typos fixed, thanks to Jordan

1 files changed, 30 insertions, 26 deletions

diff --git a/spectr.tex b/spectr.tex
index 8b08b11..78e6d64 100644
--- a/spectr.tex
+++ b/spectr.tex
@@ -25,7 +25,8 @@
 
     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
+bound states - in fact you will do that in your Quantum Mechanics course.
+From these 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}
@@ -39,7 +40,7 @@ where  $n$  is the  {\bf principal quantum number} of the atomic level, and
 is a fundamental physical constant called the {\bf Rydberg constant} (here $m_e$ is the electron
 mass).  Numerically, ${R_y} = 1.0974 \times  10^5 cm^{-1}$ and $hc{R_y} = 13.605 eV$.
 
-Because the allowed energies of an electron in a hydrogen atom, the electron can change its state
+Because of the allowed energies of an electron in a hydrogen atom, the electron can change its state
 only by making a transition (``jump'') from 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
@@ -81,8 +82,8 @@ state $n_2$ as follows:
 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
+Balmer series, corresponding to $n_2$ = 2, has its transitions in the visible part of the spectrum.
+Moreover, from all the possible transitions, 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 $3 \rightarrow 2$.
 
@@ -127,8 +128,8 @@ levels with different values of $l$ are shifted with respect to each other. This
 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
+more time closer to the nucleus, and ``feels'' stronger bounding electrostatic potential. As a result
+the corresponding energy levels are pulled down compared 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).
 
@@ -147,10 +148,10 @@ principle quantum number $n$ to take into account the level shifts. This correct
 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
+description for atomic spectra had not been developed yet. Back then spectroscopists had categorized
 atomic spectral lines corresponding to their appearance: 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
+transitions from $s$-orbital ($l=0$) always appeared \textbf{S}harp on a photographic film, while
+those with initial electron states of $d$-orbital ( $l=2$) always appeared \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
@@ -200,7 +201,7 @@ comparison.}
 
 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 $1$,
+However, not all electronic transitions 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}
@@ -222,7 +223,7 @@ $P_{3/2}$. The small difference between their energy levels is the ``fine struct
 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
+the 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
@@ -235,9 +236,9 @@ 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
+collimator slit and then gets 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
+of the collimator slit to get a 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}
@@ -262,7 +263,7 @@ should align the position of the telescope tube again (with the help of the inst
 
 \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
+of this manual. If the grating is not already in place, put it back on 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
@@ -337,7 +338,7 @@ angle reading within a half of the degree. Then use the upper scale for more acc
 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.
+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
@@ -380,12 +381,14 @@ spectrometer telescope through all first-order lines, and then discuss which lin
 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
+After that, carefully measure the left and right angles for as many
+spectral lines in the first and second 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.
+measured mean wavelengths to the accepted values given in Fig.~\ref{natrns}
+and in the table~\ref{tab:sodium}.
 Identify  at least seven of the lines with a particular transition, e.g. $\lambda = 4494.3${\AA}
 corresponds to $8d \rightarrow 3p$ transition.
 
@@ -405,7 +408,7 @@ 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
+Line$_1$ and Line$_2$ correspond to transitions to two fine-splitted $3p$ 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
@@ -418,7 +421,7 @@ can barely see any light going through. In this case you should be able to see t
 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?
+Measure the splitting between two lines in the first and the second orders. 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.
 
@@ -437,7 +440,7 @@ or
 \begin{equation}
 \frac{1}{\lambda}=\frac{{R_y}}{(3-\Delta_p)^2}-\frac{{R_y}}{n_d^2},
 \end{equation}
- where $n_d$ is the principle number of the initial $d$ state. To verify this
+ where $n_d$ is the principle number of the initial $d$ state. 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 ${R_y}$. The value of the intercept in this case is
 $\frac{{R_y}}{(3-\Delta_p)^2}$, so use it to find the quantum defect $\Delta_p$.
@@ -538,9 +541,10 @@ 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 grating'' is somewhat misleading, since we are used to
+talking about
 diffraction with regard to the modification of light intensity distribution  to
-finite size of a single aperture.
+the finite size of a single aperture.
 \begin{figure}[h]
 \includegraphics[width=\linewidth]{./pdf_figs/grating}
 \caption{\label{grating}Intensity Pattern for Fraunhofer Diffraction}
@@ -551,7 +555,7 @@ 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
+resulting intensity distribution on the screen is given by the 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
@@ -579,7 +583,7 @@ I(\theta)=\left|E_0
 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
+ Here we again neglect the diffraction from 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
@@ -599,11 +603,11 @@ 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
+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 $a+b$,
+The path length difference between two rays 1 and 2 passing through the consequential slits is $a+b$,
 where:
 \begin{equation}
 a=h\sin \theta_i;\,\, b=h\sin \theta_R