halfway done with accel lab

accelerator/report/report.tex

@@ -29,10 +29,11 @@
 
 
 \title{Lab 4: Nuclear Reactions and Nucleosynthesis}
-\author{Otho Ulrich, Eugene Kopf, Asghar Kayani, Mike Pirkola, }
+\author{Otho Ulrich, Eugene Kopf, Asghar Kayani, Mike Pirkola, Jacob, }
 \maketitle
 
 \begin{abstract}
+A 2 MeV proton beam is used to annihilate lithium and flourine atoms from a LiF foil. Alpha particle detection is used to verify the nuclear reactions involved. The probabilty cross-sections are computed for the nuclear reactions \FpaO\ and \LipaHe.
 \end{abstract}
 
 %─────────────
@@ -40,24 +41,12 @@
     \label{sec:intro}
     The nuclear properties of lithium are of interest, especially its interaction probability cross-section. Studies of the Sun's photosphere show the abundance of lithium relative to hydrogen and helium less nearly an order of magnitude. Lithium's large cross-section is believed responsible; it predicts that ionized hydrogen will readily collide with lithium, transforming the lithium to helium, and emitting an alpha particle. This is one transition that takes place in stellar nucleosynthesis, and maintained as fresh lithium is carried toward a star's core by convective currents, but even with this process and reaction in mind, modern astrophysicists have yet to completely explain the lack of abundant lithium in our sun's photosphere. Studies of the nuclear properties of lithium could elucidate a better stellar structure model, but this is outside the scope of our study. \cite{Carroll&Ostlie}
 
-    Fluorine can be annihilated in a similar fashion. It is one of the rarest elements observed by astronomers, and thought to be for the same reasons: it is readily annihilated by a proton to produce oxygen and an alpha particle. Figure~\ref{fig:abundances} shows relative abundances of many elements. To judge whether the high-availability explanation is plausible, we will determine the probability cross-sections of the fluorine-proton reaction {\FpaO} and the lithium-proton reaction {\LipaHe}. By observing the alpha particle products of each reaction, these cross-sections can be computed.\\
-
-    \begin{table}
-        \center
-        \begin{tabular}{rl}
-            M1,E1: & Incident Particle &
-            M2,E2: & Target Nucleus &
-            M3,E3: & Emitted Particle &
-            M4,E4: & Residual Nucleus &
-        \end{tabular}
-        \caption{These variables represent the mass and kinetic energies of the particles involved in the collision outlined in Figure~\ref{fig:kinematics}.}
-        \label{tab:MEpairs}
-    \end{table}
+    Fluorine can be annihilated in a similar fashion. It is one of the rarest elements observed by astronomers, and thought to be for the same reasons: it is readily annihilated by a proton to produce oxygen and an alpha particle. Figure~\ref{fig:abundances} shows relative abundances of many elements. To judge whether the high-availability explanation is plausible, we will determine the probability cross-sections of the fluorine-proton reaction \FpaO\ and the lithium-proton reaction \LipaHe. By observing the alpha particle products of each reaction, these cross-sections can be computed.\\
 
     \begin{figure}
         \center
         \includegraphics[width=4.5in]{abundances.png}
-        \caption{Relative elemental abundances in the sun's photosphere. Lithium and fluorine are much less abundant than their neighbour elements. It is thought this is due to large probability cross-sections for a proton collision with these nuclei. \cite{Carroll&Ostlie}}
+        \caption{Relative elemental abundances in the sun's photosphere. Lithium and fluorine are much less abundant than their neighbor elements. It is thought this is due to large probability cross-sections for a proton collision with these nuclei. \cite{Carroll&Ostlie}}
         \label{fig:abundances}
     \end{figure}
 
