The following content is provided by MIT OpenCourseWare
under a Creative Commons license.
Additional information about our license and MIT
OpenCourseWare in general is available at ocw.mit.edu.
Great. Well, let's get going.
Last time we ended up by discovering the electron.
We discovered the fact that the atom was not the most basic
constituent of matter. But in 1911 there was another
discovery concerning the atom, and this is by Ernest
Rutherford in England. And what Rutherford was
interested in doing was studying the emission from the newly
discovered radioactive elements such as radium.
And so he borrowed, or he got, from Marie Curie,
some radium bromide. And radium bromide was known to
emit something called alpha particles.
And they didn't really know what these alpha particles were.
Now, they did know that the alpha particles were heavy,
they were charged and that they were pretty energetic.
That is what was known. Of course, today we know these
alpha particles to be nothing other than helium with two
electrons removed from the helium, helium double plus.
Rutherford is in the lab and has this radium bromide,
alpha particles being emitted and has some kind of detector
out here to detect those alpha particles.
And he measures a rate at which the alpha particles touch his
detector. And it is about 132,000 alpha
particles per minute. That's nice.
Then what he does is takes a piece of gold foil and puts it
in between the radium bromide emitter and the detector.
And that gold foil is actually really very thin.
It is 2x10^-5 inches. Two orders of magnitude thinner
than the diameter of your hair. I often wonder how he handled
that, but he did it. He put it in the middle here
and then went to count the count rate as a result of putting this
foil there, and the count rate is 132,000 alpha particles per
minute. It didn't seem like that gold
foil did anything. The alpha particles were just
going right through to the detector.
It didn't even seem to matter that there was that gold foil.
The post-doc that was working on it, Geiger,
of the Geiger Counter, was actually disappointed.
Gee, that is a boring experiment.
But Geiger was even a little bit more unhappy because he had
this undergraduate hanging around the lab,
this undergraduate named Marsden.
And Marsden was really enthusiastic about doing
something in the lab. He really wanted to do
something. And Geiger, you know,
what am I going to do with this kid?
Geiger goes to Rutherford, look, this kid really wants to
do something. What should I have him do?
And Rutherford said, well, what you should have him
do is take this detector and have him build it so that it can
be swung around. So that it can be positioned
here. So that we can check to see
whether or not any of these alpha particles are
backscattered, scattered back into the
direction from which they came. And Geiger went away and
thought, good, this is something to give the
undergraduate. This is a ridiculous
experiment. We know all the particles are
going right through the detector.
Okay. But Marsden was real happy.
He gets to build this detector. He swings it around and gets
Geiger there to do the first experiment.
He puts the radium bromide and they listen and hear tick,
tick, tick, tick, tick, tick.
Geiger says, "Oh, it must just be
background. Let me do a control experiment.
Let me take the gold foil out of here so that all the
particles have to be going in this direction."
They take the gold foil out of there and listen,
and they hear nothing. They put the gold foil back and
they hear tick, tick, tick, tick,
tick, tick. And they put a platinum foil in
there and they hear tick, tick, tick, tick,
tick, tick. Whatever metal they put in
there, there were some particles coming off.
And they got Rutherford down in the lab.
Rutherford looks them over their shoulder.
They do this again and again. Hey, it is real.
It is real. And what is coming off?
Well, the count rate is about 20 particles per minute.
Not large but not zero. And the probability here of
this backscattering is simply the number of particles
backscattered, which is 20,
over the total number of particles, or actually the count
rate that the particles backscattered over the total
incident count rate. That is 2x10^-4.
That is not zero. Wow.
Rutherford was excited. Rutherford later wrote,
"It was quite the most incredible event that has ever
happened to me in my life. It was almost as incredible as
if you fired a 15 inch shell at a piece of tissue paper and it
came back and hit you." What was the interpretation?
The interpretation was the gold atoms that make up this foil,
they must be mostly empty. Now, they knew that those atoms
had some electrons in it because the electron had already been
discovered. But these alpha particles seem
to be going right through those gold atoms, for the most part.
The atom, which he knew to be a diameter of about 10^-10 meters,
most of that atom must be empty was the conclusion.
But occasionally these helium double plus ions,
these alpha particles, hit something massive.
And that something massive then scatters those helium ions into
the direction from which they came.
And since that probability is small, well, the size of this
massive part has to be really pretty small.
