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PROFESSOR: All right.
Let's get started.
Why doesn't everyone go ahead and take ten more seconds on
the clicker question.
All right, and let's see how we did.
Alright, excellent job, 86% of you, that's right.
What we had just done a clicker question on is
discussing light as a particle and the photoelectric effect,
so we're going to finish up with a few points about the
photoelectric effect today.
And then we're going to try a demo to see if we can convince
ourselves that the kind of calculations we make work out
perfectly, and we'll do a test up here about
half way through class.
And we'll also talk about photon momentum as another
example of light behaving up as a particle.
After that, we'll move on to matter as a wave, and then the
Schrodinger equation, which is actually a wave equation that
describes the behavior of particles by taking into
account the fact that matter also has these wave-like
properties.
So, starting back with the photoelectric effect -- yes.
STUDENT: [INAUDIBLE] class last time [INAUDIBLE] notes.
PROFESSOR: Oh, sure.
Can one of the TAs maybe come up and hand around anyone that
didn't get notes?
We have not yet perfected the art of entering and exiting
this classroom yet, we're still working on that.
Raise your hand if you need notes and we'll make sure we
get those to you.
All right, so where we left off with the photoelectric
effect was when we first introduced the effect, we were
talking about it in terms of frequencies.
So, for example, we were talking about a threshold
frequency as in a minimum frequency of light that you
need in order to eject an electron from a metal surface.
What Einstein then clarified for us was that we could also
be talking about energies, and he described the relationship
between frequency and energy that they're proportional, if
you want to know the energy, you just multiply the
frequency by Planck's constant.
So, now we can talk about it in different terms, for
example, talking about e sub i, which is the incident
energy or the energy of the light that comes in, or
talking about work function here, and that's just another
way to say threshold energy.
So, the work function's the minimum amount of energy
that's required in order to eject an electron, and most of
you understand this relationship here, which is a
little bit cut off, but it is all the way on in your notes,
and that is what you saw the clicker question on -- how you
can figure out, for example, the kinetic energy of the
ejected electron by looking at the difference between how
much energy you put in and how much energy is required to
eject that electron in the first place.
So, in this class we'll be talking about energy a lot,
and it's often useful to draw some sort of energy diagram to
visualize the differences in energy that we're discussing.
So, we do this here for the photoelectric effect, and in
terms of the photoelectric effect, what we know the
important point is that the incoming photon has to be
equal or greater in energy then the work
function of the metal.
So here we have energy increasing on the y-axis, and
you see this straight line at the bottom here is lower down
on the graph, and that's the energy of a bound electron, so
that's going to be a low stable energy.
But we see if we have a free electron, as we do in this
dotted line here, that's going to be a higher energy that's
less stable.
So, if we want to go from that stable state to that less
stable state, we need to put in a certain amount of energy
to our system, and that's what we define as the work function
here -- that difference between the free electron and
the electron bound to the metal.
So, the most basic case to understand, which is what we
just saw is a case where we have the incident energy
coming in, and that incident energy is greater than the
work function, and in that case what we see is that we
have an electron that is ejected.
That makes sense and it also makes sense that this little
extra bit here, that's the amount of energy that we have
that goes into the kinetic energy of the electrons.
So, that's how we could also graph figuring
out the kinetic energy.
So, in the second case what we have is what happens if we
have the incident energy at some amount that's less than
the work function, and in this case we're showing 1/2 of the
work function.
So in this case, we don't have enough energy to eject an
electron, so, an electron is not ejected.
And that's pretty clear, too, and the question I want to
pose to you is instead the third case here.
So in the third case what I'm showing is that we have -- now
we're not just talking about 1 photon, we're talking about 3
photons -- let's say we shoot them all at the same time at
our metal, each of them having some energy that's let's say
1/2 the work function.
So, just to take a little bit of an informal survey, who
thinks here that we will have an electron that is
ejected in this case?
So a couple hands, all right.
And what about who thinks that we will not have
enough energy here?
All right.
We've got a big majority, and both are logical ways of
thinking, but it turns out that the majority is correct,
which is not always the case, but the electron is not
ejected in this case.
