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diff --git a/spectr.tex b/spectr.tex new file mode 100644 index 0000000..6ae7624 --- /dev/null +++ b/spectr.tex @@ -0,0 +1,629 @@ +\documentclass[./manual.tex]{subfiles} +\begin{document} + +\chapter{Atomic Spectroscopy} + + \textbf{Experiment objectives}: test a diffraction grating-based spectrometer, study the energy spectrum of atomic hydrogen (H) and a hydrogen-like atomic sodium (Na), determine values of quantum defects of low angular momentum states of Na, and measure fine splitting using Na yellow doublet. + +\subsection*{History} + + The observation of discrete lines in the emission spectra of + atomic gases gives insight into the quantum nature of + atoms. Classical electrodynamics cannot explain the existence + of these discrete lines, whose energy (or wavelengths) are + given by characteristic values for specific atoms. These + emission lines are so fundamental that they are used to + identify atomic elements in objects, such as in identifying + the constituents of stars in our universe. When Niels Bohr + postulated that electrons can exist only in orbits of discrete + energies, the explanation for the discrete atomic lines became + clear. In this laboratory you will measure the wavelengths of + the discrete emission lines from two elements - hydrogen and sodium - + to determine the energy levels in the hydrogen-like atoms. + +\section*{Hydrogen Spectrum} + + The hydrogen atom is the simplest atom: it consists of a single proton and a single +electron. Since it is so simple, it is possible to calculate the energy spectrum of the electron +bound states - in fact you will do that in your Quantum Mechanics course. From this calculations it +follows that the electron energy is quantized, and these energies are given by the expression: +\begin{equation}\label{Hlevels_inf} +E_n=- \frac{2\pi^2m_ee^4}{(4\pi\epsilon_0)^2h^2n^2} + = -hcRy\frac{1}{n^2} +\end{equation} +% +where $n$ is the {\bf principal quantum number} of the atomic level, and +\begin{equation} \label{Ry} +Ry=\frac{2\pi m_ee^4}{(4\pi\epsilon_0)^2ch^3} +\end{equation} +is a fundamental physical constant called the {\bf Rydberg constant} (here $m_e$ is the electron +mass). Numerically, $Ry = 1.0974 \times 10^5 cm^{-1}$ and $hcRy = 13.605 eV$. + +Because the allowed energies of an electron in a hydrogen atom, the electron can change its state +only by making a transition ("jump") from an one state of energy $E_1$ to another state of lower +energy $E_2$ by emitting a photon of energy $h\nu = E_1 - E_2$ that carries away the excess energy. +Thus, by exciting atoms into high-energy states using a discharge and then measuring the frequencies +of emission one can figure out the energy separation between various energy levels. Since it is +more convenient to use a wavelength of light $\lambda$ rather than its frequency $\nu$, and using the +standard connection between the wavelength and the frequency $h\nu = hc/\lambda$, we can write the +wavelength of a photon emitted by electron jumping between the initial state $n_1$ and the final +state $n_2$ as follows: +\begin{equation} \label{Hlines_inf} +\frac{1}{\lambda_{12}}=\frac{2\pi^2m_ee^4}{(4\pi\epsilon_0)^2ch^3} +\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right]= Ry \left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] +\end{equation} +%Based on this description it is clear that by measuring the frequencies (or +%wavelengths) of photons emitted by an excited atomic system, we can glean +%important information about allowed electron energies in atoms. + +%To make more accurate calculation of the Hydrogen spectrum, we need to take +%into account that a hydrogen nucleus has a large, but finite mass, M=AMp (mass +%number A=1 and Mp = mass of proton)\footnote{This might give you the notion +%that the mass of any nucleus of mass number $A$ is equal to $AM_p$. This is not +%very accurate, but it is a good first order approximation.