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[Slide 1] Greetings everyone and welcome to the
last lecture of the first week of atoms
to materials. Today we're going to finish with
the hydrogen atom. We're going to describe excited
states and we're going to use the orbitals
for hydrogen to discuss multi-electron atoms, so let's
get to it. [Slide 2] Last class, last
lecture we talked about the spherically symmetric states
in hydrogen. We talked about how the ground
state had no nodes, there was an exponentially
decaying function. That wave function is called n
equals one, n is the quantum principle number
and then as we move up in energies
to n equals two, this is the first
excited state, then n equals three we increased
the number of nodes in the wave functions.
just like what happened in the particle in
a box. So for the ground state, the
sign of the wave function is positive everywhere
and no nodes. The wave function goes down
as you move away from the proton. For
the n equals two, these are wave functions
with one node in three dimensions, so that's
going to be a plain- a planar surface.
And in the case of this spherically symmetric
functions the plane is a sphere. So the
wave function starts off as positive and becomes
negative. And the second excited state has one
more node and again goes positive, then negative
and then positive again. But not all solutions
of hydrogen, of course, are spherically symmetric but
there is angular dependence. There are wave functions
where the electrons have an angular momentum around
the proton as opposed being spherically symmetric and
what we're going to do is we're going
to is think about this excited state- states
in terms of the number of nodes and
the requirement that the wave functions be normal
to one another, okay, remember that because the
Hamiltonian is a Hermitian operator, the orbitals- the
Eigenfunctions need to be orthogonal to each other.
That means that the product of the two
functions integrated over all of space needs to
add up to zero, okay. So if you
think about the 1 S and the 2
S wave functions, the first two that we're
seeing here- if you do the product of
these two, we're going to have positive near
the proton, where both wave functions are positive
and then the product is going to turn
negative as the n equals two wave function
turns negative and that this happens in a
way that if you integrate the product over
all of space you get zero. So the
two n equals one, n equals two wave
functions are orthogonal to each other. Now let's
think about functions that have one node and
we come to the conclusion that there's more
solutions than a spherical node, right, so there
are other ways to have- other matters in
which I can place a node in space
so that I have positive and negative wave
functions that satisfies orthogonally with all of these
other functions. So if you think about it
a little bit, it will be very clear
that the wave function where- that has a
planar mode, normal to one of the Cartesian
directions would satisfy these conditions. So if I'm
looking for wave functions with a single node
in three dimensions I have multiple options. I
have a node that's a sphere, but I
can also have a node that's a plane
that passes- cuts through the proton. So the
proton would be here in the middle. if
you think about doing the product between this
new wave function and the previous ones, you're
going to convince yourself very easily that that
the products are zero, okay. So let's do
an example, let's estimate the product of the
new wave function that we came up with
and the spherically symmetric wave function with one
node. So let's say this is the node
of the spherically symmetric wave function and I'm
going to compute- calculate the sign in each
of these regions. So this region here is
going to be positive because both wave functions
are positive and here I'm going to have
negative because the new wave function is negative
and the spherically symmetric function is positive, and
then I have positive over here and negative
over here. So very clearly this new wave
function that is positive on one side, let's
say plus x and negative on the other
side, say plus negative X, this wave function
is normal to the wave functions we had
before and it also has one node. And
now you can think that if I have
a wave function that has a node along
the X I can also have a wave
function that has a node along z, along
y or along z. It turns out that
there are three such wave functions, okay, that
are going to be two n=2 wave functions
that need a single node and they're going
to be on a line along x,y and
z. Okay and I can only put three
because these wave functions need to be also
orthogonal to one another so I cannot have
more than three and they have to be
orthogonal to one another. So just with our
knowledge of the number of nodes and the
fact that we live in 3D we come
to the conclusion that that for n= 2
wave function, that's wave functions with one node,
I have one possibility of a spherical node
and then I have three additional wave functions
that I have that are going to have
nodes normal to the x direction, y direction,
z direction. Let's give names to these wave
functions. The spherically symmetric family of wave functions
are called S, these are called S states.