And the reason for this, and this is a very important point
about the photoelectric effect, and the point here is
that the electrons here are acting as particles, you can't
just add those energies together.
One individual particle is being absorbed by the metal
and exciting an electron.
So, having other particles around that have the same
energy that you could technically add up if you were
adding them up like a wave, you can't do the same thing
with particles, they're all separate.
So, the take-home message is whether you have three photons
or 3,000,000 photons that you're shooting at your metal,
if you're not at that minimum frequency or that minimum
energy that you need, nothing is going to happen.
So, you might ask then well what is the significance of
shooting different amounts of photons at a metal?
Is there any significance at all, for example, in the
number of photons that are hitting the metal or being
absorbed by the metal.
And there is a relationship here, and that is that the
number of photons absorbed by the metal are related to the
number of electrons ejected from the metal.
So, in this figure here what I'm actually showing is these
little sunshines, which let's say are each
one individual photon.
So we had six photons going in, so the maximum number of
electrons that we're going to have coming out is also six
because the maximum scenario that we could have that would
maximize the number of electrons is that each one of
those photons comes in, excites an electron, ejects it
from the surface of the metal.
It's important to note, of course though, it's not just
the number, it's really important that the energy of
each one of these individual photons is, of course, greater
than the work function of the metal.
So, that's, in fact, it's that number of photons that we're
talking about when we refer to the intensity of light, and
the intensity of light is proportional to that number,
because when we talk about intensity, really we're
talking about the amount of energy that a stream of
particles, a stream of photons, has per second.
So, if we have a high intensity, we're talking about
having more photons per second, and it's important to
know also what that does not mean.
So it does not mean that we have more energy per photon.
This is a really important difference.
Intensity, if we increase the intensity, we're not
increasing the energy in each photon, we're just increasing
the number of photons that we're shooting out of our
laser, whatever our light source is.
And when we talk about intensity in terms of units,
we usually talk about watts, so if you change your
lightbulb, usually you see the intensity in terms of watts.
But in terms of SI units, which become much more useful
if you're actually trying to use intensity in a problem and
cancel out your units, we're just talking about joules per
second is what intensity is.
So at this point, you should be able to have all the
background you need on the photoelectric effect to solve
any type of problem that we throw at you, and you see
three on this problem set, and we'll probably give you one
more on your next problem set, and the reason we ask you so
many questions about the photoelectric effect is
because it actually is very similar to ionization energy
that we'll talk about later, also problems dealing with
photoelectron spectroscopy.
So, we want to make sure that this is something the entire
class is 100% solid on.
Sometimes the questions are worded quite differently, so I
just want to sum up here the different ways
they could be worded.
For example, if we talk about photons, of course, we also
just mean light, sometimes we refer to this as
electromagnetic radiation, and there's several ways that you
might be asked this in a problem or that you might be
asked to answer.
Sometimes we might just directly tell you the energy
of the photon -- that's probably the easiest scenario,
because when we think about work functions those are
usually reported in energy.
So since that's the easiest scenario, you can probably be
sure it's not going to be too frequently that you're just
given the energy, right, that might be too easy.
So really what we'll probably do is instead either give you
the wavelength or the frequency and you'll go ahead
and calculate the energy from there.
In terms of talking about the electrons, I wanted to point
out that in the book and other places you might see electrons
referred to as photoelectrons.
That's sometimes confusing for people, because it seems like
okay, is it a photon or is it an electron.
I just want to clarify that it is an electron.
It's called this just because it's an electron that results
when an electron absorbs a photon's worth of energy, so
thus it's a photoelectron.
And if we talk about electrons or photoelectrons, again we
can describe it in terms of energy, we can talk about
velocity, and from there, of course, you can figure out the
energy from 1/2 m v squared, and actually we can also
describe the electron in terms of wavelength.
So you don't actually know this yet from this class,
you'll know it by the end of the class that electrons can,
in fact, have a wavelength.
So once we cover it, it will then be fair game to ask these
photoelectron spectroscopy or these photoelectric effect
questions using the wavelength of the electron.
