Prof: Okay,
so we have the promise of the
knowledge of everything,
at least everything within a
certain category,
and we saw how we could see
quantized energy last time,
and structure,
and structure-related dynamics,
how nuclei vibrate and how it's
different for heavier atoms.
 
Today we're going to look again
at a different kind of dynamics,
and also at bonding.
 
So in the one-dimensional world
this is the payoff.
We have to go to the
three-dimensional world.
Okay, so you did problem sets
for today and you're now
familiar with things like this.
 
What do you know about the blue
curve and the red curve here?
Anybody got a suggestion of
what to do, or what they tell
you about what the true solution
is?
Yes Angela?
 
Student:  It looks like
the true solution is somewhere
between them.
 
Prof: I can't hear very
well.
Student:  It looks like
the true solution will be
somewhere between them,
because it looks like --
Prof: Yes,
the true solution should be
between them.
 
The red one is too hot and the
blue one is too cold.
Right?
 
So there'll be,
in between them,
there'll be one that'll just
come back down to the baseline
for Hooke's law.
 
Now the problem is that this
isn't Hooke's law,
because I didn't show you the
whole picture.
It's in fact a double minimum.
 
Right?
 
And those are both true
solutions.
So the correct
single minimum
energy would lie between those
two,
but for the double minimum
there's one,
there's a solution that's lower
in energy and a solution that's
higher in energy.
 
Now what would happen if we
moved the wells further apart?
Here you see the separation of
the two wells as 0.6 angstroms.
Suppose we moved it apart to be
1.3 angstroms.
Right?
 
Now we have blue and red
solutions.
But now you notice the blue
solution (or the red solution,
they're the same in the left),
they're both the same as the
single minimum.
 
So if you get the wells far
enough apart they look like
single minima with respect to
their wavefunction.
Okay?
 
Now, closer wells.
 
If we move the wells closer
together, we get a lower minimum
energy than for the single well
solution.
Okay?
 
And we get a higher energy for
the next higher one.
But they both look like the
single minimum --
but they both started when the
wells were far apart,
looking like the single minimum
solution,
one with no nodes,
one with one node.
When the wells were far apart
they had the same energy as the
single minimum,
but when the wells come
together, the one with no nodes
has less curvature,
less kinetic energy and is
lower energy than the single
minimum,
and the other one has one node,
more curvature than the single
minimum,
and a higher energy;
that is to say,
there's a splitting between the
energy of these two.
If the wells are far apart,
they're the same as the single
minimum,
both of them,
the no-nodes and the one-node
solution,
are the same as the single
minimum,
but if you bring them together
the no-nodes is lower in energy
and the one-node is higher in
energy.
So the energies split about the
original single minimum
structure.
 
Now that's really important
because that's what makes bonds.
Right?
 
So if we have two atoms or
one-dimensional wells far apart,
say A and B,
and we get a solution in A and
a solution in B,
and if we want the solution for
the whole thing,
we just put them together that
way,
no nodes.
Or we could flip B upside down,
multiply it by minus one,
add them together then and we
get one node.
But they have the same energy
as the single ones would do.
Right?
 
But if we bring them together
the energy gets lower,
less curvature.
 
Right?
 
And that consists,
or gives rise,
to a stabilization of the
particle.
The particle is more stable
when the wells are closer
together.
 
Right?
 
What would you call that?
 
Student:  A bond.
 
Prof: That's a bond.
 
The energy is lower when
they're close together.
So that holds A and B together
because the energy of the
particle is lower when the wells
are close together.
So that's bonding.
 
And part and parcel of this,
the flipside of this,
is that you have this
wavefunction,
which is higher.
 
What will we call that?
 
Student:  Antibonding.
 
Prof: Antibonding.
 
So you get a combination of
bonding and an antibonding
combination because the kinetic
energy changes when the wells
come closer together.
 
You get one that's less curved,
one that's more curved.
Okay, now that's bonding.
 
How about dynamics?
 
Well, suppose we have this
double minimum and suppose we
put a particle in and stop it
there so it has zero --
that's our zero of energy --
and then we're going to let it
go.
 
What will it do if it's
classical?
It'll roll back and forth,
right?
Bingo, let it go.
 
Roll across,
to the same height,
and roll back again,
and then roll across again to
the same height.
 
With no friction it goes on
forever, right?
But here's what happens
special, in real quantum
mechanical systems --
oops, it went too far -- and
now it's going to roll back and
forth on the right;
but every once in awhile it
goes across, from one well to
the other.
 
Now, this is called tunneling,
for obvious reasons.
Right?
 
But the word tunneling I hate.
 
