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\title{Physics 54 \\
Modern Physics Laboratory}

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\experiment{Rutherford Scattering}

By 1911 general agreement existed that atoms contain a small number
of electrons with most of the atomic mass associated with positive
charge. The problem was to determine how the positive charge and mass
are distributed. Two extreme views were proposed by J.\ J.\ Thomson and
Ernest Rutherford. Thomson considered the atom to be made of a space
filling sphere of positive charge in which the electrons were
embedded--the ``plum pudding'' model. Rutherford considered the
positive charge and mass to be contained within a central, very dense
nucleus---the ``nuclear atom'' model.

The test of these views was suggested by Rutherford and carried out
by H.\ Geiger and E.\ Marsden in 1913. The experiment is the prototype
for a great many contemporary ``particle experiments'' of the
so-called ``scattering'' type. Experiments by Hofstadter,
\textit{et al.}, on the special distribution of charge within the nucleus
itself are of this type. The experimental procedure is to send known
particles (known mass, charge, etc.) with a given momentum into a
thin target of the material under investigation and to observe the
scattering (the change of momentum) of the emergent beam. Given any
model of the target such that the forces arising between the particle
and the target are known, the expected scattering can be calculated.
The observed scattering then serves to eliminate those models for
which the predictions disagree with experiment. Rutherford's
particles were alpha particles of relatively low energy arising in
natural radioactive decay. Since only electromagnetic forces are
significant in this case, the experiments served to eliminate models
of the positive charge distribution in an atom. The plum pudding
model was definitely crossed off. The nuclear atom model, on the
other hand, predicted results in very good agreement with the data. 


\begin{enumerate}
  \item The Rutherford model with which the results of this
experiment are compared is that of a positive charge distribution
which is represented as a point charge of magnitude $Ze$, where $Z$ is
the atomic number of the target material. The mass distribution was
considered to be the same as that of the charge or, at any rate, the
center of mass was assumed rigidly attached to the point charge. The
predicted angular distribution of particles of mass $m$ and charge $Z'e$
scattered from an incident beam of particles with velocity $\vv$ by atoms
of atomic number $Z$ and mass $M$ initially at rest is 
\begin{equation}
	\label{eq:Ruth1}
	\sigma(\theta) = \left( \frac{Z Z' e^{2}}{8 \pi \epsilon_{0} \mu
        \nu_{0}^{2}} \right)^{2}
    \frac{1}{\sin^{4}(\theta/2)} \qquad\qquad
    \mu = \frac{m M}{m + M} 
\end{equation}
The ``cross section,'' $\sigma(\theta)$, is a measure of the
probability that an incident particle will be scattered in a single
collision into the angle $\theta$ to $\theta + \DD\theta$ measured
with respect to the direction of the initial velocity.

\begin{figure}[t]
  \centering
  \igw[5in]{Rutherford}
  \caption{Scattering geometry.}
  \label{fig:Rutherford}
\end{figure}

A sketch of the relation of the source, scatterer, and detector of the
alpha particles in the laboratory apparatus is shown in
\fref{Rutherford}.  The apparatus may be disassembled at the flanged
end by removing the four knurled nuts. \scream{First}, however, read
the following description.  The source and scatterer are mounted
together in a movable cage such that the angle $\beta$ is fixed. The
source is radioactive americium 241, which decays primarily by
emitting a \hval{5.29}{MeV} alpha particle.%
\footnote{A thin cover over the radioactive source reduces the energy
  of the alpha particle somewhat.} The scatterer is an annulus of gold
foil about \val{3.5}{\mu m} thick. Neither the americium source nor the
gold foil may be touched, for obvious reasons.

The detector is a solid-state device consisting of a silicon wafer
with a thin (\val{0.02}{\mu m}) gold surface covering on one side and
an aluminum surface on the other side. A potential difference of
\val{30}{V} is placed across this ``sandwich.'' When an ionizing
particle passes through the silicon, electrons are ejected by
collision with the particle from the filled band to the empty
conduction band of the silicon semiconductor. Both the electrons in
the conduction band and the ``holes'' left in the valence band move
under the applied field: the electron to the gold surface, the holes
to the aluminum. Hence a pulse of charge is collected, with size
proportional to the number of electrons injected into the conduction
band, and thus to the energy loss of the ionizing particle in the
silicon. You will count this pulse of charge with a scaler after it is
amplified.  Further details are given in the appendix to this
experiment.

