\experiment{Gamma Radiation Interactions}

\section{Introduction}

Electromagnetic radiation of wavelength greater than $\val{1}{pm}$
interacts with matter in just two ways: the photoelectric effect and
Compton scattering. Both are nonclassical and most simply described as
the interaction of a particle---the photon, or gamma ray---with an
atom.

Both interactions remove an electron from the atom, and each is
observed by detecting this electron. The identifying feature of the
photoelectric interaction is that the electron emerges with a single
energy $E = h c/\lambda$, since the photon is destroyed in the
interaction then, for energy conservation, $E$ is the photon energy.
Compton scattering, on the other hand, is interpreted as elastic
scattering of the photon of energy $E$ and momentum $p = h/\lambda =
E/c$ from an atomic electron. If the photon is scattered by an angle
$\theta$ from its original direction, then its new wavelength
$\lambda'$ is related to its original wavelength $\theta$ by the
Compton formula
\begin{equation}
	\label{eq:Gamma1}
	\lambda' - \lambda = \frac{h}{mc}(1 - \cos\theta) 
\end{equation}
You may readily show, then, that the electron acquires the energy 
\begin{equation}
	\label{eq:Gamma2}
	E_{e} = \frac{E}{1 + \frac{mc^{2}}{E(1-\cos\theta)} } 
\end{equation}
where $mc^{2}$ for an electron is \val{511}{keV}.

The experiment is performed with radiation from a radioactive source
such as cesium 137 incident upon the atoms in a 2-inch-diameter
by 2-inch-thick crystal of sodium iodide. The radiation from
\iso{Cs}{137} consists of gamma rays of energy \val{662}{keV}. Compton
and photoelectrons freed from atoms of the crystal by this radiation
rapidly come to rest within the crystal, converting their energy into
that of low energy photons.  In sodium iodide, these secondary photons
are in the visible spectrum and are detected by a photomultiplier
tube. The photomultiplier converts them into an electrical pulse of
amplitude proportional to the total energy of the secondary
photons---hence also proportional to the initial electron energy. By
observing the amplitude distribution of these pulses, we determine the
energy distribution of electrons freed from atoms of the crystal by
interactions with the incident radiation.

\section{Experiment}

The apparatus is shown schematically in \fref{Gamma1}.
\begin{figure}[h]
  \centering
  \igw[4in]{Gamma1}
  \caption{Schematic representation of the apparatus.}
  \label{fig:Gamma1}
\end{figure}

The source S is taped to the front of the sodium iodide crystal. The
enlargement illustrates photoelectric and Compton interactions within
it. Secondary photons are generated as the electrons $e^{-}$ come to
rest. These photons are detected by the photomultiplier tube PMT
mounted in a common housing with the crystal. The preamplifier
generates a negative output pulse as shown in the inset with amplitude
proportional to the energy of the Compton or photoelectron feed within
the crystal. The amplifier then inverts, shapes and amplifies the
pulse as shown.

% \begin{wrapfigure}[nlines]{<rl>}[overhang into margin]{width or 0pt}
\begin{wrapfigure}{r}{0pt}
	\igw[2in]{Gamma2}
	\caption{Discriminator levels.}
	\label{fig:Gamma2}
\end{wrapfigure}

The amplifier output is directed to a single channel analyzer SCA.
This is the instrument with which the pulse amplitude distribution is
determined. You will control two settings of the SCA---its lower level
discriminator LLD and its window. The function of these controls is 
shown in \fref{Gamma2}.
A, B, and C are
amplitude three pulses from the amplifier to the SCA. The controls
set the height of the line LLD and the width of the gap labeled
``window.'' As set, only pulse B will trigger an output window pulse
from the SCA. The maximum amplitudes of A and C lie outside the
window and neither pulse is sent on to the 
scaler. Were the LLD lowered so that the window embraced only pulse
A, then only pulses of this maximum amplitude would be registered by
the scaler. Were the window 
opened to include the maxima of both Band C, then both of these
pulses would be registered. The scaler is started and stopped by a
timer and counts the number of pulses it receives from the SCA within
this time which satisfy the condition: 
\begin{equation*}
	\mathrm{LLD} \le \text{maximum pulse amplitude} \le
    \mathrm{LLD + window}
\end{equation*}
We wish to observe the amplitude distribution of pulses generated by
electrons from gamma interactions within the sodium iodide crystal.
First, connect an oscilloscope to the output of the amplifier. You
will see a broad amplitude distribution, with a bright band of pulses
of nearly the same maximum amplitude. These are generated by
photoelectric interactions in the crystal. Set the amplifier gain so
that these pulses have $\sim$5--8 volts maximum amplitude. Remove the
scope and reconnect the amplifier to the SCA.  Set the SCA window%
\footnote{In ``window'' mode, the Window pot goes from 0--1~V, while
  the Lower Level pot goes from 0--10~V.} to 0.2~volts and the LLD
somewhat below the maximum voltage of the bright band observed on the
scope. Set the timer for $\val{10}{s}$ or so and count the number of pulses
received by the scaler at this setting of the SCA. Advance the LLD by
\val{0.2}{V}, leaving the window fixed, and count again. Proceed in this
way through the ``photopeak.'' You now know how to use the apparatus.

For \iso{Cs}{137}, the photopeak observed above occurs at \val{662}{keV}. Since
each component of the apparatus is linear, this number calibrates the
entire voltage range of the LLD in electron-volts. You are now
prepared to study the entire energy spectrum of electrons resulting
from photoelectric and Compton interactions of gamma radiation from
\iso{Cs}{137} with atoms of the sodium iodide crystal. Start with the LLD at
zero volts and
make three or four measurements at gradually increasing LLD. When you
have completed this task and understand how the apparatus works, you
have a choice: you may either proceed through the photopeak, one
channel at a time. Or, you may get checked out by your instructor and
switch to using the Maestro 32 multichannel data acquisition system,
which will allow you to record the entire spectrum
simultaneously. Instructions on using Maestro 32 are available at the
apparatus.

If you choose the manual approach, be sure to plot as you go! Where
the distribution is quite flat, you should increase the increments by
which the LLD is advanced. Where it changes rapidly, you may wish to
decrease them. Set the timer to obtain good statistics. Remember that
the uncertainty in the number $N$ of pulses counted in a given time
interval is $\pm \sqrt{N}$.
 
\subsection{Things to observe}

Classically, one would expect no structure in the pulse amplitude
distribution at all. The prominent peak you observe attests to the
nonclassical photoelectric interaction first accounted for by
Einstein. 

The Compton interaction produces electrons of energy from zero,
$\theta = 0$, to a maximum for $\theta = \pi$ (\eref{Gamma1}). The
maximum defines the ``Compton edge.'' Calculate where you expect the
edge to appear and locate it in your experimental distribution. You
will probably also observe a rather broad peak at the energy $E
-E_{e}$. This is the energy of a gamma ray which has scattered at
$\theta \cong \pi$. This occurs for gammas emitted by the source
away from the crystal, which are then scattered from the table, floor,
walls, etc. back into the crystal. The peak may be enhanced by placing
an aluminum plate immediately behind the source.  You should also
observe a narrow peak at an energy of \val{\sim\! 30}{keV}. The radiation
responsible is an x-ray from barium, the decay product of
cesium within the source. Look up the known energy of this x-ray and
compare it to your result. You may wish to place a lead brick behind
the source and observe x-rays from lead (\val{\sim\! 70}{keV}) which
arise with each photoelectric interaction in the lead. 



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