It's one of my pet peeves.
 
It's misleading and
mischievous, because it suggests
there's something weird about
the potential energy,
that you can tunnel through and
have a potential energy lower
than you would've guessed from
the potential energy curve.
But that's not right.
 
What happens is when it's in
the middle you have negative
kinetic energy.
 
You have the potential energy
that the curve shows,
but the kinetic energy is
negative.
The potential energy can be
higher than the total.
And this happens in every bond
and wavefunction,
as you saw in your problem set,
that the waves always go out
into that forbidden region of
negative kinetic energy where
they curve away from the
baseline.
So there's the forbidden
region, out to the left,
to the right,
and also in the middle for that
energy.
 
Right?
 
And what really happens is that
you go to negative kinetic
energy and high positive energy
to get over that hump,
and then you get across to the
other side.
Now how often does that happen?
 
It turns out that the time to
get from one well to the other,
which I won't be able to prove
to you because it requires
time-dependent quantum
mechanics,
and we're talking about
time-independent quantum
mechanics,
but let me just tell you that
the answer is,
if you know the energy
difference between the blue and
the red,
that the rate of getting across
is 5*10^-14 seconds,
divided by whatever that energy
is,
expressed in kilocalories/mole.
 
Right?
 
This is just an assertion,
and it's based on
time-dependent quantum
mechanics.
It's true, but I'm not telling
you why.
So you won't be satisfied,
I hope.
Okay?
 
So the energy difference here
is 1.4 kilocalories/mole.
So how rapidly does a particle
then get from -- if it started
out in the left,
how long do you have to wait,
on average, until it's in the
right one?
You have to wait 5*10^-14
divided by 1.4,
which means about 4*10^-14
seconds it takes to,
quote, "tunnel";
although it really doesn't bore
through, it goes over with
negative kinetic energy.
Okay, so there's something else
about dynamics.
And we're hoping soon to get to
reactivity, but that'll come
after the exam,
after we talk about atoms and
molecules.
 
So now it looks like we're in a
pretty powerful position,
at least with respect to one
dimension,
because with this Erwin
program we can find satisfactory
wavefunctions for any
complicated potential.
We could make it anything we
would like and just follow the
recipe,
find out allowed energies,
shapes of wavefunctions,
probability density,
rates of tunneling,
all this kind of stuff,
and we can rank all the
wavefunctions by their energy,
or their curvature - the number
of nodes they have.
Here's an unsatisfactory
wavefunction.
What do you notice that's bad
about it, just to show that
you've learned something?
 
Yes?
 
Student:  It doesn't
start at the baseline.
Prof: It doesn't start
at the baseline,
on the left.
 
How about on the right,
does it look okay on the right?
>
Prof: Lucas?
 
Student:  It tends to be
going down.
Prof: Ah,
it crosses the baseline,
in the forbidden region.
 
If it crosses the baseline
while curving away,
it's bound to go to negative
infinity, or positive infinity
if it crosses in the other
direction.
Right?
 
And on the left,
because it didn't get to the
baseline when it became flat,
it's going to keep curving and
go up to infinity.
 
So you've internalized a lot of
this stuff now.
Congratulations on doing well
on the problem set.
Okay?
 
So that was a bad one.
 
Okay, so it looks like we're in
pretty good shape.
We even can handle multiple
minima and understand tunneling
and so on.
 
The unfortunate thing is this
curve tracing recipe doesn't
work if you have more
dimensions.
Remember, the reason it would
work is if you knew the
potential and the total energy,
then you knew the kinetic
energy, and you could assign it
to curvature.
Right?
 
But if you have two dimensions,
then there are two different
curvatures you could assign it
to, and you don't know how much
to put in each of them.
 
So it's no longer such a
clear-cut problem if you have
more dimensions,
especially 3n
dimensions.
 
So when there are many
curvatures, it's not clear how
to partition the kinetic energy.
 
But Schrödinger had no
trouble.
The Schrödinger equation
is what's called a differential
equation, it involves
derivatives.
Right?
 
And how many people have had
differential equations?
Okay, maybe a fifth of you I
guess, or fewer.
Right?
 
So the rest of you are feeling,
"Gee I wish I knew
differential equations because
then I could do this."
But you don't have to,
because people already have
done it.
 
You don't have to develop this
from ground zero,
right?
 
And Schrödinger didn't
solve the differential equations
either.
 
Right?
 
He knew what the solutions
were, because people had been
studying them forever,
and the reason they studied it
was because of studying acoustic
waves,
and light also.
 