Also to point out, a lot of times you'll see electron
volts instead of joules, this is the conversion factor here
just so you all have it in your notes.
All right.
So let's test what we, in fact, know about the
photoelectric effect, and before we do that actually,
we're going to calculate what we would predict, so when we
do the demo it will be meaningful and we can tell
whether we're successful or not.
So hopefully we will be successful.
And as I point this out, we now know how to do any kind of
photoelectric effect problem, also this means you should be
able to go back to Monday's notes where we filled in all
those graphs, which were what different scientists were
observing when they were measuring either the frequency
or the intensity of light that was irradiating different
types of metals, and also the number of electrons ejected,
and the kinetic energy of those electrons ejected.
You should be able to maybe print out a blank copy of
those notes from the website and fill in all those graphs
-- not for memorizing them, but now just understanding how
the photoelectric effect works.
All right.
So let's do an in-class problem, and this will be done
with zinc. We have a zinc plate up here, and we're going
to -- in a minute I'll describe how we can probe if
electrons are coming off of it.
But we're going to irradiate it with two
different light sources.
We have a UV lamp right here, which is centered at a
wavelength of 254 nanometers.
And then since we have my red laser pointer, we will also
try with the red laser pointer, which is centered at
wavelength of 700 nanometers.
So, there are a few questions that we need to answer first.
So we want to see, do we expect to eject electrons off
of this metal surface, or do we expect that we don't have
enough energy?
So that means we're going to need to figure out what is the
energy per photon that's emitted by that UV light.
Also, what's the energy per photon of this red laser
pointer, and then it's also worth trying a calculation
dealing with intensity.
So let's also try calculating the numbers of photons that
would be emitted by this laser pointer, if, for example, we
were to use it for 60 seconds and this were a
one milliwatt laser.
So, let's do some of these calculations starting first
with what is the energy per photon, and let's start with
the UV lamp.
So we know that energy is equal to Planck's constant
times nu, but what we know about the lamp is its
wavelength, or the light that's emitted.
We know that nu is equal to c over wavelength.
So we can figure out the energy of each photon emitted
by our UV lamp by saying e is equal to h c over wavelength.
So let's just plug in these numbers here.
That means our energy is equal to 6.626 times 10 to the -34
joules times seconds.
And then we have c, the speed of light, 2.998 times 10 to
the 8 meters per second.
And we want to divide all of that by our wavelength, and to
keep our units the same we'll do meters.
So that's 254 times 10 to the -9 meters.
So hopefully if some of you have your calculators with
you, you can confirm the answer that I got, which is
that the energy is 7.82 times 10 to the -19 joules.
So, remember what we're talking about here is the
amount of energy that's in each photon.
So if we think about the work function for zinc, and the
work function for zinc is 6.9 times 10 to the -19 joules, do
we expect that when we shine our UV light on the zinc,
we'll be able to eject electrons?
What do you think?
Yes.
Good.
OK, anyone disagree?
No, OK and that's correct, because each photon of light
actually has more energy than is
needed to eject an electron.
So, we would expect to see electrons ejected with the UV
light source.
So let's now think about using instead the amount of energy
per photon in that red laser pointer.
So again, we know that energy is equal to h c divided by
wavelength, and energy is equal to -- you have written
out in your notes what the actual value for h c is, but
now our wavelength is 700 times 10 to the -9 meters.
And what we end up with for the energy then is 2.84 times
10 to the -19 joules.
All right.
So please raise your hand now if you think there'll be
sufficient energy to eject electrons
from the metal surface?
And raise your hands if you think there won't be.
OK.
Good hand raising technique.
Yes.
In fact, there is not enough energy in a single photon to
go ahead and eject an electron from this zinc surface.
So our last question we ask is what's the total number of
photons emitted if we give this given
intensity for 60 seconds?
So, keep in mind that one milliwatt is just the same as
saying 1 times 10 to the -3 joules per second.
So we have 1 times 10 to the -3 joules per second, and we
want to multiply that by -- or cancel out how much energy we
have per photon, first of all, so how much energy do we have
per photon if we're talking about the red laser pointer?
