a reminder about the celebration of learning on Wednesday, October 2nd, Test 2. It will have coverage spanning lectures 7 through 17. The new material that I am going to introduce today will not be covered, but certainly this little review is worthwhile And going back to 7 with the beginning of ionics What I wanted to do today was just set the stage and briefly review where we were last day where we looked at the emission spectrum of a target in an x-ray tube Here I have plotted intensity versus wavelength Long wavelength is low energy Energy is increasing from right to left. And what we see is a family Low voltage gives us low intensity High voltage gives us high intensity. We see this whale-shaped background which we refer to as the continuous spectrum or “bremsstrahlung And then once we reach a critical voltage beyond which we start to see this second set of lines here, which I have indicated in fuchsia And this is the characteristic spectrum, and it is quantized It is not continuous. We have discrete values of wavelength And those transitions are prompted by the ejection of inner shell electrons. And that does not occur until you get to a certain critical plate voltage which translates into a certain critical ballistic energy on the part of the incident electrons. And there are a few features here that we can point out The various discrete wavelengths of the characteristic spectrum are calculable, the k alpha and the l alpha Lines are calculable by Mosley’s law, which is given here, where the nf and ni are associated with a particular transition For k lines it is the transition from two to one And screening factor is one for the k alpha transition For the l alpha transition it is from three to two And the screening factor in that And the other point that we can calculate on this curve is the continuous spectrum is designated here as the shortest wavelength, transfer of incident ballistic energy of the electron to photon and given by the Duane-Hunt law which is 12,400 over the plate voltage giving I think we are ready now to probe atomic arrangement with x-rays And, in order to do so, we are going to rely on two important factors. First of all, we are going to model the crystal as an array of mirrors In other words, we are going to look at the atoms in the crystal and consider each atom as acting as a mirror This will give us features that we can exploit in a very simple model Let’s model atoms as mirrors. And that is going to allow us to use the laws of specular reflection And the one that we most critically rely on is that the angle of That is the first thing that we use And then the second thing that we use is interference criteria We are going to apply interference criteria. Let’s take a look at what we mean by that We have two possibilities We can have multiple beams coming in phase or we can have multiple beams coming out of phase And so here is what we talk about I am going to show a little sketch where I am plotting amplitude This is some electromagnetic signal Amplitude versus time Let’s look at two rays Here is one. This is amplitude versus time. And the second one, in phase. I am going to draw it directly beneath And so the crests line up, the troughs line up. If I take ray

one plus ray two, I am going to add them And I will get the intensification, the additive sum of amplitude because they are in phase This is called in phase radiation summing. Or the other term applied to this is constructive That is the one extreme The other extreme is totally out of phase. In that case, ray one I will draw as I did before, but ray two I will draw symmetrically opposite so that in ray two the crest of ray two aligns with the trough of ray one The trough of ray two aligns with the crest of ray one And, for argument’s sake, let’s say that these are equal in amplitude. If I have two rays equal in amplitude and they interfere but they are perfectly out of phase, if I add these two signals what I will get is nothing They will cancel. Complete And those are the only two principles we need And we can describe at a very reasonable level what is going on inside a crystal which is being irradiated by electromagnetic radiation in the x region of the spectrum. That is all I am going to use are these two ideas Let’s now go to a simple model here of a crystal. We will draw some atoms here Here is the top plane Let’s go down a second And I will make this cubic Here are two planes of atoms And let’s say I have incident light coming in at some angle theta One ray I am showing striking this atom And if I observe the laws of specular reflection, this will reflect from the mirror at the identical angle The angle of reflection equals the angle of incidence And let’s put a second ray coming down here. And it, too, will reflect at the same angle and come out like so Oh, there is a cell phone Somebody is leaving the room Let’s go. I want the cell phones off. No compromises Show some respect for your classmates. It is rude It is disrespectful to me It’s disrespectful to the class You don’t get it. If that happens on Wednesday, I will tear up your paper in front of your eyes and give I want to look at what happens when I have light coming in at a particular angle And, furthermore, I want to specify that I have in phase radiation which is also termed coherent. I have in phase incident coherent radiation And, furthermore, I want it monochromatic These are x-rays that are coherent and monochromatic Monochromatic means a single wavelength. Here is the incident radiation on the left, and this is the reflected radiation on the right. And what do I want to see here? I want to see what the conditions are to give me constructive interference because I don’t want destructive interference If I have destructive interference I won’t see anything Let’s mark this by showing some kind of a waveform What I will do is show just the first sinusoid, and then I will show the same on the