And Chladni is the guy who
originated this.
And here we have a book by
Chladni called Acoustics.
I got it out of the library.
 
This is in fact the new,
unchanged edition from 1830.
The original edition was
something like 1805 or something
like that;
actually we'll see it here I
think.
 
There's Chladni,
Ernst Florens Friedrich
Chladni, and this is the title
page of that book,
Acoustics;
the original one was 1803,
ours is 1830.
 
Okay, now what he did was take
plates of different -- he also
did violin strings and timpani
heads and things like that.
He was interested in all kinds
of acoustics.
But the particular experiment
we're interested in was he would
take a plate and suspend it in
the middle and touch it in
different places,
while he was bowing it with a
violin bow to make it vibrate.
 
Okay?
 
So you bow it in different
places,
and he put sand on the surface
of this thing to see what
patterns would be formed when
the thing vibrated --
this is why violins have the
shape they have,
so that they'll vibrate in
certain ways when you bow to
make them vibrate --
and the sand collects where the
plate isn't vibrating,
because when it's vibrating it
shakes it away.
 
So you get a picture of the
pattern of vibration.
Now, I'm going to try to show
this to you.
But it might not work,
and just in case I can go to
the Web.
 
So what we have here is a -- I
can't play a violin,
but as you know I can play a
bell.
In fact, I can play a bell in
several ways;
not only that,
>
I can do it this way too.
 
>
 
Now, that's not a very pleasing
sound.
Why?
 
Why is it -- what's noise?
 
I'm not very good at playing
the bell.
>
 
Why does dry ice make a bell
ring?
This was discovered in London
by an ice-cream vendor who had a
bell on his cart,
and he asked a physics teacher,
a woman in London,
why his bell rang when the
dry-ice from his ice-cream hit
it.
>
 
See, when you touch metal like
that,
it gives heat to the dry ice,
which causes it to give off the
pressure of gas,
but when it gives off the
pressure and the metal moves
away,
then heat's not being
transferred, so it stops.
So you get rapid pulses of
CO_2, which make the thing ring.
But depending on how you touch
it,
you get noise,
because there are many
different patterns of vibration
at different frequencies that
are being generated.
 
If I could do it just right,
I would generate just one pure
tone and then you'd say,
"What a great player of
bells he is."
 
Okay, now I'm going to try
this, and I may fail in the same
way, to do an experiment like
Chladni's.
But instead of using a violin
bow -- I'm going to put sand on
here -- instead of using a
violin bow I'm going to use this
piece of dry ice.
 
Okay?
 
And I have to -- if I just
touch it, it won't do anything
because I'll be touching sand
and I won't get heat transfer.
So I have to brush a little bit
to get a clean part of the
brass, and then let's see what
happens here.
>
 
Now I'm doing the same thing I
did with the bell,
right?
 
It's just noise.
 
But let me try someplace else
here.
This is an empirical science.
 
>
 
>
  Try something else
here.
Spread some around here.
 
If you're lucky,
this is really great.
Hear that pure one?
 
Maybe it's not sharp enough.
 
We don't want to spend too much
time doing this.
>
 
I want to try one more and then
we're going to -- you can go to
the website and see a much
better once.
Once I played just the lost
chord, a really beautiful tone;
and it's on,
you can see it on the Web.
>
 
I'm trying to find where and
how hard to touch so that I get
a really pure tone.
 
>
 
Well, you can get patterns that
way.
Okay, you can get much prettier
patterns and prettier tones than
I'm able to coax out of the
thing right now,
and you can look at it on the
Web to see it.
>
Okay, so these are some crude
Chladni figures that I've made
in the past, three rings;
two rings with a line through
the middle, vertically.
 
That's the one we just made
now, I think,
isn't it?
 
Isn't it the same number?
 
A ring and three diameters.
 
And there's a circle and four
diameters.
I actually made that in class
last year.
Okay, these are ones that
Chladni did back in the late
eighteenth century,
taken from that book,
right?
 
And here, the first ones are
just diameters,
and the next ones are rings,
together with a certain number
of diameters,
or two or more rings.
Now let's see what that picture
means.
Below you see a vertical
diameter and two rings.
On top you see color coded as
to whether the plane,
at an instant of vibration,
whether it's toward you or away
from you, as it's vibrating;
that's the pattern.
Because those lines don't move,
but either side of the line,
one side moves up,
the other side moves down.
Okay?
 
Or you could look at it this
way.
The dotted patterns and the
circles and the line don't move
as the rest of the thing
deforms.
This would be like a drum head
on a timpani.
Got the idea?
 
So that's the pattern of motion.
 
