+\usepackage{tabularx,amsmath,boxedminipage,epsfig}
+    \oddsidemargin  0.0in
+    \evensidemargin 0.0in
+    \textwidth      6.5in
+    \headheight     0.0in
+    \topmargin      0.0in
+    \textheight=9.0in
+
+\begin{document}
+\title{Spectrum of Sodium}
+\date {}
+\maketitle \noindent
+ \textbf{Experiment objectives}: measure the energy spectrum of sodium (Na),
+ determine values of quantum defects of low angular momentum states, and measure fine splitting
+using Na yellow doublet.
+
+\section*{Theory}
+Sodium (Na) belongs to the chemical group of \emph{alkali metals}, together
+with lithium (Li), potassium (K), rubidium (Rb), cesium (Cs) and Francium (Fr).
+All elements of this group have a closed electron shell with one extra unbound
+electron. This makes energy level structure for this free electron to be very
+similar to that of hydrogen, as shown in Fig.~\ref{nae}.
+
+For example, a Na atom has 11 electrons, and its electronic configuration is
+$1s^22s^22p^63s$, as determined from the Pauli exclusion principle.  Ten
+closed-shell electrons effectively screen the nuclear charge number ($Z=11$) to
+an effective charge $Z^*\approx 1$, so that the $3s$ valent electron experience
+the electric field potential similar to that of a hydrogen atom. As a result,
+the electron spectrum of all alkali metal atoms is quite similar to that of
+hydrogen:
+\begin{equation}\label{Hlevels_Naexp}
+E_n=-hcRy\frac{1}{n^2}
+\end{equation}
+%
+where $n$ is the principle quantum number, and $Ry=\frac{2\pi
+m_ee^4}{(4\pi\epsilon_0)^2ch^3}$   is  the  Rydberg constant ($Ry = 1.0974
+\times 10^5 cm^{-1}$ and $hcRy = 13.605 eV$). For each particular value of
+angular momentum $l$ the energy spectrum follows the same scaling as hydrogen
+atom. However, the absolute values of energies obey Eq.(\ref{Hlevels_Naexp})
+only for electron energy states with orbits far above closed shell - the ones
+with large value of an angular momentum $l$. Electron with smaller $l$ spends
+more time closer to the nuclear, and ``feels'' stronger bounding electrostatic
+potential. As a result the corresponding energy levels are pulled down compare
+to those of hydrogen, and the states with the same principle number $n$ but
+different angular momenta $l$ are split (\emph{i.e.} have different energies).
+\begin{figure}
+\includegraphics[height=\columnwidth]{nae.eps}
+\caption{\label{nae}Energy spectrum of Na. The energy states of H are shown in
+far right for comparison.}
+\end{figure}
+
+For each particular value of angular momentum $l$ the energy spectrum follows
+the same scaling as hydrogen atom, but with an effective charge $Z^*$:
+\begin{equation}\label{heq}
+E_n=-\frac{1}{2}\frac{Z^{*2}e^4}{(4\pi\epsilon_0)^2}\frac{mc^2}{\hbar^2c^2}
+\frac{1}{n^2}=-Z^{*2}\frac{hcRy}{n^2}
+\end{equation}
+The value of the effective charge $Z^*$ depends on the angular momentum $l$,
+and does not vary much between states with different principle quantum numbers
+$n$ but same $l$\footnote{The accepted notation for different electron angular
+momentum states is historical, and originates from the days when the proper
+quantum mechanical description for atomic spectra has not been developed yet.
