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\pagestyle{empty}
\rightline{(revised 3/19/09)}
\vspace{1in}
\begin{center}
\bf Compton Scattering\\[1cm]
\rm Advanced Laboratory, Physics 407\\
University of Wisconsin\\
Madison, Wisconsin 53706
\end{center}
\vskip 2in
\begin{center}
\bf\underline{Abstract}
\end{center}

The Compton scattering of the 662 keV gamma rays from the decay of Cs$^{137}$
is measured using a Sodium Iodide detector. The scattered energy and the
differential cross section are both measured as a function of scattering angle,
and the results are compared to the full relativistic quantum theory of
radiation.

\newpage
\pagestyle{plain}
\leftline{\underline{\bf Theory}}
\medskip
The Compton effect is the elastic scattering of photons from electrons. As a
reaction, the process is:
\[ \gamma +e^-\rightarrow\gamma +e^-\quad.\]

Since this is a two body elastic scattering process, the angle of the scattered
photon is completely correlated with the energy of the scattered photon by
energy and momentum conservation. This relation is usually written as:
\begin{equation}
\Delta\lambda=\lambda'-\lambda=\frac{h}{mc}(1-\cos\theta)
\label{1}
\end{equation}
where $\lambda$ and $\lambda'$ are the wavelengths of the incident and
scattered photon respectively, and $\theta$ is the photon scattering angle.

The energy of a photon is related to its frequency and wavelength as:
\begin{equation}
E=h\nu=hc/\lambda
\label{2}
\end{equation}
where $c$ is the velocity of light. Combining Eq.\ (\ref{1}) and (\ref{2}) the
energy of the scattered photon is:
\begin{equation}
E'=\frac{E}{1+\frac{E}{mc^2}(1-\cos\theta)}\qquad.
\label{3}
\end{equation}
The kinetic energy of the recoil electron is:
\begin{equation}
T_e=E-E'=E\frac{\gamma (1-\cos\theta)}{1+\gamma (1-\cos\theta)}
\label{4}
\end{equation}
where $\gamma =h\nu/mc^2$.

The quantity $h/mc$ in Eq.\ (\ref{1}) is called the Compton wavelength and has
the value:
\[ h/mc=2.426\times10^{-10} {\rm cm}= .02426\ \rm \AA.\]
For low energy photons with $\lambda\gg .02\ \rm \AA$, the Compton shift is
very small, whereas for high energy photons with $\lambda\ll 0.02\ \rm \AA$,
the wavelength of the {\em scattered} radiation is always of the order of
0.02 \AA, the Compton wavelength.

As an example, in this experiment gamma rays from Cs$^{137}$ are scattered from
an aluminum target; since $E=0.662$ MeV, we have $\gamma =1.29$ so that gamma
rays scattered at $\theta=180^\circ$ will have $E'=E/3.6$, which is less than
1/3 of their original energy. It thus becomes quite easy to observe the Compton
energy shift. This would not be the case for X-ray energies.

Another useful kinematic relation is the electron scattering angle in terms of
the photon scattering angle:
\[ \cot\varphi=(1+\gamma)\tan\theta/2\]
where $\varphi$ is the electron scattering angle relative to the incident
photon direction (See Fig.\ \ref{kinematics})
\begin{figure}
\begin{center}
\includegraphics[height=2.0in]{compton_1.pdf}
\end{center}
\caption{\label{kinematics}Kinematics of Compton Scattering}
\end{figure}

The above kinematic relations as well as the following discussion on cross
sections may be found in Melissinos pp.\ 252--265 which is required reading for
this experiment.

