HENRY SHACKLETON: Be a long 15 minutes.
There you go.
All right, hello everybody.
My name's Henry and today, I'll be presenting my findings
on an experiment that I conducted with my lab partner
Adin over there.
Thank you, Adin
During our experiment, we determined the angular
dependency of the scattering rate of alpha particles
through gold foil.
Using this relation, we are able to determine
that Rutherford's scattering presents
an accurate model for these scattering of effects.
In doing so, we are able to extract some information
about the actual nature of atoms and their atomic makeup.
So to give a bit of background, I'm
going to go through two different historical models
of the atom and talk about the different experimental
predictions that they predict.
So back in the early 19th century,
people weren't too sure what made up an atom.
They knew there were some electrons
and some positive stuff going around
somewhere, but other than that, they weren't too sure.
Now, the predominant theory at the time
was JJ Thomson's Plum Pudding Model, shown here.
In this model, you have these negatively charged electrons
that sort of float around in this soup of positive charge.
I apologize for the plum pudding and the soup.
If you're getting hungry from all these metaphors,
there is food back there.
Now, the experimental predictions
that we want to get out of this model
is what it says about scattering.
So if we take one of these atoms and we shoot it through
a material made up of other atoms,
what kind of scattering do we expect to see ?
And you can see in this model here,
we don't really expect that much scattering.
So the atom will go through.
It might pass through a slightly more negatively charged region
here and move a bit that way, deflect a little bit that way.
But you don't expect that much scattering, overall.
Doing out all the math, we see that this predicts small angle
scattering that dies off exponentially
as a function of angle.
This dying off is controlled by this theta sub
m called the mean multiple scattering rate.
For gold, we predict this is about 1.
So Rutherford comes along and he looks at this and says,
I don't know about this.
Let me try something else.
This right here is the Rutherford model
that he proposes.
In this model we have our negatively charged electrons.
But instead of this soup of positive charge,
we have a densely packed nucleus in the middle.
Now because this nucleus is so dense and so tightly packed,
it allows for much stronger interactions between atoms,
when it comes to scattering.
For example, you can imagine if two atoms collided head on.
The nuclei would interact and they'd just
bounce straight off.
So doing out all the math again, we
see that this larger scattering angle dies off as 1 over sine
to the fourth of theta over 2.
Now, this predicts much higher rate at larger angles than this
exponential fall off, right here .
A small point to clarify about this equation
here is that this equation is derived assuming large angles.
So for theta, roughly, larger than or equal to 10.
This is not a limit of the Rutherford
theory in and of itself.
You can derive a Rutherford model
for scattering for any angle, but it
starts to get really gross at small angles.
You can actually see this diverges
for theta equals to 0, which is not good.
So this equation right here is only
valid for theta roughly greater than or equal to 10.
So for the purposes of this experiment,
we will restrict our viewing to theta greater than
or equal to 10.
And these two models give us two different predictions
that we can compare against actual data.
So we experimentally measure scattering rates
at various angles.
And by taking this data we can fit it
to both the predicted Rutherford and Thomsom model,
and come to a conclusion about which one more
accurately predicts our model.
Now let's talk a bit about the apparatus that lets us do this.
The first part of our efforts is an alpha particle howitzer.
This howitzer contains an Americium-241 source
which emits alpha particles-- which
are two protons and two neutrons-- at, approximately,
a constant energy.
These alpha particles shoot out of the howitzer
right here in a sort of concentrated beam directed out
of the howitzer.
These awful particles hit a gold foil where something happens,
we don't know yet.
And it scatters off at an angle theta, which is then
detected by our state detector.
This detector registers the count, as well as
the energy of the incoming particle,
and sends it to an MCA to be analyzed.
Now, there's a bit of a complication
in the geometry of our set up.
You see, what we want to measure is
this angle, theta, here, our scattering rate.
What we actually control and what
we measure in our experiments is this angle,
phe which is the angle of our howitzer
relative to our detector.
So at phe equals to 0, the howitzer
points dead on at our detector.
Now, in a sort of theoretical world
where this solid state detector is infinitesimally small
and our beam is also this infinitesimally point source,
we would expect phe to correspond exactly to theta.
So at phe equals to 0, the howitzer pointed dead on,
we would expect any non-zero scattering
to cause the particles to shoot off in a direction
and not be detected by our detector.
However, in practice, this is not actually the case.
Our solid state detector is a few centimeters wide.
And because of this, it allows for a range of angles theta
that are detected given a certain howitzer angle.
In addition, our beam has some width,
which allows for further angles that can be detected given
a certain howitzer angle phe.
So this is a bunch of stuff going on.
