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Great.
Well, let's get going.

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Last time we ended up by
discovering the electron.

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We discovered the fact that the
atom was not the most basic

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constituent of matter.
But in 1911 there was another

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discovery concerning the atom,
and this is by Ernest

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Rutherford in England.
And what Rutherford was

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interested in doing was studying
the emission from the newly

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discovered radioactive elements
such as radium.

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And so he borrowed,
or he got, from Marie Curie,

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some radium bromide.
And radium bromide was known to

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emit something called alpha
particles.

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And they didn't really know
what these alpha particles were.

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Now, they did know that the
alpha particles were heavy,

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they were charged and that they
were pretty energetic.

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That is what was known.
Of course, today we know these

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alpha particles to be nothing
other than helium with two

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electrons removed from the
helium, helium double plus.

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Rutherford is in the lab and
has this radium bromide,

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alpha particles being emitted
and has some kind of detector

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out here to detect those alpha
particles.

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And he measures a rate at which
the alpha particles touch his

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detector.
And it is about 132,000 alpha

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particles per minute.
That's nice.

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Then what he does is takes a
piece of gold foil and puts it

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in between the radium bromide
emitter and the detector.

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And that gold foil is actually
really very thin.

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It is 2x10^-5 inches.
Two orders of magnitude thinner

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than the diameter of your hair.
I often wonder how he handled

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that, but he did it.
He put it in the middle here

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and then went to count the count
rate as a result of putting this

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foil there, and the count rate
is 132,000 alpha particles per

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minute.
It didn't seem like that gold

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foil did anything.
The alpha particles were just

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going right through to the
detector.

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It didn't even seem to matter
that there was that gold foil.

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The post-doc that was working
on it, Geiger,

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of the Geiger Counter,
was actually disappointed.

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Gee, that is a boring
experiment.

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But Geiger was even a little
bit more unhappy because he had

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this undergraduate hanging
around the lab,

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this undergraduate named
Marsden.

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And Marsden was really
enthusiastic about doing

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something in the lab.
He really wanted to do

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something.
And Geiger, you know,

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what am I going to do with this
kid?

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Geiger goes to Rutherford,
look, this kid really wants to

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do something.
What should I have him do?

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And Rutherford said,
well, what you should have him

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do is take this detector and
have him build it so that it can

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be swung around.
So that it can be positioned

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here.
So that we can check to see

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whether or not any of these
alpha particles are

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backscattered,
scattered back into the

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direction from which they came.
And Geiger went away and

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thought, good,
this is something to give the

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undergraduate.
This is a ridiculous

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experiment.
We know all the particles are

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going right through the
detector.

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Okay.
But Marsden was real happy.

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He gets to build this detector.
He swings it around and gets

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Geiger there to do the first
experiment.

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He puts the radium bromide and
they listen and hear tick,

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tick, tick, tick,
tick, tick.

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Geiger says,
"Oh, it must just be

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background.
Let me do a control experiment.

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Let me take the gold foil out
of here so that all the

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particles have to be going in
this direction."

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They take the gold foil out of
there and listen,

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and they hear nothing.
They put the gold foil back and

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they hear tick,
tick, tick, tick,

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tick, tick.
And they put a platinum foil in

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there and they hear tick,
tick, tick, tick,

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tick, tick.
Whatever metal they put in

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there, there were some particles
coming off.

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And they got Rutherford down in
the lab.

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Rutherford looks them over
their shoulder.

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They do this again and again.
Hey, it is real.

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It is real.
And what is coming off?

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Well, the count rate is about
20 particles per minute.

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Not large but not zero.
And the probability here of

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this backscattering is simply
the number of particles

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backscattered,
which is 20,

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over the total number of
particles, or actually the count

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rate that the particles
backscattered over the total

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incident count rate.
That is 2x10^-4.

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That is not zero.
Wow.