The wave functions that have like the one
we just discussed have an angular momentum of
one and they're called the P states. So
let's do the same exercise once more now
for n= 3 and again you can play
the same game and come up with wave
functions that have nodes in addition to it's
spherical nodes, now I have to have two
nodes. The S wave function that would be
the 3 S wave function has two spherical
nodes and the three P wave functions have
a spherical node that you can see here,
plus a plane, okay. And now given that
I have two nodes, then I can also
put the nodes along different planes and those
are called D functions. And now remember for
P states I can pick either one of
the three planes, x, y and z, so
I have the 3 P states. Okay and
we did the d, counting the number of
states is a little bit more difficult but
the solution is that I have five d
states, okay, so again the fact that we
live in three dimensions and the fact that
these wave functions need to be all orthogonal
to one another and that the number of
nodes essentially tells me the energy, I can
understand the ground states and the excited states
for hydrogen. [Slide 3] So let's formalize this
a little bit more. In order to the
general solution of the hydrogen atom I need
three quantum numbers to fully distinguish these wave
functions. And the first quantum numbers- the principal
quantum number n. This quantum number gives me
the energy of the wave function. And so
all wave functions with the same principal quantum
number would have exactly the same energy. The
next wave function is called- I'm sorry, the
next quantum number is called l. I'ts the
angular momentum quantum number and the values for
l go from 0 to n-1, okay. So
if the principal quantum number is one then
I only have l equals zero. That is
the S state and we call that S.
N equals one, we called that p and
n equals two, we call that d. Okay
and finally the last quantum number is the
projection of this magnetic quantum number along the
Z direction and this quantum number is called
z, I'm sorry, m. And m can take
values from negative l to plus l, changing
by integers okay. So again for p states
I can have -1,0 and 1 and remember
that the p state we said I have
three possibilities, x, y and z. And the
number that are correlated, they are not exactly
the same orbitals but they're correlated to one
another. So essentially with this information we can
classify this, all the excited states for hydrogen
[Slide 4] and what I'm doing here in
this slide is summarizing those results and putting
them in a position that is correlated with
their energy, okay, so the ground state is
here at the bottom , the ground state
energy iss -13.6 eV, this is called 1s
orbital in spectroscopy so n=1 and then the
angular momentum s. On their first excited states
I have two s orbitals okay, n equals
two and zero angular momentum and I also
have 2 p orbitals okay. And these two
p orbitals, I have three of them x,
y and z. The next shell up I
have three s, remember those are about 1.5
electron volts or so. I have 3 s
principle quantum number three, s meaning l equals
zero. Then I have 3 p and I
have three d states. And we could keep
going to fourth shell and then you have
f states and whatnot. Something to keep in
mind is that for hydrogen and only for
hydrogen, the energy of the states are only
governed by the principle quantum number. Alright. And
this concludes our very, very quick introduction to
the hydrogen atom in the additional reading materials
that I mentioned in lecture two you can
learn quite a bit more about what's going
on about these orbitals, and look at the
formal derivations but I think we have information
now we need to move on and think
about multi-electron atoms. [Slide 5] In this slide
I'm just adding- just listing the functional forms
for all of these orbitals that we just
discussed so that you have them for the
homework assignments. All of these functions are- can
be obtained analytically and we know the solutions
for all of these orbitals that we discussed
in terms of geometry. This table that I'm
showing here is taken from this book by
Carplus and Porter, and I recommended this book
where you can learn additional details and find
full derivations of these wave functions. [Slide 6]
Okay so let's move on and let's talk
about multi-electron atoms. So what happens as we
move from hydrogen to helium and lithium is
that the shape of these orbitals that we
found for hydrogen more or less stay the
same. And the ordering in energy of these
orbitals also stayed more or less as we
discussed, although we will talk about slight changes
that occur. So we're going to use these
states for hydrogen to talk about multi-electron systems
and, okay, so we're going to talk about
the electronic conflagration of these simple atoms. So
we're going to start with hydrogen, okay, the
ground state of hydrogen is a single electron
that I'm going to describe with this little
arrow pointing up and what I would say-the
way I would describe its electronic structure is
1s1 okay, that means one electron is the
superscript in the one s orbital. And the
next item in the periodic table is helium.
Helium has two protons in the nucleus and
two electrons so I need to put one
additional electron so I'm going to pick the
lowest possible state for my second electron. And
the electronic configuration of helium is 1s two,
at least two electrons in the 1s orbital.