When $h\nu\ll mc^2$ the probability for Compton scattering can be regarded as a
classical process and is given by the Thompson cross section which is the
classical limit of the exact Compton scattering cross section formula.
\begin{equation}
\left.\frac{d\sigma }{d\Omega }\right|_{\rm Thompson}=
r_0^2\left(\frac{1+\cos^2\theta}{2}\right)
\label{5}
\end{equation}
where $r_0=\frac{e^2}{4\pi\epsilon_0mc^2}$ is the ``classical electron radius"
and has the value\\
$r_0=2.818\times10^{-13}$ cm. When integrated over all
scattering angles, Eq.\ (\ref{5}) yields the total Thompson cross section:
\begin{equation}
\sigma_T=\frac{8}{3}\pi r_0^2\quad.
\label{6}
\end{equation}
This simple cross section has several failings:
\begin{enumerate}
\item
It does not depend on the photon energy, a fact not supported by experiment;
\item
the electron, although free, is assumed not to recoil;
\item
the treatment is nonrelativistic;
\item
quantum effects are not taken into account.
\end{enumerate}
The problem was solved by Klein and Nishina in 1928 giving the correct
quantum-mechanical calculation for Compton scattering, the so called
Klein-Nishina formula:
\begin{eqnarray}
\frac{d\sigma}{d\Omega}&=&r_0^2\left(\frac{1+\cos^2\theta}{2}\right)
\frac{1}{\left(1+\gamma(1-\cos\theta)\right)^2}\nonumber\\
&&\times
\left[\frac{\gamma^2(1-\cos\theta)^2}{(1+\cos^2\theta)(1+\gamma(1-\cos\theta))}
+1\right]\quad.
\label{7}
\end{eqnarray}
This result is for the cross section averaged over all incoming photon
polarizations. By integrating Eq.\ (\ref{7}) over all angles, the total cross
section can be obtained. While the expression for the total cross section is a
lengthy formula, two asymptotic expressions for the total cross section
$\sigma_c$ in the low energy and high energy case are more simple.
\begin{eqnarray}
\mbox{Low energies ($\gamma\ll 1$)}\quad \sigma_c &=& \sigma_T\left\{
1-2\gamma+\frac{26}{5}\gamma^2+\cdots\right\}\nonumber\\
\mbox{and High energies ($\gamma\gg 1$)}\quad \sigma_c &=&
\frac{3}{8}\sigma_T\frac{1}{\gamma}\left(\ln 2\gamma+\frac{1}{2}\right)\quad.
\label{8}
\end{eqnarray}
Note that Eq.\ (\ref{5}) for the Thompson cross section gives a symmetric
angular distribution of scattered photons ({\em i.e.}, the angular distribution
is symmetric about $90^\circ$). The Klein-Nishina formula (Eq.\ (\ref{7})) on
the other hand predicts a strongly forward peaked cross section as $\gamma$
increases. 

The following table and graph show the Klein-Nishina cross section as a
function of photon scattering angle. The table is for the .662 MeV gamma ray
energy of Cs$^{137}$, while Fig.\ \ref{fig2}
shows the angular distribution for a
range of incident photon energies.

\def\p{$\phantom{0}$}
\begin{figure}
\begin{center}
\includegraphics[width=3 in]{compton_fig2.pdf}
\end{center}
\caption{\label{fig2}\small Angular distribution of the Compton scattering as a function
of the angle of scattering for various primary frequencies $\gamma=h\nu/mc^2$}
\end{figure}
\begin{table}
\begin{minipage}{\textwidth}
\small
\renewcommand{\arraystretch}{.9}
\begin{tabular}{ccccc}
\parbox{0.2\textwidth}{\centering $\theta$\\ (Degrees)}
&$\cos\theta$
&\parbox{0.2\textwidth}{\centering Classical $E$\\ (keV)}
&\parbox{.2\textwidth}{\centering Relativistic $E$\\ (keV)}
&\parbox{.2\textwidth}{\centering Klein-Nishina\\ ($10^{-30}$ m$^2$)}\\[10pt]
0 &1.00000 &662.00 &662.00 &7.941\\
5 &\p.99619  & 658.74  & 658.75 & 7.834\\
10  & \p.98481  & 649.06  & 649.22  & 7.524\\
15 &  \p.96593  & 633.24  & 634.01  & 7.047\\
20  & \p.93969  & 611.73  & 614.03  &  6.452\\
25  & \p.90631  & 585.20  & 590.35  &  5.793\\
30 &  \p.86603  & 554.45  & 564.09  &  5.119\\
35  & \p.81915  & 520.46  & 536.34  &  4.471\\
40  & \p.76604  & 484.32  & 508.02  &  3.875\\
45  & \p.70711  & 447.17  & 479.90  &  3.348\\
50  & \p.64279  & 410.13  & 452.57  & 2.894\\
55  & \p.57358  & 374.23  & 426.43  & 2.513\\
60  & \p.50000  & 340.31  & 401.76  & 2.200\\
65  & \p.42262  & 308.95  & 378.72  & 1.948\\
70  & \p.34202  & 280.49  & 357.37  & 1.746\\
75  & \p.25882  & 255.06  & 337.72  & 1.589\\
80  & \p.17365  & 232.58  & 319.72  & 1.467\\
85  & \p.08716  & 212.88  & 303.31  & 1.374\\
90  & \p.00000  & 195.71  & 288.39  & 1.304\\
95  & $-$.08716  & 180.79  & 274.87  & 1.254\\
100 & $-$.17365  & 167.86  & 262.65  & 1.217\\
105  & $-$.25882  & 156.65  & 251.63  & 1.192\\
110  & $-$.34202  & 146.95  & 241.73  & 1.176\\
115  & $-$.42262  & 138.55  & 232.85  & 1.166\\
120  &$-$.50000  & 131.29  & 224.92  & 1.161\\
125  & $-$.57358  & 125.01  & 217.87  & 1.160\\
130  & $-$.64279  & 119.60  & 211.62  & 1.161\\
135  & $-$.70711  & 114.95  & 206.13  & 1.164\\
140  & $-$.76604  & 110.99  & 201.34  & 1.168\\
145  & $-$.81915  & 107.63  & 197.22  & 1.172\\
150  & $-$.86603  & 104.83  & 193.71  & 1.176\\
155  & $-$.90631  & 102.53  & 190.80  & 1.181\\
160  & $-$.93969  & 100.69  & 188.45  & 1.184\\
165  & $-$.96593  & \p99.29  & 186.64  & 1.187\\
170  & $-$.98481  & \p98.31  & 185.37  & 1.190\\
175  & $-$.99619  & \p97.72  & 184.60  & 1.191\\
180  & $-$1.00000 & \p97.53  & 184.35  & 1.191
\end{tabular}
\end{minipage}
\begin{center}
Table 1
\end{center}
\label{table1}
\end{table}