Now Chladni was interested in
sound, right?
So he found what pitch each of
these different patterns
corresponded to.
 
So this table shows the number
of diameters that are nodes,
the number of circular nodes,
and what pitch it corresponds
to,
and the little lines may mean
it's a little sharp or a little
flat,
something like that.
 
For example,
the one with two diameters and
no rings is the pitch C.
 
Got the idea?
 
So then he tried to figure out
if there's a mathematical
relationship among these things.
 
These were fourty-seven
patterns that he observed there.
So here he says:
"The pitch relationships
agree approximately with the
squares of the following
numbers."
 
So how many diameters,
how many circles,
and what he sees is the
frequency is roughly the number
of diameter nodes,
plus twice the number of
circular nodes,
squared.
This is his empirical
observation.
So, for example,
you could have two diameters or
one circle,
both would give the number two,
which you then square,
to get something proportional
to the frequency.
 
Right?
 
So there are two different ways
to get the same pitch -- that's
what's crucial -- either rings
or lines.
Or you can have a combination
of rings and lines.
So for higher frequencies you
can get more and more
combinations that give the same
pitch.
For example,
the number eight,
you can have four circles,
three circles and two lines;
two circles and four lines;
a circle and six lines;
or eight lines.
 
All of them give the number
eight and all give the same
pitch, at least approximately.
 
Okay?
 
So this is what he -- but the
lesson we want to take is that
there are different ways of
getting the same frequency,
by combinations of these nodes.
 
Okay, now Chladni didn't solve
his problem mathematically about
how plates or strings or
whatever vibrate.
But there were great
mathematicians working on this,
like Bernoulli,
Lagrange, Euler,
who not only made a Communist
Germany stamp,
he also made a Swiss ten-franc
note.
Okay?
 
Okay, so the Î¨s for
one-electron atoms --
now we're going from one
dimension into three dimensions,
for a real atom,
electron and an atom --
so for one-electron atom
there's going to be three
variables,
x, y,
z for the electron,
and the solutions,
the waves we're going to get,
involve what are called
spherical harmonics,
and they're 3D analogues of
Chladni's 2D figures.
 
These mathematicians could work
in three dimensions as well as
two.
 
Okay, so a 3-dimensional H-atom
wavefunction,
Î¨,
can be written as a product of
three functions,
an R function,
a Î˜ function,
and a Î¦ function.
And these were available from
other old-time mathematicians.
The R function is called
the Associated Laguerre
Function, and it's named for
Edmond Laguerre.
And the Î˜ function
is called the normalized
Associated Legendre Polynomial,
after Adrien-Marie Legendre.
Okay?
 
Now here are the solutions.
 
Schrödinger didn't find
these, he just looked them up.
These guys had already done it,
from acoustics,
for 3-dimensional things
vibrating.
Right?
 
So here's a table that
you can use,
the same way Schrödinger
would've used it,
to find out what wavefunctions
for electrons in one-electron
atoms actually look like.
 
In all your books you've seen
pictures of these things but
you've never seen --
my guess is -- you've never
seen the real thing,
and you should wonder,
"Are these pictures that
people have shown me
right?"
 
>
Now how do you understand?
 
This looks pretty complicated,
this table, and in a sense it
is.
 
But if you look at it the right
way it's really pretty simple.
It relates to the position of
an electron, shown on the right
there, relative to the nucleus.
 
And there's only one electron;
it's a one-electron atom,
otherwise we have more
dimensions.
But the nucleus can have any
charge you want it to.
So it's a one-electron atom of
any nuclear charge,
and we have the coordinates,
x, y, z, and we have a
potential.
 
What's the potential law that's
given to us, whose law?
Student:  Coulomb's.
 
Prof: Coulomb's law.
 
So it's 1/r^2.
 
And how do you write
r^2?;
r^2 is x^2 + y^2 +
z^2.
Right, everybody know that?
 
Everybody know that?
 
Good, nod, yes;
good, okay we're on the same
page here.
 
Okay, but that's a pretty
complicated function,
right?
 
So the equations that involve
that are going to be really,
really complicated.
 
But there's a clever way to get
around this.
Do you know what that way is?
 
What's the clever way of
getting around this?
The clever way of getting
around it is to change the
coordinates of the system that
you're using to one that's more
natural for the problem.
 
And the one that's natural for
the problem is spherical polar
coordinates.
 
So the three dimensions are
r,
the distance,
and Î¸,
the angle down from the z axis,
and Ï†,
the angle rotated around from
the x axis.
Student:  Isn't it the
other way around?
Prof: Well you can
define it any way you want to.
This is the way it's defined
for these purposes.
Okay?
 