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Rutherford was excited.
Rutherford later wrote,

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"It was quite the most
incredible event that has ever

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happened to me in my life.
It was almost as incredible as

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if you fired a 15 inch shell at
a piece of tissue paper and it

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came back and hit you." What was
the interpretation?

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The interpretation was the gold
atoms that make up this foil,

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they must be mostly empty.
Now, they knew that those atoms

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had some electrons in it because
the electron had already been

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discovered.
But these alpha particles seem

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to be going right through those
gold atoms, for the most part.

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The atom, which he knew to be a
diameter of about 10^-10 meters,

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most of that atom must be empty
was the conclusion.

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But occasionally these helium
double plus ions,

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these alpha particles,
hit something massive.

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And that something massive then
scatters those helium ions into

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the direction from which they
came.

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And since that probability is
small, well, the size of this

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massive part has to be really
pretty small.

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And from knowing the
probabilities and knowing

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roughly what the diameter of the
atoms were and how many layers

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of atoms he had,
he was able to back out of

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those experiments a diameter for
this massive part of 10^-14

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meters.
And he called this massive part

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the nucleus.
He called it the nucleus in

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analogy to the nucleus of a
living cell.

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The heavy part,
the dense part in a living

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cell-- that is where the name
"nucleus" comes from.

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Now, Rutherford also realized
that this nucleus here has to be

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positively charged.
He knew about electrons and

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knew the atoms then were
neutral, and so he reasoned this

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nucleus had to be positively
charged.

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And then he did a bunch more
experiments, more sophisticated

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experiments in which he actually
measured here the angular

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distribution of the helium ion
scattered from the nucleus.

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And from those very detailed
measurements of the angular

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distribution,
he was able to back out the

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fact that this nucleus,
the charge on it was actually

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plus Z times e.
Z is the atomic number.

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e is the unit charge.
He did a bunch of different

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metals and was able to establish
that the nucleus had a charge of

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plus Z times e.
His model is that there is a

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very dense center,
10^-14 meters.

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This diameter of the nucleus is
something that every MIT

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undergraduate should know.
And he realized that then the

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electrons have to fill out the
rest of this volume.

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That was his interpretation
from these results.

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And think about Marsden,
what a great UROP experiment.

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He discovered the nucleus.
Isn't that great?

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Marsden had a long and
successful career as a scientist

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also after that.
Now, I should also tell you

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that this backscattering
experiment is really the essence

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of how a quark was discovered.
Quark are the fundamental

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elementary particles in protons
and neutrons.

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Essentially,
they took a high energy

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particle, scattered it through
the proton or the neutron,

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and it backscatters.
And, in that way,

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they discovered the quark and
measured the diameter of the

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quark.
And this was done by a couple

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of my colleagues in the Physics
Department.

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Jerry Friedman and Henry
Kendall, who has since passed

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away.
Jerry Friedman is still around.

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He loves to talk to
undergraduates,

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and many of you will get that
opportunity.

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Now it is time for us to do our
own Rutherford backscattering

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experiment.
Yeah.

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[APPLAUSE]
Here is our gold lattice.

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These Styrofoam balls are the
gold nuclei.

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The space around them are the
electrons.

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These things in the center here
are just the posts on this

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frame.
[LAUGHTER] This is a piece of

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equipment from my lab that I
pressed into service,

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and so I couldn't cut these
posts away because I would have

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trouble taking my manipulator
out of my machine at a later

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time.
So they are just there.

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But this is our one monolayer
of gold nuclei.

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And so what are we going to do?
Well, what we are going to do

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is try to measure the diameter
of these Styrofoam balls in the

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same way that Rutherford did.
And so we are going to need

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some alpha particles.
What are we going to use for an

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alpha particle?
Well, we have some ping-pong

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balls for alpha particles.
Let's do that.

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We have 287 alpha particles,
or ping-pong balls,

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and we are going to measure the
probability of backscattering.