second ray. What I am trying to indicate is that I have in phase radiation coming down from the left And then over here I want to indicate that I have in phase radiation leaving There is an issue here You can see ray one travels a shorter distance than ray two Ray two has to go farther. I can actually calculate how far that is If I drop a normal from here and a normal from here, I can label this A, B and C. And we can see that ray two has to travel a farther distance, AB plus BC. And what is the condition of AB plus BC to ensure that is a whole number of wavelengths. Otherwise, I will get some amount of destructive interference And n, of course, is a whole number I don’t care It’s one, two and so on Actually, we can call this the order of the reflection And if I go through the math, we know what theta is. What is this distance between successive planes? If I am showing you adjacent hkl planes then this distance here is d, as we have defined it before. This is the distance between adjacent hkl planes. And, if I go through the trigonometry here, n lambda will equal two times d sine of theta. We’ve got the angle theta If n lambda equals 2 d sine theta, this path length will be such that as such, and this is called Bragg’s Law. It is the condition for And, in 3.091, what we are going to do is neglect higher order reflections In 3.091, we neglect higher order reflections. Let’s just put n equals one. Bragg’s Law for 3 91 will simply be lambda equals 2 d sine theta. That is the Bragg condition Suppose I have a new system and I find that I have now atoms here Let’s say I go from a simple cubic to a body-centered cubic Well, in body-centered cubic there are atoms here I know they are on a different plane relative to the board, but relative to the horizontal there is a plane of atoms at this location Let’s see what happens in this case In this case, if I have a third ray that comes in, let’s call it ray three, and ray three also is in phase with ray one and ray two, this is the reflected ray three, I can define the additional path length of ray three as DEF And what we see is that it turns out then that DE plus EF, if n equals one, this is going to be one-half a wavelength If AB plus BC is one wavelength, this is one-half wavelength. And what that means is that the reflected ray three is perfectly out of phase with rays one and two And that means, if we park a detector right here, we will measure zero intensity. We measure zero intensity at that angle Now what I want to do is generalize what we have observed here I want to generalize this analysis And, if I generalize this analysis, what I get is that if I take the

interference criterion and combine it with the crystal structure, And so, in essence, this is a fingerprint of the crystal structure. It is not a fingerprint of the chemical identity It does not say anything about identity, but any crystal structure of this description will give the Any crystal structure of this description will give no reflections at the same set of angles This, for example, will show you what would happen in the case of {001}. {001} we saw reflects in simple cubic at the given angle theta 001 But it does not reflect in body-centered cubic at theta 001 Because in body-centered cubic there are atoms at {002} And atoms at {002} will be totally out of phase with the reflection Likewise, with FCC. FCC also has atoms at the midpoint on each face It does not reflect in BCC and FCC, both of them. Now I can start to distinguish which are which And now let’s take a look at some opportunity to make sense of this Here is just a figure that I took from one of the texts just showing what I have drawn on the board I think we can zoom through this Now this is on the handout. This is the selection rules for reflection in cubic crystals What has happened here is that the geometric analysis along these lines has been performed taking into account the atom positions reflections at every plane, at every index. Body-centered cubic is not reflecting the {100}, because we have shown already that the reflection from {200} at that angle is perfectly out of phase The computations have been done And, likewise, FCC. This is indicating that you won’t see a {100} reflection in face-centered cubic or body-centered cubic Then we go down to {011}, simple cubic reflects, body-centered cubic reflects, but face-centered cubic does not reflect. You cannot look at any single line. You have to look at the entire pattern and so on And there is a simple way of codifying all of this Simple cubic, all of the indices are reflecting In body-centered cubic, if you go through this pattern, what we are looking at is the sum of h squared plus k squared plus l squared And that gives you a nice sequence, one, two, three, four, five and so on. And what you see with BCC is you have the sequence two, four, six, eight, ten. The sum of h plus k plus l is even That is a way of capturing what is going on in this series And for face-centered cubic, it turns out that the planes that reflect are either all even numbers or all odd numbers For example, face-centered cubic, you see {111}. Those are all odd {200}. Zero is termed in this calculation as being an even number Two is even, zero is even and zero is even. Three and four reflect Five and six do not. Eight does That is {220} and so on, eight, eleven, twelve, sixteen. This is a set of selection criteria that have been tabulated for your convenience And now what we can do is go into some example using this technique in order to index a crystal And so I want to show you two different techniques that allow us