Now, so that simplifies the
expression for potential
enormously.
 
Instead of being
1/âˆšx^2 + y^2/c^2,
it's 1/r,
or some constant over r.
Okay?
 
And what that means is -- and
this is the mathematical payoff
--
that the wavefunction can be
written as a product of three
functions,
each of which is the function
of only one variable.
Instead of having x^2^(
)and y^2^( )and^( )
z^2^(2) all mixed
together in this complicated
function,
you have a function only of
r,
a function only of Î¸,
and a function only of Ï†,
and you multiply them together
and you get the solution to the
Schrödinger equation for a
one-electron atom.
 
Now, that -- this table --
gives you those functions,
the R function,
the Î˜ function,
and the Î¦ function.
 
It's like a restaurant where
you get to choose one from
column A, one from column B,
one from column C -- right?
-- here's your appetizer,
your main course and your
dessert, right?
 
So you choose one of these,
one of these,
one of these,
multiply them together,
and you have the true function.
 
Right?
 
Now they look very,
very complicated.
You can't choose any
combination you want to.
Once you've chosen one -- once
you've chosen your appetizer --
you can only have certain main
courses,
and once you've chosen that,
you can only have certain
desserts,
to get solutions. Right?
This is what these
mathematicians had all figured
out.
 
Okay?
 
And so how do we go about doing
it?
Well we can name the
Î¨ we're going to get
by a nickname,
like 1s or 2p_x
or something like that,
but you can also name it by the
numbers that name these
individual functions,
and those are n,
l and m,
which I think you've
encountered before probably.
So let's make a try at it.
 
No, actually notice something
about the complicated form of
the functions.
 
First, every one of them has
Z/a_0 to the 3/2 power.
Z is the nuclear charge;
a_0 is a unit of length
of about half an angstrom about.
 
Okay, so to the 3/2 power;
that's pretty weird.
But remember what you're going
to do with the wavefunction?
What do you do with
wavefunctions,
other than find energy?
 
Students:  Probability.
 
Prof: Get probability
density, and what do you do to
do that Sam?
 
Student:  Square it.
 
Prof: You square it.
 
So when you square it,
it's going to be
(Z/a_0)^3,
when you square it.
Everybody with me on that?
 
Okay, so what that means is
you're going to get units that
are Z^3 per unit volume;
because remember a_0 is
a distance.
 
So a_0^3 is a volume.
 
So it's going to get a number
-- the nuclear charge is a
number Z,
the atomic number is a number
-- so it's going to get a number
per unit volume,
which is the right units to
have for probability density.
What is it per unit volume?
 
Okay?
 
So this is just the bit that
scales it right so that you get
the right density and the right
units.
So you can drop that out,
that's not something very
fundamentally interesting.
 
Okay, let's try making up a
1s orbital.
Where do you start?
 
Can somebody tell me where to
choose?
It's not hard.
 
They're labeled with red, right?
 
If you want 1s,
you choose the top one.
Okay?
 
And now if you've done that,
n is one and l is
zero.
 
So when you come to the next
one, l has to be zero.
Now m can be either zero
or one;
or no, actually wait a second,
if it's 1 -- no,
if l is zero,
then m has to be zero,
it says in the second column.
 
Does everybody see that?
 
See up at the top?
 
It says if l is zero,
then m can be zero;
lm.
 
Okay?
 
For m,
pardon me, to be non-zero,
like here, you have to have an
l that's greater than
zero.
 
Okay?
 
So to get a solution,
if you chose n one and
l had to be zero --
because remember that's the
only choice here,
you can't have a higher
l -- then m is
also going to be zero.
So you've got to take this
times this times this.
Okay?
 
Now how complicated are those
functions?
Okay, âˆš2/2,
that's just a constant,
that's a nothing function.
 
Or 1/âˆš(2Ï€),
big deal.
Why do you have these constants
in there?
Why do you care what constant
you multiply the wavefunction
by?
 
Student:  No idea.
 
Prof: For what purpose?
 
Student:  Normalization.
 
Prof: For normalization,
right?
If you want to get -- but if
all you want is the shape,
forget constants.
 
So what's the real working part
of the 1s wavefunction?
What's the part that varies?
 
We have here the
(Z/a_0)^3/2;
forget that;
two, forget that,
âˆš2/2 forget that,
in fact the 2s would
cancel;
âˆš2 here would cancel that
âˆš2;
so it's 1/âˆšÏ€
times this, times what?;
e^(-Ï /2).
That's the real working part of
it.
Right?
 
When you square it,
what are you going to get,
for e^(-Ï /2)?
 
