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HENRY SHACKLETON: Be
a long 15 minutes.

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There you go.

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All right, hello everybody.

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My name's Henry and today,
I'll be presenting my findings

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on an experiment that I
conducted with my lab partner

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Adin over there.

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Thank you, Adin

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During our experiment,
we determined the angular

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dependency of the scattering
rate of alpha particles

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through gold foil.

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Using this relation, we
are able to determine

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that Rutherford's
scattering presents

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an accurate model for these
scattering of effects.

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In doing so, we are able
to extract some information

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about the actual nature of
atoms and their atomic makeup.

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So to give a bit
of background, I'm

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going to go through two
different historical models

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of the atom and talk about
the different experimental

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predictions that they predict.

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So back in the
early 19th century,

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people weren't too sure
what made up an atom.

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They knew there
were some electrons

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and some positive
stuff going around

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somewhere, but other than
that, they weren't too sure.

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Now, the predominant
theory at the time

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was JJ Thomson's Plum
Pudding Model, shown here.

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In this model, you have these
negatively charged electrons

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that sort of float around in
this soup of positive charge.

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I apologize for the plum
pudding and the soup.

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If you're getting hungry
from all these metaphors,

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there is food back there.

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Now, the experimental
predictions

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that we want to get
out of this model

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is what it says
about scattering.

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So if we take one of these
atoms and we shoot it through

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a material made
up of other atoms,

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what kind of scattering
do we expect to see ?

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And you can see in
this model here,

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we don't really expect
that much scattering.

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So the atom will go through.

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It might pass through a slightly
more negatively charged region

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here and move a bit that way,
deflect a little bit that way.

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But you don't expect that
much scattering, overall.

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Doing out all the math, we see
that this predicts small angle

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scattering that dies
off exponentially

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as a function of angle.

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This dying off is
controlled by this theta sub

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m called the mean
multiple scattering rate.

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For gold, we predict
this is about 1.

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So Rutherford comes along and
he looks at this and says,

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I don't know about this.

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Let me try something else.

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This right here is
the Rutherford model

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that he proposes.

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In this model we have our
negatively charged electrons.

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But instead of this
soup of positive charge,

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we have a densely packed
nucleus in the middle.

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Now because this nucleus is so
dense and so tightly packed,

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it allows for much stronger
interactions between atoms,

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when it comes to scattering.

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For example, you can imagine
if two atoms collided head on.

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The nuclei would
interact and they'd just

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bounce straight off.

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So doing out all
the math again, we

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see that this larger scattering
angle dies off as 1 over sine

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to the fourth of theta over 2.

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Now, this predicts much higher
rate at larger angles than this

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exponential fall
off, right here .

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A small point to clarify
about this equation

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here is that this equation is
derived assuming large angles.

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So for theta, roughly,
larger than or equal to 10.

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This is not a limit
of the Rutherford

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theory in and of itself.

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You can derive a
Rutherford model

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for scattering for
any angle, but it

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starts to get really
gross at small angles.

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You can actually
see this diverges

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for theta equals to
0, which is not good.

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So this equation
right here is only

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valid for theta roughly
greater than or equal to 10.

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So for the purposes
of this experiment,

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we will restrict our viewing
to theta greater than

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or equal to 10.

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And these two models give
us two different predictions

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that we can compare
against actual data.

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So we experimentally
measure scattering rates

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at various angles.

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And by taking this
data we can fit it

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to both the predicted
Rutherford and Thomsom model,

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and come to a conclusion
about which one more

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accurately predicts our model.

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Now let's talk a bit about the
apparatus that lets us do this.

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The first part of our efforts
is an alpha particle howitzer.

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This howitzer contains
an Americium-241 source

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which emits alpha
particles-- which

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are two protons and two
neutrons-- at, approximately,

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a constant energy.

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These alpha particles
shoot out of the howitzer

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right here in a sort of
concentrated beam directed out

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of the howitzer.

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These awful particles hit a gold
foil where something happens,

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we don't know yet.

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And it scatters off at an
angle theta, which is then

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detected by our state detector.

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This detector registers
the count, as well as

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the energy of the
incoming particle,

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and sends it to an
MCA to be analyzed.

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Now, there's a bit
of a complication

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in the geometry of our set up.

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You see, what we
want to measure is

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this angle, theta, here,
our scattering rate.

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What we actually
control and what

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we measure in our
experiments is this angle,

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phe which is the
angle of our howitzer

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relative to our detector.

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So at phe equals
to 0, the howitzer

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points dead on at our detector.