+Back then spectroscopists had categorized atomic spectral lines corresponding
+to their appearend: for example any spectral lines from electron transitions
+from $s$-orbital ($l=0$) appeared always \textbf{S}harp on a photographic film,
+while those with initial electron states of $d$-orbital ( $l=2$) appeared
+always \textbf{D}iffuse. Also spectral lines of \textbf{P}rinciple  series
+(initial state is $p$-orbital, $l=1$) reproduced the hydrogen spectrum most
+accurately (even though at shifted frequencies), while the \textbf{F}undamental
+(initial electron state is $f$-orbital, $l=3$) series matched the absolute
+energies of the hydrogen most precisely. The orbitals with higher value of the
+angular momentum are denoted in an alphabetic order ($g$,
+$h$, \textit{etc}.) }:\\
+\begin{tabular}{ll}
+States&$Z^*$\\
+s~($l=0$)&$\approx$ 11/9.6\\
+p~($l=1$)&$\approx$ 11/10.1\\
+d~($l=2$)&$\approx$ 1\\
+f~($l=3$)&$\approx $ 1\\
+\end{tabular}
+\\These numbers mean that two states with the lowest angular momentum ($s$ and
+$p$) are noticeably affected by the more complicated electron structure of Na,
+while the energy levels of the states with the higher values of angular
+momentum ($d$, $f$) are identical to the hydrogen energy spectrum.
+
+An alternative (but equivalent) procedure is to assign a {\it quantum defect}
+to the principle quantum  $n$ instead of introducing an effective nuclei
+charge. In this case Eq.(\ref{heq}) can be written as:
+\begin{equation}\label{qdef}
+E_n=-\frac{hcRy}{(n^*)^2}=-\frac{hcRy}{(n-\Delta_l)^2}
+\end{equation}
+where $n*=n-\Delta_l$, and $\Delta_l$  is the corresponding quantum defect.
+Fig. \ref{nadell} shows values of quantum defects which work approximately for
+the alkalis. One sees that there is one value for each value of the angular
+momentum $l$. This is not exactly true for all alkali metals, but for Na there
+is very little variation in $\Delta_l$ with $n$ for a given $l$.
+
+\begin{figure}
+\includegraphics[width=0.5\columnwidth]{nadell.eps}
+\caption{\label{nadell}Quantum Defect $\Delta_l$ versus $l$ for different
+alkali metals. Taken from Condon and Shortley p. 143}
+\end{figure}
+%\begin{figure}
+%\includegraphics[height=3in]{nadel.eps}
+%\caption{\label{nadel}Quantum Defect $\Delta_l$ variation with $n$. The
+%difference between the quantum defect of each term and that of the lowest term
+%of the series to which it belongs is plotted against the difference between
+%the total quantum numbers of these terms. Again from Condon and Shortley p. 144.}
+%\end{figure}
+
+The spectrum of Na is shown in Fig. \ref{nae}. One can immediately see that
+there are many more optical transitions because of the lifted degeneracy of
+energy states with different angular momenta. However, not all electronic
+transition are allowed: since the angular momentum of a photon is $1$, then the
+electron angular momentum cannot change by more than one while emitting one
+spontaneous photon. Thus, it is important to remember the following
+\emph{selection rule} for atomic transitions:
+\begin{equation}\label{selrules}
+\Delta l = \pm 1.
+\end{equation}
+According to that rule, only transitions between two ``adjacent'' series are
+possible: for example $p \rightarrow s$ or $d \rightarrow p$ are allowed, while
+$s \rightarrow s$ or $s \rightarrow d$ are forbidden.
+
+The strongest allowed optical transitions are shown in Fig. \ref{natrns}.
+\begin{figure}
+\includegraphics[height=\columnwidth]{natrans.eps}
+\caption{\label{natrns}Transitions for Na. The wavelengths of selected
+transition are shown in {\AA}. Note, that $p$ state is now shown in two
+columns, one referred to as $P_{1/2}$ and the other as $P_{3/2}$. The small
+difference between their energy levels is the ``fine structure''.}
+\end{figure}
+%\begin{figure}
+%\includegraphics[height=4in]{series.eps}
+%\caption{\label{series}Series for Hydrogen, Alkalis are similar.}
+%\end{figure}
+Note that each level for given $n$ and $l$ is split into two because of the
+\emph{fine structure splitting}. This splitting is due to the effect of
+electron \emph{spin} and its coupling with the angular momentum. Proper
+treatment of spin requires knowledge of quantum electrodynamics and solving
+Dirac equation; for now spin can be treated as an additional quantum number
+associated with any particle. The spin of electron is $1/2$, and it may be
+oriented either along or against the non-zero electron's angular momentum.
+Because of the weak coupling between the angular momentum and spin, these two
+possible orientation results in small difference in energy for corresponding
+electron states.
