Freshman Organic Chemistry I

CHEM 125a - Lecture 5 - X-Ray Diffraction

Chapter 1. Introduction: Focusing Lux [00:00:00]

Professor Michael McBride: So as we ended last time, we said despite Earnshaw, which says you can’t have these Lewis structures, might there really be shared-pair bonds and lone-pairs, and how do we know; we have to look, or feel. So last time we tried feeling with scanning probe microscopy — AFM, STM, SNOM. They’re really great. You can see atoms; you can see molecules, but you can’t see bonds. Okay, so what’s the key word here?

Students: Lux.

Professor Michael McBride: Lux, right? So maybe we’ll see it, if we can’t feel it. Now this is the entrance to this building, the old delivery entrance out on Prospect Street. And you know there’s interesting stuff all over the building. Up here on top there’s a fluted filter paper and a funnel, but hidden back in the shadows is something that’s a little surprising to you, back there, which is a microscope. What science do you associate that with?

Students: Biology.

Professor Michael McBride: Biology, not chemistry, right? What’s it doing? Well at least it’s back in the shadows there. But maybe the eyeball up there says we can see the things we can’t feel. Okay? This is a picture — does anybody recognize this?

Student: Flea.

Professor Michael McBride: You know what it’s from?

Student: Bubonic plague?

Professor Michael McBride: It’s from Robert Hooke’s great book, Micrographia, from 1665. It’s a wonderful book. The page is about this big. And imagine the person who had fleas, seeing that for the first time. It’s exquisite, this drawing. And he said in the book:

“But Nature is not to be limited by our narrow comprehension; future improvements of glasses may yet further enlighten our understanding, and ocular inspection may demonstrate that which as yet we may think too extravagant either to feign or to suppose.”

So if he had a microscope that could do that in 1665, we must have a microscope now that’s powerful enough to see bonds, right? And in fact we do. In fact, there’s a brand new one that they promise will come online next Tuesday in a room about 100 yards over that way, and the trick it uses is the same one Newton used to measure the distance of the air gap for Newton’s rings, or that Hooke explained by the pulses, right? And it’s interference that comes from scattering. So you’ve seen oil on water, thin layers of oil on water, and you see rainbows in them. Why? Okay, so light comes in and scatters from two surfaces, that let’s say are 200 nanometers apart. Okay? So the path difference between the one that reflects from the top and the one that reflects from the bottom, the difference is 400 nanometers; one goes 400 nanometers further than the other. Now suppose that happened to be one wavelength. Right? Then if that purple light that came in, one going further than the other, they would come out exactly one wave apart. So it would be as if it were a single wave and they would reinforce one another and you’d see the purple light strongly. Okay? But if the wavelength difference were half — if the path difference were half the wavelength, like for red light —

Oh-oh. What do we do here, Elaine? Better get it quick, before something comes up. You’re a good sport.

[Technical adjustments]

OK. Thank you.

Professor Michael McBride: If it’s one wavelength path difference, then it reinforces and we get a nice strong wave coming out. Okay? But if it’s half-a-wavelength difference, like for the red light, then they cancel. One is a maximum when the other is a minimum, and they’re zero. Okay? So you get no scattering. So that’s why you get different colors, because the oil is different thicknesses in different parts of the slick. Okay, so here’s the new machine that’s right over here that’s supposed to come on — new machines actually; there are two of them; it was a package deal. And that’s Chris Incarvito who’s the director of the Chemical Instrumentation Center and the proud owner of these two new machines. So the one he’s looking at there is a user — is to be operated by users. So it’s just you sort of push buttons and you get where the atoms are in the molecule — at least that’s the hope — and it costs about $200,000. Okay, so there’s a little thread there that has a crystal on, but you need a magnifying glass to see it. You won’t see it with your eyes; it’s a tiny, tiny crystal. Okay? So here’s an x-ray tube. X-rays come down the pipe there, hit the crystal and get scattered, and are detected by this CCD detector. Okay? And then from that information, those scattered rays, you get where the atoms are in the molecule; more precisely, as you’ll see, where the electrons are in the molecule, or in the crystal.