And from knowing the probabilities and knowing
roughly what the diameter of the atoms were and how many layers
of atoms he had, he was able to back out of
those experiments a diameter for this massive part of 10^-14
meters. And he called this massive part
the nucleus. He called it the nucleus in
analogy to the nucleus of a living cell.
The heavy part, the dense part in a living
cell-- that is where the name "nucleus" comes from.
Now, Rutherford also realized that this nucleus here has to be
positively charged. He knew about electrons and
knew the atoms then were neutral, and so he reasoned this
nucleus had to be positively charged.
And then he did a bunch more experiments, more sophisticated
experiments in which he actually measured here the angular
distribution of the helium ion scattered from the nucleus.
And from those very detailed measurements of the angular
distribution, he was able to back out the
fact that this nucleus, the charge on it was actually
plus Z times e. Z is the atomic number.
e is the unit charge. He did a bunch of different
metals and was able to establish that the nucleus had a charge of
plus Z times e. His model is that there is a
very dense center, 10^-14 meters.
This diameter of the nucleus is something that every MIT
undergraduate should know. And he realized that then the
electrons have to fill out the rest of this volume.
That was his interpretation from these results.
And think about Marsden, what a great UROP experiment.
He discovered the nucleus. Isn't that great?
Marsden had a long and successful career as a scientist
also after that. Now, I should also tell you
that this backscattering experiment is really the essence
of how a quark was discovered. Quark are the fundamental
elementary particles in protons and neutrons.
Essentially, they took a high energy
particle, scattered it through the proton or the neutron,
and it backscatters. And, in that way,
they discovered the quark and measured the diameter of the
quark. And this was done by a couple
of my colleagues in the Physics Department.
Jerry Friedman and Henry Kendall, who has since passed
away. Jerry Friedman is still around.
He loves to talk to undergraduates,
and many of you will get that opportunity.
Now it is time for us to do our own Rutherford backscattering
experiment. Yeah.
[APPLAUSE] Here is our gold lattice.
These Styrofoam balls are the gold nuclei.
The space around them are the electrons.
These things in the center here are just the posts on this
frame. [LAUGHTER] This is a piece of
equipment from my lab that I pressed into service,
and so I couldn't cut these posts away because I would have
trouble taking my manipulator out of my machine at a later
time. So they are just there.
But this is our one monolayer of gold nuclei.
And so what are we going to do? Well, what we are going to do
is try to measure the diameter of these Styrofoam balls in the
same way that Rutherford did. And so we are going to need
some alpha particles. What are we going to use for an
alpha particle? Well, we have some ping-pong
balls for alpha particles. Let's do that.
We have 287 alpha particles, or ping-pong balls,
and we are going to measure the probability of backscattering.
The probability of backscattering will be the
number that actually backscatter divided by the number that we
throw, or the total number. That is what we are going to
measure. But now I have to take this
probability and I have to relate it to the diameter of these
nuclei. How am I going to do that?
Well, that probability is going to be equal to the total surface
area of the crystal here. I have already measured the
total area. I know that the total area is
2,148 square inches. That is in the denominator,
but now the numerator is simply the total area of the nuclei.
The total area of the nuclei is the area of one nucleus,
A sub i, summed over the total number of
nuclei, which I have already counted as 119.
And so the total area is times the cross-sectional area
here of any one of these nuclei. And that is pi d squared over 4.
I can solve that equation,
for the diameter, in terms of the probability.
And when I solve that equation, d is equal to 4.79 times the
probability to the one-half power.
What we are going to do is measure this probability by
throwing the ping-pong balls and calculating and determining how
many backscatter. And then we are going to use
that to get this diameter of the nuclei.
The same experiment that was done to actually measure the
diameter of the nucleus. Now you are going to do this
experiment. Every one of you are going to
get a ping-pong ball from the TAs.
TAs, why don't you give out the ping-pong balls,
and then I will give you some instructions.
All right. The pi d squared over 4
is the cross-sectional area in terms of the diameter of
these balls. I just wrote it in terms of d
instead of r. Yes?
That is correct. Good point.
That balls that we are throwing actually have size compared to
in the case of the Rutherford backscattering experiment where
the projectile was almost a point compared to the size of
the nucleus. In our experiment,
you are quite right, our balls are about the
diameter there. And so, if we were doing a more
exact experiment, we would do a little different
calculation. We would take into
consideration the size of the actual ball that we were
throwing. But we are not going to do
that. Because we are not throwing
that many balls, we don't really have the
statistics to do a more exacting kind of calculation.