The first one is called diffractometry And since it is done with x-rays, it is also known as x-ray diffraction, XRD And the characteristics of x-ray diffraction are we use radiation of fixed wavelength and we vary the If we have fixed wavelength and we are varying the angle, variable angle What I am saying is variable angle of incidence. In other words, I want to probe all of the incident angles and build a library of responses. For a given incidence angle, what is the intensity of the reflected radiation? The way I do that is to park the beam and rotate the specimen Ultimately, I end up with a plot that looks like this Now we are going to decide if the pattern we observe conforms to this set of rules up here And, by making that comparison, we can deduce which of these three crystal structures the crystal belongs to. And I am keeping this as a very narrow set I am saying this is 3 91 so we confine ourselves to the universe of cubic crystals If you want to get more general then it could be any one of the crystals that comes out of the 14 There is a huge database out there that has values of intensity versus angle for thousands and thousands of metals and compounds A huge amount of mineralogy in this database. And so, when you bring a specimen of unknown provenance, you are able to use this technique and infer what the constituents are First of all, let’s take a look at what the specimen looks like We grind the material into a fine powder and we put it in a silicate It is a very narrow tube This is definitely not to scale This dimension here is about a tenth of a millimeter It is a very narrow tube, a tenth of a millimeter made of silicate glass It is an oxide glass, a cousin of window glass Why do we choose it? Not because it is transparent visible light We don’t care. We are shooting x-rays through this thing, for heaven’s sakes. Why do we choose this? Because it is amorphous and it doesn’t have a crystal structure of its own. If we are trying to irradiate the contents of this and the tube itself has a crystal structure, how are we going to sort out what is the contents from the container. This is amorphous, no long range order. And inside we put little grains of powder The powder dimension is on the order of about 50 micrometers Oh, the wall thickness, I didn’t mention that. That is one one-hundredth of a millimeter Saying it is paper thin is an exaggeration. It is one-hundredth of a millimeter And this distance is about two centimeters. You grind in a mortar and pestle, drop the powder in here, and this goes into the diffractometer and is irradiated And we are going to turn this in order to present as many different angles as possible We have many grains in here Each grain, the theory is if we grind these fine enough, each one of these can act as a single crystal Now light is coming in and reflecting off this, but its cousin is sitting at a slightly different angle And so I am going to try to, through the mix of powder and the rotation of the specimen, try to get all of the different incident angles so that I will get as many reflected angles as possible I worked with this It was my PhD thesis The material was sensitive to moisture in the air so I had to work in a glove box I had these big thick rubber gloves on holding something like this Only in real life it is about this

big. And you are inside the glove box and it’s Toronto in August and perspiration is pouring You are supposed to be crying now The perspiration is pouring down my face. I have to get the powder to fall in here. And I have got to flick this a little bit You laugh now It’s not funny That’s the way. That’s how science goes. Anyway What happens now? We take this. Now I am going to give you the top view of the diffractometer It is essentially the combination of an x-ray generator, This is monochromatic x-ray, some lambda fixed, coming in Top view has this sitting in a carousel, and this carousel spins And then I have a detector. And the detector itself moves through a variety of angles And so here we have an incidence angle and here we have a reflectance angle. Ultimately, we get a plot of intensity versus reflected angle from this specimen That is how this thing works And here is what the output could look like. You will have nothing and then you will have peaks at discrete lines Why? At certain angles the Bragg condition is satisfied At other angles the Bragg condition is not satisfied These are the Bragg peaks associated with magical angles And what are the magical angles? They are the angles associated with those planes This is how we are trying to make sense of what is going on here I wanted to give you the thrill of discovery, so we are going to do a virtual lab here What I have for you is