The other machine that he’s especially proud of costs $350K. And why is it more expensive? Because it collects more. Instead of using that little disc of a CCD, it has a curved image plate. Right? So when the x-rays come down here and hit the tiny crystal, many more of them get collected by the plate. So it’s more efficient. So those will be coming soon. And if all you have to do is press a button and the structure comes out, that seems fine. But that’s not the goal of this course, to press a button. We want to understand how such a thing works. Seeing individual molecules, atoms, and maybe bonds, there’s a problem, and the problem is wavelength. Because you know in principle you can’t resolve things that are closer together than the wavelength of light, and the wavelength of light we’ve been talking about here is 400-800 nanometers. Whereas what’s the distance of a carbon-carbon bond? Anybody remember?

Student: 1.86.

Professor Michael McBride: One and a half angstroms, right? So three orders of magnitude smaller than the wavelength of light. So this is a problem, and it’s why, as you’ll see, you use x-rays, because they have short wavelengths.

Chapter 2. Defining and Scattering Light to See: X-Ray Crystallography and Diffraction [00:07:11]

Now to understand this we’d have to know what light is. So what is light? You people have studied physics and so on. So what’s light? Wilson?

Student: Who knows?

Professor Michael McBride: Who knows, okay.

Student: An electromagnetic wave.

Professor Michael McBride: It’s an electromagnetic wave. Okay. Now I’ve seen — indeed it is — I’ve seen waves on the ocean, right? Does it mean that electromagnetic waves are like those waves? In what way is an electromagnetic wave a wave? What’s wavy about it? Well you can make a graph that’s a wave, that involves light. So here what we’re going to plot is the force on a charge. The charge is at a particular position. We’ll have the charge over there on the right; it’s fixed there, and we’re going to measure the force on it that’ll make it accelerate up or down. Okay? Oops. It doesn’t work quite the way it does on mine. Please let me know when that happens. No maybe it’s going to work. [It does not.]

What that first thing did was went up and down, up and down, up and down. Okay? But if you plot it then as a function of time, it’s a wave. Right? The field, at a particular point, as a function of time, is a wave. It goes up and down and up and down. Okay, now so we plot the field, or the force on a charge at one position as a function of time.

But there’s another way of making a plot that also shows light as a wave, which is to show the field at different places at the same time. Right? So we’re going to change it and now the horizontal axis is the position but it’s at a particular instant. So as the thing goes along there are different forces at different positions, okay? And again, that will be a wave. So that’s the sense in which light is a wave.

And what is it that we’re plotting? We’re plotting the electric field. So light is associated with an electric field that goes up and down. Right? And you can plot it in time or in space and get a wave. So that’s the sense in which it’s a wave. It’s also, incidentally, a magnetic field; it’s electromagnetic. Maxwell dealt with things like that. And there’s a magnetic field perpendicular to the electric field, but we don’t care about it, at least not until next semester when we talk about nuclear magnetic resonance, because the electric field is so much stronger in its effect on molecules. Now, its effect is on charges; on electrons, on protons, therefore on nuclei. Now, accelerated electrons scatter light. So here comes the light in. We’ll see if this works. Okay, so it makes the electron go up and down, as the light goes by. But up and down moving of electrons, accelerating electrons, is what creates electric fields. Right? That’s what an antenna is, is electrons moving back and forth.

So when the original light comes through and hits the electron and makes it vibrate up and down, that electron vibrating emits radiation, electromagnetic radiation, in all directions; all directions except this direction, and stronger the more you’re perpendicular to the direction it’s going up and down. But anyhow it scatters the light. So most of the light still goes through the direct beam, but a little bit of it gets scattered; much less would be scattered — a single electron wouldn’t scatter very much, you’d need lots of electrons to scatter a lot, and they have to be cooperating. Okay? Now you tell me, why are we interested in electrons scattering light? There are just as many protons in a sample as there are electrons. Why don’t we worry about the protons scattering light?

Student: Electrons are a lot less massive.