What's e^(-Ï /2)
squared?
Yes Alex?
 
Student: 
e^-Ï .
Prof: e^-Ï .
 
Okay, so the probability
density is going to be
e^-Ï ,
times a constant.
Okay, so here we got it.
 
It's a constant times
e^(-Ï /2);
when we square it,
it's e^-Ï .
Now, there's something -- there
are two things
that are interesting here.
 
One, we wanted to get a
function of r,
Î¸ and
Ï†,
but we don't have r in
it.
Right?
 
We have Ï  instead.
 
Why do we substitute r
by Ï ?
There's a good reason for that.
 
Ï  is defined --
which is, indeed,
the Greek version of r is
Ï  -- it's defined as
r times a constant.
 
The constant is twice the
atomic number,
divided by n,
the quantum level n,
times that distance a_0.
 
And why do you do it that way?
 
It's because if you do it that
way, then the same Ï 
works, no matter what the
nuclear charge is,
and no matter what n is.
 
So you get the same -- so it
makes the table very compact,
because you use the same thing
no matter what atom you're
dealing with,
if you work in units of
Ï  rather than
r;
that is scale r,
to take into account the fact
that you can have various
nuclear charges.
Okay, allows using the same
e^(-Ï /2) for any
nuclear charge (Z) and
any n.
And notice, every one of these
has e^(-Ï /2)^( )on
it.
 
Now, why does that make sense?
 
Why is it no surprise that
these things have e^-r?
Have we ever seen that before,
e^-x as a wavefunction?
What kind of wavefunction --
what does that mean when you
have e^-r?
 
>
Prof: That's a constant
negative kinetic energy,
constant negative kinetic
energy, which is going to be the
situation for any nucleus and an
electron.
Once the electron gets pretty
far away, the energy stops
changing.
 
Right?
 
So the potential energy is
constant, the kinetic energy is
constant.
 
Right?
 
If the electron is bound to the
nucleus, that means it can't
just fly off to infinity,
right?
So it's below the ultimate
potential energy.
Then you have constant negative
kinetic energy,
when you're far from the
nucleus, right?
So e^-Ï ^(.
 
) So it doesn't surprise you
that every one of these has
e^-Ï  in it.
 
Okay, now let's just look at
this scaling quickly.
Okay, so here's
e^-Ï ^( )and Ï 
-- as a function of
Ï .
Okay?
 
Now suppose -- so we can
rearrange this.
r is that.
 
If we wanted to make a new
plot, which has the horizontal
axis being r,
being distance instead of
Ï .
 
Okay now, so r for a
hydrogen atom,
the 1s orbital of a
hydrogen atom,
is 0.53 -- so n is one,
a_0 is half an angstrom,
Z is one;
so it's .53/2 times
Ï .
 
So we could just put a new
scale on there.
So here's half an angstrom for
a hydrogen, here's one angstrom
for a hydrogen.
 
Everybody with me on this so
far, how I did that?
Right?
 
I just went through and found
out for any given Ï 
here,
that I already had,
one, two, three,
four, five, e^-Ï .
For any give Ï ,
what would the r be if I
were talking about a hydrogen
atom?
Okay?
 
And I just put a new scale on
there.
Now if I were talking about a
carbon atom,
which has a nuclear charge of
plus six,
then this number Z is
going to be much bigger,
it's going to be six times as
big.
What effect will that have on
the scale?
So suppose we're at five,
for Ï .
Instead of dividing by one,
for Z here,
I'm going to be dividing by
six.
So I'm going to get much
shorter distances,
real distances,
rather than Ï .
Right?
 
Does that surprise you -- that
the function is going to squash
in if I have a plus six nuclear
charge?
Does it surprise you or not?
 
It's what you expect.
 
A bigger charge is going to
suck the electrons in more.
Okay?
 
So if I do it for carbon,
I get .53/12 instead of .53/2,
and the scale is going to look
like that.
Instead of this being 0.5 here
for hydrogen,
it's 0.1, right?
 
So then if I squash it in so
that they're on the same scale
--
those are different angstrom
scales for carbon and hydrogen
--
if I make them the same scale,
the bigger nuclear charge sucks
the 1s function in by a factor
of six,
and it looks like that.
 
But now, of course,
if I want the real,
the normalized function,
it's got to be higher,
so that the total area is the
same.
I have the squared function
here, e^-Ï ;
remember that the wavefunction
was e^(-Ï /2);
the density is e^-Ï .
 
To get the same thing I'm going
to have to multiply that by six.
So it's going to look like that.
 
