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\begin{flushright}
(Revised 1/21/98)
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\begin{center}
{\bf INTERACTIONS OF GAMMAS}\\
\vspace{.625in}
{Advanced Laboratory, Physics 407 } \\
{University of Wisconsin} \\
{Madison, Wisconsin 53706} \\
\vspace{.8in}
\end{center}
\begin{abstract}  
In this experiment a Sodium Iodide Scintillator is used to both detect 
and measure gammas in the range of 22 keV to 2.6 MeV.  The experiment is to 
study the interactions of gammas in various materials and in the detector 
itself.
\end{abstract}

\section*{\bf\underbar{Theory}}
\setcounter{section}{1}
A photon can interact with matter by a number of competing mechanisms.  The 
interaction can be with the entire atom, as in the photoelectric effect, or 
with one electron in the atom, as in the Compton effect, or with the atomic 
nucleus (as in pair production).  The probability for each of these competing 
independent processes can be expressed as a collision cross section per atom, 
per electron, or per nucleus in the absorber.  The sum of all these cross 
sections, normalized to a per atom basis, is then the probability that the 
incident photon will have an interaction of some kind while passing through a 
very thin absorber which contains one atom per cm$^2$ of area normal to the 
path of the incident photon.

The total collision cross section, $\sigma$, per atom when multiplied by the 
number, $n$, of atoms per cm$^3$ of absorber is then the linear attenuation 
coefficient, $\mu_0$ per centimeter of travel in the absorber. The fraction of 
incident photons which can pass through a thickness $x$ of absorber without 
having an interaction of any kind is given by 
%%
$$I_0 = e^{-\mu_0x} = e^{-\sigma nx}\quad.$$
%%
We sometimes wish to express the absorption in terms of the equivalent matter 
traversed, namely $\xi$ = 
gm/cm$^2$.  Then the thickness of the material can be 
expressed by $d\xi$, where:
$$d\xi = \rho dx .$$
The \underline{mass} absorption coefficient is defined by:
$$\mu_m = {\mu_0\over \rho},$$
so that the fraction of the beam not absorbed is:
$${I\over I_0} = e^{-\mu_m\xi}\quad.$$

In this experiment we will measure the mass absorption coefficient of materials 
of different $Z$ for gamma rays of a large range of energies.  We will try to 
understand these results in terms of what we know about the interactions of 
gamma rays with matter.

As mentioned above, there are three mechanisms of interaction which are 
important at the gamma ray energies we are interested in:  photoelectric effect, 
pair production, and Compton effect.  These are described briefly below.  You 
should also read Eisberg and Resnick, p. 53--56 and/or Evans, Chapters 23 and 
24.
\begin{itemize}
\item
\underline{Photoelectric Effect}\\
In this interaction the photon ejects an electron from an atom (generally from 
the $K$ or $L$ shells).  The photon is completely absorbed and all its energy 
is transferred to the atomic electron.  The atom then emits characteristic 
X--rays and Auger electrons as it returns to normal.
\item
\underline{Pair Production}\\
In pair production a photon of sufficiently high energy is annihilated and an 
electron--positron pair is created.  For a free photon conservation of energy 
and momentum would not be possible, so pair production must 
take place in the field of a nucleus (or of another electron) which will take 
up the balance of momentum.  The energy threshold for this process is $2mc^2$ 
or 1.02 MeV.
\item
\underline{Compton Effect}\\
In the Compton effect the gamma ray scatters off of a loosely 
bound electron and 
loses only part of its energy.  The electron recoils in one direction and  the 
gamma goes off in another direction with a reduced energy.
\end{itemize}

The following figure (\ref{evans712}) (from Evans, p. 712) provides a guide to the relative 
importance of these three principal interactions over a wide range of energy, 
$h\nu$, of the incident photons and atomic numbers, $Z$ of the attenuating
material.