Professor Michael McBride: So what?

Student: They move a lot more easily.

Professor Michael McBride: Ah, the electrons are what moves. You need the moving charge to generate the scattered light. The lightest positive charged things are — of normal particles — are a thousand times heavier than the electrons. So they don’t move very much. It’s the electrons that scatter the light. So with x-rays, you see electrons, not nuclei. Okay, they’re too heavy. Now then, but you need a lot of electrons to scatter enough to see, and they have to cooperate. Right? So here we have ripples on a pond; you’ve all seen that kind of thing with a little bit of rain falling. And so suppose you had two spots that are generating ripples, two electrons let’s say, and they give off circular waves here, in two dimensions, and they interfere with one another. But you can see a pattern is emerging, and when you’re very, very far away from these, compared to the distance between the electrons, as indeed you are in these samples we’re talking about where you have an infinitesimal crystal and the detector plate is a substantial distance away, right? The electrons are really close to one another. So you’re way, way, way far away on the scale of the distance between the electrons. Okay?

And now you can see what pattern there is for high points and low points, for waves. Okay? Because you can see here that there’s a pattern like that, that goes along the maxima. Everybody see that? And halfway in between those dashed yellow lines, there’s no change at all; nothing’s happening there. Right? And if you go out very, very far away, you can see that asymptotically these approach these straight lines that originate between the electrons, or on the crystal in a real case. So what you have coming out of the crystal are straight rays of light, x-ray light. Okay? And the angles at which they come out depend on where the electrons are. If the electrons are closer together or further apart or displaced in this direction, you’ll get different patterns of the ripples. So somehow it must be possible, or there must be a connection between the positions of the electrons and where these rays are coming out. Right? And if you can go backwards, to go from where the rays come out, to where the electrons were, then you’ve solved the problem. Right?

You’re not creating an image, the way you would with a normal microscope. You’re trying to interpret the scattering. So the angular intensity distribution, at great distance, depends on the scatterer distribution at the origin; that is, in the crystal. Now, if you have normal light, and a lens like Hooke’s microscope, then you can use these lenses to refocus it. So the same information is coming out. The sample there is emitting these rays, but a lens collects them and refocuses them to make an enlarged image that you then observe. The problem is that you can’t do that with x-rays. People are making efforts now, with nanofabrication kind of things, to make things that will act like lenses for x-rays, and they’ve made some steps, but nothing like you would need to actually observe electrons in a crystal. Okay? Be sure to read — there’s a webpage — have some of you read it so far? — that has to do with what I’m lecturing on now, which will help you a lot I think; so look for that. So we’re interested in seeing molecules, atoms, bonds, collectively, by x-ray crystallography. That is, we’re not seeing the image of individual electrons, we’re seeing the scattering that comes from all the electrons acting together. So we’re not seeing them one at a time, we’re seeing something collective about them. So that, for example, a real sample, or a sort of fake sample, of benzene, which had a bunch of — six carbon atoms in a ring, might look like this at any given time. It’s sort of regular but things are a little bit one way or the other, vibrating and so on.

Now you wouldn’t see this in x-ray, because everything is cooperating in what you’re getting. You’d see some sort of average of all of them, and it would be a little bit smeared because of that. Right? You don’t see — with scanning probe microscopy, with these sharp tips, you actually can feel individual atoms, and if one atom is someplace else, you’ll see it, or feel it there. That’s not true in x-ray. You see an averaged structure. And it’s averaged in two ways. There’s blurring, from motion and from defects. There’s one benzene molecule missing there. It’s time-averaged because it took you time to collect these, this information. With synchrotrons and really, really intense beams, people are trying to get faster and faster data collection, but still what you see is time-averaged on the scale of how fast atoms are moving in your sample. So it’s time-averaged, what you get. And it’s also space-averaged, right? It’s as if you put them all on top of one another, that’s what you would see, but some are displaced a little one way, the other, and so on. So it’s a little fuzzy. Okay? And this is an advantage for scanning probe microscopy, which operates in real space, actually feels individual things.