\pagestyle{plain}
\leftline{\underline{\bf Radiation Units}}

Milliroentgen per hour (mR/h) are units of radiation exposure. Exposure
indicates the production of ions in a material by radiation, and it is defined
as the amount of ionization produced in a unit mass of dry air at standard
temperature and pressure. The roentgen is the conventional unit for exposure,
where
\[ \mbox{radiation exposure unit: 1 roentgen = 1 R = }
2.58\times10^{-4}\mbox{ coulomb per kilogram.} \]
Thus, 1 R of radiation produces $2.58\times10^{-4}$ C of positive ions in a
kilogram of air at standard temperature and pressure, and an equal charge of
negative ions.

Radiation safety standards are expressed in units of {\em roentgen equivalent
mammal} per year (rem/yr). We now relate roentgens to rems via the {\em rad}
unit and the {\em RBE} or {\em QF}, discussed below.

The absorbed dose is the radiation energy absorbed per kilogram of absorbing
material. It is measured in rads, where
\begin{center}
absorbed dose unit: 1 rad = 0.01 joule per kilogram.
\end{center}
In animal tissue it takes about 30 eV or $4.8\times10^{-18}$ J to produce an
ion pair, and assuming the magnitude of the charge of each ion is
$1.60\times10^{-19}$C, then for animal tissue
\[ \rm 1\ R\times\frac{4.8\times10^{-18}\,J}{1.6\times10^{-19}\,C}
=2.58\times10^{-4}\frac{C}{kg}\times30\frac{J}{C}=0.008\frac{J}{kg}\simeq1\
rad.
\]
Hence, a 1-R exposure to x rays or $\gamma$ rays produces an animal tissue
absorbed dose of approximately 1 rad.

The effects of radiation on biological systems depend on the type of radiation
and its energy. The {\em relative biological effectiveness} (RBE) or {\em
quality factor} (QF) of a particular radiation is defined by comparing its
effects to those of a standard kind of radiation, which is usually taken to be
200-keV x rays. The RBE or QF is the ratio of the dose in rads of a particular
kind of radiation to a 1-rad dose of 200-keV x rays, where the particular
radiation produces the same biological effect as the x rays. Note that RBE or
QF is dimensionless.
\[
\mbox{RBE or QF:}\ \frac{\mbox{Number of rads of a particular kind of
radiation}}{\mbox{1 rad of 200-keV x rays}}\]
In animal tissue the RBE is about 1.0 for x rays, $\gamma$ rays, and $\beta$
rays, and it ranges from about 2 to 20 for protons, neutrons, and $\alpha$
particles.