Knowledge of these interactions will also be important in understanding the 
detection of gammas with a scintillation detector.

\begin{figure}
\epsfysize=.25\textheight \epsfbox{gamma_fig1.eps}
\caption{Relative importance of the three major types of 
$\gamma $-ray interaction. The lines show the values of $Z$ and 
$h\nu $ for which the two neighboring effects are just equal.\label{evans712}}
\end{figure}

\medskip
\subsubsection*{\underbar{Detection of Gammas -- Scintillation Detectors}}

Gamma rays may be detected in a number of ways, including gas--filled counters, 
sold--state detectors, and scintillation detectors.  This experiment makes us 
of a scintillation detector.

In a scintillation detector the gamma ray passes into an organic or inorganic 
crystal where it interacts with the atoms of the crystal by one of the above 
interactions.  The result of these interactions are the production of charged 
particles (electrons and gammas) and scattered gammas.  The scattered gammas 
travel a distance through the crystal which depends on their energy and they 
may or may not interact again before leaving the crystal.  The charged 
particles travel a relatively short distance through the crystal leaving behind 
a trail of exited atoms.  A few materials will emit light in the UV and visible 
wave lengths as the excited atoms return to the ground state.  These 
``scintillations'' are detected by a photomultiplier attached to the crystal. 
Materials to be used as scintillators must of course be transparent to 
their own
light so that it can be seen by the photomultiplier.  A discussion of organic 
and inorganic materials suitable for scintillators is given in an appendix.  In 
this experiment we use an inorganic crystal, NaI, as a scintillator.

If monoenergetic gammas interact in the NaI crystal the pulses produced by the 
photomultiplier will not all be if the same height but will show a certain 
distribution in height.  A ``pulse height spectrum'' is obtained by plotting a 
graph of the number of 
pulses
per unit time versus pulse height.  The feature of these pulse 
height spectra can be understood in terms of the interaction the gamma 
undergoes in the crystal.

In the photoelectric effect and pair production all of the gamma ray energy in 
converted to electron (or positron) energy.  Path lengths of charged particles 
are short so there is a good chance for all of the energy being absorbed in the 
crystal.  In Compton scattering, however, part of the energy is converted to 
electron energy and part remains in the scattered photon.  The scattered photon 
may interact again in the crystal by a second Compton scattering
or by photoelectric absorption, but in some cases the scattered gamma ray 
escapes the crystal, carrying its energy with it.

If all the gamma ray energy is captured in the crystal there will still be a 
pulse height  distribution of finite width
due to a number of effects which depend on the 
location of the events in the crystal:
\begin{enumerate}
\item
Inhomogeneities in the crystal.
\item
Variations in reflections from the walls.
\item
Variations in photocell sensitivity.
\item
Statistical fluctuation due to the moderate number of photoelectrons in the 
photomultiplier.
The pulse height spectrum in this case would look like:\\
\epsfysize=.825in \epsfbox{gamma_fig2.epsf}
%\special{epsf=gamma_fig2 scale=0.9}
\end{enumerate}

In practice, however, because of the finite size of the crystal some of the 
Compton scattered gammas will escape the crystal and in this case the spectra 
will look like these:\\
\epsfxsize=\textwidth \epsfbox{gamma_fig3.epsf}
%\special{epsf=gamma_fig3}

The peak A corresponds to interactions in which no gamma escapes.  The peak B 
and the plateau C correspond to Compton interactions in 
which the scattered gamma 
escapes the crystal.  B is called the ``Compton Edge'' and corresponds to a 
Compton scattering in which the gamma is scattered in the backward direction 
and then escapes the crystal.  In this case the photon transfers a maximum 
amount of energy to the electron so that a minimum of energy escapes.  Small 
peaks and bumps like that at D may be caused in various ways:

\begin{list}{(D\arabic{enumi})}{\usecounter{enumi}}
\item  Photoelectric interactions in the source itself or in the shielding can 
give K--shell X--rays.  These are then detected in the usual manner by the 
detector at energies around 20 to 50 keV.\label{D1}

\item  Gammas may Compton scatter (even backwards) from the shielding or 
photomultiplier glass and enter the crystal and be detected.   The energy is 
dependent upon the angle and the energy of the original gamma.

\item  A small group between B and A can be caused by a K--shell X--ray escaping 
from the crystal after a photoelectric interaction in the crystal.
\end{list}

All of the above defects (except \ref{D1}) are reduced by using larger (and more 
expensive) crystals which have a higher probability of both capturing the 
original gammas and also capturing scattered gammas which might otherwise 
escape.