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The probability of
backscattering will be the

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number that actually backscatter
divided by the number that we

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throw, or the total number.
That is what we are going to

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measure.
But now I have to take this

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probability and I have to relate
it to the diameter of these

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nuclei.
How am I going to do that?

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Well, that probability is going
to be equal to the total surface

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area of the crystal here.
I have already measured the

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total area.
I know that the total area is

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2,148 square inches.
That is in the denominator,

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but now the numerator is simply
the total area of the nuclei.

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The total area of the nuclei is
the area of one nucleus,

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A sub i,
summed over the total number of

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nuclei, which I have already
counted as 119.

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And so the total area is
times the cross-sectional area

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here of any one of these nuclei.
And that is pi d squared over 4.

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I can solve that equation,

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for the diameter,
in terms of the probability.

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And when I solve that equation,
d is equal to 4.79 times the

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probability to the one-half
power.

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What we are going to do is
measure this probability by

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throwing the ping-pong balls and
calculating and determining how

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many backscatter.
And then we are going to use

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that to get this diameter of the
nuclei.

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The same experiment that was
done to actually measure the

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diameter of the nucleus.
Now you are going to do this

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experiment.
Every one of you are going to

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get a ping-pong ball from the
TAs.

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TAs, why don't you give out the
ping-pong balls,

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and then I will give you some
instructions.

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All right.
The pi d squared over 4

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is the cross-sectional
area in terms of the diameter of

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these balls.
I just wrote it in terms of d

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instead of r.
Yes?

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That is correct.
Good point.

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That balls that we are throwing
actually have size compared to

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in the case of the Rutherford
backscattering experiment where

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the projectile was almost a
point compared to the size of

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the nucleus.
In our experiment,

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you are quite right,
our balls are about the

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diameter there.
And so, if we were doing a more

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00:16:24,000 --> 00:16:27,000
exact experiment,
we would do a little different

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calculation.
We would take into

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consideration the size of the
actual ball that we were

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throwing.
But we are not going to do

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that.
Because we are not throwing

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that many balls,
we don't really have the

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statistics to do a more exacting
kind of calculation.

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00:16:53,000 --> 00:16:57,000
But you are quite right.
Yes?

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Well, he didn't know.
Although, he knew the fact that

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it was backscattering,
that it had to be much,

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much less massive than the
nucleus.

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I think that he also measured
the energy of the backscattered

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particle.
And from that,

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you can back out the fact that
it is much less massive than the

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nucleus.
There are a few other details,

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00:17:26,000 --> 00:17:29,000
you are quite right,
that I have left out in this

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discussion that he had to know
in order to get this number.

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Here is the thing.
You have to aim your alpha

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particles at this lattice.
And then you have to watch your

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00:17:46,000 --> 00:17:50,000
ball.
[LAUGHTER] You have to watch to

235
00:17:50,000 --> 00:17:56,000
see if it scatters back at you,
because at the end I am going

236
00:17:56,000 --> 00:18:02,000
to ask you if your ball
backscattered.

237
00:18:02,000 --> 00:18:07,000
And we need an accurate count.
Now, if you hit one of these

238
00:18:07,000 --> 00:18:11,000
things and it backscatters,
that doesn't count.

239
00:18:11,000 --> 00:18:16,000
Only if it hits the Styrofoam
ball does it count.

240
00:18:16,000 --> 00:18:20,000
If it hits the Styrofoam ball
and goes through,

241
00:18:20,000 --> 00:18:25,000
that doesn't count.
It literally has to backscatter

242
00:18:25,000 --> 00:18:30,000
at you.
Was there a question over here?

243
00:18:30,000 --> 00:18:32,000
If you miss you miss.
[LAUGHTER] Now,

244
00:18:32,000 --> 00:18:37,000
I do invite you to come a
little closer so that you can at

245
00:18:37,000 --> 00:18:40,000
least hit the crystal.
Yes?

246
00:18:47,000 --> 00:18:50,000
That is correct.
Well, you have got a defect.