a data set If you flip over your sheet, this is all you get, the output. Now, for reasons I don’t want to go into, the angles are often reported as two theta instead of theta And it has to do with the geometry of this specimen and blah, blah, blah But that is not important We have a copper target in the x-ray generator, which means we will have monochromatic light lambda 1 418 angstroms. We know that. And, remember, we said we had to fix our lambda. We have done that We have rotated the specimen variable theta And these are the peaks I don’t care about the intensities All I am doing right now is telling you these are the peaks that I got How do we take that and infer what the crystal is? Well, you can go to the library and get all these self-help books I have not had the time to write a full self-help book, but I got this one self-help sheet here. This is my program You won’t find this in any books This I am giving to you special. I want to walk through this and show you how to take that dataset and index the crystal The question we are trying to answer is what is the crystal structure? And you only have three answers. It’s either going to be And that is the dataset That is all you have, plus your mind. It says start with the two theta values and generate a set of sine squared theta values Oh, I should perhaps tell you how I came up with this, what is the underlying principle Here is the underlying principle To solve the puzzle, all I did was I took these two equations I took the Bragg’s Law, which is lambda equals 2 d sine theta, and I took the formula for the d spacing between planes, which you saw two days ago, which is the ratio of the lattice constant divided by the sum of the squares of the Miller Indices of the plane If I take those two equations and Lambda squared over 4 a squared equals sine squared theta over h squared plus k squared plus l squared. Now, look at the left side of this equation. Lambda is the wavelength of the incident radiation, and that is fixed. We fixed that in the experiment, so this is a constant It is fixed. A is the lattice constant of the crystal That is the crystal property, so that is fixed. That means the ratio of lambda squared over 4 a squared is a constant Well, when I went to school, what they taught me in math was that

if the left-hand side of the equation is constant, the right-hand side of the equation is constant. That means even though theta is varying with hkl, this ratio varies in such a way as Whatever hkl is, at that theta, the sine squared of that angle over the h squared plus k squared plus l squared remains constant as you go through all of the lines. This is the underlying principle. Let’s go into this The first thing I want to do is build sine squared theta There they are. Those are the theta values. And now I have taken them, divided them by two and then And then it says normalize by dividing sine squared theta by the first member of the column All I did was took everything and divided it by 0 43. This 0.143 divided by 0 43 is one. This is, as you can see, 0.190 divided by 0.140 is about 1.34 and so on. Why? Because, look, I am going to this What is the next thing? Clear the fractions from the normalized column 0, 1.34 and so on 1.34 to me, allowing for some experimental error, is roughly one and a third It is roughly four-thirds. And that is 2.67. That is roughly eight-thirds. Why not multiply the whole darn column by 3 and throw away the little residuals And, sure enough, you get a set of whole numbers, 3, 4, 8, 11, 12 and so on Now it says speculate, this is where you get to participate, on the hkl values that if expressed as h squared plus k squared plus l squared would generate the sequence of the clear fractions column And remember what you have to go on You have that set of simple cubic, face-centered cubic, body-centered cubic. What set of hkl’s would give you this? Well, what sum of integer squares would give you three? I wonder. Maybe 1, ,1. How about 4? And so you can see what is going on here. This I have gotten from the theta, and this is my speculation, just trying. And now here is the punch line. Compute for each theta the value of this The sine squared theta over h squared plus k squared plus l squared. And, if you get a constant, bingo, you must have hit the jackpot Sine squared theta is this H squared plus k squared plus l You do it for the next one, the next one and next one allowing for experimental error I have a constant. What does that say? It says that my speculation of what the hkl’s are is good It is self-consistent. And what do I know? Those are either all even or all odd. That is the sequence for face-centered cubic So, on this basis, I can say that I have something that is But, furthermore, what else can I get? Well, I know the lambda value and I know this thing This is that constant there What I can do is take this constant and plug it into this equation And, furthermore, I can determine that the a, whatever this crystal is, is going to be lambda of copper k alpha divided by 2 times the square root of that constant And, if I plug in the numbers, I get 3.53 angstroms. It is FCC It is 3.53 angstroms. If I look at it and it looks metallic, I am going to say it is nickel, because I happen to know that nickel has a lattice constant of 3 3 angstroms. I know I have to do some other things, but certainly that makes me think about whether I have nickel present Here is the gambit That is how you are going to solve these problems You are going to go through this set of operations Just to recap, if you are going to look at this, simple cubic, the hkl sequence is 1, 2, 3, 4, 5, 6, 8, 9, because there is no sum of integers that will give you seven if you sum squares of integers It will give you all the lines, Body-centered cubic has to be 2, 4, 6, 8, 10, 12, 14, 16. But, when you first do this, you don’t know because you divide by