Okay, so x-rays were discovered in 1895 by Röntgen, using Crookes’ tube that we talked about. Okay? And he took a picture of his wife’s hand there, Frau Röentgen’s hand, in 1896. But what he sees is not a picture that you can blow up, like with a microscope, because it’s just a shadow. All the bones do is stop the x-rays that are going through. So you don’t get something that’s enlarged. Right? You’re not going to be able to use it the way you use medical x-rays. Okay. But in 1912 Max von Laue invented x-ray diffraction, which is this scattering and detect — not trying to focus things, not using a shadow, but looking at these rays that come out and trying to figure out from them, something about what the atoms were. And that’s his diffraction picture. That’s what you’d see on that — remember there was that round CCD plate that’s on this new machine?

You might get a pattern that looked sort of like that. That was copper sulfate in 1912. But the real breakthrough by was William Lawrence Bragg, who was 22 years old. He had just graduated from Cambridge University when he determined the structure of a crystal using Laue’s x-ray diffraction pattern. So he figured out how to go the other way, to go from the x-ray’s scattered pattern to what the atoms were that were doing it. Right? And he did this in 1912 when he was 22-years-old, and he got the Nobel Prize in 1915. He’s still the youngest Nobel Laureate. Right? His son gave me permission to use this picture. He’s a very nice guy, lives in Cambridge. Okay, then of course we’ve come a long way from that. I think you probably recognize this picture, right? What is it?

[Students speak over one another]

Professor Michael McBride: It’s the scattering from DNA. Is it a picture of DNA?

Students: No.

Professor Michael McBride: No it’s not a double helix. What it is, is the way x-rays come out when they hit the double helix. Okay? And you have to figure backwards, and that’s what Crick was able to do; and we’ll discuss how he did that. Okay, so that’s 1952, 40 years after Bragg’s discovery. And then just in 2000, not so long ago, this has gone to the complete structure of the ribosome at 2.4 angstroms resolution, which was done here at Yale, in this building and the next building. Okay, and that’s what it looks like. It has twenty-five nanometers across, 250 angstroms, as long as whatever number that is; lots of carbon-carbon bonds. And you see all those atoms. There are greater than 100,000 atoms, not counting the hydrogens. Incidentally, why is it easier to see other atoms than to see hydrogens?

Student: They have never counted them.

Professor Michael McBride: Because that’s where the electrons are. Hydrogens come with only one electron. Other atoms come with lots of electrons. So it’s hard to see hydrogens; much easier to see the other things. Okay, now what can electron diffraction show, x-ray diffraction show? That’s what we want to know. Can it show molecules? Yes. Can it show atoms? Apparently; I just showed you a picture. But what we really want to know is whether it can show bonds and whether Lewis was right. So to understand this, we have to know how diffraction works. I mean, you could just be like the people that will go and punch the button on that machine, but that’s not what I think you would be satisfied with. So I’ll help you out. So like all light, x-rays are waves; they just are very short. So now I’m going to demonstrate with a machine here, which was designed for an overhead projector, but I believe it’s going to work here.

Okay, so here are our waves. Okay, so here’s a wave coming in. Okay? And when it gets to this position, it hits an electron here and an electron here; forget this one for now. Okay, everybody got me? At a given time, these two electrons are being pushed up; then they’re going to be pushed down, as this wave passes by. Okay, now as those two get pushed up and down, they give off waves in almost all directions; all directions for our purposes. So they give off waves in all different directions. Right? Now notice the one, the part of that scattered x-ray that goes straight forward, in the same direction the original light came in, is bound to be in-phase with one another. Right? And we can test that with this line here, because if they’re going right straight ahead, this one is a maximum when this one is a maximum. Everybody with me on that? And it’ll keep that. So you’ll get scattering by both of them cooperating, coming out of it straight ahead. But how about if it’s at an angle? Okay, so I can push this up and change the angle. How about at that angle, how much light is going to be coming out from these two electrons?

Student: None.