\subsection*{\underline{A Typical Spectrum}}

The following figure shows a typical spectrum obtained with a 
5.1 cm $\times$ 5.1 cm
NaI scintillator and a Cs$^{137}$ source.\\
\epsfxsize=\textwidth \epsfbox{gamma_fig4.eps}
% \special{epsf=gamma_fig4.eps}
 
The Cesium--137 nucleus decays by the following scheme:\\
\epsfxsize=\textwidth \epsfbox{gamma_fig5.epsf}\\
%\special{epsf=gamma_fig5}\\
The Cs$^{137}$ emits an electron ($\beta$) plus a neutrino 
thereby increasing the 
nuclear charge from 55 to 56 by changing a neutron to a proton.
$${\rm Cs}^{137} \rightarrow {\rm Ba}^{137} + e^- + \bar\nu_e$$


In 92\% of the decays only 514 keV is transferred to 
the electron and neutrino
and the Ba$^{137}$ is left in a metastable, excited state.   
This decays ($T_{{1\over 2}}$ = 2.6 min.)
by either emitting a \underline{662 keV} gamma (90\%) 
or ``internally converting'' the gamma (10\%) before it leaves the 
Ba atom and ejecting a K shell electron instead.  In effect, the gamma 
interacts with the K shell electron as the gamma is created by the nucleus.  
In this case a fast electron is thrown out and then followed by a \underline{32 
keV} x--ray as the K shell of the Ba is refilled.

The spectrum shows the following features:
\begin{list}{\Alph{enumi}.}{\usecounter{enumi}}
\item
This is due to the complete absorption of 662 keV gamma, usually a Compton 
scatter followed by a second scatter and/or a photoelectric interaction.  The 
lead shielding collimates the incoming gammas so that all strike near the 
center of the NaI crystal and so are unlikely to escape without a second 
interaction.  The NaI absorption of 662 keV gammas follows the equation
$$N(x) = N_0 e^{-x/d}\quad.$$
%%
where d = 3.7 cm.  However scattered gammas are much more 
likely to interact.  For example, if the energy of the scattered gamma is 300 
keV then d = 1.6 cm. 
\item
``Compton Edge''.
The gammas which are scattered once  and then escape can transfer 
a maximum of 478 keV by being scattered backwards ($\theta =\pi$).
\item
This region is due to single Compton scatters with $0<\theta < \pi$.  The 
region extends from the Compton edge to zero energy.
\item
This peak is due to gammas which have been scattered through $0 < \theta < \pi$ 
outside the crystal.  For example some 662 keV gammas leave the source and 
strike the lead at the back of the cavity holding the source.  The 662 keV can 
be scattered forward with an energy of 184 keV and pass into the detector where 
it will be detected.  Alternatively a 662 keV gamma may pass through the 
crystal without interaction and then be backscattered from the front face of 
the photomultiplier back into the detector crystal.  
\item
The lead shielding can produce X--rays by electrons returning
to the  K--shell 
after they have been ejected by a photoelectric interaction with a gamma ray.  
The $K_\alpha$ X--rays from lead have an energy of 75 keV.
\item
This peak is due to the 32 keV X--rays produced by Ba$^{139}$ after internal 
conversion.
\item
Pulses smaller than 20 keV can be produced by photomultiplier noise and so the 
LLD of the detector has been set to remove these.
\end{list}