the first sine squared And so you don’t know it is 2, 4, 6, 8. You could end up with 1, 2, 3, 4. Now, don’t we have a problem? The simple cubic is going 1, 2, 3, 4 and body-centered cubic is going 1, 2, 3, 4 But watch this, perseverance. Half of 10 is 5 Half of 12 is 6. Half of 14 is 7 Bingo. If you guessed that something when you divide through you get a sequence that gives you 1, 2, 3, 4, 5, 6, 7, you know that it cannot be simple cubic because there is no sum of integer squares that gives you 7. This 7, in fact, must be 14 divided by 2 This is body-centered cubic Face-centered cubic is the easiest one to pick off because it gives you 3, 4, 8, 11, 12 I mean, I just need to see 3, 4 and I am done. I always check for FCC first. Anyway, that is how diffractometry works And it allows you to index crystals and determine their crystal structure. And, more broadly speaking, Let’s look at the second technique which is called Laue diffraction named after Max von Laue who got the Nobel. He was the first person to demonstrate x-ray identification of crystal structure He got the Nobel in 1912 for this And in Laue we fix the angle, the specimen is stationary and variable wavelength radiation How do we get variable wavelength radiation? Look at that spectrum up there. You see the bremsstrahlung? The bremsstrahlung gives you a range of wavelengths How do we get monochromatic radiation? How do we get monochromatic x-rays? And don’t tell me we buy an x-ray laser. How do we get monochromatic x-rays? Well, it is really cool What you do is take this bremsstrahlung with everything on it And you present it to a single crystal. And then you park over here, this is the entrance slit to your diffractometer What will go through that entrance slit? There is only one wavelength that will go through that entrance slit, and that is the wavelength that satisfies the Bragg condition for this lattice spacing at this angle. What I can do is make a single crystal monochromator that basically takes this characteristic Sends that in I present a single crystal And there is only one wavelength of light that will come out of here that satisfies the Bragg condition for this d spacing and this angle The single crystal acts as a monochromator Just use Bragg’s Law That is all. Well, in this case, we take the whole mess, everything And here is what the specimen looks like. The specimen goes into a darkened box called a “camera obscura.” And what happens is I can put the specimen on the front here This is some crystal, relatively thin And I put film on the back I have film that is sensitive to x-ray. Now I send the beam in This is white x-ray meaning it is variable wavelength It is the whole spectrum And what happens is the white x-ray interacts with the specimen and shoots off at different angles because I have different at its own special angle What happens is we end up with a spot pattern on the film Spot pattern is the output And the symmetry of the spot pattern. I am not going to take you into k space and so on, but let me simply tell you that all you need to appreciate at this point is that the symmetry in the spot pattern is imitative of the symmetry in the crystal Let me say something about atomic

symmetry. And all we are talking about here is rotational symmetry because we are looking at something on edge. For example, if I gave you the 001 plane, in a cubic crystal it looks like this Now what I want to do is look at rotation about a normal axis I am going to put a normal axis right here at the center, and I am going to rotate this I am going to ask you the question through what angle do I have to rotate this in order to get this back? If you close your eyes and I start rotating, suppose I rotate it 45 degrees, you open your eyes, you are going to see that and know something has moved How many degrees do I have to rotate before I get this back? 90. If I rotate 90 degrees, I will end up with this. These are indistinguishable 90 degrees is what I need to rotate And the crystallographers, instead of talking about the angle of rotation to restore what you are looking at, they talk about fold And fold is simply 360 degrees rotational symmetry Suppose I look at (011) (011) is the diagonal. This is a and this is square root of 2a And now, if I put an axis in the center, through what angle do I have to rotate before I get this back? That is 180. This has two-fold symmetry Finally, let’s do (111) The (111) plane looks like this And so, obviously, this you would have to go through 120 And this would exhibit three-fold symmetry. For example, remember I showed you that crystal of the silicon, single crystal silicon? The single crystal silicon, let’s say we have cut it and we want have cut it so that the 111 plane is parallel to the surface One way of determining this would be to put the single crystal of silicon into a Laue unit and irradiate it. And, if indeed we have the 111 plane, we will see a spot pattern that exhibits