phy-4660/accelerator/report/report.tex

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\newcommand{\FpaO}{$^{19}\textrm{F(p,}\alpha)^{16}\textrm{O}$}
\newcommand{\LipaHe}{$^7\textrm{Li(p,}\alpha)^4\textrm{He}$}

%\newcommand{}$^7\textrm{Li(p,}\alpha)^4\textrm{He}$ reaction.\\






\title{Lab 4: Nuclear Reactions and Nucleosynthesis}
\author{Otho Ulrich, Eugene Kopf, Mike Pirkola, Jacob Burke, Andrew Messecar, Spencer Henning, Asghar Kayani}
\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}

%─────────────
\section{Introduction: Lithium and Fluorine}
    \label{sec:intro}
    The nuclear properties of lithium are of interest, especially the reaction differential cross-section for \LipaHe. Studies of the Sun's photosphere show the abundance of lithium relative to hydrogen and helium less nearly an order of magnitude. A large cross-section for \LipaHe\ 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 is 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 rearranged 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 rearranged 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-probability explanation is plausible, we will determine the reaction differential cross-sections of the fluorine-proton reaction \FpaO\ and the lithium-proton reaction \LipaHe. In this study, we performed a prompt radiation analysis by observing the alpha particle products of each reaction, and from the kinetic energy spectrum of these products, the differential 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 neighbor elements. It is thought this is due to large reaction differential cross-sections for a proton collision with these nuclei. \cite{Carroll&Ostlie}}
        \label{fig:abundances}
    \end{figure}




%─────────────

\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 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
        \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}
    
    The measurement apparatus serves to create a kinetic energy spectrum of charged particles. An alpha particle or proton (or any other charged particle) incident on the detector creates a current pulse which is converted to a voltage pulse across a high-impendence conductor. The voltage signal is then sent by way of a pre-amplifier to the receiving amplifier in the control room. A multi-channel analyzer receives voltage signals from the second amplifier, binning counts as a function of voltage. The amplifier is adjustable, allowing the voltage range to fit properly within the MCA's detection domain, and the voltage received at the MCA is directly proportional to the kinetic energy of the alpha particle. \cite{ADVLABACCEL} In Section~\ref{sec:reactions:calibration}, a kinetic energy scale is calibrated to the voltage scale, thus providing the charged-particle spectrum.


%─────────────

\section{Nuclear Reactions and Detection Plan}
    \label{sec:reactions}
    
    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.

    \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. In our experiments, the detection angle $\theta = 149.95\degree\pm0.05\degree$ \cite{ADVLABACCEL}.}
        \label{fig:kinematics}
    \end{figure}

    \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{table}
        \center
        \begin{tabular}{lrllr}
            Isotope  &Mass (amu)&Abundance&Kinematic Factor&\\
            \hline 
            p/$^1$H$^+$&1.00727646688(13)&&&\cite{Carroll&Ostlie} \\
            $\alpha$/$^4$He$^{++}$&4.001506179127(63)&&&\cite{codatarecommended} \\
            $^4$He   &4.002603 &1.000 &&\cite{Chuetal} \\
            $^7$Li   &7.016004 &&&\cite{Chuetal} \\
            $^{16}$O &15.994915&0.9976&&\cite{Chuetal} \\
            $^{19}$F &18.998405&&&\cite{Chuetal} \\
            % From Kayani?
            %$^{28}$Si&28.086   &1.0000&0.1808&\cite{tesmer1995handbook} \\
            %$^{63}$Cu&62.930   &0.6917&0.4897&\cite{tesmer1995handbook} \\
            %$^{65}$Cu&64.928   &0.3083&0.4975&\cite{tesmer1995handbook} \\
            $^{28}$Si&28.086   &1.0000&0.8746&\cite{tesmer1995handbook} \\
            $^{63}$Cu&62.930   &0.6917&0.9420&\cite{tesmer1995handbook} \\
            $^{65}$Cu&64.928   &0.3083&0.9437&\cite{tesmer1995handbook} \\
        \end{tabular}
        \caption{Properties of nuclei and particles important in our reactions. Li and F nuclei are constituent to LiF foil and are expected to undergo nuclear reactions, with residual nuclei $^{16}$O and $^4$He. The Cu and Si nuclei are involved in Rutherford scattering. p and $\alpha$ refer to a proton and an alpha particle, respectively. Abundances are relative to unit probability for that species, and kinematic factors are included for target nuclei involved in Rutherford scattering of a proton at $150\degree$. Numbers reported without uncertainty are assumed to have uncertainty $\pm$1 on the most precise digit.}
        \label{tab:nuclei} 
    \end{table}