247
00:18:50,000 --> 00:18:55,000
These are a little bit lighter.
Oh, you have some more here.

248
00:18:55,000 --> 00:18:57,000
Oh, okay.
You can have a regular one.

249
00:18:57,000 --> 00:19:01,000
Anybody need one yet?
I have a couple.

250
00:19:01,000 --> 00:19:03,000
Oh, all right.
You need one?

251
00:19:03,000 --> 00:19:08,000
Because I need them all thrown.
Did you have a question?

252
00:19:13,000 --> 00:19:15,000
What is the mean free path?

253
00:19:20,000 --> 00:19:25,000
That I am going to have to give
you an expression for at some

254
00:19:25,000 --> 00:19:30,000
other time, but there is
certainly a decay pathway.

255
00:19:30,000 --> 00:19:35,000
I have another ball here.
Now, are you ready?

256
00:19:35,000 --> 00:19:43,000
You can come down closer,
but now I have one piece of

257
00:19:43,000 --> 00:19:48,000
advice for you.
That is, only fools aim for

258
00:19:48,000 --> 00:19:56,000
their chemistry professor.
[LAUGHTER] Go to it.

259
00:20:53,000 --> 00:21:00,000
Did you throw your balls?
You missed the crystal.

260
00:21:00,000 --> 00:21:05,000
All right.
Has our supply of alpha

261
00:21:05,000 --> 00:21:11,000
particles been exhausted?
All done?

262
00:21:11,000 --> 00:21:18,000
All right.
[APPLAUSE] Now comes the big

263
00:21:18,000 --> 00:21:23,000
test.
How many of you had an alpha

264
00:21:23,000 --> 00:21:33,000
particle that backscattered?
Let's keep your hand high

265
00:21:33,000 --> 00:21:37,000
because I have to count
accurately.

266
00:21:37,000 --> 00:21:42,000
In this section I see one.
Two?

267
00:21:42,000 --> 00:21:43,000
Cheater.
No.

268
00:21:43,000 --> 00:21:48,000
Two, three, four,
five, six, seven,

269
00:21:48,000 --> 00:21:51,000
eight, nine,
ten, eleven,

270
00:21:51,000 --> 00:21:58,000
twelve, thirteen.
Did I get everybody?

271
00:22:03,000 --> 00:22:04,000
I got everybody?
13?

272
00:22:04,000 --> 00:22:09,000
Right, not deflection.
If it hit and went through,

273
00:22:09,000 --> 00:22:15,000
that does not count.
It has to come back at you.

274
00:22:20,000 --> 00:22:23,000
Yes.
[LAUGHTER] That is right.

275
00:22:23,000 --> 00:22:27,000
All right.
Does anybody want to change

276
00:22:27,000 --> 00:22:30,000
their count?
13 balls?

277
00:22:30,000 --> 00:22:34,000
I am sorry?
If it just hit it and moved but

278
00:22:34,000 --> 00:22:38,000
did not backscatter,
it does not count.

279
00:22:38,000 --> 00:22:43,000
The nuclei will move.
They will move,

280
00:22:43,000 --> 00:22:48,000
certainly, because there is a
momentum transfer.

281
00:22:48,000 --> 00:22:51,000
Well, not quite like that.
No.

282
00:22:51,000 --> 00:22:55,000
We have 13 balls that
backscattered?

283
00:22:55,000 --> 00:23:00,000
Okay.
Let's see what we got.

284
00:23:05,000 --> 00:23:13,000
The probability,
here, then, is 13 over 287.

285
00:23:13,000 --> 00:23:16,000
That probability is equal to

286
00:23:19,000 --> 00:23:29,000
If I now that this probability
and plug it into here,

287
00:23:29,000 --> 00:23:40,000
what we are going to get is a
diameter of 1.0 inches.