@@ -68,11 +57,14 @@
 
 \section{Proton Beam and Detector}
     \label{sec:detector}
-    The Tandem Van de Graff Accelerator Lab provided a $1.95\pm0.05$ MeV proton beam incident on a LiF foil. Under these conditions, we expected the nuclear reactions described in Section~\ref{sec:reactions} to occur, and Rutherford scattering. We used a circular normal-faced charged-particle detector to observe the alpha particle products of \FpaO and \LipaHe and the protons from Rutherford scattering. The detector was placed in a ``backscattering'' position, at $149.95\pm0.05$. Figure
+    The Tandem Van de Graff Accelerator Lab provided a $1.95\pm0.05$ MeV proton beam incident on a LiF foil. Under these conditions, we expected the nuclear reactions described in Section~\ref{sec:reactions} to occur, and Rutherford scattering. We used a circular normal-faced charged-particle detector to observe the alpha particle products of \FpaO\ and \LipaHe\ and the protons from Rutherford scattering. The detector was placed in a ``backscattering'' position, at $149.95\degree\pm0.05\degree$. 
 
     \begin{figure}
         \center
-        \includegraphics[width=4.5in]{detector.jpg}
+        \includegraphics[width=3.2in]{detector.jpg}\\
+        \includegraphics[width=3.2in,angle=180]{detector_angle.jpg}
+        \caption{The detector was positioned at $149.95\degree\pm0.05\degree$ relative to the direction of the proton beam. This angle was maintained through both experiments.}
+        \label{fig:detector}
     \end{figure}
     
 
@@ -83,14 +75,31 @@
 \section{Nuclear Reactions and Detection Plan}
     \label{sec:reactions}
     
-    The nuclear reactions and \FpaO and \LipaHe are of interest, and both can be analyzed using the kinematic diagram in Figure~\ref{fig:kinematics}. In this diagram, each M\#,E\# pair refers to the mass and kinetic energy of a particle involved in the collision, with associations defined in Table~\ref{tab:MEpairs}. In the case where the incident particle does not have sufficient kinetic energy to overcome the electric potential barrier of the target nucleus, Rutherford scattering will occur, and the incident particle will backscatter without annihilating the target nucleus. The maximum expected kinetic energy resulting in Rutherford scattering can be computed, and by observing this value, the energy scale of the detector can be calibrated. When it can overcome the potential barrier, a nuclear reaction may occur.
+    The nuclear reactions \FpaO\ and \LipaHe\ can be analyzed in terms of the kinematic diagram in Figure~\ref{fig:kinematics}. In this diagram, each M\#,E\# pair refers to the mass and kinetic energy of a particle involved in the collision, with associations defined in Table~\ref{tab:MEpairs}. In the case where the incident particle does not have sufficient kinetic energy to overcome the electric potential barrier of the target nucleus, Rutherford scattering will occur. When it does overcome the potential barrier, a nuclear reaction may occur. We ignore tunneling in this analysis, which is a potential source of error. 
+
+    %\subsection{Rutherford Scattering}
+
+    The maximum expected kinetic energy resulting in Rutherford scattering can be computed, and by observing this value, the energy scale of the detector can be calibrated. 
+
+
+    \begin{table}
+        \center
+        \begin{tabular}{rl}
+            M1,E1: & Incident Particle \\
+            M2,E2: & Target Nucleus \\
+            M3,E3: & Emitted Particle \\
+            M4,E4: & Residual Nucleus \\
+        \end{tabular}
+        \caption{These variables represent the mass and kinetic energies of the particles involved in the collision described in Figure~\ref{fig:kinematics}.}
+        \label{tab:MEpairs}
+    \end{table}
 
 
 
     \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 $\theta = 149\pm0.05\degree$ \cite{ADVLABACCEL}.}
+        \caption{In general, an incident particle collides with a target nucleus, resulting in an emitted particle and a residual nucleus. The backscattering experiments in this study have $\theta = 149.95\degree\pm0.05\degree$ \cite{ADVLABACCEL}.}
         \label{fig:kinematics}
     \end{figure}