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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK, let's just
take 10 more seconds. OK. Does someone want to
explain the answer here? I'm not sure if this is on. Give it a try. AUDIENCE: So it
says in the problem that the X-rays have
the same wavelength, so you know that they also
have the same frequency, so that discounts
1, 2, and 5 and 6. So then it's just a
choice between 3 and 4. And in the video
the other day, you said in order to
image these proteins, you need high intensity light. So, 3. PROFESSOR: Yup. That's a great explanation. Here, I don't really
know what these-- these might come in handy today. I don't know. OK. Yeah, so the trick was
better quality data, so you probably figured
out that that, then, was the higher intensity. So this is a true thing. So we have data collection. There's equipment here
at home, and a lot of universities have what
they call home data collection equipment, but we often travel
to synchrotrons where we have higher intensity, i.e.
more photons per second, and then you get
better quality data. And so sometimes people
do these things remotely where you ship your samples
and someone else collects it, but my lab likes to go. And you stay up all night
and collect great data, and it's a bonding experience. You saw a little bit
of that on the video. OK. We ended last time looking
at the Schroedinger equation and seeing that the
Schroedinger equation could be solved for a hydrogen
atom, giving information about binding
energy, the binding of the electron to
its nucleus, and also a wave function, which we
haven't talked about yet. So we're going to
continue talking about this binding energy,
and then next week we're going to move into wave
functions, or orbitals. So the binding
energy that comes out of the Schroedinger
equation, no one should ever just believe things. It looks fancy, but
does it really work? Is it really doing
this right estimation? And again, it just came
out of Schroedinger's mind, so it's always nice to
verify that this equation is working pretty well. So today we're going
to talk about how we were able to verify
that the binding energy that the Schroedinger equation
was predicting actually agrees with experiment. So we're going to continue
talking about binding energies, then go on to the
verification, with a demo, of how people were
able to show that there was good agreement here. All right. So let's continue
with binding energies. So we're still talking about
the hydrogen atom and energy levels, and we saw the last time
that the Schroedinger equation could be derived
for a hydrogen atom such that the binding
energy, or e to the n, was equal to minus this Rydberg
constant, RH, over n squared were n is the principle
quantum number. And so this is what
we saw last time, and now we have a graphical
depiction of this. And you'll note that this is
a negative value over here. So if n is 1 and we have
the principle quantum number of one, we have
minus RH over one squared. And so we just have
the negative value for the Rydberg constant,
2.18 times 10 to the minus 18th joules. And as we go up
here in energy, we would get to an energy of zero. And if energy here is zero,
what must be true about n? What kind of number is n here? Infinity, right. So if this is infinity,
that number goes to zero. And so if the electron
is infinitely far away from the nucleus, it's
basically a free electron. It doesn't feel any
kind of attraction. It's infinitely far away. Then your binding energy
would be zero, i.e. it's not bound,
and that would be true with this infinitely
far away distance. And then in between the n
equals 1, 2n equals infinity, we can use this equation
for the hydrogen atom to figure out what
these energy levels are. So when we have the
n equals 2 state, it would be minus RH
over 2 squared, or 4, and so we can calculate
what that number is here, minus 0.545 times 10 to
the minus 18th joules. N equals 3 so we have
RH over 3 squared. We can do the math over here. 4-- you get the idea--
minus RH over 4 squared, and we have then over 5
squared, over 6 squared, and you can see the energy, and
you can calculate the energy levels here. All right. So when you have an electron in
this n equals 1 state, that's the lowest energy, it's
the most negative number, and that's known as
the ground state. And when you have an electron
in this ground state, that's the most stable
state for the hydrogen atom. So again, from these lower
ground state up to this state here. Now we're going to
introduce another term which you'll hear a lot, and
this is ionization energy. So the ionization energy,
the amount of energy you need to put in to ionize
an atom or release an electron. So the ionization energy of a
hydrogen atom in the nth state is going to be equal
to the binding energy, but the signs of these
are going to be different. So we have this equation where
binding energy equals minus IE, the ionization energy. So we talked about the fact that
the binding energy is negative, and the ionization energy
is always positive. So for the binding energy, when
the binding energy is zero, it means the
electron isn't bound. So a negative value
for binding energy means that the
electron's being held by the nucleus-- the