The rem is defined as
\[ \mbox{rem}\equiv\mbox{dose in rads}\times\mbox{RBE}\ .\]
For animal tissue a 1-rad dose of $\gamma$ rays is equivalent to an exposure of
1 R, and the RBE is about 1 for $\gamma$ rays in animal tissue; therefore, the
dose in rems and the exposure in roentgens are equivalent. 

Radiation standards adopted by the United States Government are the following:
\begin{enumerate}
\item
For workers employed around nuclear facilities: 5 rem/yr, which would be 2.5
mrem/h continuously while at work for those on a 40-h week. For comparison, 300
to 600 rem of acute whole-body radiation is fatal to humans.
\item
For the general population living near a nuclear facility: 0.5 rem/yr.
\item
For worldwide population: 5 rem total up to age 30 (0.17 rem/yr), in addition
to natural background radiation, which is about the same intensity. Primary
concern is for genetic damage. It is estimated, rather uncertainly, that 0.17
rem/yr may produce 5000 extra deaths and 5000 birth defects in the United
States per year.
\end{enumerate}
\par
\bigskip
\leftline{\underline{\bf Shielding Calculations}}
 
The calculations consist of two parts.  First we calculate the attenuation of 
gammas which is desirable.  Then we calculate the thickness of lead to give 
this attenuation.  
\renewcommand{\theenumii}{\Alph{enumii}.}
\renewcommand{\labelenumii}{\Alph{enumii}}
\begin{enumerate}
\item
\begin{enumerate}
\item
A source of 100 millicuries (100 mCi) of Cs$^{137}$ gives an 
exposure of 0.039 Roentgen/hour (R/hr) at a distance of 1 meter.  
By the inverse square law, the exposure at the distance of 
0.3 meters is
\[
 x = 0.43\ {\rm R/hr}\qquad\mbox{(with no shielding)}\]

\item
This corresponds to a dose equivalent $D$ for tissue of .415 rads/hr.

\item
Converting this to the Dose Equivalent in rems.
\[ {\rm DE}=D\times{\rm QF}\]
where QF=1 for gamma rays. 

Hence we have
\[ \rm DE=0.415\ rems/hr\qquad \mbox{(if no shielding)}\]

\item
Assume that a student may spend 16 hours/month standing at this 
short distance (0.3 meters) from the source.  The Dose equivalent/year with 
this occupancy factor is then 
\begin{center}
79.7 rem/year \qquad (if no shielding)
\end{center}
\item
Radiation standards adopted by the U.S. (1990) for worldwide population: 5 rem
total up to age 30 (0.17 rem/yr), in addition to natural background radiation,
which is about the same intensity. We will assume that $1\times10^{-3}$ rem/yr
is satisfactory for the shielding of our Cs$^{137}$ source. We have therefore
designed the lead shielding to give an attenuation factor of:
\[ f=\frac{1\times10^{-3}\ {\rm rem/yr}}{79.7\ {\rm rem/yr}}=12.5\times10^{-6}
\quad.\]
A shielding calculation (not shown here) for 662 keV gamma rays gives the
result that 10.5 cm of lead will accomplish the above attenuation factor.

\end{enumerate}
\end{enumerate}




\leftline{\underline{\bf Apparatus}}
\medskip
\leftline{\underline{\bf Gamma Source}}

The gamma source is Cs$^{137}$. The present source had a nominal strength of
100 millicuries when it was purchased in 1975, although the actual strength
appears to have been somewhat smaller. A careful measurement of the source
strength in May, 1991, yielded a value of $42\pm2$ millicuries.

This is a strong source and 
so there are 3 very important rules.  The decay scheme of Cs$^{137}$ is shown
in Fig.\ \ref{fig3}. The details of the source are shown in Fig.\ \ref{fig4}.
\begin{enumerate}
\item
Do not attempt to remove the source from inside its lead container.
\item
The container barrel points towards a concrete wall.  Do not try to move 
the container or table.
\item
Do not place any part of your body within 15$^{\circ}$ of the barrel 
axis unless the stopper is in the position to block the barrel.
\end{enumerate}

The lead container is designed conservatively so that the dose you would 
receive if you leaned against the container for 16 hours each month would be 
still less than 10\%\ of the allowable dose for students and 2\%\ of the 
allowable dose for industrial workers.  The dose is approximately 1\%\ of the 
natural background dose.  As a result, the lead container weighs about 250 kg.