\caution{Do not touch the detector! The gold coating is fragile, the
silicon can be ruined by contamination, and static electricity could
damage the detector irreversibly.}

The distance from the scattering foil to the detector may be varied
from about 1 to 20~cm by means of the vacuum sealed plunger attached
to the source cage and extending outside the apparatus. The scattering
angle $\theta$ may thus be varied from about \dg{27} to \dg{90}.

Since the range of the alpha particles in air at normal pressure is
only a few centimeters, it is necessary to evacuate the entire
apparatus. The brass vacuum chamber is closed at one end
by the sliding plunger and flange. The other end is closed by the
mounting bracket of the detector seated against an O-ring seal.

Current pulses from the silicon detector generate voltage pulses in
the amplifier circuit. These pulses are counted by a scaler. The
experiment consists in determining the number of counts registered by
the scaler in a measured time interval as the source cage plunger is
moved in or out to vary the scattering angle $\theta$.

\item  Carefully study the apparatus prior to its evacuation. You will be
given the minimum value of $d$ (i.e., when the plunger is in as far as
possible) for your apparatus. You will need this value together with
your measurements of the external position of the plunger to compute
the scattering angle $\theta$ and to correct for changes in the
detector solid angle (see below).  Begin collection of data with the
plunger withdrawn as far as possible to measure the counting rate for
the smallest scattering angle. Record the time necessary to accumulate
at least 100 counts at all scattering angles.\footnote{In taking data,
  choose intermediate plunger positions in light of the plot you will
  be making. (See below)} The standard deviation for $N$ counts is
$\sqrt{N}$ so that 10\% statistics are obtained with 100 counts. The
counting rate at minimum scattering angle will probably be of the
order of 30 counts per minute, falling to some 4 counts per minute at
the largest angles.

\begin{figure}[h]
  \centering
  \igw[2.5in]{Rutherford3}
\end{figure}

The counting rate must be corrected for the change in the solid angle
subtended by the detector at the gold annulus. The apparent size of
the detector as seen from the annulus is a function of their
separation, $d$. Ignoring the finite size of the detector and the
annulus width, this correction consists of two factors.  First, the
detector size would vary as $1/d^{2}$ were it viewed ``head on'' from
the annulus.  This is very nearly the case when $d$ is much greater
than the radius of the annulus. For small separations, however, the
projection of the detector into the line of sight from the annulus
must be taken into account. The projected area goes as $\cos \alpha$,
or as $d/D$. Combining these two factors, the apparent size of the
detector varies as $\displaystyle \frac{d}{D} \, \frac{1}{D^{2}}$.

The counting rate is multiplied by the reciprocal of this factor to
obtain a counting rate proportional to that which would have been
measured with a detector whose size appeared always the same to the
scattering annulus. The counting rate corrected for solid angle is
proportional to the cross section $\sigma(\theta)$. 

To compare your results with the predictions of the Rutherford model,
plot the logarithm of the corrected counting rate vs. the logarithm
of $\sin (\theta/2)$. (What should this plot look like according to the
Rutherford model?) Enter your data in this plot with bars to indicate
the standard deviations resulting from counting statistics. 
\end{enumerate}


\section[Detector]{The EGG Ortec Silicon Charged Particle Detector}

Silicon is a semiconductor with a gap of 1.1 electron volts between
the top of the filled band and the bottom of the (nearly empty)
conduction band. At any temperature above absolute zero, some
electrons will have enough thermal energy to reach the conduction
band; for the detector you use, with 30 volts potential difference
across the silicon wafer, this gives rise to a ``dark current'' of
about $\val{300}{nA}$ or $\val{1 \times
10^{12}}{electrons/second}$. (Incidentally, since the silicon wafer
is about $\val{150}{\mu m}$ thick, the electric field is
$\val{30}{V}/\val{1.5 \times 10^{-4}}{m} = \val{200,000}{V/m}$.)