three-fold symmetry This allows us to do indexing The other thing we can do is put the specimen on the back and have the film on the front And we end up with what is known as Laue backscatter Let’s take a look at a few more symmetries. This is going back to some Escher prints I don’t think I showed you this one the other day What is that crystal structure? It is either body-centered cubic or face-centered cubic The two are the same in two-dimensions This would exhibit four-fold Symmetry now. Three-fold symmetry 120. I even tried it. I rotated this. I actually programmed Photoshop. I said rotate 120 degrees Bingo. Now you see Now, these are some Laue patterns of various crystals This is what you see in the Laue unit. Three different crystals here Let’s look. What do we see? How about this one? You won’t be able to tell the crystal structure All you can see is the pattern All you can ask from this is what is the pattern? Which of these symmetries do we have? Is this four-fold, three-fold or two-fold? This looks like four-fold, 90 degrees, pretty good This one here, well, you’re right You can say six. It looks like 60 but, in fact, you need to go the full and there are other reasons you have to go into that This is two-fold Remember this? This one is, you can see, a 60 degree angle, but you can see this has three-fold symmetry This gets back to somebody who said 60 degrees. You can see this pattern requires because of the basis here, in fact, it requires 120 degrees What about this one? Exactly It is one-fold. This has translational symmetry but no rotational symmetry Obviously, if you go 90 degrees, the dogs are going to be pointing up and down If you go 180 degrees, it is not going to work either This has rotational one-fold. I am going to jump over this, I think. I see we are getting close

to the end. Here is another symmetry. This is the inverse of the dogs. The dogs have no rotational symmetry but translational symmetry This pattern here has rotational It is called Penrose Tile And, in particular, there is the rotational symmetry And how many turns do we go here? It has five-fold symmetry. Five That is very weird Five-fold symmetry Now, this does not conform to any rules of crystallography Such a pattern is called aperiodic We saw that you couldn’t make bathroom tile of this shape You would have to have other shapes that would fit into the interstices and so on. There is another one See, it fills to space, but you have to add some extra But it has five-fold symmetry If you take this whole pattern and rotate it 72 degrees, Now I am going to show you something that was really, really unexpected So far we are over here in the top left. We are studying ordered solids which they have a unit cell, they are periodic, and we call them a crystal. And pretty soon we are going to start studying disordered solids which form glasses In 1982, there was a fellow by the name of Dan Schechtman who was on sabbatical from the Technion in Israel. He was working down at the It is now known as the National Institute of Standards and Technology. Sometimes you will see this acronym NIST And they were down in Gaithersburg, Maryland. And Schechtman and John Kahn and coworkers were looking at a set of alloys from the aluminum manganese system, and they were very highly ordered Over the range of composition, pure aluminum is FCC The question is what happens when you take something that is not face-centered cubic and alloy it with something that is face-centered cubic? Well, initially aluminum retains its face-centered cubic crystal structure But, after a while, you get to such a high manganese level that the crystal structure changes. And over certain ranges of composition these alloys are amorphous. They are glass formers But then came something totally unexpected. When Schechtman prepared the alloy that contained 25% manganese he got something extremely weird in the x-ray x-ray diffraction pattern Five-fold symmetry. This lax translational symmetry, it is aperiodic, these were termed quasicrystals And this caused a storm in the material science community In the mid ‘80s, all kinds of NSF programs, Department of Defense programs were funding research into quasicrystals to try to figure out they might have some unique properties Maybe they have conductivities Maybe they have ductilities. Maybe they have corrosion resistance that is unknown in classical face-centered, body-centered, simple cubic alloys Here is one of Schechtman’s Laue patterns. Look at that That is so weird. There were jokes about this. If you know the geography, Gaithersburg is not far from Washington Some people joked that this had five-fold symmetry because the research was funded by the Pentagon And this is really something These are the pages from Schechtman’s lab notebook April 8, 1982, aluminum 25% manganese. And he is analyzing his diffraction patterns and is going And it was real It was very, very exciting That is what quasicrystals are about. What you can see is how using Laue you can infer that And then you have to go in and determine what the lattice constant is and so on. The last thing, before you go. I ran out of time last day. What is the connected between 7-Up and lithium? Before Mosley people referred to things according to their atomic mass. The atomic mass of lithium is 7. That is why it is 7 I have this on good authority from people in the lithium industry All right. Have a good weekend