In practice, of course, you will spend less time than 16 hours and will be at a 
distance which reduces the background intensity 50 fold.

\begin{figure}[h]
\begin{center}
\includegraphics[height=2.0in]{compton_fig3.pdf}
\end{center}
\caption{\label{fig3}Cs$^{137}$ Decay Scheme}
\end{figure}\nopagebreak
\begin{figure}[h]
\begin{center}
\includegraphics[height=2.0in]{compton_fig4.pdf}
\end{center}
\caption{\label{fig4}Cs$^{137}$ Source Details}
\end{figure}


\leftline{\underline{\bf Source Container and Barrel}}

The container is made of lead and so the 662 KeV gammas have a nearly equal 
probability of interacting via the Photo-Electric effect and via the Compton 
Effect.  The most important angle of scattering is in the forward $\pm 45 
^{\circ}$ cone because
\begin{itemize}
\item[(a)] The differential cross section is larger in the forward direction, 
and
\item[(b)] The minimum energy is lost by the gamma when it is scattered in the 
forward direction.
\end{itemize}
As we require a jet of well collimated 662 KeV gammas with as little 
contamination as possible of gammas with lower energies, the container barrel 
has been designed to greatly reduce the chance of gammas escaping after a 
forward scatter ($\theta < 90^{\circ}$).

{\samepage
If a 662 KeV photon is scattered through more than 45$^{\circ}$ then it is 
likely to be absorbed by a small thickness of lead.
\[ N=N_0 e^{-x/\lambda}\]

\nopagebreak
\begin{center}
\begin{tabular}{ccc}\hline
Angle of Scattering  & Energy   & $\lambda$ in Lead \\
\hline
\makebox[0pt][r]{$0^{\circ}$}    & 662 keV  & 0.88 cm \\
\makebox[0pt][r]{$45^{\circ}$}  & 479 keV  & 0.57 cm \\
\makebox[0pt][r]{$90^{\circ}$}   & 288 keV  & 0.22 cm \\
\makebox[0pt][r]{$135^{\circ}$}  & 206 keV  & 0.11 cm \\
\makebox[0pt][r]{$180^{\circ}$}  & 184 keV  & 0.09 cm \\
\hline
\end{tabular}
\end{center}

\nopagebreak
\begin{figure}[h]
\begin{center}
\includegraphics[height=1.5in]{compton_5.pdf}
\end{center}
\caption{\label{fig5}Source Collimation}
\end{figure}}


The gammas which do not pass out through the aperture are incident upon the 
surfaces A, B, C, D. (See Fig.\ \ref{fig5}.)

\underline{\bf A and B} A few gammas will Compton Scatter to give 184 MeV 
gammas which scatter to the right.  However, the 662 gammas will have 
penetrated about 0.88 cm into the lead and so only a few of the 184 MeV gammas 
will escape from inside the AB surfaces.  Some gammas will interact via the 
photoelectric effect.  Both will cause 75 keV X-rays (Lead K shell) but again 
only a few X-rays will escape from inside the AB surfaces.

\underline{\bf C} The aperture is designed so that scattered gammas and X-rays 
cannot escape directly from C.

\underline{\bf{D}} Again scattered gammas and X-rays cannot escape directly.  
Although the Compton scatters of 90$^{\circ}$ to 45$^{\circ}$ may have enough 
energy for a second Compton scatter on the opposite surface E, the distance 
they must travel from the interaction point in D to leave D is greater than the 
penetration into D.  Note that a few gammas will Compton Scatter near the DE 
corner and will escape into the main beam.  Obviously if we had a ring of dense 
HIGH $Z$ material such as uranium, at the corner, this background could be 
reduced. (See Fig.\  \ref{fig6})
\begin{figure}[h]
\begin{center}
\includegraphics[height=2.0cm]{compton_6.pdf}
\end{center}
\caption{\label{fig6}Source Collimator Scattering}
\end{figure}

\underline{\bf{E}} This is a conical surface designed so that it cannot be 
directly illuminated by gammas from the source.