288
00:23:40,000 --> 00:23:48,000
And the diameter on the average
of those particles is about 0.85

289
00:23:48,000 --> 00:23:52,000
inches.
You did a really pretty good

290
00:23:52,000 --> 00:23:56,000
job.
You got the diameter of the

291
00:23:56,000 --> 00:24:00,000
nucleus.
[APPLAUSE]

292
00:24:00,000 --> 00:24:04,000
That is great.
And that is the way the nuclear

293
00:24:04,000 --> 00:24:07,000
diameter was,
in fact, measured and

294
00:24:07,000 --> 00:24:11,000
discovered.
But now we have the problem

295
00:24:11,000 --> 00:24:16,000
that the scientists had in 1912,
and that is what is the

296
00:24:16,000 --> 00:24:21,000
structure of the atom?
We now know it has a nucleus.

297
00:24:21,000 --> 00:24:25,000
It has an electron.
How do they hang together?

298
00:24:25,000 --> 00:24:31,000
Where are they in the atom?
We are going to talk about the

299
00:24:31,000 --> 00:24:35,000
classical description here of
the atom.

300
00:24:35,000 --> 00:24:39,000
And the first question that we
have to ask is,

301
00:24:39,000 --> 00:24:45,000
what is the force that keeps
the electron and the nucleus

302
00:24:45,000 --> 00:24:49,000
together?
What are the four fundamental

303
00:24:49,000 --> 00:24:51,000
forces?
Gravity is one.

304
00:24:51,000 --> 00:24:55,000
And that is the strongest or
the weakest?

305
00:24:55,000 --> 00:24:56,000
Weakest.
Gravity.

306
00:24:56,000 --> 00:25:01,000
Next stronger force?
Electromagnetic.

307
00:25:01,000 --> 00:25:05,000
I will just abbreviate it EM.
Next stronger force?

308
00:25:05,000 --> 00:25:06,000
Weak force.
And the next?

309
00:25:06,000 --> 00:25:08,000
Strong.
Weak and strong are

310
00:25:08,000 --> 00:25:12,000
intranuclear forces.
They are operable between the

311
00:25:12,000 --> 00:25:17,000
protons, the neutrons and the
other elementary particles that

312
00:25:17,000 --> 00:25:20,000
make up the nucleus.
It does not have a lot of

313
00:25:20,000 --> 00:25:24,000
effect, the weak and the strong
force, on chemistry,

314
00:25:24,000 --> 00:25:30,000
except for beta emission for
the radioactive elements.

315
00:25:30,000 --> 00:25:36,000
Gravity actually does have no
known chemical significance to

316
00:25:36,000 --> 00:25:40,000
chemistry.
And so all of chemistry is tied

317
00:25:40,000 --> 00:25:45,000
up here in the electromagnetic
force, which I am,

318
00:25:45,000 --> 00:25:50,000
at the moment,
going to simplify and just call

319
00:25:50,000 --> 00:25:55,000
the Coulomb force.
Now, we know how to describe

320
00:25:55,000 --> 00:26:01,000
the Coulomb force between
charged particles.

321
00:26:01,000 --> 00:26:03,000
We know what expression to
write down.

322
00:26:03,000 --> 00:26:06,000
Let's do that.
If we have the nucleus,

323
00:26:06,000 --> 00:26:09,000
which is positively charged,
and the electron here,

324
00:26:09,000 --> 00:26:13,000
which is negatively charged,
and they are at some distance r

325
00:26:13,000 --> 00:26:16,000
between each other,
the expression that describes

326
00:26:16,000 --> 00:26:20,000
how that force of interaction
changes with distance,

327
00:26:20,000 --> 00:26:24,000
this Coulomb's force law,
it is just the magnitude of the

328
00:26:24,000 --> 00:26:28,000
charge of the electron times the
magnitude of the charge on the

329
00:26:28,000 --> 00:26:31,000
nucleus over 4 pi epsilon nought
times r squared.