Right.
So this value right here.
So for every photon we have 2.84 times
10 to the -19 joules.
We're saying let's do this for 60 seconds.
So what we end up with for the number of photons in this
laser beam of light is 2.1 times 10 to the 17 photons.
So this gives you a little bit of an idea of just how many
individual photons there are in a laser beam of light.
This is a huge number of photons.
So the question is does this matter?
How about if we shoot this many photons?
Does it make any difference at all in terms of whether we can
eject an electron?
No, it actually doesn't.
It is an impressive number, it is very, very large, but it
doesn't make a difference.
So we see that we do not eject electrons in the case of the
laser pointer, even if we have this intensity, even for 60
seconds -- it is still not related to the energy of an
individual photon, so we won't see an effect.
All right.
So let's hope that we can confirm our predictions here
by actually doing it, and Professor Drennen well help me
out by loading up our device with electrons, and I'll
explain exactly what our set up here is as she does that.
So basically what we have is this zinc plate here.
So that's what we want to load up with electrons, and then
see if we can remove some.
But that's a little bit hard, we aren't all that good at
seeing electrons with our eyes, so we need to think of a
way to do this.
So what she's going to do is start loading up the
electrons, and you see this wand here move slowly, and it
takes a while to do it, start become perpendicular.
The reason for that is because all of this is connected, so
we're moving electrons everywhere in the system.
And since we have two bars that are together like this,
once they're both loaded up with electrons there's going
to be negative charges that repel, so the electrons will
want to get as far away as possible, and they're on their
slow way to doing that, to getting as far away from each
other as possible.
And if we do, in fact, hit it with light to get the
electrons off, it will go back to the straight up in
position, or if it gets knocked hard enough
it does that, too.
Sometimes it's easier actually not touch it to the metal, I
should have--
TA: It's hard to see if it's moving or not.
PROFESSOR: So, our technology TA is also our paper TA.
Darcy will hold up the yellow paper.
Right, there we go, now we're making a little progress.
TA: It was moving before, you just couldn't see it.
PROFESSOR: So, does anyone have any questions about the
set up here, does it make sense what we're looking for
the bar to go back once we make some progress.
This demo works wonderfully in the winter months in Boston
when we will all be full of static at all times.
We're still close enough to the summer that the air is not
just filling us up with extra static electricity, so it's a
little more challenging here.
We'll try to make this happen only once.
I think that's probably, if we can get one more.
So, it works, I think it's just getting too much
[INAUDIBLE].
Sometimes it helps to not actually hit the metal, just
put it next to the -- there we go.
I wonder if there's some UV light out of this new lighting
set up in our classroom here.
That would be a little tricky.
All right.
I think this -- if this sticks.
Yeah, it's the pressure of the paper.
I think that's good enough, we'll be able to see.
If you can keep showing that, though, Darcy, we'll try
different scenarios and I'll try not to put
laser in your eye.
Actually you can look down as well as an added precaution.
OK, let's try it with that.
That's enough then.
So, the first thing we're going to try is with the red
laser pointer, because that we are expecting not to have an
effect, and that will prevent Professor Drennen from having
to charge up our apparatus again.
So, Darcy will look down at this moment and we will hit
this with the laser pointer, and what we see is nothing is
happening at all.
OK, good.
Control one working.
So now very carefully take our UV light source --
Darcy again will divert her eyes and her skin.
Let me make sure this is actually on.
OK, so we've got UV light here, and let's see what we
can see, and we lose electrons, if
that's what's happening.
And it often doesn't go all the way, because actually this
device gets stuck right there.
So let's charge it up again and see if we can check again.
But did you see movement?
Are you buying our story here?
This is actually very representative of when you do
research in the laboratory, you will find often things do
not work quite exactly as they worked 20 minutes ago when you
just checked it in your office, for example.
And sometimes it's a matter of factors that you need to
figure out what it is, and maybe it's that there's extra
light in the room we don't know about.
It might just be -- so, we did get it back to
the starting position.
Next time maybe we'll charge it up before class.
All right.