electron's bound. For ionization energy,
that's the energy you need to add to the system
to release the electron, and you're always going to
need to add some energy, so that's a positive number. So when you think about
ionization energy, it's a positive
number that you're going to be expecting there. Now we can consider
this same diagram, and we already talked
about these energy levels and now we can think about these
in terms of ionization energies as well. So the difference
from this state, where energy is 0, to the
ground state down here, the ionization energy--
the energy that's needed to ionize an electron
that's in n equals 1 here, is going to be equal
to minus the binding energy of that electron in
that n equals one state. So again here, it's
not too hard if you know this information
and this equation to figure out what the
ionization energy is. So that's just, then, going
to be the positive value of the binding energy. So binding energy
minus Rydberg's constant here, 2.18 times
10 to the minus 18th joules. So the ionization energy,
then, for a hydrogen atom in the ground state is positive
2.180 times 10 to the minus 18th. And I'm just going to try
to use the same number of significant figures. I always try to pay attention
to my significant figures. All right. So we can do this again
for the n equals 2 state, or the first excited state. So here's the n equals 2 state. So now we're going to be talking
about this differential energy here. So the ionization
energy for an electron in this first excited state. Again, that will be ionization
energy equals minus the binding energy for that state,
and so that's going to be, then, the positive value here. So the binding energy was minus,
if we change this to-- this is 18.5 to the 18, or 5.0
to the minus 19th joules. Try to keep the significant
figures the same. All right. So why don't you
give this a try now and we'll have a
clicker question. OK. 10 more seconds. A very long one second. OK. Interesting. So maybe you can talk to
your neighbor and someone can tell me what the trick is here. OK. We have someone who is going
to tell us what the trick is. AUDIENCE: The ground
state is n equals one, and from there, the
excited states go up one. So the first excited
state is n equals 2, and then the third one
will be n equals 4. And since you're looking
for the ionization energy, you go to the energy
for n equals 4, and you multiply by negative
one, which is 40.14 10 to the negative 18th. PROFESSOR: Great. Yeah, so let's see what I have. I need to get some more stuff. Sorry. OK. So the trick. It's not a hard problem. You just had to figure out what
the third excited state meant, so once you've figured
that out, it was pretty easy to get it right. OK. So keep that in mind. Now that you've potentially
made that mistake, you will not make
that one again. All right. So now we can think about this
also in more general terms-- only slightly more
general terms, frankly-- which is to consider,
for other one electron ions, we can have a more
general equation. So we had, for
the hydrogen atom, the binding energy, En, is minus
Rydberg's constant, RH, over n squared. And now I've added z squared,
which is the atomic number. And for hydrogen, it's
1, so it wasn't around. We didn't need it before. But we can consider other ions
that also have one electron. They will also work
with this equation. So there's a couple of things
that kind of fall out of this. One, that an electron is
going to be bound more weakly when n is a big number here, and
so that makes sense from what we were looking at before. And that an electron
is also going to be bound more
tightly when z is big, and we haven't really talked
about that because we've just been talking about
the hydrogen atom, and so it always has the same z. But if you have a
different z, you're going to have a bigger
positively charged nucleus, and so it makes sense that
you would then have a tighter bonding electron. All right. So you might think, what
are other things that have just one electron that
this is going to apply to? And so far, of
course, we've just been talking about
our friend, hydrogen, that has its one
electron and z equals 1. But we have ions that can
also have one electron. So helium plus, and it has a z
of 2, but when it's helium plus it only has one electron. Lithium, plus 2, also
only has one electron, and it has a z of three. And then what about something
that's a one electron system with a plus 64. Without looking at your
periodic table, what does the z have to be here? Yeah. So, 65. So in working these
kinds of problems, if you're talking about a
one electron ion or atom and it's not hydrogen,
don't forget about z. And we need to have
that in our equation. All right. So talking about these
binding energies now, out of the
Schroedinger equation, you can calculate
ionization energies if you know the binding energy. All of this is
good, but how do we know that we can trust
the Schroedinger equation, that equations
really are working? So the way that they
figured this out is from experiment, and
particularly, experimentally figuring out what the energy
levels were, and thinking, does this match with the
Schroedinger equation? So they were able to use photon
emission to be able to do this. So let's consider what