+
+\section*{Procedure and Data Analysis}
+Align a diffraction-grating based spectrometer as described in ``Atomic
+Spectroscopy of Hydrogen Atoms'' experimental procedure.
+
+Then determine the left and right angles for as many spectral lines  and
+diffraction orders as possible. Each lab partner should measure the postilions
+of all lines at least once.
+
+Reduce the data using Eq. \ref{nlambda} to determine wavelengths for each
+spectral line (here $m$ is the order number):
+\begin{equation}\label{nlambda}
+m\lambda=\frac{d}{2}(\sin\theta_r+\sin\theta_l)
+\end{equation}
+Determine  the wavelengths of eight Na spectral lines measured in both first
+and second order. Combining first and second order results obtain the mean and
+standard deviation (error) of the mean value of the wavelength for each line.
+Compare these measured mean wavelengths to the accepted values given in
+Fig.~\ref{natrns} and in the table below:
+
+\begin{tabular}{lll}
+  Color&Line$_1$(\AA)&Line$_2$(\AA)\\
+Red&6154.3&6160.7\\
+Yellow & 5890.0&5895.9\\
+Green & 5682.7&5688.2\\
+&5149.1&5153.6\\
+& 4978.6&4982.9\\
+Blue&4748.0&4751.9\\
+&4664.9&4668.6\\
+Blue-Violet&4494.3&4497.7\\
+\end{tabular}
+
+Line$_1$ and Line$_2$ corresponds to transitions to two fine-spitted $3p$
+states $P_{1/2}$ and $P_{3/2}$. These two transition frequencies  are very
+close to each other, and to resolve them with the spectrometer the width of the
+slit should be very narrow. However, you may not be able to see some weaker
+lines then. In this case you should open the slit wider to let more light in
+when searching for a line. If you can see a spectral line but cannot resolve
+the doublet, record the reading for the center of the spectrometer line, and
+use the average of two wavelengthes given above.
+
+ Identify  at least seven of the lines with a particular transition, e.g.
+$\lambda = 4494.3${\AA} corresponds to $8d \rightarrow 3p$ transition.
+
+\subsection*{Calculation of a quantum defect for $n=3, p$ state}
+Identify spectral lines which corresponds to optical transitions from $d$ to
+$n=3,p$ states. Since the energy states of $d$ series follows the hydrogen
+spectra almost exactly, the wavelength of emitted light $\lambda$ is given by:
+\begin{equation}
+\frac{hc}{\lambda}=E_{nd}-E_{3p}=-\frac{hcRy}{n^2}+\frac{hcRy}{(3-\Delta_p)^2},
+\end{equation}
+or
+\begin{equation}
+\frac{1}{\lambda}=\frac{Ry}{(3-\Delta_p)^2}-\frac{Ry}{n^2},
+\end{equation}
+ where $n$ is the principle number of the initial $d$ state. To verify this
+expression by plotting $1/\lambda$ versus $1/n^2$ for the $n$= 4,5, and 6. From
+the slope of this curve determine the value of the Rydberg constant $Ry$. From
+the intercept determine the energy $E_{3p}$ of the $n=3,p$ state, and calculate
+its quantum defect $\Delta_p$.
+\subsection*{Calculation of a quantum defect for $s$ states}
+Now consider the transition from the $s$-states ($n=5,6,7$) to to the $n=3, p$
+state. Using $hc/\lambda=E_{ns}-E_{3p}$ and the results of your previous
+calculations, determine the energies $E_{sn}$ for different $s$ states with
+$n=5,6,7$  and calculate $\Delta_s$. Does the value of the quantum defect
+depends on $n$?
+
+Compare the results of your calculations for the quantum defects $\Delta_s$ and
+$\Delta_p$ with the accepted values given in Fig. \ref{nadell}.
+
+\subsection*{Calculations of fine structure splitting}
+For the Na D doublet measure the splitting between two lines
+$\Delta\lambda=\lambda_{3/2}-\lambda_{1/2}$ in the second diffraction order
+(why the second order is better than the first one?). Compare to the accepted
+value: $\Delta\lambda=$5.9\AA . Compare this approach to the use of the
+Fabry-Perot interferometer.
+
+\end{document}
+\newpage