So we kind of saw what was happening here, you saw it
move a little bit.
They'll keep trying to get it going, but maybe we should
move on with our lives here while this is happening, and
we'll click it back at the end, and if we have a nice set
up at any point, I'll just stop and we'll go back and
we'll look at it again.
Since, I think that's just not going to happen right now.
So let's switch, actually, back to our notes.
So, ideally what we did see was, in fact, it does have
enough energy with the UV lamp, it wasn't a dramatic
shift you saw because we didn't start very high and
then it went to that stuck point.
But luckily we had the control of the red laser pointer where
nothing moved at all.
So hopefully you're convinced that your predictions worked
well and you are able to predict what's going on when
you're looking at the photoelectric effect.
So, it turns out that the photoelectric effect is not
the only evidence for the fact that light has these
particle-like characteristics.
And one thing that Einstein put forth is he figured if
well, what we're saying is that light is, in fact, a
stream of particles, each one of those particles or photons
must, therefore, have a momentum.
And that's really neat to think about, because photons,
of course, are massless particles, they have no mass,
so it's neat to think about something that has no mass,
but that actually does have a momentum.
And the relationship that he put forth is that the momentum
is equal to Planck's constant times nu divided by the speed
of light, or it's often more useful for us to think about
it in terms of wavelength.
So, since the speed of light equals lambda nu, we can say
that momentum is equal to h divided by lambda.
And there was experimental evidence that came along that
supported this, and this is called the Compton scattering
experiment, and this was done by Arthur Compton, and
basically what he did was he took x-ray light, which had
some frequency, which was a very high frequency because it
was x-rays, and he shot it at a stationary electron.
And what he was able to observe was that the electrons
scattered and now had some momentum, and that both the
frequency, and therefore, the momentum of the light that he
shot in, went down once it was scattered.
So what he's showing here is, first of all, that the light
has some momentum and when it hits an electron it can
actually transfer some of that momentum to the electron.
So the transfer of momentum from a photon to an electron
is what was being observed, and it was seen as completely
separate evidence to the photoelectric effect that,
yes, in fact, light is behaving in these
particle-like ways.
So up to this point, before it was really established that
yes, light is like a particle sometimes, there was this very
strong distinction between what is
light and what is matter.
And the distinction was when we're talking about light,
light is a wave, and when we're talking about matter,
well, matter is a particle.
And these behave completely separate, they don't overlap
at all in terms of behavior, but then, of course, with the
photoelectric effect with Compton's scattering, what we
see is that, oh actually, sometimes photons behave as if
they're particles.
So now this relationship's beginning to get a little bit
fuzzy in terms of what is the difference between how we
treat light and matter.
And actually, this was taken a step further by Louis de
Broglie who in his PhD thesis, as part of his work as a
graduate student, put forth the idea that, OK, Einstein
says, and everyone agrees that, in fact, light is
particle-like at times, and light, in fact, of course has
a wavelength, and if it has a wavelength we're saying that
it can have momentum.
And what de Broglie said is well, if it's true that light,
which has a wavelength can have momentum, then it must
also be true that matter, which has momentum, also has a
wavelength.
And you can look at this in two different ways.
One is that he's just re-arranged an equation here
and gotten both his PhD thesis and a Nobel Prize, but I think
the more representative way to think about this is the real
revolutionary idea that he put forth, which is that matter
can actually behave as a wave. And in terms of equations that
we use, it's sometimes easier to plug in the fact, since
momentum is equal to mass times velocity.
We can know the wavelength of any matter -- and he's not
limiting this, for example, to electrons.
What de Broglie is saying we can know the wavelength of any
matter at all, as long as we know its
mass and it's velocity.
And Einstein credited de Broglie, which is a fair
statement of lifting a corner of the great veil, because
really there was this fundamental misunderstanding
about what the difference was between matter and light, and
the reality is that they can both be like-particles and
they can both show characteristics of waves.
So I mentioned, however, that in terms of de Broglie's work.
This was Nobel Prize worthy, absolutely, but it was also
his PhD thesis.