But you are quite right. Yes?
Well, he didn't know. Although, he knew the fact that
it was backscattering, that it had to be much,
much less massive than the nucleus.
I think that he also measured the energy of the backscattered
particle. And from that,
you can back out the fact that it is much less massive than the
nucleus. There are a few other details,
you are quite right, that I have left out in this
discussion that he had to know in order to get this number.
Here is the thing. You have to aim your alpha
particles at this lattice. And then you have to watch your
ball. [LAUGHTER] You have to watch to
see if it scatters back at you, because at the end I am going
to ask you if your ball backscattered.
And we need an accurate count. Now, if you hit one of these
things and it backscatters, that doesn't count.
Only if it hits the Styrofoam ball does it count.
If it hits the Styrofoam ball and goes through,
that doesn't count. It literally has to backscatter
at you. Was there a question over here?
If you miss you miss. [LAUGHTER] Now,
I do invite you to come a little closer so that you can at
least hit the crystal. Yes?
That is correct. Well, you have got a defect.
These are a little bit lighter. Oh, you have some more here.
Oh, okay. You can have a regular one.
Anybody need one yet? I have a couple.
Oh, all right. You need one?
Because I need them all thrown. Did you have a question?
What is the mean free path?
That I am going to have to give you an expression for at some
other time, but there is certainly a decay pathway.
I have another ball here. Now, are you ready?
You can come down closer, but now I have one piece of
advice for you. That is, only fools aim for
their chemistry professor. [LAUGHTER] Go to it.
Did you throw your balls? You missed the crystal.
All right. Has our supply of alpha
particles been exhausted? All done?
All right. [APPLAUSE] Now comes the big
test. How many of you had an alpha
particle that backscattered? Let's keep your hand high
because I have to count accurately.
In this section I see one. Two?
Cheater. No.
Two, three, four, five, six, seven,
eight, nine, ten, eleven,
twelve, thirteen. Did I get everybody?
I got everybody? 13?
Right, not deflection. If it hit and went through,
that does not count. It has to come back at you.
Yes. [LAUGHTER] That is right.
All right. Does anybody want to change
their count? 13 balls?
I am sorry? If it just hit it and moved but
did not backscatter, it does not count.
The nuclei will move. They will move,
certainly, because there is a momentum transfer.
Well, not quite like that. No.
We have 13 balls that backscattered?
Okay. Let's see what we got.
The probability, here, then, is 13 over 287.
That probability is equal to
If I now that this probability and plug it into here,
what we are going to get is a diameter of 1.0 inches.
And the diameter on the average of those particles is about 0.85
inches. You did a really pretty good
job. You got the diameter of the
nucleus. [APPLAUSE]
That is great. And that is the way the nuclear
diameter was, in fact, measured and
discovered. But now we have the problem
that the scientists had in 1912, and that is what is the
structure of the atom? We now know it has a nucleus.
It has an electron. How do they hang together?
Where are they in the atom? We are going to talk about the
classical description here of the atom.
And the first question that we have to ask is,
what is the force that keeps the electron and the nucleus
together? What are the four fundamental
forces? Gravity is one.
And that is the strongest or the weakest?
Weakest. Gravity.
Next stronger force? Electromagnetic.
I will just abbreviate it EM. Next stronger force?
Weak force. And the next?
Strong. Weak and strong are
intranuclear forces. They are operable between the
protons, the neutrons and the other elementary particles that
make up the nucleus. It does not have a lot of
effect, the weak and the strong force, on chemistry,
except for beta emission for the radioactive elements.
Gravity actually does have no known chemical significance to
chemistry. And so all of chemistry is tied
up here in the electromagnetic force, which I am,
at the moment, going to simplify and just call
the Coulomb force. Now, we know how to describe
the Coulomb force between charged particles.
We know what expression to write down.
Let's do that. If we have the nucleus,
which is positively charged, and the electron here,
which is negatively charged, and they are at some distance r
between each other, the expression that describes
how that force of interaction changes with distance,
this Coulomb's force law, it is just the magnitude of the
charge of the electron times the magnitude of the charge on the
nucleus over 4 pi epsilon nought times r squared.
I am going to just treat the force as a scalar,
just for simplicity purposes here.
Epsilon nought is the permittivity of vacuum.
It is a factor in there for unit conversation.
r, then, is the distance between the electron and the
nucleus. What does this say?
Well, this says that when r goes to infinity,
what is the force? Zero.