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Now, in a sort of
theoretical world

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where this solid state detector
is infinitesimally small

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and our beam is also this
infinitesimally point source,

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we would expect phe to
correspond exactly to theta.

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So at phe equals to 0, the
howitzer pointed dead on,

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we would expect any
non-zero scattering

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to cause the particles to
shoot off in a direction

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and not be detected
by our detector.

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However, in practice, this
is not actually the case.

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Our solid state detector
is a few centimeters wide.

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And because of this, it allows
for a range of angles theta

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that are detected given
a certain howitzer angle.

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In addition, our
beam has some width,

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which allows for further angles
that can be detected given

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a certain howitzer angle phe.

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So this is a bunch
of stuff going on.

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We've got a lot of different
things contributing

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to uncertainty.

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But we can account for
this by, essentially,

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asking the simple
question, all right, we

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have a howitzer at an angle phe.

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What is the probability
of a detection

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that we see having come from
a particle being scattered

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at an angle theta?

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Again, in this
theoretical case, where

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we have point sources
and point detectors,

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this probably would just
correspond to a delta function

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centered at phe.

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However this is not
practically the case.

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And from simple
geometric considerations,

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which I can go into
more detail later,

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we expect as a
first approximation

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a sort of triangle shaped
distribution around phe.

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This is called the
angular response function.

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And we determine it by
removing the gold foil

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from our detectors, so
we just have our howitzer

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and our detector, and we see
how deviations in our phe

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affect our counting rates.

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Now, again, to reference back
in this theoretical scenario

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with point sources
and point detectors,

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we would expect to only
see counts at phe equals

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to 0, with our howitzer
pointed dead on the detector.

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And as soon as we move it a
little bit, the counts go away.

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However, this is not
actually the case.

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And by measuring
these deviations,

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we can determine our
angular response function.

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So we move or
howitzer around, we

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determine how our angle
phe affects our counts,

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and we get something like this.

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This is fitted to a
triangle distribution.

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Now, we can take a few
things out of this.

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I see a fellow in the crowd
laughing at my chi squared.

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The chi squared
is not that good.

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This is because we acknowledge
that a triangle is only

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a first approximation.

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We don't expect our data
to actually perfectly model

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this triangle distribution.

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However, it simplifies
our math greatly.

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So we use this
triangle distribution,

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and try to extract
some data from this.

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The second thing that
we can see from this

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is that this distribution
is actually a bit off center

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at theta equals to negative 2.

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Now, normally, we would expect
the highest count rate when

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our howitzer is pointed dead
on at our detector at theta

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equals to 0.

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This offset indicates
that the protractor

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that we used to
measure our angles

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is actually a little bit off.

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And putting that on actually
corresponds to theta

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equals to 2.

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Finally, this gives us our
angular response function.

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If we replace zero with
our howitzer angle phe,

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this gives us the spread
around phe of scattering

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angles theta that we detect.

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We accommodate this in our
equations via a convolution.

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Now, this can
intuitively be thought

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as taking these two rates
which are given Thomson

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and Rutherford the probability
of having a scattering at angle

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theta.

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We then multiply it
by the probability

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of a scattering
angle theta being

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detected by our howitzer
positioned at an angle phe.

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And then we integrate
over all theta.

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This gives us two
different counting rates

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as a function of our
howitzer angle phe,

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one for our Rutherford theory,
and one for our Thomson theory.

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This gives us two equations that
we can use to compare our data,

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given geometric considerations
about our apparatus.

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So after that, we are
finally ready to go.

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So we take our gold
foil, we put it back in.

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We measure at different
angles between phe

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equals to 10 degrees
and 60 degrees.

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We let a detector
detect for a while.

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And over the period of time at
any one of our measuring rates,

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we get something that
looks a bit like this.

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Now, this MCA bin number
down here corresponds

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to the energy of the particles
that we're detecting.

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And on the y-axis,
here, is our count rate.

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Now, the first thing
that you might think,

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and the first thing that we
thought as well, is well,

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this is about a
Gaussian right here.

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That's our particle, and
everything below that

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is just noise.

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However, as we found out, the
energy loss through a material

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is actually described by a
Landau distribution, which

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is a Gaussian, but with
a slightly longer tail

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on one side.

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What you see here is a Gaussian
with a slightly longer tail

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on one side.

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So by fitting this
Landau distri--

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We fit this to a
Landau distribution

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and get chi squared's
between 0.5 and 2,

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which suggests that all
this did actually come

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from a Landau distribution.

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And all of these are valid
scattering data points, and not

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just noise.