So that's how the radial
distribution,
probability density
distribution,
looks different,
for a one-electron atom with a
plus one or a plus six nuclear
charge.
Carbon holds its core electrons
in much more tightly than
hydrogen does;
no big surprise there.
Okay.
 
So here's something to think
about.
How would it differ if instead
of talking about the 1s
orbital of carbon,
I was talking about the
2s orbital of carbon?
 
Right?
 
Now, Ï  is going to
be different,
and here, instead of one,
it's going to be twp.
So it's going to change the
scale by a factor of two.
So the 2s are further
out.
Okay, so for Wednesday I want
you to do these problems --
you can do them in groups if
you want to --
some about Chladni Figures,
and then some things about
energy and some atomic orbital
problems.
Okay?
 
Now we already looked at the
1s.
Let's look at some other atomic
orbitals.
Incidentally,
this function,
the Coulombic function,
is simpler in three dimensions
than it is in one dimension.
 
It was that complicated thing
in one-dimension,
remember that had the cusp on
it.
It was a really complicated
function, but it's really simple
in three dimensions,
it's just exponential in
r.
 
How does it vary with
Î¸?
As you come down from the axis,
how does the wavefunction vary,
the 1s?
 
What does it say?
 
How does it depend on
Î¸?
Student:  It doesn't.
 
Prof: It doesn't.
 
How does it depend on
Ï†?
Student:  It doesn't.
 
Prof: It doesn't.
 
So what shape does it have?
 
Student:  A sphere.
 
Prof: It's spherically
symmetric, it depends only on
r.
 
And the dependence on r
is as easy as pie,
it's just e^-r^( )(or
Ï  --
it depends on what units you
measure r in).
Okay, now let's look at
2s.
Now tell me what to do.
 
How do I write a 2s
function from this table?
First I have to choose the
appetizer.
Where do I go?
 
Student:  Second one.
 
Prof: Second one down is
2s.
Okay?
 
So I take this one,
and now 2s,
that means l is zero.
 
So where do I go;
what direction do I go?
Here.
 
And then what direction do I go?
 
Student:  Straight
across.
Prof: Straight across,
because m is zero.
Okay.
 
So here's -- there's 2s;
multiply those together.
Now what's interesting,
there are a bunch of constants
that are going to give me --
make it normalized.
So that's a constant,
that's a constant,
just like they were before;
this is a constant,
that's a constant.
 
But this part is interesting:
(2-Ï ) times
e^(-Ï /2).
 
That's the function that we're
interested in,
(2-Ï ) times
e^(-Ï /2).
That has an interesting thing.
 
What happens when Ï 
has the value two?
It has the e^(-Ï /2);
that's just decaying,
all of them have that.
 
But how about the
2-Ï ?
That gives an interesting thing.
 
What's its value when Ï 
is two?
Student:  Zero.
 
Prof: Zero.
 
So there's a node.
 
What shape is the node?
 
Is it a point,
is it a line,
is it a wiggly thing?
 
Student:  A sphere.
 
Prof: It's a sphere,
it's spherically symmetric
again.
 
So there's a spherical node.
 
So inside -- and at a node,
the wavefunction changes sign.
So inside it's one sign,
outside it's the opposite sign.
Okay?
 
Now how about a 2p_z
orbital?
So where are we going to go?
 
2p, we'll start with
this, and for p_z,
we do this one,
and for z we do this
one.
 
So it's this times this times
this.
I think I have that circled
here;
let's see, yeah.
 
Is this one interesting?
 
Student:  No.
 
Prof: No.
 
Is this one interesting?
 
Student:  Yes.
 
Prof: What part of it is
interesting?
Student:  Cosine.
 
Prof:
cos(Î¸) is
interesting, right?
 
That's real variation.
 
And what part's interesting
here?
It's Ï  times
e^(-Ï /2).
Okay, so it's some constant
times Ï  times the
cosine of Î¸ times
e^(-Ï /2).
You always have the
e^(-Ï /2);
forget that.
 
It just means it decays as you
go out.
Now Ï  times
cos(Î¸).
What does that mean?;
Ï ,
which is this distance,
times the cos(Î¸),
what is that?
 
Russell?
 
It's z.
 
So I can simplify this a lot,
if I mix my metaphors between
polar and Cartesian coordinates,
it's z times
e^(-Ï /2).
 
That's the 2p_z orbital.
 
Now, can you guess what the
2p_x and the 2p_y
look like?
 
Can somebody guess the
2p_y?
Josh?
 
Student:  You replace
x with --
Prof: Replace what?
 