photon emission is, and then we're going to prove
that this equation that I've been showing you actually holds. So photon emission,
this is a situation that occurs when
you have an electron going from a higher energy
initial state going to a lower energy state. And as it goes from
this high energy state to the low energy state, there's
a difference between these two energy states,
delta E, and that's going to be equal
to the higher energy initial state minus the
energy in the final state. So there's this difference in
energy between the two states, and the photon that gets
emitted when this energy transition happens has the
same energy as the difference between those. So the energy of the emitted
photon is also delta E. So you emit all of that energy
as you have that change. So the difference here-- we
can consider an actual case where we're going from
an energy difference of n equals 6 to an energy
level of an equals 2, and we can think about what
the energy difference is between these two and we can
just write that equation out. So the initial energy, the
electron started at n equals 6, the energy of the
n equals 6 state, and it goes to the energy
of the n equals 2 state. So energy n equals 6
minus energy n equals 2. All right. So of course if you
know energy, you can know a lot of other
things about the photon. So you can calculate frequency
of that emitted photon. So again, we have our
energy difference here and we can then solve
for the frequency of the emitted photon, which is
equal to the energy difference. That energy, divided
by Planck's constant, and you could also write
it out, the initial energy minus the final energy over h. All of these are
equivalent things. And when you know
frequency, we're talking about light here, so you
can calculate the wavelength. So let's think now about
what we might expect in terms of frequencies
and wavelengths, depending on the energy
difference between the two different states. So here, if we think
first about this electron with the purple line, we have
a large energy difference here between this
state and this state down here, between n
equals 5 to n equals 1. So when we have a large
difference in energy, what do we expect
about the frequency of the emitted photon? Is it going to be a high
frequency or low frequency? High. Yep. So large energy, high frequency. And so then what would be
true about the wavelength of that emitted photon? Short. Right. Now if we had a small
difference, say n equals 3 to n equals 1, which is a smaller
difference in energy, what's true about the frequency here? Right. Low frequency. And wavelength? Right, long wavelength. Right. So now we're actually going to
see some photons being emitted, and let me just build in to
this experiment a little bit. So we have an evacuated glass
tube filled with hydrogen. And if you have negative
and positive electrodes, you can emit light from
this and then analyze the different wavelengths. So we are not going to be
the first people to see this, but we're going
to try this and we should observe these different
wavelengths coming off. And after we observe
them, we will try to calculate
what they're due to, and then if the experimental
results of the wavelength and frequency observed can be
explained by the Schroedinger equation. But first, let's actually
see the visible spectra that is created by hydrogen.
And so we have our demo TAs, and actually if all our
TAs can help pass out some little glasses to
help everyone see this. And when we're ready, we're
going to do lights down. But let's get everything
handed out first. All right. I got the lights. TA: So this is a hydrogen
lamp, and when you turn it on, the electricity excites
all the hydrogen inside and then you see this glow from
the electromagnetic radiation being emitted by these excited
hydrogens relaxing down to the ground state. PROFESSOR: Want
to try the light? TA: So we're going to
try this for those of you who don't have the glasses,
but let's see if this works. It was kind of there. Is that what we're
supposed to see? You're supposed to
see all of them. I guess you can't, really. I guess depending on
how you move this thing, maybe you end up
seeing all of them. Is it working? PROFESSOR: It's not working. TA: It's not working? No? Yeah, I know. It kind of works, depending
on how I move this thing. PROFESSOR: Sometimes
it works really well. Should we just try walking? Can we hold it up and see
whether people also can see it? TA: All rights. So we're going to hold it up. See if you guys-- PROFESSOR: Can just see
it without the camera. TA: OK. So what you should
be able to see, for those of you that
have your glasses, is a continuous spectrum
with the various colors. PROFESSOR: You might have
to get the angle right. Are people in the middle
of the room able to see it? Can anyone see it? Yeah? People on the edge of
the room, can you see it? I think it's harder
from-- and the camera's blocking people a little bit. TA: It works. I can see it. PROFESSOR: Do you want to move
it up farther, like in front of that? Try turning it around. All right. Maybe we'll turn it
slightly, and then we can come down, maybe,
and try it after class if it's not working very well. When it's tilted, are you
having better luck over here? All right. I guess we'll bring