\underline{\bf{F}} The forward outer surface is positioned so that 662 keV
gammas 
leaving the source must travel through a distance of 12$\lambda$, or 10.56 cm.
The attenuation factor is then approximately $e^{-12} = 6 \times 10^{-6}$.  The 
total gammas intensity is greater since the scattered gammas are not all 
absorbed.  The ``Build Up Factor" for gammas of ${1 \over 2}$ MeV in lead is 
about 2 and so the overall attenuation is about $12\times 10^{-6}$.

\underline{\bf{G}} The backward outer surface has an extra $2\lambda$ added as 
an extra (and unnecessary) safety factor.
\begin{eqnarray*}
H &=& 14 \times 0.88\ \rm cm \\
             &=& 12.32\rm\ cm \\
             &=& 4.85\ \mbox{inches}
\end{eqnarray*}
The overall attenuation factor is then approximately 
\[ 2 \times e^{-14} = 1.6 \times 10^{-6}\]
The jet of gammas is intended to be
\begin{itemize}
\item[(a)] wider than the scatterer so that calculations of the cross section 
may assume the scatterer was uniformly illuminated, and
\begin{figure}[h]
\begin{center}
\includegraphics[width=2.5in]{compton_7.pdf}
\end{center}
\caption{\label{fig7}Gamma Beam Geometry}
\end{figure}
\item[(b)] sufficiently narrow that the detector may measure scattered 
intensifies at small angles from the scatterer. See Fig.\ \ref{fig7}.
\end{itemize}
\noindent
The approximate angles are:

Angle of scatterer subtended at the source $\pm 5^{\circ}$

Angle of Uniform $\gamma$ jet from the source $\pm 10^{\circ}$

Angle of outside of $\gamma$ jet from the source $\pm 20^{\circ}$

Angle of scintillator subtended at the scatterer $\pm 8^{\circ}$

Minimum angle of scattering which is outside the jet: 15$^{\circ}$.


\leftline{\underline{\bf Tapered Plug}}

The plug is intended to block the beam.  The taper is made from brass and so must 
be longer than if it had been made from lead.  We except a few scattered gammas 
to sneak along the taper since the fit cannot be perfect.  For this reason an 
extra 3.8 cm of steel is used to block those escaping along the taper.

\smallskip
\leftline{\underline{\bf{Detector}} (Harshaw Chemical Type 858 Serial 6V230 
with voltage divider)}

The gamma detector is a NaI crystal scintillation cylinder with a diameter of 2 inches and a 
thickness of 2 inches.  The crystal is hermetically sealed and in good optical 
contact with the photocathode of an RCA 6342A phototube with 10 stages.  The 
Harshaw Quality Assurance Report states that the detector has a resolution when 
measuring 662 keV gammas of 
\[\frac{\mbox{full width at 1/2 maximum counting rate}}{\mbox{pulse 
height}} = 8.4\%. \]
Read the description of the experiment on Interactions of Gammas for further 
information.

The photomultiplier requires a positive high voltage via the white cable and 
High Voltage BNC connector.  The dynode potentials are controlled by the built 
in voltage divider with a total resistance of 6.2 Mohms.  The anode is at the 
positive high voltage and so a 1 nf capacitor is used to pick off the 1 
$\mu$sec 100 millivolt negative signals.  The maximum HV rating is +1500V but 
the experiment needs only $\sim1100$--1200 V.

The NaI-PM assembly is mounted inside a tapered lead cylinder.  The lead weighs
about  23 kg and is intended to serve as a collimator so that only gamma rays
from the target region are detected. See Fig.\ \ref{fig8}.
\par
\vfill
\begin{figure}[h]
\begin{center}
\includegraphics[width=2.5in]{compton_8.pdf}
\end{center}
\caption{\label{fig8}Detector Geometry}
\end{figure}
\noindent
The NaI crystal diameter is 2 inches but only 1 3/4 inches are exposed to the 
gammas to give the greatest probability of full absorption of each gamma 
entering the NaI.

The lead cylinder is mounted on rollers and rotates about a center which can 
support a scatterer.\par
\bigskip
\leftline{\underline{\bf{Scatterer}}}

The scatterer is an aluminum disc (1.27 cm thick, 3 cm diameter).  The Compton 
effect depends only on the electrons but does assume that the electrons are 
loosely held.  For this reason, the Z of the scatterer should not be too high.  
The K shell electrons in aluminum have a level of $-1560$ eV and the L shell 
electrons have levels of $-118$, $-74$ and $-73$ eV. 