330
00:26:36,000 --> 00:26:40,000
I am going to just treat the
force as a scalar,

331
00:26:40,000 --> 00:26:43,000
just for simplicity purposes
here.

332
00:26:43,000 --> 00:26:47,000
Epsilon nought is the
permittivity of vacuum.

333
00:26:47,000 --> 00:26:51,000
It is a factor in there for
unit conversation.

334
00:26:51,000 --> 00:26:56,000
r, then, is the distance
between the electron and the

335
00:26:56,000 --> 00:27:00,000
nucleus.
What does this say?

336
00:27:00,000 --> 00:27:03,000
Well, this says that when r
goes to infinity,

337
00:27:03,000 --> 00:27:06,000
what is the force?
Zero.

338
00:27:06,000 --> 00:27:09,000
The particles are infinitely
far apart.

339
00:27:09,000 --> 00:27:13,000
There is no force between them.
In this case,

340
00:27:13,000 --> 00:27:16,000
an attractive force between
them.

341
00:27:16,000 --> 00:27:20,000
When r is equal to zero,
what is the force?

342
00:27:20,000 --> 00:27:23,000
Infinite.
And anywhere in between,

343
00:27:23,000 --> 00:27:28,000
that force is described by this
one over r squared

344
00:27:28,000 --> 00:27:33,000
dependence.
You can see that as the

345
00:27:33,000 --> 00:27:39,000
particles come closer and closer
together, the force between them

346
00:27:39,000 --> 00:27:42,000
gets larger and larger.
The closer they get,

347
00:27:42,000 --> 00:27:47,000
the larger the force,
the more they want to be

348
00:27:47,000 --> 00:27:50,000
together.
This expression is just telling

349
00:27:50,000 --> 00:27:56,000
me, if I held one particle and
the other particle in my hand,

350
00:27:56,000 --> 00:28:02,000
and I held them at some
distance from each other --

351
00:28:02,000 --> 00:28:06,000
That expression is just telling
me the force with which I have

352
00:28:06,000 --> 00:28:09,000
to kind of exert to keep them
apart.

353
00:28:09,000 --> 00:28:14,000
But now, if I let them go,
you know what is going to

354
00:28:14,000 --> 00:28:16,000
happen.
They are going to come

355
00:28:16,000 --> 00:28:19,000
together.
They are going to want to come

356
00:28:19,000 --> 00:28:23,000
together because of that force.
And what is not in this

357
00:28:23,000 --> 00:28:27,000
expression?
What is not in that expression

358
00:28:27,000 --> 00:28:31,000
is any information about how
those particles move under

359
00:28:31,000 --> 00:28:37,000
influence of that force.
Nowhere in this expression is

360
00:28:37,000 --> 00:28:42,000
there an r of t,
how that distance changes with

361
00:28:42,000 --> 00:28:45,000
time.
And so what we need to describe

362
00:28:45,000 --> 00:28:48,000
that is a force law.
And in 1911,

363
00:28:48,000 --> 00:28:52,000
the force law that seemed to
describe the motion of all

364
00:28:52,000 --> 00:28:56,000
bodies, including astronomical
ones, of course,

365
00:28:56,000 --> 00:29:01,000
the equation of motion that
described how bodies move are

366
00:29:01,000 --> 00:29:06,000
Newton's equations of motion.
And, in particular,

367
00:29:06,000 --> 00:29:09,000
F equals ma.
And, of course,

368
00:29:09,000 --> 00:29:13,000
I can write that acceleration
as a time derivative of the

369
00:29:13,000 --> 00:29:16,000
velocity, dv over dt.

370
00:29:16,000 --> 00:29:19,000
And that velocity,
of course, itself is a change

371
00:29:19,000 --> 00:29:22,000
in the position with respect to
time.

372
00:29:22,000 --> 00:29:26,000
This is m, the second
derivative of r with respect to

373
00:29:26,000 --> 00:29:30,000
time.