So, we can think about what would happen if we're on his
thesis defense, we're on his thesis committee, we would
need to think of some pretty mean, hard, nasty questions to
be asking de Broglie about this theory -- that's what
happens when you defend your thesis.
This is necessary, it's hard to find holes in a Nobel Prize
worthy idea.
But let's just try maybe one of the basic questions they
could ask, and they can say, all right, de Broglie, so you
say that all matter, absolutely all matter has
wave-like behavior.
Why is it that we're never observing this, for example,
why is it the table doesn't defract as we bring
it through the door?
Why don't we see the influence of the wave-like behavior on
every day matter?
So it turns out that he could have picked anything to
explain this, and hopefully done out the calculation, and
we'll do this ourselves.
And the example we'll pick is considering, for example, a
Matsuzaka fastball.
So, many of you are new to the Boston area, now I still
realize, and I want to let you know it's not required that
you be a Red Sox fan to be at MIT.
We do encourage it, however, and in general, I find you
don't have to give up that old team, you can keep your old
team, even if it's teams I won't name, just
keep them to the side.
And you can join on to the Red Sox nation on top of that, and
part of being a good Red Sox fan is knowing the statistics
of your team.
For example, if we're talking about a pitcher, like
Matsuzaka, we might want to know the speed
of his average fastball.
We might want to know his ERA.
If you're really into it and you're at MIT, maybe you want
to know the wavelength of these average fastballs.
So, let's go ahead and look at that.
So, if we're trying to figure out the wavelength of a
Matsuzaka fastball, we need to consider the velocity first,
which is 42 miles per hour.
We don't usually do our chemistry calculations in
miles per hour, so let's switch that to 42 meters per
second, so it's -- sorry, it's 94 miles per hour.
And we can use the de Broglie relationship that wavelength
should be equal to h over mass times volume.
And we can put up here Planck's constant -- and I
want to make note that instead of writing joules per second,
I actually wrote out with a joule is.
A joule is a kilogram meter squared per second squared.
Occasionally, you'll find you need to cancel out units,
because, of course, you're always doing unit analysis as
you solve your problems, and sometimes you'll need to
convert joules to kilogram meters
square per second squared.
We divide that by the mass, so 0.12 kilograms, that's the
mass of a regulation baseball for the major leagues, and the
velocity of the baseball is 42 meters per second.
So, we can cross out our units doing our unit analysis.
The seconds cross out, the kilograms cross out, one of
the meters crosses out from the top, so we're left with an
answer in meters.
It's always good when we're looking for a wavelength that
our answer is in a unit of length,
that's a good sign already.
And what we find out is the wavelength of a Matsuzaka
fastball is 1.1 times 10 to the -31 meters.
So, this is really small, this is undetectably small.
And especially when we consider it, what tends to be
important is the size of wavelength in relationship to
its environment.
So 1.1 times 10 to the -31 meters is not, in fact, a
significant number when we're comparing it, for example, to
the length of a ball, or the size of the baseball field.
So that would probably be de Broglie's answer for why, in
fact, we're not observing the wavelength behavior of
material on a day-to-day life.
So, that's for Matsuzaka, and even if you don't memorize all
the wavelengths for all the pitchers.
I would expect, whether you're a Red Sox fan or not, you to
be able to look at a list of different pitchers and their
average velocity for their fastball, and tell me who has
the longest or the shortest wavelength.
You should all be able to know that relationship.
So why don't we go to a clicker question here, and see
if you can tell us this.
So we have 4 different pitchers we're showing here --
they all have different strengths.
It's not always how fast you throw the fastball, sometimes
it's your different styles or the different ways that you
decide when to throw what.
So, first we have Matsuzaka at 94 miles per hour.
So, click one if you think that he's going to have the
longest wavelength.
Tim Wakefield on the DL right now throws a lot slower,
because he has that tricky knuckle ball, he doesn't need
to throw as fast. Then we have Beckett who can get up 96 just
on a regular old day.
And Timlin who is about 91 miles per
hour, one of our relievers.
So, why don't you take ten seconds to do that.
If you can't decide, Timlin is my favorite ever, so that
would be a good back up choice if you forgot the relationship
between wavelength and the relationship between speed.