} such that the +%electron and the nucleus orbit a common center of mass. For this two-mass +%system the reduced mass is given by $\mu=m_e/(1+m_e/AM_p)$. We can take this +%into account by modifying the above expression (\ref{Hlines_inf}) for +%1/$\lambda$ as follows: +%\begin{equation}\label{Hlines_arb} +%\frac{1}{\lambda_A}=R_AZ^2\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] \mbox{ +%where } R_A=\frac{R_{\infty}}{1+\frac{m_e}{AM_p}} +%\end{equation} +%In particular, for the hydrogen case of ($Z=1$; $M=M_p$) we have: +%\begin{equation}\label{Hlines_H} +%\frac{1}{\lambda_H}=R_H\left[\frac{1}{n_2^2}-\frac{1}{n_1^2}\right] +%\end{equation} +%Notice that the value of the Rydberg constant will change slightly for +%different elements. However, these corrections are small since nucleus is +%typically several orders of magnitude heavier then the electron. + + +Fig.~\ref{Hspec.fig} shows the energy levels of hydrogen, and indicates a large number of observed +transitions, grouped into series with the common electron final state. Emitted photon frequencies +span the spectrum from the UV (UltraViolet) to the IR (InfraRed). Among all the series only the +Balmer series, corresponding to $n_2$ = 2, has its transitions in visible part of the spectrum. +Moreover, from all the possible transition, we will be able to to observe and measure only the +following four lines: $n_1=6 \rightarrow 2$, $5 \rightarrow 2$, $4 \rightarrow 2$, + and $3 \rightarrow 2$. + +\begin{figure} +\includegraphics[width=0.7\linewidth]{./pdf_figs/spec} +\caption{\label{Hspec.fig}Spectrum of Hydrogen. The numbers on the left show the energies of the +hydrogen levels with different principle quantum numbers $n$ in $eV$. The wavelength of emitted +photon in {\AA} are shown next to each electron transition. } +\end{figure} + +%In this lab, the light from the hydrogen gas is broken up into its spectral +%components by a diffraction grating. You will measure the angle at which each +%line occurs on the left ($\theta_L$) and ($\theta_R$) right sides for as many +%diffraction orders $m$ as possible, and use Eq.(\ref{mlambda}) to calculate +%$\lambda$, using the following expression, derived in the Appendix. +%\begin{equation}\label{mlambda} +%m\lambda = \frac{h}{2}\left(\sin\theta_L+\sin\theta_R\right) +%\end{equation} +% Then the same +%expression will be used to check/calibrate the groove spacing $h$ by making +%similar measurements for a sodium spectral lines with known wavelengths. +% +%We will approach the data in this experiment both with an eye to confirming +% Bohr's theory and from Balmer's early perspective of someone +% trying to establish an integer power series linking the +% wavelength of these four lines. + +\section*{Sodium spectrum} +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 these elements consist of a closed +electron shell with one extra unbound electron. Not surprisingly, the energy level structure for this +free electron is very similar to that of hydrogen. For example, a Na atom has 11 electrons, and its +electronic configuration is $1s^22s^22p^63s$. 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 experiences the electric field potential similar to that of a hydrogen atom, given by the +Eq.~\ref{Hlevels_inf}. + +However, there is an important variation of the energy spectrum of alkali metals, related to the +electron angular momentum $l$. In hydrogen the energy levels with same principle quantum number $n$ +but different electron angular momentum $l=0, 1, \cdots (n-1)$ are degenerate. For Na and others the +levels with different values of $l$ are shifted with respect to each other. This is mainly due to the +interaction of the unpaired electrons with the electrons of the closed shells. For example, the +orbits of the electron with large angular momentum value $l$ is far above closed shell, and thus +their energies are basically the same as for the hydrogen atom. An 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). + + +To take onto account the modification of the atomic spectra while still using the same basic +equations as for the hydrogen, it is convenient to introduce a small correction $\Delta_l$ to the +principle quantum number $n$ to take into account the level shifts. This correction is often called a +{\it quantum defect}, and its value % an effective nuclei charge $Z^*$ keeping the 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{equation}\label{qdef} +E_{nl}=-\frac{hcRy}{(n-\Delta_l)^2}%=-\frac{hcRy}{(n-\Delta_l)^2} +\end{equation} + +%\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} +In particular, the energies of 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: +% +%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[height=\columnwidth]{./pdf_figs/nae} +\caption{\label{nae}Energy spectrum of Na. The energy states of H are shown in far right for +comparison.