From b606cc565f91c5ba2176893957a1cd7e86d728bc Mon Sep 17 00:00:00 2001
From: "Eugeniy E. Mikhailov" <evgmik@gmail.com>
Date: Wed, 28 Aug 2019 16:37:37 -0400
Subject: Synchronized manual with one of Justin

Updated source was located at
located at https://bitbucket.org/jrsteven/phys251_manual/src/master/
---
 ediffract.tex | 30 +++++++++++++++---------------
 1 file changed, 15 insertions(+), 15 deletions(-)

(limited to 'ediffract.tex')

diff --git a/ediffract.tex b/ediffract.tex
index e1da926..8ace7bf 100644
--- a/ediffract.tex
+++ b/ediffract.tex
@@ -35,7 +35,7 @@
 
 \begin{figure}[h]
 \centering
-\includegraphics[width=\textwidth]{./pdf_figs/ed1} \caption{\label{ed1}Electron
+\includegraphics[width=\textwidth]{./pdf_figs/ed1_new} \caption{\label{ed1}Electron
 Diffraction from atomic layers in a crystal.}
 \end{figure}
 \section*{Theory}
@@ -48,7 +48,7 @@ Fabry-Perot or Michelson interferometers. The de Broglie wavelength for the
 electron is given by $\lambda=h/p$, where $p$ can be calculated by knowing the
 energy of the electrons when they leave the ``electron gun'':
 \begin{equation}\label{Va}
-\frac{p^2}{2m}=eV_a,
+\frac{1}{2}mv^2=\frac{p^2}{2m}=eV_a,
 \end{equation}
  where $V_a$ is the accelerating potential.  The condition for constructive
 interference is that the path length difference for the two waves
@@ -77,7 +77,7 @@ electron's mass, and $V_a$ is the accelerating voltage.
 
 \section*{Experimental Procedure}
 
-\textbf{Equipment needed}: Electron diffraction apparatus and power supply, tape, ruler.
+\textbf{Equipment needed}: Electron diffraction apparatus and power supply, ruler and thin receipt paper or masking tape.
 
 \subsection*{Safety}
 You will be working with high voltage. The power supply will be connected for you,
@@ -127,8 +127,8 @@ Anode Current & $I_A$& 0.15 mA at 4000 V ( 0.20 mA max.)
 Slowly change the voltage and determine the highest achievable accelerating
 voltage, and the lowest voltage when the rings are visible.
 \item  Measure the diffraction angle $\theta$ for both inner and outer rings for 5-10 voltages from that range,
-using the same masking tape (see procedure below). Each lab partner should
-repeat these measurements (using an individual length of the masking tape).
+using the same thin receipt paper (see procedure below). Each lab partner should
+repeat these measurements (using an individual length of the thin paper).
 \item Calculate the average value of $\theta$ from the individual measurements for each
 voltage $V_a$. Calculate the uncertainties for each $\theta$.
 \end{enumerate}
@@ -137,7 +137,7 @@ voltage $V_a$. Calculate the uncertainties for each $\theta$.
 
 To determine the crystalline structure of the target, one needs to carefully
 measure the diffraction angle $\theta$. It is easy to see (for example, from
-Fig.~\ref{ed1} ) that the diffraction angle $\theta$ is 1/2 of the angle
+Fig.~\ref{ed1}) that the diffraction angle $\theta$ is 1/2 of the angle
 between the beam incident on the target and the diffracted beam to a ring,
 hence the $2\theta$ appearing in Fig.~\ref{ed4}. You are going to determine
 the diffraction angle $\theta$ for a given accelerated voltage from the
@@ -147,13 +147,13 @@ L\sin{2\theta} = R\sin\phi,
 \end{equation}
 where the distance between the target and the screen $L = 0.130$~m is controlled during the production process to have an accuracy better than 2\%. {\it Note, this means that the electron tubes are not quite spherical.}
 
-The ratio between the arc length and the distance between the target and the
-radius of the curvature of the screen $R = 0.066$~m gives the angle $\phi$ in
-radian:  $\phi = s/2R$. To measure $\phi$ carefully place a piece of
-masking tape on the tube so that it crosses the ring along the diameter.
+The ratio between the arc length $s$ and the
+radius of the curvature for the screen $R = 0.066$~m gives the angle $\phi$ in
+radians:  $\phi = s/2R$. To measure $\phi$ carefully place a piece of
+thin receipt paper on the tube so that it crosses the ring along the diameter.
 Mark the position of the ring for each accelerating voltage, and then
-remove the masking tape and measure the arc length $s$ corresponding to
-each ring. You can also make these markings by using the thin paper on which cash register receipts are printed.
+remove the paper and measure the arc length $s$ corresponding to
+each ring. You can also make these markings on masking tape placed gently on the tube.
 
 
 \begin{figure}
@@ -170,10 +170,10 @@ is proportional to $d$, one needs to substitute the expression for the
 electron's velocity  Eq.(\ref{Va}) into the diffraction condition given by
 Eq.(\ref{bragg}):
 \begin{equation}\label{bragg.analysis}
-2d\sin\theta=\lambda=\frac{h}{\sqrt{2m_ee}}\frac{1}{\sqrt{V_a}}
+\sin\theta=\frac{h}{2\sqrt{2m_ee}}\frac{1}{d}\frac{1}{\sqrt{V_a}}
 \end{equation}
 
-Make a plot of $1/\sqrt{V_a}$ versus $\sin\theta$ for the inner and outer rings
+Make a plot of $\sin\theta$ (y-axis, with uncertainty) vs $1/\sqrt{V_a}$ (x-axis) for the inner and outer rings
 (both curves can be on the same graph). Fit the linear dependence and measure
 the slope for both lines. From the values of the slopes find the distances
 between atomic layers $d_{inner}$ and $d_{outer}$.
@@ -181,7 +181,7 @@ between atomic layers $d_{inner}$ and $d_{outer}$.
 Compare your measurements to the accepted values: $d_{inner}=d_{10} = .213$~nm
 and $d_{outer}=d_{11}=0.123$~nm.
 
-\section*{Looking with Electrons}
+\section*{Seeing with Electrons}
 	
 The resolution of ordinary optical microscopes
 is limited (the diffraction limit) by the wavelength of light ($\approx$ 400
-- 
cgit v1.2.3