It looks like, in fact, people did not forget that
relationship, and only 1% of you humored me.
So, let's see what the correct answer is, and it is, in fact,
Wakefield, right, because there's an inverse
relationship between how fast a particle is going and what
its wavelength is.
So, in terms of wavelength, Wakefield has the largest
wavelength, but in terms of being significant, we're still
not even close.
It's still undetectably small.
Yes.
STUDENT: Why doesn't wavelength go to infinity as
it stops, like a standing [INAUDIBLE].
PROFESSOR: As it stops.
So, let's think.
I would think that it would approach inifinity, and I
would need to think about it and get back to you in terms
of why we don't actually hit it and see something with an
infinite wavelength.
I'm sure there's some upper limit as there are to most
things, like if we think of wavelengths and different
types of light, there is so large that you can get, but
you would be approaching that level.
All right.
So we can switch back actually to our notes here --
oh, do we have--?
OK.
We're going to just try this one more time just
so you can see it.
It'll still likely get stuck in that spot, but we'll just
show you one more time the effects of the UV light, and
actually we'll throw in an extra trick here, too.
We know that UV light gets absorbed by glass, so it
shouldn't be able to go through the glass.
So first if Professor Drennen can try it through the glass,
and we see nothing's happening.
Let's move the glass away.
All right.
[APPLAUSE]
PROFESSOR: All right.
Good.
So we can fully believe what our calculations were now,
which is a nice thing to do.
Let's go back to considering the wavelengths
of different objects.
We considered a baseball, but let's also think
about now an electron.
And an electron is something where, in fact, we might be
able to, if we calculate it and see how that works out,
actually observe some of its wave-like properties.
So, if we do this calculation for an electron, saying it
moves at 10 to the 5 meters per second, then what we end
up with for a wavelength is 7 times 10 to the -9 meters.
A lot of times we talk about these kind of distances either
in nanometers or in angstroms so we can say this is 70
angstroms. So this is, first of all, even just on an
absolute scale, this is way, way larger than the
wavelengths we're talking about for baseball.
In addition, if we compare this to the diameter of an
atom, which is on the order of somewhere between one and ten
angstroms, now we're seeing that, in fact, this wavelength
is significantly larger than its environment.
So certainly we would expect to see that it has an effect
in terms of seeing its wave-like properties.
And this was experimentally validated, hopefully, even
more clearly than our experiment here.
And at first this was done by Davidson and Germer, and they
were American scientists who tried defracting electrons
from a nickel crystal.
They did this in Bell Laboratories, and they found
that, in fact, the electronis did defract.
And G.P. Thompson showed a similar thing.
What he did was he defracted electrons through a very thin
gold foil, and this is a picture -- oops, that is not.
OK.
It is a picture from your book here showing the defraction
pattern of an electron going through that gold foil.
So, you can see that, in fact, it's confirmed that an
electron can have both wavelength and
particle-like behavior.
And it turns out that Davidson and Thompson shared a Nobel
Prize for this discovery of seeing the wave-like behavior
of electrons.
So, this is actually kind of neat to point out, because we
all remember J.J. Thomson from our second lecture, and J.J.
Thomson got a Nobel Prize in 1906 for showing that
electrons exist in that they are particles.
And it turns out that G.P. Thompson, well, that's his
son, so we can actually think of this -- and I'm sure this
wasn't the case, but I like to think of it as a little bit of
child rebelling against the father.
So, the father gets a Nobel Prize for showing that an
electron is a particle, and the son says, well, what can I
do to top that?
I'm going to show the exact opposite.
I'm going to say that an electron's a wave no matter
how much my father says differently, and I'm going to
get a Nobel Prize for that, and he does.
But the nice part of the story is, it turns out
they're both right.
An electron is a particle, but an electron's also a wave. So,
father and son, happy ending, they both
have their Noble prizes.
So, what happens now that we, in fact, know
that matter is a wave?
Well, this allows us to try to go back and explain some
phenomena that over the years, mounting evidence was building
that couldn't be explained.