    \subsection{Calibration with Rutherford Scattering}
        \label{sec:reactions:calibration}
        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. Rutherford scattering can be analyzed in terms of figure~\ref{fig:kinematics}, where M1 = M3, and M2 = M4. The scattering angle $\theta = 149.95\degree\pm0.05\degree$. When we assume the collision is elastic, the kinetic energy of the scattered Proton $E3 \propto E1$. We define a kinematic factor K as the constant of proportionality. This factor can be computed as \cite{ADVLABACCEL} 

        \begin{equation}
            K = \left(\frac{M1 cos(\theta) + \left(M2^2 - M1^2sin^2(\theta)\right)^{1/2}}{M1 + M2}\right)^2.
            \label{eq:kinematicfactor}
         \end{equation}
         Thus,
        \begin{equation}
            E3 = K \times E1.
            \label{eq:rutherford}
        \end{equation}
          

        We observe Rutherford scattering of a proton by copper and silicon atoms by performing two experiments, one where the proton beam is incident on a copper foil, and then a silicon foil. Kinematic factors are tabulated in table~\ref{tab:nuclei}. The proton may deposit energy in the target material, so the elastic case is the maximum energy case. A spectrum of energy from scattered protons should therefore be observed, a sum of approximately Gaussian peaks that drops to noise after this highest-energy peak. The expected peaks are tabulated in table~\ref{tab:predictions}.

        \begin{table}
            \center
            \begin{tabular}{rl}
                Reaction&Energy (MeV) \\
                \hline
                % From Kayani's(?) kinematic factors
                %$^{28}$Si(p)&0.352$\pm$0.009\\
                %$^{63}$Cu(p)&0.95$\pm$0.02\\
                %$^{65}$Cu(p)&0.97$\pm$0.03\\
                $^{28}$Si(p)&1.706$\pm$0.05\\
                $^{63}$Cu(p)&1.837$\pm$0.05\\
                $^{65}$Cu(p)&1.840$\pm$0.05\\
                \FpaO&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}.}
            \label{tab:predictions}
        \end{table}


    \subsection{Nuclear Reaction Peaks}
    \label{sec:reactions:peaks}
        The kinetic energy of the alpha particle products from \FpaO and \LipaHe are characteristic of those reactions. Binding energies of the involved nuclei are significant compared to their rest masses, so the relativistic equivalence must be considered when computing the energy Q released or absorbed during the reaction, \cite{ADVLABACCEL} so

        \begin{equation}
             Q = \delta(mc^2) = (M1+M2)c^2 - (M3+M4)c^2.
             \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} 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}, 
        \end{equation} where
        \begin{equation}
            A = [(M1{\times}M3{\times}E1)^{1/2}/(M3+M4)]cos\theta
        \end{equation} and
        \begin{equation}
            B = [M4{\times}Q+E1(M4-M1)]/(M3+M4). \cite{Meisel1996}
        \end{equation}

        %\begin{equation}
        %    E3 = \frac{E1 M1 M3}{(M3 + M4)^2}\Bigg\{2cos^2\theta + \frac{M4(M3+M4)}{M1 M3}\left(\frac{Q}{E1} + 1 - \frac{M1}{M3}\right)
        %\end{equation}
        %\begin{equation}
        %    + 2cos\theta\left[cos^2\theta + \frac{M4(M3+M4)}{M1 M3}\left(\frac{Q}{E1} + 1 - \frac{M1}{M4}\right)\right\]^{\frac{1}{2}}\Bigg\}
        %    \label{eq:E3}
        %\end{equation}
             

%─────────────    
\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=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=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. 


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