We've got a lot of different things contributing
to uncertainty.
But we can account for this by, essentially,
asking the simple question, all right, we
have a howitzer at an angle phe.
What is the probability of a detection
that we see having come from a particle being scattered
at an angle theta?
Again, in this theoretical case, where
we have point sources and point detectors,
this probably would just correspond to a delta function
centered at phe.
However this is not practically the case.
And from simple geometric considerations,
which I can go into more detail later,
we expect as a first approximation
a sort of triangle shaped distribution around phe.
This is called the angular response function.
And we determine it by removing the gold foil
from our detectors, so we just have our howitzer
and our detector, and we see how deviations in our phe
affect our counting rates.
Now, again, to reference back in this theoretical scenario
with point sources and point detectors,
we would expect to only see counts at phe equals
to 0, with our howitzer pointed dead on the detector.
And as soon as we move it a little bit, the counts go away.
However, this is not actually the case.
And by measuring these deviations,
we can determine our angular response function.
So we move or howitzer around, we
determine how our angle phe affects our counts,
and we get something like this.
This is fitted to a triangle distribution.
Now, we can take a few things out of this.
I see a fellow in the crowd laughing at my chi squared.
The chi squared is not that good.
This is because we acknowledge that a triangle is only
a first approximation.
We don't expect our data to actually perfectly model
this triangle distribution.
However, it simplifies our math greatly.
So we use this triangle distribution,
and try to extract some data from this.
The second thing that we can see from this
is that this distribution is actually a bit off center
at theta equals to negative 2.
Now, normally, we would expect the highest count rate when
our howitzer is pointed dead on at our detector at theta
equals to 0.
This offset indicates that the protractor
that we used to measure our angles
is actually a little bit off.
And putting that on actually corresponds to theta
equals to 2.
Finally, this gives us our angular response function.
If we replace zero with our howitzer angle phe,
this gives us the spread around phe of scattering
angles theta that we detect.
We accommodate this in our equations via a convolution.
Now, this can intuitively be thought
as taking these two rates which are given Thomson
and Rutherford the probability of having a scattering at angle
theta.
We then multiply it by the probability
of a scattering angle theta being
detected by our howitzer positioned at an angle phe.
And then we integrate over all theta.
This gives us two different counting rates
as a function of our howitzer angle phe,
one for our Rutherford theory, and one for our Thomson theory.
This gives us two equations that we can use to compare our data,
given geometric considerations about our apparatus.
So after that, we are finally ready to go.
So we take our gold foil, we put it back in.
We measure at different angles between phe
equals to 10 degrees and 60 degrees.
We let a detector detect for a while.
And over the period of time at any one of our measuring rates,
we get something that looks a bit like this.
Now, this MCA bin number down here corresponds
to the energy of the particles that we're detecting.
And on the y-axis, here, is our count rate.
Now, the first thing that you might think,
and the first thing that we thought as well, is well,
this is about a Gaussian right here.
That's our particle, and everything below that
is just noise.
However, as we found out, the energy loss through a material
is actually described by a Landau distribution, which
is a Gaussian, but with a slightly longer tail
on one side.
What you see here is a Gaussian with a slightly longer tail
on one side.
So by fitting this Landau distri--
We fit this to a Landau distribution
and get chi squared's between 0.5 and 2,
which suggests that all this did actually come
from a Landau distribution.
And all of these are valid scattering data points, and not
just noise.
We further confirm this by measuring our noise,
by taking our howitzer pointed away from our detector,
and just letting our detector collect.
We ultimately determined that the noise within our energy
range of interest is very small, much smaller than any count
rate that we care about in our experiment.
Therefore, because of this and the fact that our distribution
models a Landau distribution, which is what we expect,
we conclude that we can use all of the points
that we detect as valid scattering data points.
But of course, it comes with some good old counting
uncertainty.
In addition, we have some uncertainty in our angles.
We read the angles of our howitzer
by eye from a little protractor that we have in our apparatus.
But our eyes aren't so good, so we
tack a plus or minus 1 degree uncertainty
to angular measurements.
My lab partner thinks 0.5.
I think 1 because I'm not as certain.
So we take all these measurements.
We divide by the amount of time that we
collected to get a count rate.
We graph this as a function of our angle
that we measure to that, and fit this to a Rutherford
and Thomson convolved fit.
And we see something like this.
And I think this really speaks for itself.
This is not an error in my code.
Thomson is really just that bad in accounting for our data.
But we see here that the Rutherford scattering
prediction more accurately describes the gradual falloff
of our scattering rates as the howitzer angle gets bigger, as
opposed to the Thomson fit, with the chi squared of 2000.