} +\end{figure} +%\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]{./pdf_figs/natrans} +\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*{Experimental setup} +\textbf{Equipment needed}: Gaertner-Peck optical spectrometer, hydrogen discharge lamp, sodium +discharge lamp. + +Fig. \ref{expspec} gives a top view of the Gaertner-Peck optical spectrometer used in this lab. The +spectrometer consists of two tubes. One tube holds a vertical collimator slit of variable width, and +should be directed toward a discharge lamp. The light from the discharge lamp passes through the +collimator slit and then get dispersed on the diffraction grating mounted in the mounting pedestal in +the middle of the apparatus. The other tube holds the telescope that allows you to magnify the image +of the collimator slit to get more accurate reading on the rotating table. This tube can be rotated +around the central point, so that you will be able to align and measure the wavelength of all visible +spectral lines of the lamp in all orders of diffraction grating. +\begin{figure} +\includegraphics[height=4in]{./pdf_figs/expspec} +\caption{\label{expspec}Gaertner-Peck Spectrometer} +\end{figure} + +It is likely that you will find a spectrometer at nearly aligned condition at the beginning on the +lab. Nevertheless, take time making sure that all elements are in order to insure good quality of +your data. + +\textit{Telescope Alignment:} Start by adjusting the telescope eyepiece in or out to bring the +crosshairs into sharp focus. Next aim the telescope out the window to view a distant object such as +leaves in a tree. If the distant object is not in focus, you will have to adjust the position of the +telescope tube - ask your instructor for directions. + +\textit{Collimator Conditions:} Swing the telescope to view the collimator which is accepting light +from the hydrogen discharge tube through a vertical slit of variable width. The slit opening should +be set to about 5-10 times the crosshair width to permit sufficient light to see the faint violet +line and to be able to see the crosshairs. If the bright column of light is not in sharp focus, you +should align the position of the telescope tube again (with the help of the instructor). + +\textit{Diffraction Grating Conditions:} In this experiment you will be using a diffraction grating +that has 600 lines per mm. A brief summary of diffraction grating operation is given in the Appendix +of this manual. If the grating is not already in place, put it back to the baseclamp and fix it +there. The table plate that holds the grating can be rotated, so try to orient the grating surface to +be maximally perpendicular to the collimator axis. However, the accurate measurement of angle does +not require the perfect grating alignment. Instead, for each spectral line in each diffraction order +you will be measuring the angles on the left ($\theta_l$) and on the right ($\theta_r$), and use both +of the measurements to figure out the optical wavelength using the following equation: +\begin{equation}\label{nlambda} +m\lambda=\frac{d}{2}(\sin\theta_r+\sin\theta_l), +\end{equation} +where $m$ is the diffraction order, and $d$ is the distance between the lines in the grating. + + % Use +% the AUTOCOLLIMATION procedure to achieve a fairly accurate +% alignment of the grating surface. This will determine how to +% adjust the three leveling screws H1, H2, and H3 and the +% rotation angle set screw for the grating table. +% +% \textbf{AUTOCOLLIMATION} is a sensitive way to align an optical +% element. First, mount a ``cross slit'' across the objective lens of +% the collimator, and direct a strong light source into the +% input end of the collimator. Some of the light exiting through +% the cross slit will reflect from the grating and return to the +% cross slit. The grating can then be manipulated till this +% reflected light retraces its path through the cross slit +% opening. With this