the lights back up, and I'll show you what
you should have seen, if it didn't work for you. So how many people were
able to see the spectra? OK. All right. So a good number of people. Great. I feel like this room is
not as perfect for this as some other rooms. But there are some rooms that
actually don't get dark at all, and then you can't
really see anything. All right. So maybe if we have a chance,
we can try again at the end. All right. So this is what you
should have seen. You should have seen these
different series of lights, or series of colors coming off,
and we're not the first people to see it. So J.J. Balmer, in 1885,
reported seeing these colors, and he wanted to calculate
the frequencies of the lights that you were seeing
emitted from this. And so he did calculate
the frequency, and then he tried to figure out
the mathematical relationship between the different
frequencies of light that he was observing. And he found that the
frequency equaled 3.29 times 10 to the 15th per second,
times 1 over 4 minus 1 over some number, n, where
n was either 3, 4, or 5. And he really didn't understand
what the significance of this was, but it was pretty. You had hydrogen in
this sealed tube, and there were colors that came
off, and they had frequencies. So that's kind of where that
stood for a little while. So now let's think about what
those different colored lights were due to. So we have here energy
levels, and the transitions that we were observing
are all going to be n equals 2 final state. And we can think about why you
didn't see any transitions to n equals one. Think about that. We'll come back to
that in a minute. But there were these
different transitions that were being observed
from 3 to 2, 4 to 2, 5 to 2, and 6 to 2. So now let's think about which
colors-- which wavelengths-- are due to which
of the transitions. So for the red one,
what do you think? 3, 4, or 5 transitioned to 2? 3. It is 3. And you could think about that
in terms of the smaller energy. That's the smallest
energy, so that would be a low frequency
and a long wavelength. So the one with the longest
wavelength-- and red is the longest
wavelength-- so that must be the transition from an
initial n of 3 to n equal 2. And then we can fill in the
rest, so this one over here must have been n equals 4 to 2. This one here, then, would
be the blue, n equals 5 to 2. And then the purple or indigo
at the end, n equals 6 to n equals 2. So they saw these
four colors, there were these different
transitions, and so then, now,
we can calculate what the frequencies
of these are and think about this, then, in terms
of Schroedinger's equation and test Schroedinger's equation
to see if it predicts this. So we can calculate
the frequency, then, of the emitted
photons, and we had frequency equals the initial
energy minus the final energy state, or this delta E,
over Planck's constant. And from the
Schroedinger equation, we know about what these energy
levels are from Schroedinger. And this is, again, for
hydrogen, so z equals 1, so z isn't shown. We have the binding energy
equals minus RH, Rydberg constant over n squared. And now we can put these
equations together. So we can substitute these
energies in using these, and so we can do that here. We'll pull out Planck's
constant, so 1 over h. And then we can substitute in
minus RH over the initial n level squared minus minus
RH over the final n squared. And we can also simplify
this a little more, pull out RH over here, and now
we just have 1 over the final-- we have minus a minus,
so we've rearranged this-- 1 over n final squared
minus 1 over n initial squared. And we have an
equation that solves for the frequency in terms of
Rydberg, Planck's constant, and what the principle quantum
numbers are, what n is. So let's look at
this a little more. Now remember, this is all
going to n final of 2, and so we can put that
equation up here again. So when this is
2, 2 squared is 4. And if you remember
back, Balmer had a 4-- had this part
of the expression-- but he had a strange
number over here that he experimentally
determined. But if you take RH and divide
by Planck's constant, Rydberg constant divided
by Planck's, you get that number that
Balmer had found back in 1885-- 3.29 times 10
to the 15th per second. So when we plug in the values
from Schroedinger's equations, you come up with the
experimentally determined values for frequencies,
or wavelength, of the emitted light. And of course,
from the frequency, you can calculate
the wavelength, and the wavelengths that
were observed experimentally agreed with the wavelength
you would calculate from the Schroedinger's
equation to one part, n times 10 to the 8th. So the agreement was
absolutely amazing. So Schroedinger's
equation, which was taking into account
the wavelike properties of the electrons,
were able to predict, for a hydrogen atom,
what wavelengths you should see emitted in
that hydrogen atom spectra. So this was really exciting. Schroedinger
equation was working, we had a way of
describing the behavior that we were observing
for these electrons, and that was really incredible. And I think Balmer should
get a lot of credit as well for all of this. And the people who were doing
these early experiments, they didn't know
what it was meaning, but they were coming up