Now, there are different ways of potentially giving Thomson
a better shot, accommodating for it,
may be giving it a better chance of describing our data.
And I can go into this in more detail
later, if people are interested.
But, ultimately, we see that Thomson is fundamentally unable
to account for this data.
Whereas Rutherford describes it quite well.
Now, there's an additional caveat
to this chi squared that we have here.
So this chi squared comes from fitting
from our Rutherford and Thomson data convolved
with our beam profile.
However, our beam profile does have
some uncertainty associated.
In that fit that we had, there is some uncertainty
in the triangle.
So to sort of estimate how this uncertainty affects our data,
we vary our beam profile within one standard deviation
of its fit parameters, and see how our goodness of fit
changes.
And we see here that the goodness of fit
for a Rutherford model, while it varies somewhat
depending on the convolution that we use,
it doesn't change a significant amount.
And it certainly doesn't change enough to warrant
considering the Thomson model.
The discrepancy between the chi squareds
and the visual discrepancy you see here very
clearly suggests that Rutherford does
a much better job of predicting our data than Thomson.
Which, again, suggests a more Rutherford make up of the atom,
as opposed to the Thomson Plum Pudding model.
So in conclusion, Thomson's Plum Pudding model
is found to fundamentally be unable to describe
the scattering rates that we observe at large angles.
The Rutherford model more accurately predicts
this gradual fall off, which suggests a more Rutherford-like
view of our atom.
And that's it.
Thank you.
[APPLAUSE]
INSTRUCTOR: All right, the paper is open for questions.
HENRY SHACKLETON: Yeah.
AUDIENCE: Did you try the fit without using the triangular
correction?
HENRY SHACKLETON: Yes I did.
And this is another interesting point, which
shows that our convolution does benefit our data,
does a much better-- not a much better, not as bad as Thomson,
relatively speaking.
But it does a noticeably better job
at describing our data than the pure Rutherford scattering, 1
over sine to the fourth fit.
AUDIENCE: So the difference between 2.14--
HENRY SHACKLETON: And, sorry, 8.18.
AUDIENCE: Oh, I see it.
HENRY SHACKLETON: That's a chi squared of eight
for just fitting with the raw Rutherford scattering
without the convolution.
Yes.
AUDIENCE: Do you have any plots that show your Landau
distribution fits to the data?
HENRY SHACKLETON: I thought I did.
You saw me switch forward and then,
realize that it wasn't there.
So I switched back.
No, so I don't have a picture of the Landau fit.
One thing that I can describe, which
is a small caveat to this, is that the Landau distribution
describes energy loss.
And this is usually a Gaussian with one tail on--
this isn't too well.
It's a Gaussian with a slightly longer tail on the right hand
side.
Now, this is a Gaussian with a slightly longer tail
on the left hand side.
However, what we see here is energy.
And energy loss is sort of the inverse of this.
So we actually fit this to a reverse Landau distribution.
And throughout all our data, we get reduced chi squareds
between 0.5 and 2, which indicates a good fit.
INSTRUCTOR: A different question for Henry?
All right, let's--
HENRY SHACKLETON: Yep, Shawn?
SHAWN: What was the largest angle you
were able to use the data from?
HENRY SHACKLETON: Sorry?
SHAWN: What was the largest angle
you were able to get your data?
HENRY SHACKLETON: 60 degrees.
60 degrees, we collected over the course of five days.
And I don't think we could have actually collected
data at much higher angles than that because our noise
rate right here is 0.0002.
Now, this is still a bit smaller than our counting
rate for 60 degrees.
But if we started to go further, we
would start to enter the regime where noise becomes much more
dominant.
SHAWN: Do you happen to know how many total counts was
that over five days?
HENRY SHACKLETON: Over five days?
Off the top of my head, I believe 400.
You can actually see--
I know the rate here, which you can very easily go
backwards in the math, I guess.
SHAWN: So if the rate--
HENRY SHACKLETON: This is somewhere down here.
SHAWN: --100 counts per day--
HENRY SHACKLETON: Yeah, something like that.
SHAWN: Or a couple counts per hour.
HENRY SHACKLETON: Yes, something like that.
SHAWN: All right, thank you.
The noise that we mentioned here is actually
restricted to our energy of interest.
So there's a fair bit of noise at much lower energy
levels, specifically between this MCA bin
number of 0 and 50, down here.
There's a fair bit of noise there.
But once you start getting up to this larger area right here
between 50 and 1500, we start to see much less noise.
INSTRUCTOR: Another question?
All right, let's thank Henry again.
[APPLAUSE]