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\section*{\underbar{Useful Gammas}}\nopagebreak
%\thicksize=.25pt
%\thinsize=.25pt
%\begin{table}[h]
\begin{tabular}{|l|l|l|l|l|}
\hline  
${\rm Energy \atop (keV)}$ & \hfill Source & \hfill Lifetime & ${\rm \% \ 
Giving \atop Gamma}$ & \multicolumn{1}{c}{Origin or Comment} \\
\hline
2614.4* & Th$^{228}$ & \?\?\?1.9 y && Gives a range of 84 through 2614 keV \\
2204.2 & Ra$^{226}$ & 1600\x y && (From daughter Bi$^{214}$) \\
1769.7 & Ra$^{226}$ & 1600\x y && (From daughter Bi$^{214}$) \\
1332.5* & Co$^{60}$ & \?\?\?5.3 y & 100 & \\
1274.5 & Na$^{22}$ & \?\?\?2.6 y & 100 & \\
1173.2* & Co$^{60}$ & \?\?\?5.3 y & 100 & \\
\?661.6* & Cs$^{137}$ & \?\?30\x y & \?85 & \\
\?511.0* & Na$^{22}$ & \?\?\?2.6 y & 180 & From positron-electron annihilation
\\ 
\?356* & Ba$^{133}$ &\?\?10\x y && \\
\?302  & Ba$^{133}$ &\?\?10\x y && \\
\?280* & Se$^{75}$ & \?120\x d & \?25 & \\
\?265 & Se$^{75}$ & \?120\x d & \?60 & Replace each 2 years \\
\?136 & Se$^{75}$ & \?120\x d & \?57 & \\
\?122* & Co$^{57}$ & \?\?23\x d & \?98 & \\
\?\?87.7* & Cd$^{109}$ & \?\?\?1.3 y & \?\?1.0 & K-X-ray from Ag$^{109}$ after K 
capture \\
\?\?81.0* & Ba$^{133}$ & \?\?10\x y && \\
\?\?75.0 &&&& K-X-ray sometimes from Pb shielding \\
\?\?59.5* & Am$^{241}$ & \?458\x y & \?36 & \\
\?\?32.1* & Cs$^{137}$ & \?\?30\x y & \?\?6.8 & K-X ray from Ba$^{137}$ after
internal conv. \\ 
\?\?31 & Ba$^{133}$ & \?\?10\x y && K-X-ray from Cs$^{133}$ after K capture \\
\?\?22.1 & Cd$^{109}$ & \?\?\?1.3 y & \?25 & K-X-ray from Ag$^{109}$ after K
capture \\
\hline
\end{tabular} 
\begin{flushleft}
\hspace*{4.4cm} y = year\\
\hspace*{4.4cm} d = day\\[2ex]
The table of available gammas gives 18 energies. However, some are
closer together and since there is little point in using energies within 5\% of
each other, you should choose about 13 energies such as those marked *.
%\end{table}
\end{flushleft}
%%
\section*{\bf\underline{Apparatus}}

\bigskip
\subsubsection*{\underbar{Gamma Ray Sources}}

A number of radioactive sources (Th$^{228}$, Ra$^{226}$, Cs$^{137}$, 
Ba$^{133}$, Cd$^{109}$, Se$^{75}$, Co$^{60}$, Co$^{57}$ and Na$^{22}$) are 
available mounted in small plastic rods 
and are kept in a lead container in Room 
2507.  Most of them are supplied by Isotope Products Labs.  The sources are not 
strong (1--100 $\mu$Ci but they could interfere with the counting in this 
experiment and for this reason they should be kept inside the shielding when 
not in use.

\subsection*{\underbar{Gamma Ray Detector}}

The gamma ray detector is  a 
scintillation detector consisting of a 
scintillation crystal, a photomultiplier tube, and a preamplifier assembled in a 
single unit supplied by a Mech--Tronics Nuclear.  The crystal is a 2 inch 
diameter, 2 inch thick NaI crystal doped with thallium.  The photomultiplier is 
an RCA 8053 with 10 dynodes and a maximum voltage rating of 1500 volts.  The 
photomultiplier anode output is connected to an amplifier by a short length 
of coaxial cable. 

During operation, the crystal absorbs energy from a gamma event and 
produces a proportional flash of light.  The light flash causes the 
photomultiplier tube cathode to emit a proportional quantity of electrons.  
These are attracted from dynode to dynode through the tube with a 
multiplication effect at each successive dynode due to secondary emission.  The 
highly intensified burst of electrons which arrives at the anode of the tube, 
still proportional to the energy of origin, is transferred to form a charge at 
the input capacitor in the preamplifier.  The amplifier responds by creating 
a positive output pulse which retains the basic proportional significance.