The scatterer was chosen as a disc so that the probability of multiple 
scattering may be minimized.  The scatterer is mounted on rigid foam plastic so 
that the support will scatter few gammas.
\par
\medskip
\leftline{\underline{\bf{Spectrum Techniques UCS-30}}}

The UCS-30 is a unit that contains a preamplifier, high voltage supply, and 
Pulse Height Analyzer (PHA). The UCS-30 software controls all the settings.   
The nominal settings are +1174 V for the HV, Coarse Gain = 32, and Fine Gain =
1.75. We normally use 512 channels for the PHA. The amplifier gain is set to match
the 0-8 V range of the PHA and is set so that the Cs$^{137}$ peak falls
at about 90\% of the maximum PHA channel. 

\par
\medskip
\leftline{\underline{\bf{Procedure}}}
\begin{enumerate}
\item
\begin{itemize}
\item[(a)] Read the section on the design of the shielding for this 
experiment.
\item[(b)] Use the Radiation Monitor to make a survey of the gamma intensity 
around the experiment with the plug on the source container both open and shut. 
We have included a considerable safety factor in the design of this experiment. 
However, if you are not completely reassured by both of the checks (a) and (b) 
above, then discuss your numbers with the instructor.  You 
should not, of course, lean into the jet of the gammas.
\item[(c)] Connect the equipment and take care that the cables will not be 
caught.
\item[(d)]
Make sure the plug is back in the source.
\end{itemize}

\item
Rotate the detector out of the beam and use sources to calibrate the linearity
of the analyzing system. Start with Cs$^{137}$ and keep the spectrum on the
display after stopping a short run. Change to Ba$^{133}$ and accumulate another
spectrum on the same display. Note the peak channels for both spectra and plot
peak channel \#\ vs energy. The plot should come fairly close to going through
the origin. More sources can be used to check the overall linearity with some
redundancy. A sample calibration using eight different photon energies is shown
in Fig.\ \ref{fig9}. Your calibration curve will allow you to convert from peak
channel number to energy. Set the amplifier gain so that the Cs$^{137}$ peak is
about  90\%\ of full scale (use 512 channels full-scale).
\end{enumerate}

\begin{figure}[h]
\includegraphics[width=\textwidth]{compton_fig9.pdf}
\caption{\label{fig9}}
\end{figure}

\begin{enumerate}
\setcounter{enumi}{2}
\item
Mount the Al scatterer, unplug the source and record the spectra due to the
scattered gammas at $15^\circ$ intervals starting at a scattering angle of
$15^\circ$, going out to $135^{\circ}$ The scatterer should be rotated so as to 
approximately bisect the
angle between the source and the detector. (Why? The gammas are not reflected
like light---why not?) Use your knowledge of counting statistics to decide how
long to count the spectra.

At each angle take a run with and without the Al target. These runs do not have
to be for equal times if they are correctly normalized.

The PHA software may be used to subtract the appropriately normalized
background. The software has a ``strip background'' feature that will
automatically subtract a background run from a target run and correctly do
the subtraction even if the data and background runs are for different times. 
Consult the UCS-30 manual for details. You can then place a region of interest (ROI)
about the subtracted spectrum and the gross counts in the ROI will be recorded
as part of the spectrum display. The spectra data can be saved on the computer hard drive, 
and the spectrum display can be sent directly to a printer. 

For each spectrum, identify the channel number of the peak of the Compton
scattering. Use the plot of (2) above to compute a scattered energy $E_2$ for
each angle. The calculation  of the differential cross section will, in addition,
require a measurement of the scattering rate at each angle, determined by evaluating
the total number of counts in the ROI and nomalizing to the time of the run. 
The rate must subsequently be corrected for the efficiency of the NaI detector 
and the photofraction. Why are these corrections different? If there is still
a significant background under the subtracted spectrum, it may be more accurate
to determine the counts that fall within the FWHM of the detected peak and correct back to
the full area. For a Gaussian shaped peak, the counts within the FWHM region
correspond approximately to $\displaystyle\frac{1}{1.314}$ of the total area.
Efficiency and photofraction data for a $2''\mbox{diam.}\times2''$ thick NaI
crystal are included as figures 10 and 11.