the grating surface is normal to the +% collimator light. + Then, with the hydrogen tube ON and in place at + the collimator slit, swing the rotating telescope slowly + through 90 degrees both on the left \& right sides of the forward + direction. You should observe diffraction maxima for most + spectral wavelength, $\lambda$, in 1st, 2nd, and 3rd order. If these + lines seem to climb uphill or drop downhill + the grating has to be adjusted in its baseclamp to + bring them all to the same elevation. + + Also, turn on the sodium lamp as soon as you arrive, since it requires 10-15 minutes to warm up + and show clear Na spectrum. + +\subsection*{Experimental studies of Hydrogen Balmer line} + +Swing the rotating telescope slowly and determine which spectral lines from Balmer series you +observe. You should be able to see three bright lines - Blue, Green and Red - in the first (m=1) and +second (m=2) diffraction orders on both left \& right sides. In the third order (m=3) only the Blue, +\& Green lines are visible, and you will not see the Red. + +One more line of the Balmer series is in the visible range - Violet, but its intensity is much lower +than for the other three line. However, you will be able to find it in the first order if you look +carefully with the collimator slit open a little wider. +% +%\emph{Lines to be measured:} +%\begin{itemize} +%\item \emph{Zero order} (m=0): All spectral lines merge. +%\item \emph{First order} (m=1): Violet, Blue, Green, \& Red on both left \& +% right sides. +%\item \emph{Second order} (m=2): Violet, Blue, Green, \& Red on +% both left \& right sides. +%\item \emph{Third order} (m=3): Blue, \& Green. +%\end{itemize} +% You might not see the Violet line due to its low +% intensity. Red will not be seen in 3rd order. + +After locating all the lines, measure the angles at which each line occurs. The spectrometer reading for each line should be measured at least \emph{twice} by \textit{different} lab partners to avoid systematic errors. \textbf{Don't forget}: for every line you need to measure the angles to the right and to the left! + +You should be able to determine the angle with accuracy of $1$ minute, but you should know how to +read angles with high precision in the spectrometer: first use the bottom scale to get the rough +angle reading within a half of the degree. Then use the upper scale for more accurate reading in +minutes. To get this reading find a tick mark of the upper scale that aligns perfectly with some tick +mark of the bottom scale - this is your minute reading. Total angle is the sum of two readings. + +To measure the frequency precisely center the crosshairs on the line as accurately as possible. +Choose the width of lines by turning the collimator slit adjustment screw. If the slit is too wide, +it is hard to judge the center of the line accurately; if the slit is too narrow, then not enough +light is available to see the crosshairs. For Violet the intensity is noticeably less than for the +other three lines, so you may not see it. However, even if you find it, a little assistance is +usually required in order to locate the crosshairs at this line. We suggest that a low intensity +flashlight be aimed toward the Telescope input, and switched ON and OFF repeatedly to reveal the +location of the vertical crosshair relative to the faint Violet line. + +%\subsubsection*{ Calibration of Diffraction Grating:} Replace the hydrogen tube with +% a sodium (Na) lamp and take readings for the following two +% lines from sodium: $568.27$~nm (green) and $589.90$~nm (yellow). Extract from +% these readings the best average value for $h$ the groove +% spacing in the diffraction grating. Compare to the statement +% that the grating has 600 lines per mm. Try using your measured value +% for $h$ versus the stated value $600$ lines per mm in +% the diffraction formula when obtaining the measured +% wavelengths of hydrogen. Determine which one provide more accurate results, and discuss the conclusion. + +\subsubsection*{ Data analysis for Hydrogen Data} +Calculate the wavelength $\lambda$ for each line observed in all orders, calculate the average +wavelength value and uncertainty for each line, and then identify the initial and final electron +states principle numbers ($n_1$ and $n_2$) for each line using Fig.