Remember Pauli allows me to put two electrons
in the same orbital as long as they
have opposite spin. Okay so the next atom
up in- the next item in the periodic
table with three electrons is lithium. So now
lithium- the third electron comes in and it
needs to decide whether it goes to the
2s or the 2p, okay. And in principle
for hydrogen those states all degenerate, they all
have the same energy. But think a little
bit more about what happens with lithium and
see whether this remains the case. So again
the 2s and 2p have exactly the same
energy for hydrogen where the electron sees a
bare proton is the potential, one over r
potential. [Slide 7] But let's think about what
happens for the lithium atom. So what I'm
showing here in the plots is the probability
density function of finding an election at distance
r from the proton. And this is the
1s state and in the second- in the
bottom panel I am showing the 2s state
and also the 2p state in green. So
let's again go back to Lithium. I have
two electrons in the 1s orbital, okay, so
I'm going to say these guys occupy- that's
the electronic density at zero here. I have
the proton and this is distance from the
proton as a function in units of the
portrayed. So now my third electron. Electrons are
indistinguishable but let's talk about this just if
it simplifies the description. The third electron has
to decide whether it goes to the 2s
orbital, that's the red shape at the bottom,
or to the 2p orbital, that's the green,
and what we know is that if all
I had was a bare proton, these two
states would have exactly the same energy, but
I don't have a bare proton, what I
have is a 3 plus nucleus, a nucleus
with the charge of 3 plus surrounded by
two electrons in the one s orbital and
so when this third electron comes by, if
it's far away, Let's say the election is
over here, that third electron only sees an
effective charge that's plus one. So if an
electron is far away it feels exactly the
same way as with hydrogen, however as it
approaches the proton and starts overlapping with the
electrons this size, of the size of about
the Boer radius or so, it starts to
realize that is not hydrogen atom and that
the effective potential is different . So this
third electron is going to be- these two
orbitals are going to be comparable in energies
when they are far away because they're like
hydrogen but that the orbital that stand spends
more time near the nucleus will have an
advantage because the nucleus is really 3 plus
so that's much better than hydrogen and the
orbital would really not like to spend time
at the Bohr radius where the other two
electrons are centered because that'll lead to repulsion.
And so to really distinguish between these two
electrons you have to do the math but
there is a way, an intuitive way of
distinguishing the two and it is that the
2s orbital has this little bump just before
the first node and there's a relatively high
probability for that electron to be very, very
close to the nucleus where the potential energy
is very, very attractive. Whereas you can see
that the 2p has a much lower probability
for the electron being close to the proton.
So it's this little peak that makes the
difference and what happens is that the 2s
orbital is actually lowering energy than the 2p
orbital, okay, for multi-electron systems and this is
a general trend, that the higher the angular
momentum, the higher the energy within the same
shell, okay, so 2s is lower than 2p,
3s is lower than 3p and lower than
3d. And all of this is caused by
shielding. By the inner core- inner electrons shielding
the nucleus from the outer- the outer shells,
okay. That causes the outer shells to break
the symmetry that they have in hydrogen and
break the degeneracy that they have in hydrogen.
[Slide 8] So this kind of shows the
how the levels split up. What we see
on the left is hydrogen, okay, so over
here we have hydrogen. All the states, the
n=1 state of course is only one and
n= 2 state, they're all the same, n=3
states, they're all the same and in multi-electron
systems on the right-hand side we see that
the 2s and 2p split the 3s 3p
3p also split and the 4s, 4d, 4p-
I'm sorry, p, d and f also split.
And between the three and the 4 the
shells start crossing so things get more complicated.
So the take-home message is that within a
shell the higher the angular momentum, the higher
the energy and this is due to the
shielding of the nuclear charge, nuclear potential by
the inner core electrons. [Slide 9] Okay so
now we have all the information we need
to go through and start putting elections on
at least the first few rows of atoms,
okay. So hydrogen I put one electron pointing
up, it's called 1s1, I'm sorry, 1s1, and
then helium one electron up, one electron going
down, 1s2. Lithium- one electron up, one electron
down and 1s, when I get to the
two shell I have 2s and 2p, I
know that the 1s is better. So lithium
is going to be 2s, so the configuration,
the electronic configuration is 1s2, 2s1, okay. And
then I know everything going down is going
to be completely full in the 1 s
orbitals. Okay, beryllium is next, look at the
periodic table down here, that's where we are.