Professor Michael McBride: None, they exactly cancel each other, right? But if I go to this angle, now they’re just as strong as they were originally, right? And then it’ll go weak, weak, weak, weaker, weakest, nothing; then stronger, stronger, stronger, stronger, very strong, weaker, weaker, weaker, weaker, weaker, zero. So there’ll be a modulation — as you go out at different angles here, going either up or down, it’ll be strong, then nothing, then strong, then nothing, strong then nothing. Right? So now what will determine, what will determine how frequently these angles recur, at which reinforcement occurs? Is that clear, the question? What determines the angles? Yeah — pardon me?

Student: The wavelength.

Professor Michael McBride: Ah, obviously the wavelength. What if the wavelength were very, very, very short?

Student: Lots of angles.

Professor Michael McBride: Then you’d get lots of them. You wouldn’t have to do it very much before the next one came in, if they were very — what else determines it, besides the wavelength. Yes John?

Student: The distance between the two of them.

Professor Michael McBride: The distance. That’s what we’re really interested in, is using this; knowing the wavelength of the x-rays, using this to measure distance. Okay? Now let’s forget the one on the bottom here. Let’s put this one in. Okay? Now — oh okay, for reference let’s look at the bottom a second. So the first reinforcement came here, between the top one and the bottom one. Everybody with me? How about for the top one and the middle one, where did the first one come? You had to go to a higher angle, there, to get reinforcement between the top and the middle. Right? So the closer things are together, the fewer the angles are. Everybody with me on that? Notice that’s a reciprocal relationship. The closer things are in what we call ‘real space,’ the distance between the electrons, the further apart are the rays that come out, in angle. Okay? So it’s backwards. Closer together, further apart, in what’s called ‘reciprocal space.’

Chapter 3. Wave Machines [00:25:07]

Okay, now suppose you had a whole row of electrons that were evenly spaced. Okay? So here’s the first, the second, the third. Now we’ll take all three of them. Right? And what you notice is that as we go out it gets weaker, weaker, weaker, weaker, weaker, weaker, weaker, weaker, weaker. Now when we get to the angle here, where the first and the third were very strong, what do we see now? The one in the middle cancels the first one, and if it were a very long row of them, the fourth one would — here the second one cancels the first one — the fourth would cancel the third, the sixth would cancel the fifth, and so on, and you wouldn’t get anything. Right? But then you’d get it again when you got to the second. The one that would’ve been the second angle, if it was just one and three, will be strong, because this one halfway in between will reinforce. Right? So closer together, further apart. And if you have a whole row of equally spaced things, they’ll all be together, according to whatever the distance between successive ones is. Yes, Shai?

Student: What are the chances that electrons are going to be spaced evenly along —

Professor Michael McBride: Ah, what could cause electrons to be spaced exactly evenly?

Student: A crystal.

Professor Michael McBride: A crystal; x-ray crystallography, right? So that’s why you use crystals. Okay, that’s what we want to say here, I think.

[Technical adjustments]

Professor Michael McBride: Okay, so that’s the wave machine. And if you want to do it in the privacy of your room, you can go to this website at Stonybrook and download something that allows you to run a Java applet that does sort of this kind of thing. This, if you blow it up in there, the waves look sort of funny there, at the atoms, and the reason is because you’re measuring the phase perpendicular to different directions of the wave. That’s just to help you out, if you try this and have — and are confused by that. Okay, now there are — so suppose you have just an arbitrary set of electrons, and the waves coming in and hitting them, what directions will you get reinforcement in? Well there are two directions that you’re guaranteed, no matter what the spacing is, and that’s this. Here the light comes in and goes out. So the direct beam, you have the same path, because one of them gets hit earlier, but has a longer path coming out, and the other one gets hit later but has a shorter path coming out. So they’re guaranteed to be exactly in-phase if they’re scattering straight ahead. Okay? But only a little of the light is getting scattered. Most of it is the direct beam. So you don’t even notice the difference, from that. So that’s not very exciting. But there’s another angle at which you’re sure that these two are going to be in-phase, and that’s this angle. Now how do we get to that angle? I’ve lost some of the — since you downloaded it there’s some more stuff in here now. But — so now I can’t quite remember, let me just try.