Student:  Replace
z with x.
Prof: Replace z
with x,
you get the 2p_x
orbital;
replace z with y,
you get the 2p_y
orbital.
 
Pretty straightforward
functions.
Now how about -- you've seen
pictures of p orbitals.
Now that you know what the
functions look like,
it's interesting to go back to
the pictures you've seen and see
what they mean.
 
How do they relate to the
actual formula?
So you could plot -- remember,
it's three parts,
an R function,
a Î˜ function,
a Î¦ function.
 
We could look at each of those
functions separately,
if we wanted to.
 
Okay?
 
We've been talking about the
R function,
how it would behave.
 
But how would the
Î˜ function behave
in this, cosine Î¸?
 
So we'll do a polar plot that
shows the value of cosine
Î¸ as a function of
Î¸.
Everybody with me?
 
Okay, so what is it when
Î¸ is zero;
what's cosine of Î¸?
 
Student:  One.
 
Prof: One.
 
Okay, so there it is,
and we'll put a point there,
okay?
 
Now let's go to + and -30°.
 
Cosine is 0.86,
and we'll put points there.
And let's go to 45Â°,
it's the âˆšÂ½,
0.71, put points there.
 
At 60°, it's ½;
put points there.
What is it if it's 90°?
 
If it's 90°,
the cosine is zero.
So we got a point at the origin.
 
Right?
 
So you can see what the shape
is.
What if you go the other
direction, what if you go beyond
90°, what happens to the
cosine?
Student:  It's negative.
 
Prof: Changes sign.
 
So you've got another circle
that's negative on the other
side.
 
So there's a p orbital.
 
But I'm not showing the whole
p orbital,
I'm just showing how the
angular part varies.
So you've seen pictures like
that of p orbitals,
which is the Î˜
function.
Or I could square it,
in order to look at probability
density, as a function of
Î¸,
and it would look like that.
 
So you've seen pictures more or
less like that.
Or I could make a picture like
that, which is a contour plot
showing how big it is,
on a slice, through the
nucleus.
 
Right?
 
So it's Ï  times
e^(-Ï /2) times
cos(Î¸);
that multiplied by a constant
is the wavefunction.
 
Okay, so you can see it's
positive up at the top,
negative at the bottom.
 
And let's look at it a little
bit.
And, notice it doesn't depend
on Ï†,
the angle of rotation around
here.
So we can take that picture and
rotate it to give a
three-dimensional dumbbell.
 
Okay?
 
We're just looking at one slice
here, in order to be explicit.
Okay, now so there's a certain
point there that has a certain
Ï  and a certain
Î¸,
and therefore it has a certain
value.
Now, let's find the maximum,
of that function.
Okay?
 
So first it depends on
cos(Î¸).
What will Î¸ be in
order for this to be a maximum?
Student:  Zero.
 
Prof: Zero.
 
So it has to be along the axis.
 
Okay?
 
Now, so Î¸ has to
be equal to zero.
Okay, now how do we find where
it's going to be maximum with
respect to Ï ?
 
We take the derivative with
respect to Ï ,
set it equal to zero.
 
So that's this,
or simplified that,
or simplified that;
Ï =2.
So there's the maximum.
 
And if we put the constants in,
we could find the true
probability density anywhere;
just plug in Î¸,
plug in Ï ,
and square it,
multiply it by the constant.
 
That's the electron density at
any given, at the point you're
talking about.
 
Now, or you could have a
computer do it for you,
and that's where we're indebted
to Dean Dauger,
who's a physicist.
 
He went to Occidental College
and actually wrote this program
while he was there,
but then he got a Ph.D.
at UCLA in Physics,
and he wrote this for
physicists, but we'll use it.
 
Okay, here's a picture of Dean
Dauger.
He also juggles.
 
And here's a picture of him at
an Apple Developer Conference,
and he's juggling,
the guy on the left there.
He also drops one of the pins,
he's not perfect.
Okay.
 
So this is his program.
 
He's at the Apple Developer
Conference.
It works only on an Apple,
so if you don't have an Apple,
borrow somebody's Apple and
look at the program,
because it's really fun.
 
This is the screen of the
program, and I don't have time
to take you through it now,
but you should try it.
So look at the PowerPoint of
this on the next few slides to
learn what all the different
information is that's being
given here,
and the kinds of questions you
can ask.
 
There's a 1s orbital,
right?
And it's as if the thing were
really --
the electron were really
colored, and it's a picture of
how it would actually look if
you could see electron density.
But all, of course,
he's done is used fancy
computer graphics to plot the
square of the wavefunction.
Right?
 
But it's a really nice program.
 
We'll talk about this a little
more next time.