374
00:29:30,000 --> 00:29:34,000
If I know the force that is
operation, which is this,

375
00:29:34,000 --> 00:29:39,000
I can take this and plug it in
here, and I am going to have a

376
00:29:39,000 --> 00:29:44,000
differential equation.
And that differential equation

377
00:29:44,000 --> 00:29:49,000
is going to allow me to solve
for what r is as a function of

378
00:29:49,000 --> 00:29:53,000
time, the distance between the
two particles.

379
00:29:53,000 --> 00:29:58,000
And it is going to allow me to
solve for that distance in a way

380
00:29:58,000 --> 00:30:03,000
that we call deterministic,
exactly.

381
00:30:03,000 --> 00:30:06,000
In other words,
if I know where the particles

382
00:30:06,000 --> 00:30:11,000
are to start with,
using this equation of motion,

383
00:30:11,000 --> 00:30:14,000
this force law,
I can tell you where those

384
00:30:14,000 --> 00:30:19,000
particles are going to be for
all future time exactly.

385
00:30:19,000 --> 00:30:23,000
It is deterministic,
the classical mechanical

386
00:30:23,000 --> 00:30:26,000
approach.
Now, in order to solve this

387
00:30:26,000 --> 00:30:31,000
differential equation,
I am going to have to develop a

388
00:30:31,000 --> 00:30:36,000
model for the atom.
All differential equations,

389
00:30:36,000 --> 00:30:40,000
for the most part,
describing physical processes

390
00:30:40,000 --> 00:30:44,000
are going to need a model.
They are going to need some

391
00:30:44,000 --> 00:30:48,000
boundary conditions or initial
conditions.

392
00:30:48,000 --> 00:30:52,000
And the model,
of course, that came to mind

393
00:30:52,000 --> 00:30:55,000
for the atom,
is one in which the nucleus is

394
00:30:55,000 --> 00:31:00,000
in the center.
And the electron moves around

395
00:31:00,000 --> 00:31:06,000
that nucleus with uniform
circular motion and with a fixed

396
00:31:06,000 --> 00:31:09,000
radius.
We are going to call that fixed

397
00:31:09,000 --> 00:31:14,000
radius r star.
It is a planetary model.

398
00:31:14,000 --> 00:31:19,000
That seems like a good guess
for the structure of the atom.

399
00:31:19,000 --> 00:31:25,000
Now, if you have a particle
undergoing uniform circular

400
00:31:25,000 --> 00:31:30,000
motion at some well-defined
radius here.

401
00:31:30,000 --> 00:31:34,000
That particle is being
constantly accelerated.

402
00:31:34,000 --> 00:31:40,000
And I can write that
acceleration a as the linear

403
00:31:40,000 --> 00:31:45,000
velocity squared over that
radius of its orbit.

404
00:31:45,000 --> 00:31:51,000
It is being
accelerated because the velocity

405
00:31:51,000 --> 00:31:55,000
vector.
The direction is changing,

406
00:31:55,000 --> 00:32:00,000
so there is a constant
acceleration.

407
00:32:00,000 --> 00:32:02,000
Now, this expression,
for many of you,

408
00:32:02,000 --> 00:32:06,000
I pulled out of the air.
Some of you have seen it

409
00:32:06,000 --> 00:32:08,000
before.
It is an 8.01 topic.

410
00:32:08,000 --> 00:32:13,000
You are going to see it this
semester, but later on and in

411
00:00:08,010 --> 00:32:15,000
You are not responsible for

412
00:32:15,000 --> 00:32:19,000
this right now here,
but you will recall later on

413
00:32:19,000 --> 00:32:23,000
this semester that you have seen
it here in 5.112.

414
00:32:23,000 --> 00:32:26,000
But, if this is the
acceleration,

415
00:32:26,000 --> 00:32:30,000
I can take this expression for
the acceleration and plug it

416
00:32:30,000 --> 00:32:36,000
into here.
Plug in my operating force law.