\item
The exact relativistic kinematics of Compton Scattering  predicts
\[ \frac{1}{E_2}=\frac{1}{E_1}+\frac{1}{m_0c^2}(1-\cos\theta)\ . \]
Plot $1/E_2$ from your measurements against $1-(\cos\theta)$.

\item
The non-relativistic theory suggests a different dependence of $E_2$ on
$\cos\theta$. The values have been calculated and tabulated for you. Plot these
on the same plot as your data.

\item
Check and comment upon:
\begin{itemize}
\item[(a)] The agreement of your data and nonrelativistic theory.
\item[(b)] Would you expect good agreement for small $\theta$?
\item[(c)] Is the experimental data in the form predicted by relativistic
theory?
\item[(d)] From the slope, find the rest energy of an electron.
\end{itemize}

\item
The differential cross section $d\sigma/d\Omega$ is defined to be proportional to the scattering 
probability and has the units of area per unit solid angle. Thus the scattering rate
(for the case where the beam area is larger than the target) will be given by:
$$ R(\theta) = I_{\gamma}\times N_e\times\frac{d\sigma}{d\Omega}\times\Delta\Omega_{det}$$
where $R(\theta)$ is the scattering rate for a 100\% efficient detector, $I_{\gamma}$
is the source intensity per unit area at
the target, $N_e$ is the number of electrons in the target, and $\Delta\Omega_{det}$ is the solid angle 
subtended by the NaI detector. For the case of our Cs$^{137}$ source:
$$ I_{\gamma} = \frac{{\rm present\ source\ intensity\ (Ci)}\times 0.94\times 0.90\times 3.7\times 10^{10}}{4\pi\ 
{\rm (source\ to\ scatterer)}^2} $$
The efficiency factors include the efficiency of the NaI at energy $E^{\prime}$, $\epsilon_{NaI}(E^{\prime})$
and the photofraction at energy $E^{\prime}$, $\epsilon_{pf}(E^{\prime})$. Thus the measured rate $G(\theta)$
is related to the true rate $R(\theta)$ by:

$$ R(\theta) = \frac{G(\theta)}{\epsilon_{NaI}(E^{\prime})\times \epsilon_{pf}(E^{\prime})}$$
and finally:

$$ \frac{d\sigma}{d\Omega} = \frac{G(\theta)}{I_{\gamma}\times N_e\times 
\Delta\Omega_{det}\times\epsilon_{NaI}(E^{\prime})\times \epsilon_{pf}(E^{\prime})}.$$

Use Figs.\ \ref{fig10} and \ref{fig11} below for the efficiencies.  
\begin{figure}[h]
\begin{center}
\includegraphics[height=3.25in]{compton_fig10.pdf}
\end{center}
\caption{\label{fig10}Calculated Efficiency for a 2 in by 2 in NaI Crystal}
\end{figure}

\item Plot your differential cross sections with the statistical error bars as a function 
of $\theta$.  On the same plot place points calculated from the Klein-Nishina 
formula (see the table).  Compare the experimental and theoretical predictions and plot the
ratio of Experiment/Theory as a function of angle. 

\item Be prepared to answer questions on all aspects of the experiment 
such as:
\begin{itemize}
\item[(a)] How does a pulse height  analyzer work?
\item[(b)] Are you sure the HV did not drift? How would you check?
\item[(c)] What fraction of the gammas would suffer multiple scattering?
\item[(d)] Why do the energy peaks have a gaussian shape?
\item[(e)] What was the radiation level where you stood?
\item[(f)] At what angle $\theta$, is the kinetic energy
given to an electron insufficient 
to knock it free from the K-shell of aluminum?
\item[(g)] Could you distinguish between the energy of a photon which suffered 
a single $\theta = 90^{\circ}$ scatter and the energy of a photon which 
suffered two $\theta = 45^{\circ}$ scatters?
\end{itemize}
\end{enumerate}

\begin{figure}[h]
\begin{center}
\includegraphics[width=\textwidth]{compton_fig11.pdf}
\end{center}
\caption{\label{fig11} Calculated Efficiencies and Photofractions for Various Size 
Thallium Activated Sodium Iodide Crystals, W.F. Miller, John 
Reynolds, W.J. Snow, Rev. Sci. Instrum. {\bf 28}, 
717 (1957)}
\end{figure}
\end{document}