So, for example, when people, and we'll talk about this next
class, were looking at different characteristics
spectra of different atoms, what they were seeing is that
it appeared to be these very discreet lines that were
allowed or not allowed for the different atoms to emit, but
they had no way to explain this using classical physics.
And it turns out that the Schrodinger equation is an
equation of motion in which you're describing a particle
by describing it as a wave. So you're basically having a wave
equation for a particle, and for our purposes we're talking
about a very particular particle.
What we're interested in is the electron.
So basically describing electrons by their wave-like
properties.
And this is Erwin Schrodinger, and this is the equation that
he put forth where we have h hat psi being equal to e psi.
So, let's explain what these are.
So this symbol here is actually what
we call a wave function.
That doesn't mean a whole lot in itself, it will mean more
in about two lectures from now.
But right now, what I want you to be thinking of a wave
function as is just some representation of an electron.
So, it's some way of describing an electron.
Specifically, we'll talk more about this, it's talking about
different orbitals, it's the spatial part of an orbital.
But before we get to that, in terms of thinking just think,
OK, this is representing my particle, this is representing
my electron that's what the wave function is.
This e term here is the energy, or in our case when we
talk about an electron in a hydrogen atom, for example,
the binding energy of that electron to the nucleus.
So, e is binding energy.
And h with the carrot or the hat here, well, that carrot or
hat tell us it must be an operator, and this is called
the Hamiltonian operator.
So when you operate on the wave function, what you end up
with is getting the binding energy of the electron, and
the wave function back out.
When we need to describe the wave function term a little
bit more specifically so we can describe, for example, the
position of the electron, and I just want to mention that we
do have two choices if we're trying to describe this, we
could use cartesian coordinates, or we could use
polar coordinates where we're either talking about x y z or
r theta and phi.
So, I just want to point out that when you look at wave
functions, we are going to be using those spherical polar
coordinates, and the reason is because a very important
interaction here is the interaction between the
electron and the nucleus, which we want to describe the
distance of in terms of r.
So, you can see, it's much easier to describe that as one
term, r here, instead of using both y and z.
Another reason I wanted to point this out in terms of the
polar coordinates that we're using, is I think they're
actually flipped from what you're used
to seeing in physics.
Sometimes different disciplines have different
conventions, which can be very confusing because the whole
point of what's happening now is there's so much interplay
between different disciplines, but still I think this might
be one remaining one where in our case theta is that
distance from z, that angle there, where phi is this
distance or angle from the x-axis.
So just keep it in mind that it's flipped.
It turns out we won't really using it, needing to identify
it on the graph so much in chemistry.
We'll be using the solutions, so you shouldn't have a
problem, but I wanted to point it out so it does not look too
strange to you.
In terms of the Schrodinger equation, we now can write it
in terms of our polar coordinates here.
So we have the operation on the wave function in terms of
r, theta, and phi and remember this e is just our binding
energy for the electron, and we get back
out this wave function.
So, you might ask, this looks pretty simple up here, right,
just with that h hat.
It turns out, we can write it out fully.
It's three different second derivatives in terms of the
three different parameters.
It's a little bit complicated.
You won't have to solve it in this class, you can wait till
you get to 18.03 to start solving these types of
differential equations, and hopefully, you'll all want the
pleasure of actually solving the Schrodinger equation at
some point.
So, just keep taking chemistry, you'll already have
had 18.03 by that point and you'll have the
opportunity to do that.
What I want to point out also is that this h hat, the
Hamiltonian operator written out for the simplest case we
can even imagine, which is a hydrogen atom where we only
have one electron that we're dealing with, and of course,
one nucleus.
So you can imagine it's just going to get more and more
complicated as we get to other types of atoms, and of course,
molecules from there.
So, we just want to appreciate that what we'll be using in
this class is, in fact, the solutions to the Schrodinger
equation, and just so you can be fully thankful for not
having to necessarily solve these as we jump into the
solutions and just knowing that they're out there and
you'll get to solve it at some point,
hopefully, in your careers.
So, we'll pick up with that, with some of the solutions and
starting to talk about energies on Friday.