Okay, so this blue line here is called the scattering vector, and that’s how different the arrow coming out is from the arrow going in. Right? That’s just a funny mathematical or geometric thing that people have defined; they call it the scattering vector. Right? Now this length is exactly the same as that length. This length is exactly the same as that length, and when you turn at a given angle, those two will — the vector between those will be perpendicular to the direction they’re going, and they’ll be in-phase again. Right? And if you draw the line that connects the two points, notice the scattering vector is perpendicular to that line. So you say that this incoming wave was scattered perpendicular to this line. That’s how a mirror works. Right? That angle is called the specular angle, because speculum is the Latin word for mirror. Right? So you’re bound to get reinforcement at the specular angle, and you know that from having looked at mirrors.

Okay, now suppose there were another electron on that same line or plane. Okay? For the same reason it’s guaranteed to be in-phase. So everything on that plane will scatter in-phase at that particular angle. They’ll all reinforce one another. Great. So all electrons on a plane perpendicular to the scattering vector, scatter in-phase at the specular angle. Now suppose you have a whole bunch of electrons and you have to figure out how they’re going to reinforce or cancel one another. There’s a trick you can use. You see if you can discern a set of planes that are evenly spaced, that contain all, or almost all, of the electrons you’re interested in. Okay, now if you look at this pattern, you can see that there’s a set of planes, equally spaced, that passes through all of them. Can you perceive it? There.

Okay, there are three electrons on the first plane, four on the second, two on the third and one on the fourth. Okay? Now why is that handy? Okay, so there’s the scattering vector, perpendicular to these planes, and all the electrons on any one plane will scatter in-phase with one another. Okay, so it’s as if there were a single electron three times bigger, or four or two or one, on those successive planes, because they’re all going to be in-phase with one another. So we could pretend they’re all at one point, on any given plane. Okay? So now all we have to worry about is the phase relationship from one plane to the next. Got that? If they were in-phase from one plane to the next, as well as on the planes, then everything will be in-phase and you’ll get really bright reflection coming out at that scattering angle. Okay? Now, so let’s think — so it’s as if we had these collected electrons all lying on the scattering vector and we want to know whether they’re in-phase with one another.

Okay, so here’s light that comes in and out at a certain angle from the first one; three electrons worth of scattering from that, the ones that were on that plane — okay? — and four from the second plane. Now are the three and the four in-phase with one another? What condition would have to apply in order for those to be in-phase with one another? They have different path lengths, but if the paths differ by an integral number of wavelengths, then they’ll be in-phase with one another and reinforce. Okay? So there’s the path difference in red. If that happens to be — suppose the wavelength of the light, and the angle of the scattering — which also determines that path difference, as you can see — suppose that the wavelength and the angle are such that that happens to be one wavelength? Okay, so those are in-phase with one another; all seven will be scattering together. Right? How about the next one, with two electrons, how about its phase? What was the condition we talked about, up here at the top?

Student: Evenly.

Professor Michael McBride: Evenly spaced. So what do you know about the next one?

Student: [inaudible]

Professor Michael McBride: It’s going to be in-phase too, two waves behind the first one. Right? It’ll be two wavelengths behind the first one, exactly, at some angle. Right? And the next one will be three wavelengths behind. So all these electrons are going to be scattering in-phase at that particular angle. So you get a really strong reflection coming out. Right? It’ll be as if it were all the electrons working together. So the net in-phase scattering is as if there were ten electrons doing the scattering. That’s great, right? Now suppose that the first path difference, instead — suppose you had a different wavelength or a different angle. So the first, between the three and the four, suppose the difference in path was half a wavelength. Then how would it differ?

[Students speak over one another]

Professor Michael McBride: How about between the first and the second. Wilson?

Student: Cancel.

Student: The first and the second?

Professor Michael McBride: The first and the second, would they exactly cancel and get zero?

Student: There’d be one left.

Student: Zero.

Professor Michael McBride: Why?