When there is no voltage across the silicon wafer, the Fermi energies
(see Eisberg and Resnick, Chapter 13, pp.\ 507 \textit{et seq.}) of the
electrons in the gold, aluminum, and silicon are equal; electrons
move between these layers to change the potential of these layers
until this equality is reached. The particular silicon wafer we use
has donor impurities (see E \& R, p.\ 507), so the Fermi energy in
the silicon lies \val{0.16}{eV} below the bottom of the
conduction band. There are, accordingly, thermally injected electrons
in the conduction band. Once the \val{30}{V} power supply is turned on,
these electrons are swept away, giving rise to the ``dark current.'' 

\begin{figure}[b!]
  \centering
  \igw[4.5in]{Rutherford2}
  \caption{Circuit}
  \label{fig:RutherfordCircuit}
\end{figure}

When an $\alpha$ particle enters the silicon, it collides with
electrons in the silicon lattice, giving many of them enough energy to
reach the conduction band. The average energy lost by the
$\alpha$-particle to create an electron-hole pair is measured to be
\val{3.6}{eV}. Thus a \val{5}{MeV} $\alpha$-particle, completely
stopped in the silicon, gives rise to $5 \times 10^{6}/3.6 = 1.4
\times 10^{6}$ electron-hole pairs, or $2.2 \times 10^{-13}$
coulombs. The capacitance of the detector is 70 picofarads (\val{7
  \times 10^{-11}}{F}), so collecting this charge causes a voltage
change $\Delta V = \pval{2 \times 10^{-13}}{C}/ \pval{7 \times
  10^{-11}}{F} = \val{3}{mV}$. The detector voltage is supplied
through a \hval{20}{M\Omega} resistor, so the recovery time is $RC =
\pval{7 \times 10^{-4}}{F} \val{2 \times 10^{7}}{\Omega} = \val{1.4}{ms}$.

\fref[f]{RutherfordCircuit} shows a circuit diagram of the detector,
its power supply, and the first (preamplifier) stage of amplification.
Here $R_{L}$ is the ``equivalent resistance'' of the silicon wafer;
since the ``dark current'' is about $\val{3 \times 10^{-7}}{A}$ for a
potential of $\val{30}{V}$, $R_{L} = \val{100}{M\Omega}$. The Model
109A preamplifier set at $10 \times$ gain gives a pulse of
$\val{150}{mV/MeV}$ for a Si detector. This preamp also reduces the
pulse width to approximately $\val{50}{\mu s}$. The amplifier
following the preamp further reduces the pulse width and
increases the peak voltage.


\section{Addendum}

Since the laboratory notes were written, the brass cylinder has been
replaced by a plastic (Lexan) cylinder, which is semitransparent and
slightly shorter. Accordingly, the angle of scattering of the alpha
particles must be calculated with the new dimensions.  The dimension
you need to know is $d$ \emph{Page 3-3} of the notes. $d$ the distance
from the plane of the gold scattering foil to the surface of the
detector. It is also the distance between the two knurled knobs, one
the handle on the plunger and the other the one the rod slides
through, less 0.08 cm. We have placed a sleeve of length 2.00 cm on
the plunger rod (to keep a bump on the can carrying the americium
source and the foil from striking the detector), and therefore the
shortest $d$ available to you is 1.92 cm.

\smallskip
Why the Lexan cylinder, and these changes? 
\smallskip

For reasons I did not understand, this experiment usually produced an
exponent in the range of $-4.3$ to $-4.5$ instead of the $-4$ which
Professor Rutherford had in mind. Mark Chalice, '94, asked me two
years ago if any of the alpha particles which go through the foil
undeflected might then strike the brass cylinder wall and be
scattered there. Indeed, most of the alpha particles that strike the
foil do go through essentially undeflected, having lost some energy
by many collisions with electrons, thus ionizing gold atoms. These
alpha particles enter the brass. Most spend out their range losing
energy in more electron collisions, but a few of them may indeed be
scattered by the copper and zinc nuclei of atoms which make up brass.
From the cross-section equation on Page 3-1 of the notes, we see
that the scattering cross section depends on $Z^{2}$. For gold, $Z = 79$; for
copper, $Z =29$, and for zinc, $Z =30$. Accordingly, these brass nuclei
are only about 14\% as effective as gold in Rutherford scattering, but
the path length in the brass can be considerable. From the same
equation on Page 3-1, we also learn that as the alpha particle slows
down, the probability of scattering increases. Thus the cylinder
walls in front of the gold foil may constitute a significant second
scatterer. The angle of scattering at which the alpha particle is
detected is greater for these brass-scattered alphas, and hence they
are no longer much detected as the foil nears the detector.
Accordingly, we are led to believe that the power law is greater than
4. 

The solution to the problem is not to use brass, but a plastic, for
which the atoms in the wall are predominately carbon, hydrogen and
oxygen. These have at most 1\% of the scattering cross section of
gold. Essentially all the alpha particles striking the wall lose
their kinetic energy through electron collisions and are not
scattered.

\begin{flushright}
  J.\ B.\ Platt, January 1994
\end{flushright}


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