There are two connectors on the photomultiplier tube base:
\begin{itemize}
\item
Anode Output:  Connects photomultiplier output pulses to the amplifier. This
output pulse is always negative and has an amplitude depending on the input 
pulse and the HV of the photomultiplier. The connection between the photomultiplier
output and the amplifier input is always made with a 50 ohm coaxial cable.
\item
HV:  Connects the high voltage power supply to the photomultiplier tube, using HV BNC 
for the white HV cable.
\end{itemize}

\subsection*{\underline{Amplifier}}

The amplifier is an ORTEC unit. Usually you will find the model 575A.
The amplifier accepts both positive or negative input pulses and provides both pulse 
shaping and amplitude expansion of these pulses. The amplifier gain is set
via a control knob setting and there are both unipolar positive and 
bipolar positive going outputs on BNC connectors. The function of the amplifier
is to produce positive output pulses of suitable amplitude and shape so
that they can be fed into the pulse height analyzer. 

\subsection*{\underline{Multi--Channel Pulse Height Analyzer}}

The multichannel analyzer is a PC containing an ORTEC model 916A pulse height
analyzer card (MCA) run by the Maestro II software package.
The MCA card contains an Analog to 
Digital Converter (ADC), single channel analyzer (SCA), multichannel scaler 
(MCS), and a dual--ported memory.  The card, along with the standard software, 
transforms the personal computer into a multichannel analyzer.  An input of 0 
to 10 volt positive pulses from a shaping amplifier is the only external signal 
necessary for pulse height analysis operation.  For further details 
consult the manual for the 916A and the manual for the Maestro II software
which runs the MCA board. The software is located in the MCA directory and
is started by typing {\bf MCA} on the command line. The MCA has been set
up for the input voltage range of 0 to 10 Volts to correspond to channel 0 (0 V)
and channel 512 (10 V). 

\subsection*{\underline{High Voltage Supply}}

The high voltage supply for the gamma ray detector is usually a Hewlett 
Packard Model 6522A using the positive HV output connector. 
The supply can furnish an output of $\pm$ 0 -- 2000 V DC. We normallly run the 
photomultiplier somewhere between 800-1000 Volts. 

\subsection*{\underline{Oscilloscope}}

Any mobile scope may be used to display the pulses being input to  the 
multi--channel pulse height analyzer.

\subsection*{\underline{Lead Collimators}}

\epsfxsize=\textwidth \epsfbox{gamma_fig6.epsf}
%\special{epsf=gamma_fig6}

The source is placed in the center of a lead cylinder so that gamma rays may 
pass freely:
\begin{itemize}
\item
through a side hole in the cylinder.
\item
through a second collimating hole in  a circular lead plate.
\item
through any absorber.
\item
through a third collimating hole in a second circular lead plate.
\item
through a fourth hole in the lead shielding the NaI detector.
\end{itemize}

The purpose of the 4 collimating holes is to insure that only
gammas which are not scattered may reach the detector.  Once a gamma ray has 
been scattered by the lead or by the absorber, then it is unlikely to reach the 
detector.  The NaI crystal has a diameter of 
5.1 cm but the entry hole in the 
lead has a diameter of only 3.2 cm.  This is to insure that the gammas 
which enter the NaI have a high probability of losing all their energy in the 
NaI.

\section*{\bf\underline{Procedure}}
\renewcommand{\thesubsection}{\Alph{subsection}.}
\subsection{%
Set up the system}
\begin{enumerate}
\item
Connect the ``Output'' connector of the detector to the DC input of the 
amplifier.  Set the input polarity switch to ``neg'' and use the ``Uni''
output. 
\item
Connect the output of the amplifier to the BNC connector labeled ``ADC'' on the 
MCA card located in the rear of the PC. 
\item
Connect the high voltage supply to the ``HV'' connector on the detector using 
the white HV cable with HV BNC connectors.  Turn it on and set to about +800 
volts to start. 
\item
Place a Cs$^{137}$ source with a strength near 100 $\mu$Ci in the cylinder and 
check the pulses from the preamplifier and amplifier
with a scope.  Note the amplitude and width of the pulses.
\end{enumerate}