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We further confirm this
by measuring our noise,

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by taking our howitzer pointed
away from our detector,

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and just letting our
detector collect.

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We ultimately determined that
the noise within our energy

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range of interest is very small,
much smaller than any count

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rate that we care about
in our experiment.

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Therefore, because of this and
the fact that our distribution

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models a Landau distribution,
which is what we expect,

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we conclude that we can
use all of the points

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that we detect as valid
scattering data points.

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But of course, it comes
with some good old counting

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uncertainty.

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In addition, we have some
uncertainty in our angles.

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We read the angles
of our howitzer

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by eye from a little protractor
that we have in our apparatus.

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00:10:36,450 --> 00:10:38,250
But our eyes aren't
so good, so we

251
00:10:38,250 --> 00:10:40,260
tack a plus or minus
1 degree uncertainty

252
00:10:40,260 --> 00:10:41,760
to angular measurements.

253
00:10:41,760 --> 00:10:43,560
My lab partner thinks 0.5.

254
00:10:43,560 --> 00:10:46,680
I think 1 because
I'm not as certain.

255
00:10:46,680 --> 00:10:48,416
So we take all
these measurements.

256
00:10:48,416 --> 00:10:50,040
We divide by the
amount of time that we

257
00:10:50,040 --> 00:10:52,110
collected to get a count rate.

258
00:10:52,110 --> 00:10:54,720
We graph this as a
function of our angle

259
00:10:54,720 --> 00:10:57,570
that we measure to that,
and fit this to a Rutherford

260
00:10:57,570 --> 00:10:59,640
and Thomson convolved fit.

261
00:10:59,640 --> 00:11:02,800
And we see something like this.

262
00:11:02,800 --> 00:11:06,450
And I think this really
speaks for itself.

263
00:11:06,450 --> 00:11:08,340
This is not an error in my code.

264
00:11:08,340 --> 00:11:12,390
Thomson is really just that
bad in accounting for our data.

265
00:11:12,390 --> 00:11:14,910
But we see here that the
Rutherford scattering

266
00:11:14,910 --> 00:11:18,690
prediction more accurately
describes the gradual falloff

267
00:11:18,690 --> 00:11:21,840
of our scattering rates as the
howitzer angle gets bigger, as

268
00:11:21,840 --> 00:11:27,070
opposed to the Thomson fit,
with the chi squared of 2000.

269
00:11:27,070 --> 00:11:29,790
Now, there are different ways
of potentially giving Thomson

270
00:11:29,790 --> 00:11:31,650
a better shot,
accommodating for it,

271
00:11:31,650 --> 00:11:34,350
may be giving it a better
chance of describing our data.

272
00:11:34,350 --> 00:11:36,210
And I can go into
this in more detail

273
00:11:36,210 --> 00:11:37,800
later, if people are interested.

274
00:11:37,800 --> 00:11:41,430
But, ultimately, we see that
Thomson is fundamentally unable

275
00:11:41,430 --> 00:11:42,950
to account for this data.

276
00:11:42,950 --> 00:11:45,870
Whereas Rutherford
describes it quite well.

277
00:11:45,870 --> 00:11:48,000
Now, there's an
additional caveat

278
00:11:48,000 --> 00:11:50,050
to this chi squared
that we have here.

279
00:11:50,050 --> 00:11:52,500
So this chi squared
comes from fitting

280
00:11:52,500 --> 00:11:56,970
from our Rutherford and
Thomson data convolved

281
00:11:56,970 --> 00:11:58,750
with our beam profile.

282
00:11:58,750 --> 00:12:00,600
However, our beam
profile does have

283
00:12:00,600 --> 00:12:01,860
some uncertainty associated.

284
00:12:01,860 --> 00:12:04,320
In that fit that we had,
there is some uncertainty

285
00:12:04,320 --> 00:12:05,830
in the triangle.

286
00:12:05,830 --> 00:12:09,570
So to sort of estimate how this
uncertainty affects our data,

287
00:12:09,570 --> 00:12:13,170
we vary our beam profile
within one standard deviation

288
00:12:13,170 --> 00:12:15,780
of its fit parameters, and
see how our goodness of fit

289
00:12:15,780 --> 00:12:16,810
changes.

290
00:12:16,810 --> 00:12:19,500
And we see here that
the goodness of fit

291
00:12:19,500 --> 00:12:22,170
for a Rutherford model,
while it varies somewhat

292
00:12:22,170 --> 00:12:25,890
depending on the
convolution that we use,

293
00:12:25,890 --> 00:12:28,320
it doesn't change a
significant amount.

