new stuff

accelerator/report/report.tex

@@ -57,7 +57,7 @@ A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF
 
 \section{Proton Beam and Detector}
     \label{sec:detector}
-    The Tandem Van de Graff Accelerator Lab provided a $1.95\pm0.05$ MeV proton beam for three experiments. When incident on a lithium-fluoride foil, we expect the nuclear reactions described in Section~\ref{sec:reactions} to occur. When the beam is incident on a silicon or copper foil, we expect Rutherford scattering. We used a circular normal-faced surface barrier detector to observe the alpha particle products of \FpaO\ and \LipaHe\ and the protons from Rutherford scattering. \cite{ADVLABACCEL} The detector was positioned at $149.95\degree\pm0.05\degree$ from the proton beam, which we define as the lab frame of reference; see figure~\ref{fig:detector}.
+    The Tandem Van de Graff Accelerator Lab provided a $1.95\pm0.05$ MeV proton beam for three experiments. When incident on a lithium-fluoride foil, we expect the nuclear reactions described in Section~\ref{sec:reactions} to occur. When the beam is incident on a silicon or copper foil, we expect Rutherford scattering. We used a circular normal-faced surface barrier detector to observe the alpha particle products of \FpaO\ and \LipaHe\ and the protons from Rutherford scattering. \cite{ADVLABACCEL} The detector was positioned at $149.95\degree\pm0.05\degree$ from the proton beam, which we define as the lab frame of reference; see figure~\ref{fig:detector}. The detector covers a solid angle $\Omega = 1.895\pm0.001$ steradians. In each experiment, the beam is run until total integrated charged reaches $2\times10^{-4}$ coulombs. 
 
     \begin{figure}
         \center
@@ -80,7 +80,7 @@ A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF
     \begin{figure}
         \center
         \includegraphics[width=4.5in]{collision.png}
-        \caption{In general, an incident particle collides with a target nucleus, resulting in an emitted particle and a residual nucleus. The backscattering experiments in this study have $\theta = 149.95\degree\pm0.05\degree$ \cite{ADVLABACCEL}.}
+        \caption{In general, an incident particle collides with a target nucleus, resulting in an emitted particle and a residual nucleus. In our experiments, the detection angle $\theta = 149.95\degree\pm0.05\degree$ \cite{ADVLABACCEL}.}
         \label{fig:kinematics}
     \end{figure}
 
@@ -148,8 +148,8 @@ A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF
                 $^{28}$Si(p)&1.706$\pm$0.05\\
                 $^{63}$Cu(p)&1.837$\pm$0.05\\
                 $^{65}$Cu(p)&1.840$\pm$0.05\\
-                \FpaO&\\
-                \LipaHe& \\
+                \FpaO&7.949\pm0.03\\
+                \LipaHe&7.688\pm0.03\\
 
             \end{tabular}
             \caption{Expected peaks for nuclear reactions and maximum energy of a Rutherford scattered proton. Rutherford scattering energies are computed using equation~\ref{eq:rutherford} with the kinematic factors from table~\ref{tab:nuclei}.}
@@ -166,7 +166,7 @@ A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF
              \label{eq:Q}
          \end{equation}
 
-        Given this method of computing Q, non-relativistic conservation of total energy and momentum provide a sufficiently useful expression for the kinetic energy of the alpha particle product E3 as a function of the kinematic quantities described in figure~\ref{fig:kinematics}. \cite{Ziegler_1975} The expectation value of E3 can be computed as 
+        Given this method of computing Q, non-relativistic conservation of total energy and momentum provide a sufficiently useful expression for the kinetic energy of the alpha particle product E3 as a function of the kinematic quantities described in figure~\ref{fig:kinematics}. \cite{Ziegler_1975} With results in table~\ref{tab:predictions}, the expectation values of E3 were computed as
 
         \begin{equation}
             E3^{1/2} = A\pm(A^2+B)^{1/2}, 
@@ -187,36 +187,45 @@ A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF
         %\end{equation}
              
 
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 %─────────────    
 \section{Results}
     \label{sec:results}
 
+        The Rutherford scattering curves used to calibrate the energy scale are shown in figures \ref{fig:rutherford} and /ref{fig:calibration}. The best fit for this linear scale is 
+        \begin{equation}
+            x = 0.009\ \textrm{MeV}\times\textrm{channel}+0.00392\ \textrm{MeV}.
+            \label{eq:scale}
+        \end{equation}
+
+        An attempt is made to locate the predicted peaks, and instead two peaks are found at .
+
         \begin{figure}
             \center
-            \includegraphics[width=4.5in]{rutherford.pdf}
+            \includegraphics[width=6in]{rutherford.pdf}
             \caption{Across channels 50 through 205, we observed the signal from Rutherford scattering of protons by copper nuclei (black) and silicon nuclei (red). The maximum count before the cutoff is estimated at 7600 and 18500 and drawn with the blue and purple line, respectively.}
+            \label{fig:rutherford}
         \end{figure}
 
         \begin{figure}
             \center
-            \includegraphics[width=4.5in]{calibration.pdf}
+            \includegraphics[width=6in]{calibration.pdf}
             \caption{The kinetic energy scale is calibrated to channel number using the Rutherford energies predicted in table~\ref{tab:predictions}. The scale is set where the predicted energy matches 20\% of the maximum before cutoff. The 20\% line is drawn in the same color as its associated maximum line, and same for the identified energy coordinate.}
+            \label{fig:calibration}
+        \end{figure}
+
+        \begin{figure}
+            \center
+            \includegraphics[width=6in]{nuclear.pdf}
+            \caption{Two peaks were observed not far from the predicted energies along the calibrated scale. The peaks do not match the expected values, but we progress under the assumption that the scale is wrong and these are in fact the alpha particles predicted as products of the \LipaHe and \FpaO reactions.}
+            \label{fig:peaks}
         \end{figure}
 
 %─────────────
 \section{Conclusion}
     \label{sec:conclusion}
 
+    The energy scale may be off due to a poor choice in locating the Rutherford scattering energy along the curve, however, look at the different between selecting the 100\% mark over the 20\% reveals a change in the scale of only about .018 MeV, less than 10\% of the scale's deviation from the observed nuclear reaction peaks. This indicates a more significant flaw in the analysis remains to be discovered. 
+
 
 \printbibliography