\subsubsection*{\underline{Take Care}}

The gamma sources used in this experiment have strengths of up to 100 
$\mu$Ci.  Although these are relatively safe sources, you should act as if 
they are stronger.
\begin{enumerate}
\item
Never leave sources lying about.  They should be taken from their lead storage 
container directly to the cylindrical lead mounting or vice--versa.
\item
Handle the sources by the clear plastic rods.
\item
Report a missing source immediately.
\item
Use the survey meter to check radiation levels around the sources.
\end{enumerate}
%\vskip .25in
\subsection{Observe some typical spectra}
\begin{enumerate}
\item
Observe the spectrum of pulses from the Cs$^{137}$ source.  Adjust the 
amplifier gain so that the peak at 662 keV is located in the upper half of
of the available range. 
Collect enough data to smooth out the statistical fluctuations.  Print out the 
spectrum and tape it in your data book.  Identify the features of the spectrum.
\item
Collect and record the spectra from a few other isotopes, such as Na$^{22}$, 
Co$^{60}$, Ba$^{133}$, and Th$^{228}$.  Always adjust the gain so that the 
significant peaks are within the range of the analyzer.
\end{enumerate}

\subsection{Measure the attenuation in Aluminum of 662 keV gamma rays}
\begin{enumerate}
\item
Measure the intensity of gammas transmitted through 7 or 8 different 
thicknesses of 
Al using the gammas in the 662 keV peak from Cs$^{137}$.  Set up 
the 662 keV peak as a ``Region of Interest'' (ROI) and use ``Sum'' counts as 
your measure of intensity.  Does this properly correct for the background 
counting rate?
\item
Plot the intensity of the 662 keV peak as a function of thickness of A1 on 
semi--log paper.  Draw a straight line through the data points and from the 
slope of the line determine the linear attenuation coefficient of A1 for 662 keV 
gammas.  Taking the density of Al to be 2.70 g/cm$^3$, calculate the mass 
attenuation coefficient.  Compare your result to that quoted in Melissinos, p. 
208.
\item
Using SigmaPlot, fit the above data to a straight line with 
a least squares fit.  Have the computer calculate the slope and the standard 
deviation of the slope.  How does this value of the slope compare with that 
which you determined graphically?
\end{enumerate}

\subsection{Measure the attenuation in Al as a function of the energy, E, of 
the gammas.}
\begin{enumerate}
\item
Some gamma ray sources available in the lab and the energy of their gammas are 
listed in the accompanying table.  The most useful of these are indicated with 
an asterisk.
Measure the attenuation in Al of these gammas.  You need only measure 
the transmission for one thickness at each energy.  Because the transmission 
varies so much with energy, the thickness required for an accurate measurement 
will vary with energy.  The optimum thickness for determining the transmission 
with maximum accuracy in a given counting time is that thickness which gives a 
transmission of 37\%.  Can you prove this?  In order to get the various peaks 
at a reasonable channel position for measurement, you will have to change the 
gain of the amplifier as you change the energy of the gammas.
\item
Calculate the mass attenuation constant, $\mu_m$, at each energy and plot your 
results as a function of energy.
\end{enumerate}

\subsection{Measure the attenuation of 662 keV gammas in materials of 
different $Z$.}
\begin{enumerate}
\item
The absorbing materials available in the lab are listed in the accompanying 
table, along with their density, atomic number, and the energy corresponding to 
their K--shell absorption edge.  Measure the transmission of each of these 
materials for the 662 keV gamma ray from Cs$^{137}$.
\item
Calculate the mass attenuation coefficient for each material and plot your 
results as a function of the atomic number $Z$, of the material.
\end{enumerate}