294
00:12:28,320 --> 00:12:30,840
And it certainly doesn't
change enough to warrant

295
00:12:30,840 --> 00:12:32,550
considering the Thomson model.

296
00:12:32,550 --> 00:12:34,260
The discrepancy between
the chi squareds

297
00:12:34,260 --> 00:12:37,080
and the visual discrepancy
you see here very

298
00:12:37,080 --> 00:12:38,910
clearly suggests
that Rutherford does

299
00:12:38,910 --> 00:12:42,250
a much better job of predicting
our data than Thomson.

300
00:12:42,250 --> 00:12:46,620
Which, again, suggests a more
Rutherford make up of the atom,

301
00:12:46,620 --> 00:12:50,020
as opposed to the Thomson
Plum Pudding model.

302
00:12:50,020 --> 00:12:52,500
So in conclusion, Thomson's
Plum Pudding model

303
00:12:52,500 --> 00:12:55,650
is found to fundamentally
be unable to describe

304
00:12:55,650 --> 00:12:58,860
the scattering rates that
we observe at large angles.

305
00:12:58,860 --> 00:13:00,990
The Rutherford model
more accurately predicts

306
00:13:00,990 --> 00:13:06,000
this gradual fall off, which
suggests a more Rutherford-like

307
00:13:06,000 --> 00:13:08,230
view of our atom.

308
00:13:08,230 --> 00:13:08,940
And that's it.

309
00:13:08,940 --> 00:13:09,717
Thank you.

310
00:13:09,717 --> 00:13:17,350
[APPLAUSE]

311
00:13:17,350 --> 00:13:22,489
INSTRUCTOR: All right, the
paper is open for questions.

312
00:13:22,489 --> 00:13:23,447
HENRY SHACKLETON: Yeah.

313
00:13:23,447 --> 00:13:27,085
AUDIENCE: Did you try the fit
without using the triangular

314
00:13:27,085 --> 00:13:27,644
correction?

315
00:13:27,644 --> 00:13:28,810
HENRY SHACKLETON: Yes I did.

316
00:13:28,810 --> 00:13:31,930
And this is another
interesting point, which

317
00:13:31,930 --> 00:13:35,350
shows that our convolution
does benefit our data,

318
00:13:35,350 --> 00:13:38,680
does a much better-- not a much
better, not as bad as Thomson,

319
00:13:38,680 --> 00:13:39,570
relatively speaking.

320
00:13:39,570 --> 00:13:41,950
But it does a
noticeably better job

321
00:13:41,950 --> 00:13:46,020
at describing our data than the
pure Rutherford scattering, 1

322
00:13:46,020 --> 00:13:49,051
over sine to the fourth fit.

323
00:13:49,051 --> 00:13:52,390
AUDIENCE: So the
difference between 2.14--

324
00:13:52,390 --> 00:13:54,660
HENRY SHACKLETON:
And, sorry, 8.18.

325
00:13:54,660 --> 00:13:56,002
AUDIENCE: Oh, I see it.

326
00:13:56,002 --> 00:13:57,960
HENRY SHACKLETON: That's
a chi squared of eight

327
00:13:57,960 --> 00:14:00,084
for just fitting with the
raw Rutherford scattering

328
00:14:00,084 --> 00:14:01,565
without the convolution.

329
00:14:04,540 --> 00:14:05,776
Yes.

330
00:14:05,776 --> 00:14:08,584
AUDIENCE: Do you have any
plots that show your Landau

331
00:14:08,584 --> 00:14:10,930
distribution fits to the data?

332
00:14:10,930 --> 00:14:12,347
HENRY SHACKLETON:
I thought I did.

333
00:14:12,347 --> 00:14:13,805
You saw me switch
forward and then,

334
00:14:13,805 --> 00:14:15,070
realize that it wasn't there.

335
00:14:15,070 --> 00:14:16,280
So I switched back.

336
00:14:16,280 --> 00:14:18,640
No, so I don't have a
picture of the Landau fit.

337
00:14:18,640 --> 00:14:20,140
One thing that I
can describe, which

338
00:14:20,140 --> 00:14:24,850
is a small caveat to this, is
that the Landau distribution

339
00:14:24,850 --> 00:14:26,160
describes energy loss.

340
00:14:26,160 --> 00:14:30,210
And this is usually a
Gaussian with one tail on--

341
00:14:30,210 --> 00:14:31,450
this isn't too well.

342
00:14:31,450 --> 00:14:34,450
It's a Gaussian with a slightly
longer tail on the right hand

343
00:14:34,450 --> 00:14:35,115
side.