The particles are infinitely far apart.
There is no force between them. In this case,
an attractive force between them.
When r is equal to zero, what is the force?
Infinite. And anywhere in between,
that force is described by this one over r squared
dependence. You can see that as the
particles come closer and closer together, the force between them
gets larger and larger. The closer they get,
the larger the force, the more they want to be
together. This expression is just telling
me, if I held one particle and the other particle in my hand,
and I held them at some distance from each other --
That expression is just telling me the force with which I have
to kind of exert to keep them apart.
But now, if I let them go, you know what is going to
happen. They are going to come
together. They are going to want to come
together because of that force. And what is not in this
expression? What is not in that expression
is any information about how those particles move under
influence of that force. Nowhere in this expression is
there an r of t, how that distance changes with
time. And so what we need to describe
that is a force law. And in 1911,
the force law that seemed to describe the motion of all
bodies, including astronomical ones, of course,
the equation of motion that described how bodies move are
Newton's equations of motion. And, in particular,
F equals ma. And, of course,
I can write that acceleration as a time derivative of the
velocity, dv over dt.
And that velocity, of course, itself is a change
in the position with respect to time.
This is m, the second derivative of r with respect to
time.
If I know the force that is operation, which is this,
I can take this and plug it in here, and I am going to have a
differential equation. And that differential equation
is going to allow me to solve for what r is as a function of
time, the distance between the two particles.
And it is going to allow me to solve for that distance in a way
that we call deterministic, exactly.
In other words, if I know where the particles
are to start with, using this equation of motion,
this force law, I can tell you where those
particles are going to be for all future time exactly.
It is deterministic, the classical mechanical
approach. Now, in order to solve this
differential equation, I am going to have to develop a
model for the atom. All differential equations,
for the most part, describing physical processes
are going to need a model. They are going to need some
boundary conditions or initial conditions.
And the model, of course, that came to mind
for the atom, is one in which the nucleus is
in the center. And the electron moves around
that nucleus with uniform circular motion and with a fixed
radius. We are going to call that fixed
radius r star. It is a planetary model.
That seems like a good guess for the structure of the atom.
Now, if you have a particle undergoing uniform circular
motion at some well-defined radius here.
That particle is being constantly accelerated.
And I can write that acceleration a as the linear
velocity squared over that radius of its orbit.
It is being accelerated because the velocity
vector. The direction is changing,
so there is a constant acceleration.
Now, this expression, for many of you,
I pulled out of the air. Some of you have seen it
before. It is an 8.01 topic.
You are going to see it this semester, but later on and in
You are not responsible for
this right now here, but you will recall later on
this semester that you have seen it here in 5.112.
But, if this is the acceleration,
I can take this expression for the acceleration and plug it
into here. Plug in my operating force law.
And, in so doing, I am going to get --
-- e squared over 4 pi epsilon nought r star squared.
That is the F. Mass times the acceleration,
m times v squared over r star. That is my equation of motion
particular to this problem of a planetary model.
And now I can solve that for v squared, the linear velocity of
that electron going around the nucleus.
That comes out to be e squared over 4 pi epsilon nought m r
star.
Now, the reason I wanted to calculate the velocity squared
here is because I want to calculate kinetic energy.
And that is easy to do. Kinetic energy,
I will call K, is one-half m times v squared.
If I plug in the v squared right in there,
I get one-half e squared over 4 pi epsilon nought r star.
So far, everything looks okay.
We have a planetary model. Coulomb's law is operable.
We know the acceleration. We just calculated the kinetic
energy of this electron going around the nucleus.
What I want to do now is I want to know the total energy of the
system. I just calculated the kinetic
energy of the system, but I want to know the total
energy of the system. And the total energy of the
system, I am going to call this capital E, total energy,
is the kinetic energy plus the potential energy.
And I want the total energy of the system for two reasons.
One is I want to show you that the system is bound,
that the total energy is going to be negative,
that it is lower than the total energy when the electron and the
nucleus are separated. I want to show you that within
this classical model, the electron and the nucleus do
look bound. To do that, I need to show you
the total energy is negative. To do that, I need to calculate
the potential energy. That is what I want to do.
Secondly, I want to get an expression for the total energy.
Because, using that expression, I am going to show you how this
classical mechanics fails. How Newton's equations of
motion won't work to describe this problem.
Now, I have run out of time. I will do that on Monday,
but that is where we are going. All right.
See you on Monday.