\subsection{Comment on the following:}
\begin{enumerate}
\item
The $Z$ dependence of the Compton scattering at about 500 keV.  
\item
Any indication of pair production.
\item
The $Z$ and $E$ dependence of the photo--electric effect.
\end{enumerate}
Be prepared to answer questions on the following:
\begin{enumerate}
\item
The effectiveness of the lead shielding.
\item
The shape of any gamma spectrum.
\item
The principle of operation of a multichannel analyzer.
\item
The principle of operation of the photomultiplier.
\item
The K absorption edge of lead is at 88 keV yet it emits X--rays at only 75 keV.
\end{enumerate}
%%
\begin{tabular}{|c|c|c|c|}\hline
Absorber & {\bf Z}  & Density (gm/cm$^2$)  & K-shell Edge (KeV) \\ \hline\hline
Wax (CH$_2)_n$ &    &   2.71  &          \\
Lucite (${\rm C_5H_8O_2}$) &    & 1.16-1.20  &          \\
Be             & 4  &   1.848       & 0.111   \\
C (graphite)   & 6  &   1.5       & 0.284   \\
Al             & 13 &   2.70   & 1.56    \\
Fe	       & 26 &   7.86   & 7.114   \\
Ni             & 28 &   8.90   & 8.333   \\
Cu             & 29 &   8.96   & 8.999   \\	   
Mo             &  42&   10.2   & 20.00   \\
Sn             & 50 &   7.29   & 29.20   \\
W              & 74 &   19.1   & 69.52   \\
Au             & 79 &   19.3   & 80.72   \\
Hg             & 80 &   13.55       & 83.10   \\
Pb             & 82 &   11.35   & 88.10   \\
U (depleted 238)& 92 &  18.9       & 115.6   \\ \hline 
\end{tabular}

\section*{\bf\underline{References}}
\begin{enumerate}
\item
R.D. Evans, ``The Atomic Nucleus'' McGraw Hill, (1955).
\item
Eisberg and Resnick, ``Quantum Physics'' Wiley, (1974).
\item
R.D. Evans ``Gamma Rays'' Chap. 8e in ``American Institute of Physics 
Handbook'', 3rd Ed., McGraw--Hill, (1972).
\end{enumerate}
\newpage
\section*{\bf\underline{Appendix}}

\subsection*{\underline{Scintillator Materials}}
\begin{enumerate}
\item
\underline{{\bf Organic Scintillators}}  These are the cheapest.  Examples are:
\begin{enumerate}
\item
Anthracene crystals.
\item
Stilbene Crystals.
\item
Polystyrene with traces of Terphenyl and POPOP.
\item
Xylene or Toluene with traces of Terphenyl and POPOP.
\item
Various commercial plastic
scintillators  such as NE102. 
\end{enumerate}
Organic scintillators give fast light pulses - about 5 to 15 nsec long.  The 
basic material usually will absorb the light which it emits and so wavelength 
shifters are dissolved in the liquid (or plastic) to absorb the short 
wavelengths and re--emit them with longer  wavelengths.  The wavelength 
shifters are usually very complex aromatic substances and are used in very 
dilute concentrations.

The wavelength shifters also absorb UV (from sunlight or the fluorescent 
lights) and emit a very pleasant and characteristic blue glow.

Organic scintillators have 3 disadvantages:
\begin{enumerate}
\item
Their efficiency of conversion (particle energy loss to light output) is lower 
than for NaI (see below).
\item
Their conversion is not as precise as for NaI.  Identical particles giving the 
same energy loss E in an organic scintillator will result in various light 
outputs with a $\pm$ 10\% variation.
\item
When compared with inorganic scintillators such as NaI or CsI, organic 
scintillators have both lower mass densities and lower average atomic numbers 
Z.  Hence the organic scintillators interact less with gammas (to produce free 
electrons and ionization) and are usually not used as gamma detectors.
\end{enumerate}
\item
\underline{{\bf Inorganic Scintillators}}

The best are single crystals of ZnS, NaI, CsI or BaF.
The light output can be 
sometimes increased by doping (or ``activating'') the crystal with an element 
such as thallium.  This experiment uses a 
5.1cm $\times$ 5.1 cm cylinder of NaI with 
about 0.1\% Tl.  The Tl increases the light output by about 15\%.

Inorganic scintillators give  a larger and more precise light pulse than 
organic scintillators.  However, the light pulse is emitted over a longer time,
about 1 $\mu$sec and so NaI is not as useful for fast coincidence work as 
organic scintillators.

Read Melissinos 194-196.

\end{enumerate}

\end{document}