~\ref{Hspec.fig}. + +Make a plot of $1/\lambda$ vs $1/n_1^2$ where $n_1$ = the principal quantum number of the electron's +initial state. Put all $\lambda$ values you measure above on this plot. You data point should form a +straight line. From Equation~(\ref{Hlines_inf}) determine the physical meaning of both slope and +intercept, and compare the data from the fit to the expected values for each of them. The slope +should be the Rydberg constant for hydrogen, $Ry$. The intercept is $Ry/(n_2)^2$. From this, +determine the value for the principal quantum number $n_2$. Compare to the accepted value in the +Balmer series. + +\subsection*{Experimental studies of Sodium Spectrum} + +Switch to Sodium lamp and make sure the lamp warms up for approximately 5-10 minutes before starting +the measurements. + +Double check that you see a sharp spectrum in the spectrometer (adjust the width of the collimator +slit if necessary). In the beginning it will be very useful for each lab partner to quickly scan the +spectrometer telescope through all first-order lines, and then discuss which line you see corresponds +to with transition in Table~\ref{tab:sodium} and Fig.~\ref{natrns}. Keep in mind that the color +names are symbolic rather than descriptive! + +After that carefully measure the left and right angles for as many spectral lines in the first and orders +as possible. The spectrometer reading for each line should be measured at least \emph{twice} by +\textit{different} lab partners to avoid systematic errors. + +Determine the wavelengths of all measured Na spectral lines using Eq. \ref{nlambda}. Compare these +measured mean wavelengths to the accepted values given in Fig.~\ref{natrns} and in the table below. +Identify at least seven of the lines with a particular transition, e.g. $\lambda = 4494.3${\AA} +corresponds to $8d \rightarrow 3p$ transition. + +\begin{table} + +\centering +\begin{tabular}{l|l|l} + Color&Line$_1$(\AA)&Line$_2$(\AA)\\ \hline +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} +\caption{\label{tab:sodium}Wavelength of the visible sodium lines.} +\end{table} +Line$_1$ and Line$_2$ correspond 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. + +\textbf{Measurements of the fine structure splitting}. Once you measured all visible spectral lines, +go back to some bright line (yellow should work well), and close the collimator slit such that you +can barely see any light going through. In this case you should be able to see the splitting of the +line because of the \emph{fine structure splitting} of states $P_{1/2}$ and $P_{3/2}$ . For the Na D +doublet the splitting between two lines $\Delta\lambda=\lambda_{3/2}-\lambda_{1/2}$. + +Measure the splitting between two lines in the first and the second order. Which one works better? +Discuss this issue in your lab report. Compare to the accepted value: $\Delta\lambda=5.9$~\AA. +Compare this approach to the use of the Fabry-Perot interferometer. + + +\subsection*{Data analysis for Sodium} + +\textbf{Calculation of a quantum defect for $n=3, p$ state. } + +Once you identified all of your measured spectral lines, choose only those that correspond to optical +transitions from any $d$ 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_d^2}+\frac{hcRy}{(3-\Delta_p)^2}, +\end{equation} +or +\begin{equation} +\frac{1}{\lambda}=\frac{Ry}{(3-\Delta_p)^2}-\frac{Ry}{n_d^2}, +\end{equation} + where $n_d$ is the principle number of the initial $d$ state. To verify this +expression by plotting $1/\lambda$ versus $1/n_d^2$ for the $n_d$= 4,5, and 6. From the slope of this +curve determine the value of the Rydberg constant $Ry$. The value of the intercept in this case is +$\frac{Ry}{(3-\Delta_p)^2}$, so use it to find the quantum defect $\Delta_p$. + +Compare the results of your calculations for the quantum defect with the accepted value +$\Delta_p=0.86$. + +%\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 $n$? +% + +%\textbf{Example data table for writing the results of the measurements}: +% +%\noindent +%\begin{tabular}{|p{1.in}|p{1.in}|p{1.in}|p{1.in}|} +%\hline +% Line &$\theta_L$&$\theta_R$&Calculated $\lambda$ \\ \hline +% m=1 Violet&&&\\ \hline +% m=1 Blue&&&\\ \hline +% m=1 Green&&&\\ \hline +% m=1 Red&&&\\ \hline +% m=2 Violet&&&\\ \hline +% \dots&&&\\ \hline +% m=3 Blue&&&\\ \hline +% \dots&&&\\\hline +%\end{tabular} + +\section*{Appendix: Operation of a diffraction grating-based optical spectrometer} + +%\subsection*{Fraunhofer Diffraction at a Single Slit} +%Let's consider a plane electromagnetic wave incident on a vertical slit of +%width $D$ as shown in Fig. \ref{frn}. \emph{Fraunhofer} diffraction is +%calculated in the far-field limit, i.e. the screen is assume to be far away +%from the slit; in this case the light beams passed through different parts of +%the slit are nearly parallel, and one needs a lens to bring them together and +%see interference. +%\begin{figure}[h] +%\includegraphics[width=0.7\linewidth]{frnhfr.eps} +%\caption{\label{frn}Single Slit Fraunhofer Diffraction} +%\end{figure} +%To calculate the total intensity on the screen we need to sum the contributions +%from different parts of the slit, taking into account phase difference acquired +%by light waves that traveled different distance to the lens. If this phase +%difference is large enough we will see an interference pattern. Let's break the +%total hight of the slit by very large number of point-like radiators with +%length $dx$, and we denote $x$ the hight of each radiator above the center of +%the slit (see Fig.~\ref{frn}). If we assume that the original incident wave is +%the real part of $E(z,t)=E_0e^{ikz-i2\pi\nu t}$, where $k=2\pi/\lambda$ is the +%wave number. Then the amplitude of each point radiator on a slit is +%$dE(z,t)=E_0e^{ikz-i2\pi\nu t}dx$. If the beam originates at a hight $x$ above +%the center of the slit then the beam must travel an extra distance $x\sin +%\theta$ to reach the plane of the lens. Then we may write a contributions at +%$P$ from a point radiator $dx$ as the real part of: +%\begin{equation} +%dE_P(z,t,x)=E_0e^{ikz-i2\pi\nu t}e^{ikx\sin\theta}dx. +%\end{equation} +%To find the overall amplitude one sums along the slit we need to add up the +%contributions from all point sources: +%\begin{equation} +%E_P=\int_{-D/2}^{D/2}dE(z,t)=E_0e^{ikz-i2\pi\nu +%t}\int_{-D/2}^{D/2}e^{ikx\sin\theta}dx = A_P e^{ikz-i2\pi\nu t}. +%\end{equation} +%Here $A_P$ is the overall amplitude of the electromagnetic field at the point +%$P$. After evaluating the integral we find that +%\begin{equation} +%A_P=\frac{1}{ik\sin\theta}\cdot +%\left(e^{ik\frac{D}{2}\sin\theta}-e^{-ik\frac{D}{2}\sin\theta}\right) +%\end{equation} +%After taking real part and choosing appropriate overall constant multiplying +%factors the amplitude of the electromagnetic field at the point $P$ is: +%\begin{equation} +%A=\frac{\sin (\frac{\pi D}{\lambda}\sin\theta)}{\frac{\pi +%D}{\lambda}\sin\theta} +%\end{equation} +%The intensity is proportional to the square of the amplitude and thus +%\begin{equation} +%I_P=\frac{(\sin (\frac{\pi D}{\lambda}\sin\theta))^2}{(\frac{\pi +%D}{\lambda}\sin\theta)^2} +%\end{equation} +%The minima of the intensity occur at the zeros of the argument of the sin. The +%maxima are near, but not exactly equal to the solution of: +%\begin{equation} +% (\frac{\pi D}{\lambda}\sin\theta)=(m+\frac{1}{2})\pi \end{equation} +%for integer $m$. +% +%The overall pattern looks like that shown in Fig. \ref{sinxox}. +%\begin{figure} +%\includegraphics[width=\linewidth]{sinxox.eps} +%\caption{\label{sinxox}Intensity Pattern for Fraunhofer Diffraction} +%\end{figure} + +%\subsection*{The Diffraction Grating} +A diffraction grating is a common optical element, which consists of a pattern +with many equidistant slits or grooves. Interference of multiple beams passing +through the slits (or reflecting off the grooves) produces sharp intensity +maxima in the output intensity distribution, which can be used to separate +different spectral components on the incoming light. In this sense the name +``diffraction grating'' is somewhat misleading, since we are used to talk about +diffraction with regard to the modification of light intensity distribution to +finite size of a single aperture. +\begin{figure}[h] +\includegraphics[width=\linewidth]{./pdf_figs/grating} +\caption{\label{grating}Intensity Pattern for Fraunhofer Diffraction} +\end{figure} + +To describe the properties of a light wave after passing through the grating, +let us first consider the case of 2 identical slits separated by the distance +$h$, as shown in Fig.