Student: There’s one —

Professor Michael McBride: Ah, there are three in the first and four in the second, right? So you got plus three but minus four. How about the next one? Speak up gang.

Students: Plus two.

Professor Michael McBride: Plus two. And the last one?

Students: Minus one.

Professor Michael McBride: Minus one. And what’s the net scattering?

Student: Nothing.

Students: Zero.

Professor Michael McBride: Zero. Okay? So you can see how this is a neat trick to work. If you can see in the pattern a bunch of planes which would contain the electrons, then you can figure out in a particular direction, how strong the scattering would be. Now, we’re going to do an experiment and it requires the room to be dark. And so I’m going to start turning the lights off and ask Filip to get into position to do stuff in the dark here. Okay, and turn this one off. Okay. Now I’ll show you first what we’re going to do. So pull out the little thing there. And there’s a laser that’s focused right here, but it’s too bright, because the other things we’re going to see are very dim, which is why we had to turn the lights off. So I’m going to put this black tube there, with a little hole in the end. Hopefully I can get it positioned so that most of the laser will go inside here and we won’t see it very well. Let me get it positioned right. It’s a moving target so it’s not so easy.

Okay now, so here’s the view from the ceiling. There’s the screen, and this laser is coming from the back of the room and hitting the screen, right here. Okay? Now what Filip is going to do is put things called diffraction masks, which are just slides, thirty-five millimeter slides, and he’s going to put them in the path of the light. And that’s, the distance from Filip to here, is 10.6 meters; I measured it. Okay? And so put in the first slide please. And see this? Those are all deflections at different angles. And what’s doing the scattering is a slide that looks like this. Okay? A jail window, right? Okay, so what it’s giving is a row of dots. Now I’m going to ask Filip to do two things here. First, rotate the slide around this axis, so rotate it like this. What do you think will happen? You see, right? The row of spots is perpendicular to the direction of the lines. Okay? And now I’m going to ask him to do something else. I’m going to ask him to twist the slide like this, which changes the effective distance between these. And notice what happens as he makes them closer? Twisting makes them look closer together, right? What happens to the spots?

Student: Farther apart.

Professor Michael McBride: Does that surprise you?

Student: That’s that reciprocal —

Professor Michael McBride: That’s that reciprocal relationship. When things get closer together the angles get bigger. Okay, now things are going to get darker here. So I’m going to do some things. Turn that out and also get the room light. Well no, I want to show you something here first. Okay, so I’m going to show you what the masks are going to be and then we’ll show you the effect from the masks. Okay, so the next — so here’s a question for you to work for homework. What is the spacing of the lines on Filip’s slide? How far are the bars in the jail window apart on that slide, in order to give a spacing here — oh I did the wrong, I was premature with that — in order to get that 10.8 centimeter spacing here, at a distance of 10.6 meters, with 63 [ correction: 633] nanometer wavelength light? Okay? You can use that to find the distance between the bars on his slide, and I want you to do that, because if you can do it then you understand how this works.

Chapter 4. Structural Information in Wave Machines: The Case of Benzene [00:39:43]

Okay, but the next one he’s going to show — not yet but he’ll put it in — is one that’s a similar spacing, but of pairs of lines. There’s a pattern being repeated, pairs of lines. Okay? And then he’s going to go to a whole bunch of hexagons of dots, which we will call benzene, but it’s not exactly like benzene would be, a benzene gas, because they’re all oriented exactly the same way. Right? So it’s oriented benzene; that’s the next one we’re going to look at. And then we’re going to look at this one, which is also oriented benzene. Can you see the difference in the pattern between the one on the top and the one on the bottom? Those red lines are a hint.

Student: They’re connected.

Professor Michael McBride: In the bottom they’re all pairs of hexagons, that are distributed randomly, where in the top they’re individual ones. Now then he’s going to show you this one. How’s that different?

Student: Quadrupled.