344
00:14:35,115 --> 00:14:37,240
Now, this is a Gaussian
with a slightly longer tail

345
00:14:37,240 --> 00:14:38,200
on the left hand side.

346
00:14:38,200 --> 00:14:40,600
However, what we
see here is energy.

347
00:14:40,600 --> 00:14:43,660
And energy loss is sort
of the inverse of this.

348
00:14:43,660 --> 00:14:46,360
So we actually fit this to a
reverse Landau distribution.

349
00:14:46,360 --> 00:14:50,770
And throughout all our data,
we get reduced chi squareds

350
00:14:50,770 --> 00:14:53,850
between 0.5 and 2, which
indicates a good fit.

351
00:14:57,746 --> 00:15:02,129
INSTRUCTOR: A different
question for Henry?

352
00:15:02,129 --> 00:15:02,930
All right, let's--

353
00:15:02,930 --> 00:15:04,301
HENRY SHACKLETON: Yep, Shawn?

354
00:15:04,301 --> 00:15:06,209
SHAWN: What was the
largest angle you

355
00:15:06,209 --> 00:15:07,500
were able to use the data from?

356
00:15:07,500 --> 00:15:07,860
HENRY SHACKLETON: Sorry?

357
00:15:07,860 --> 00:15:09,234
SHAWN: What was
the largest angle

358
00:15:09,234 --> 00:15:10,732
you were able to get your data?

359
00:15:10,732 --> 00:15:11,940
HENRY SHACKLETON: 60 degrees.

360
00:15:11,940 --> 00:15:15,847
60 degrees, we collected
over the course of five days.

361
00:15:15,847 --> 00:15:17,930
And I don't think we could
have actually collected

362
00:15:17,930 --> 00:15:23,150
data at much higher angles
than that because our noise

363
00:15:23,150 --> 00:15:26,390
rate right here is 0.0002.

364
00:15:26,390 --> 00:15:29,830
Now, this is still a bit
smaller than our counting

365
00:15:29,830 --> 00:15:30,902
rate for 60 degrees.

366
00:15:30,902 --> 00:15:32,360
But if we started
to go further, we

367
00:15:32,360 --> 00:15:35,570
would start to enter the regime
where noise becomes much more

368
00:15:35,570 --> 00:15:36,869
dominant.

369
00:15:36,869 --> 00:15:39,119
SHAWN: Do you happen to know
how many total counts was

370
00:15:39,119 --> 00:15:40,766
that over five days?

371
00:15:40,766 --> 00:15:42,140
HENRY SHACKLETON:
Over five days?

372
00:15:42,140 --> 00:15:44,150
Off the top of my
head, I believe 400.

373
00:15:44,150 --> 00:15:45,637
You can actually see--

374
00:15:45,637 --> 00:15:47,720
I know the rate here, which
you can very easily go

375
00:15:47,720 --> 00:15:49,462
backwards in the math, I guess.

376
00:15:49,462 --> 00:15:50,420
SHAWN: So if the rate--

377
00:15:50,420 --> 00:15:51,425
HENRY SHACKLETON: This
is somewhere down here.

378
00:15:51,425 --> 00:15:52,730
SHAWN: --100 counts per day--

379
00:15:52,730 --> 00:15:53,540
HENRY SHACKLETON: Yeah,
something like that.

380
00:15:53,540 --> 00:15:55,249
SHAWN: Or a couple
counts per hour.

381
00:15:55,249 --> 00:15:57,040
HENRY SHACKLETON: Yes,
something like that.

382
00:15:57,040 --> 00:15:59,557
SHAWN: All right, thank you.

383
00:15:59,557 --> 00:16:01,390
The noise that we
mentioned here is actually

384
00:16:01,390 --> 00:16:03,070
restricted to our
energy of interest.

385
00:16:03,070 --> 00:16:05,560
So there's a fair bit of
noise at much lower energy

386
00:16:05,560 --> 00:16:10,240
levels, specifically
between this MCA bin

387
00:16:10,240 --> 00:16:12,610
number of 0 and 50, down here.

388
00:16:12,610 --> 00:16:14,080
There's a fair bit
of noise there.

389
00:16:14,080 --> 00:16:16,580
But once you start getting up
to this larger area right here

390
00:16:16,580 --> 00:16:20,050
between 50 and 1500, we
start to see much less noise.

391
00:16:24,550 --> 00:16:27,550
INSTRUCTOR: Another question?

392
00:16:27,550 --> 00:16:29,050
All right, let's
thank Henry again.

393
00:16:29,050 --> 00:16:30,898
[APPLAUSE]