with the data that allowed to test theories later on. OK. So this was a series
going to a final n of 2, and so we have
the Balmer series. That was the visible
series that we were seeing. So what about, why wasn't
anything going to n equals 1? And that's a clicker question. So at the end, I'm going to
ask you to put up the winners. Is my number good? OK. 10 more seconds. OK. So 71%. So the trick here is to
think about-- well, actually, someone can tell me, maybe. What was the trick
here to think about? I'll get a little exercise. I'll come up. How's everyone doing up here? AUDIENCE: So for
the Lyman series, there's a more difference
from the Balmer series, so there's more energy in
the transition when it goes down back to the ground state. So for that, with
more energy, it's going to be a
shorter wavelength, and that's ultraviolet. PROFESSOR: So here it would
be convenient to remember your orders of what are short
and long wavelength kinds of light. OK. So we have the UV range, then. And so that's why you
didn't observe it. It was happening,
but you didn't see it because it was in the UV. All right. So then we can go on and
look at the other things that can happen here. So we can have n
final of 3, and you don't need to know the
names of these series, but that would be near IR. N equals 4 would
be in the IR range. So only some of what's happening
is actually visible to us. We see beautiful colors
from transitions, but there's other
things happening, too, that are not visible to us. So at this point, we're feeling
pretty good about those energy levels, about the
Schroedinger equation being able to successfully predict
what kind of energy levels you have, that binding energy. And now, this
verification was good, from your photon emission. But there's another
property that you can have, which is photon absorption. So why don't we do yet
another clicker question as a competition, after
all, about absorption. OK, 10 seconds. OK. So now we're talking
about a different process. We were talking before
about electrons. They're starting up in a higher
energy level, going lower. But with photon
absorption, we're going the other direction, so
we're going from a lower state. We're being excited. They're absorbing energy
and being excited, so we can have a final
state that is higher, and the energy is gained in this
process, so it's being excited. All right. So we can think about
the same things, then, in terms of absorption. So if we have a big
energy difference, if it's absorbing a lot of
energy, big energy difference, it's going to be absorbing
light with a high frequency and a short wavelength. If there's a small
energy difference, it'll absorb a photon with a low
frequency or a long wavelength. And we'll come back to
some of these ideas, actually, well into the
course, and we'll actually look at some pretty colors. So in this case now, we can
calculate frequency again, but our equation is a
little bit different. So we have the Rydberg constant
and Planck's constant again, but now we have 1 over initial
n squared minus final n squared, and so this term should
be a positive term. We should be getting out
a positive frequency. So if I take this again
and put it up here, you want to think about
whether, if you're talking about
absorption or emission, that's what's telling you if the
energy is being gained or lost. So you're not going to have
negative frequencies in one case. You're going to be
absorbing the light of a particular frequency, or
emitting light of a frequency. So pay attention
to your equations, and think about whether
your answer actually make sense when you do them. And again, all
these equations are going to be provided to
you in an equation sheet. OK. So let's consider the summary
of both of these things now. So we're talking about
admission versus absorption. And so we have this
Rydberg formula, which is what this
is called, and it can be used to
calculate the frequency of either emitted photons
or absorbed photons, so from either process. And if we want to
make it more general, again, it's just a
one electron case, but we can put our z in
for any one electron ion. So frequency equals Z squared,
Rydberg constant over Planck's constant, and we have 1 over
the final minus initial, or initial minus final,
depending on which process you're talking about. And over here, we'd be
talking, then, about emission. So our initial energy is
higher going to lower, and when that
happens, you're going to be releasing light with
the energy difference that is due to the difference
between these states. So we're going to
have our electron is going to emit this energy. In the absorption process
for this equation, we're going to go from
an initial state that's lower to a higher state. So we'll have a final state
that's higher than initial. That's absorption. It's absorbing energy. It's getting excited,
and so the electron is absorbing that energy. So this really summarizes,
now, what we need to know about binding energies. Schroedinger equation
also tells us about wave functions, which is
what we're moving into next. So that's all for
today, except we have a very important
announcement, which is congratulation to recitation 6. Lisa, you are the first winner
of the clicker competition. Have a great weekend, everybody.