Professor Michael McBride: It’s quartets of hexagons. And finally he’s going to show you this one, which is a crystal of hexagons, and in a crystal they truly would be oriented. Okay? And then we’re going to — the pièce de résistance, at the end of this sound and light show, is going to be that. You know what that is? It’s a light bulb filament, that he’s going to hold up in the path of this; so a helix. Okay, now we’re ready to do the trick. So we’ll mute that and do this, and if you have a laptop open you’ll need to close it because we got to — it’s not very bright, so we have to make it as dark as possible, and I have to find my way back there and close my laptop. Okay you ready Filip?

Teaching Fellow: Yes.

Professor Michael McBride: Okay, so first it’s going to be pairs of jail bars. So how does it look different from what we saw before?

Student: Like triplets.

Student: Like different lights in there.

Professor Michael McBride: It’s the same kind of dots but their intensity is modulated. They’re not just evenly — they’re not just all the same or slowly dying away as you move out to right or left. They come in a pattern. Okay, now let’s do the next one. This is the one that’s benzene, but randomly, a random collection of benzene. So how would you describe that?

Student: A snowflake.

Professor Michael McBride: It looks like a snowflake. Oh, my laptop has come on. [Laughter] Go figure. I’m putting things to cover the lights up here. Okay, so it’s a snowflake. Now we’re going to look at pairs of benzenes. How is it different? It’s still the snowflake.

[Students speak over one another]

Professor Michael McBride: But there are bars across it. Everybody see that?

Student: Yes.

Professor Michael McBride: Can we stop the green blink back there or put something across it? That’s good. Okay, now we’re going to do quartets of benzenes. How does that look? How would you describe that?

[Students speak over one another]

Professor Michael McBride: It’s now got bars going in both directions, not just horizontaloid but also verticaloid; they’re actually vertical I think, but not quite. Okay, now a lattice of benzenes. Isn’t that great? So it takes the same — it’s the same underlying pattern, the snowflake, but it concentrates all the light that would’ve been spread out, into very, very fine points. Remember when it was just pairs, we put bars across, but it was the same amount of light coming out, so that it got focused sort of, and when we have a whole lattice it gets fabulously focused. I can see things quite far out from the middle here, because I’m so close to the screen. I’ll hold my hand up but you can’t see my hand. [Laughter] They go way out. So that’s what a crystal is. The crystal takes the same underlying pattern, that comes from the molecule, and focuses it into intense points at many, many angles. So it’s like looking at the scattering from a single molecule but looking at it through a pegboard. You know what a pegboard is? In a tool shop you used to put a pegboard up. It’s a piece of masonite or something that had holes in it, regularly, and you could put little hooks in it to hang your hammer and your saw and whatnot. So it’s as if you have, you’re looking through such a pegboard at that underlying pattern. And now we’re ready for the pièce de résistance, which is the light bulb filament, which is hard to see. This is why we really needed the lights out. We needed it for the snowflake too because it was so — so what do you see for the light bulb filament?

Students: An X.

Professor Michael McBride: You see an X, right? Can people in the back see the X? And there are dots along the arms of the X. Okay? So that does it, you can open your laptops again now and we’ll try to get some lights back on. Sorry for straining your eyes.

[Technical adjustments]

Professor Michael McBride: Okay, so there’s the scattering from these things and we can — now so let’s try to understand it. So this was the slide with randomly positioned but oriented benzenes. It turns out that random positioning generates the same diffraction as a single pattern but gives it more intensity. If we’d just put one hexagon there you wouldn’t have seen enough from it. Okay, so here’s the pattern you got from isolated benzenes, which is that snowflake pattern. Right? And when we had a lattice made of those, you saw the same underlying pattern, but only at little — only at regularly spaced spots that had to do with how far apart the hexagons were from one another, and much more intense, because all the light that would be scattered all over the screen here is focused in those little dots. Okay, now the thing that generated this snowflake was benzene, or what we’re calling benzene, the hexagon. Okay, now can you see what we need in order to understand? What do you look for in order to see what directions will give you scattering? Ah sorry, I was having so much fun we went over. So we’ll talk about this next week. Thanks.

[end of transcript]

Back to Top