417
00:32:36,000 --> 00:32:42,000
And, in so doing,
I am going to get --

418
00:32:50,000 --> 00:32:56,000
-- e squared over 4 pi epsilon
nought r star squared.

419
00:33:00,000 --> 00:33:04,000
That is the F.
Mass times the acceleration,

420
00:33:04,000 --> 00:33:09,000
m times v squared over r star.
That is my equation of motion

421
00:33:09,000 --> 00:33:14,000
particular to this problem of a
planetary model.

422
00:33:14,000 --> 00:33:20,000
And now I can solve that for v
squared, the linear velocity of

423
00:33:20,000 --> 00:33:23,000
that electron going around the
nucleus.

424
00:33:23,000 --> 00:33:29,000
That comes out to be e squared
over 4 pi epsilon nought m r

425
00:33:29,000 --> 00:33:34,000
star.

426
00:33:34,000 --> 00:33:39,000
Now, the reason I wanted to
calculate the velocity squared

427
00:33:39,000 --> 00:33:44,000
here is because I want to
calculate kinetic energy.

428
00:33:44,000 --> 00:33:47,000
And that is easy to do.
Kinetic energy,

429
00:33:47,000 --> 00:33:51,000
I will call K,
is one-half m times v squared.

430
00:33:51,000 --> 00:33:57,000
If I plug in
the v squared right in there,

431
00:33:57,000 --> 00:34:03,000
I get one-half e squared over 4
pi epsilon nought r star.

432
00:34:03,000 --> 00:34:10,000
So far, everything looks okay.

433
00:34:10,000 --> 00:34:15,000
We have a planetary model.
Coulomb's law is operable.

434
00:34:15,000 --> 00:34:21,000
We know the acceleration.
We just calculated the kinetic

435
00:34:21,000 --> 00:34:26,000
energy of this electron going
around the nucleus.

436
00:34:26,000 --> 00:34:33,000
What I want to do now is I want
to know the total energy of the

437
00:34:33,000 --> 00:34:37,000
system.
I just calculated the kinetic

438
00:34:37,000 --> 00:34:42,000
energy of the system,
but I want to know the total

439
00:34:42,000 --> 00:34:47,000
energy of the system.
And the total energy of the

440
00:34:47,000 --> 00:34:52,000
system, I am going to call this
capital E, total energy,

441
00:34:52,000 --> 00:34:58,000
is the kinetic energy plus the
potential energy.

442
00:34:58,000 --> 00:35:02,000
And I want the total energy of
the system for two reasons.

443
00:35:02,000 --> 00:35:06,000
One is I want to show you that
the system is bound,

444
00:35:06,000 --> 00:35:10,000
that the total energy is going
to be negative,

445
00:35:10,000 --> 00:35:15,000
that it is lower than the total
energy when the electron and the

446
00:35:15,000 --> 00:35:19,000
nucleus are separated.
I want to show you that within

447
00:35:19,000 --> 00:35:23,000
this classical model,
the electron and the nucleus do

448
00:35:23,000 --> 00:35:26,000
look bound.
To do that, I need to show you

449
00:35:26,000 --> 00:35:32,000
the total energy is negative.
To do that, I need to calculate

450
00:35:32,000 --> 00:35:36,000
the potential energy.
That is what I want to do.

451
00:35:36,000 --> 00:35:41,000
Secondly, I want to get an
expression for the total energy.

452
00:35:41,000 --> 00:35:47,000
Because, using that expression,
I am going to show you how this

453
00:35:47,000 --> 00:35:51,000
classical mechanics fails.
How Newton's equations of

454
00:35:51,000 --> 00:35:55,000
motion won't work to describe
this problem.

455
00:35:55,000 --> 00:36:01,000
Now, I have run out of time.
I will do that on Monday,

456
00:36:01,000 --> 00:36:04,000
but that is where we are going.
All right.

457
00:36:04,892 --> 00:36:07,000
See you on Monday.