So beryllium I'm going to do one up
and one down. Boron- the 2s are going
to be full okay, from down- from now
on and boron I need to put one
more electron and of course it doesn't matter
in which P states whether Px, Py or
Pz they all have the same energy so
I can put it everywhere I can put
it any spin that I want. Now carbon.
So carbon has one electron in the P
states and I need to put another electron.
Now, so I have say one electron in
Px, so the electron is there, the second
election is not going to want to be
in Px because of electron electron repulsion so
the second electron that I put in going
to be either in Py or Pz, okay.
So that's a relatively simple to understand it's
just pure electrostatic repulsion. Now the question is-
for the second electron, what is the spin?
So do I put- I have one electron
with spin up along the x, do I
put the second electron with spin up- this
is one electron along x, is the electron
along y also also with the same spin
or do I put them spin up and
spin down? And the answer is a purely
quantum mechanical phenomena that we're going to discuss
now, only in an intuitive way, but in
the homework assignment you're going to do an
exercise to think about these two electron systems
and how these occur. Let me tell you
what the answer is, if I have an
electronic Px with spin up, the other electron
in Py is going to have- is going
to want the same spin and so the
answer is that the spins- it's going to
be parallel spins. Okay, so the two electrons
are going to want to have the same
spin and the reason that why they want
to have the same spin is kind of
complicated but they have correlations- it's called exchange
interaction. You know from the Pauli exclusion principle
that if two particles have the same spin
they cannot be on the same orbital. I
can put them in the same orbital only
if they have different spin, so if they
have the same spin they have this knowledge
of each other that's called exchange, that they
have to avoid one another and what happens
is when you do the math if two
spins in two different orbitals have- if two
electrons in two different orbitals have the same
spin, they are going to avoid each other
more, it's a correlation, than if they have
a different spin so Pauili doesn't really care.
So two electrons with the same state try
to avoid each other a little bit more
than if they have different spins. That rule
is summed up in in what's called Hahn's
rule, and so and what Hahn's rule says
is that given set of orbitals are going
to place the electrons in a way that
maximizes the total spin. These rules are a
bit more complex than we just discussed but
essentially these rules that we applied allow you
to understand the electronic structure of atoms, okay.
So one interesting result is that carbon is
actually- has a nonzero spin, it's magnetic, the
carbon atom. Nitrogen all spins up. Oxygen, I'm
going to add one more electron and of
course now I don't have a choice, I'm
going to put in the same orbital and
with opposite spin so also we would predict
that oxygen up has a total spin of
one, okay, each electron contributes one half of
magnetic moment and so the two have electrons
with parallel spin that contribute to one. Florine
is up down, okay. And then neon, of
course, at the end of the second row,
full shell. Okay, so with this we've finished
week one. There's a lot of information here,
we started in lecture two to discuss about
quantum mechanics and why we need quantum mechanics.
We discussed a couple of postulates that we'll
be using to describe quantum mechanics and then
we solved a couple of simple problems, starting
from the particle in a box then going
to hydrogen and then using hydrogen's wave functions
to talk about the electronic structure of multi-electron
systems. So we've come a long way and
what we're going to do in week two
is to use our knowledge to talk about
bonding. What happens when I bring two items
together to form molecules and also to form
solids? So homework one, the homework associated with
week one will be available to everyone who
registered in the course and it's going to
help solidify some of these concepts that we
have been discussing, get you a little bit
of practice thinking about quantum mechanics and thinking
about the orders of magnitude and the implications
of what we have discussed, and also you're
going to do your first density functional theory
calculation to look at the very simple calculation
of an oxygen atom to estimate these exchange
correlations, the importance of putting electrons with the
same spin versus electrons with different spins and
that cannot be done on the back of
the envelope, so you're going to use a
research grade simulation code to quantify that number
and develop a little bit of an intuition
for these energy terms. Thank you very much
and I'll see you in week two. Bye.