~\ref{grating}a. We will assume that the size of the slits +is much smaller than the distance between them, so that the effect of +Fraunhofer diffraction on each individual slit is negligible. Then the +resulting intensity distribution on the screen is given my familiar Young +formula: +\begin{equation} \label{2slit_noDif} +I(\theta)=\left|E_0 +E_0e^{ikh\sin\theta} \right|^2 = 4I_0\cos^2\left(\frac{\pi +h}{\lambda}\sin\theta \right), +\end{equation} +where $k=2\pi/\lambda$, $I_0$ = $|E_0|^2$, and the angle $\theta$ is measured +with respect to the normal to the plane containing the slits. +%If we now include the Fraunhofer diffraction on each slit +%same way as we did it in the previous section, Eq.(\ref{2slit_noDif}) becomes: +%\begin{equation} \label{2slit_wDif} +%I(\theta)=4I_0\cos^2\left(\frac{\pi h}{\lambda}\sin\theta +%\right)\left[\frac{\sin (\frac{\pi D}{\lambda}\sin\theta)}{\frac{\pi +%D}{\lambda}\sin\theta} \right]^2. +%\end{equation} + +An interference on $N$ equidistant slits illuminated by a plane wave +(Fig.~\ref{grating}b) produces much sharper maxima. To find light intensity on +a screen, the contributions from all N slits must be summarized taking into +account their acquired phase difference, so that the optical field intensity +distribution becomes: +\begin{equation} \label{Nslit_wDif} +I(\theta)=\left|E_0 ++E_0e^{ikh\sin\theta}+E_0e^{2ikh\sin\theta}+\dots+E_0e^{(N-1)ikh\sin\theta} +\right|^2 = I_0\left[\frac{sin\left(N\frac{\pi +h}{\lambda}\sin\theta\right)}{sin\left(\frac{\pi h}{\lambda}\sin\theta\right)} +\right]^2. +\end{equation} + Here we again neglect the diffraction form each individual slit, assuming that the + size of the slit is much smaller than the separation $h$ between the slits. + +The intensity distributions from a diffraction grating with illuminated + $N=2,5$ and $10$ slits are shown in Fig.~\ref{grating}c. The tallest (\emph{principle}) maxima occur when the denominator + of Eq.(~\ref{Nslit_wDif}) becomes zero: $h\sin\theta=\pm m\lambda$ where + $m=1,2,3,\dots$ is the diffraction order. The heights of the principle maxima are + $I_{\mathrm{max}}=N^2I_0$, and their widths are $\Delta \theta = + 2\lambda/(Nh)$. + Notice that the more slits are illuminated, the narrower diffraction peaks + are, and the better the resolution of the system is: + \begin{equation} +\frac{ \Delta\lambda}{\lambda}=\frac{\Delta\theta}{\theta} \simeq \frac{1}{Nm} +\end{equation} +For that reason in any spectroscopic equipment a light beam is usually expanded +to cover the maximum surface of a diffraction grating. + +\subsection*{Diffraction Grating Equation when the Incident Rays are +not Normal} + +Up to now we assumed that the incident optical wavefront is normal to the pane of a grating. Let's +now consider more general case when the angle of incidence $\theta_i$ of the incoming wave is +different from the normal to the grating, as shown in Fig. \ref{DGnotnormal}(a). Rather then +calculating the whole intensity distribution, we will determine the positions of principle maxima. +The path length difference between two rays 1 and 2 passing through the consequential slits us $a+b$, +where: +\begin{equation} +a=h\sin \theta_i;\,\, b=h\sin \theta_R +\end{equation} +Constructive interference occurs for order $m$ when $a+b=m\lambda$, or: +\begin{equation} +h\sin \theta_i + h\sin\theta_R=m\lambda +\end{equation} + +\begin{figure} +\includegraphics[width=\columnwidth]{./pdf_figs/pic4i} +%\includegraphics[height=3in]{dn.eps} +\caption{\label{DGnotnormal}Diagram of the light beams diffracted to the right +(a) and to the left (b).} +\end{figure} + +Now consider the case shown in Fig. \ref{DGnotnormal}(b). The path length between two beams is now +$b-a$ where $b=h\sin\theta_L$. Combining both cases we have: +\begin{eqnarray} \label{angles} +h\sin\theta_L-h\sin\theta_i&=&m\lambda\\ +h\sin\theta_R+h\sin\theta_i&=&m\lambda \nonumber +\end{eqnarray} +Adding these equations and dividing the result by 2 yields the following expression connecting the +right and left diffraction angles: +\begin{equation}m\lambda=\frac{h}{2}\left(\sin\theta_L+\sin\theta_R\right) +\end{equation} + +\end{document} + |