 You can follow along with this presentation using printed slides from the Nano Hub. Visit www.nanohub.org and download the PDF file containing the slides for this presentation. Print them out and turn each page when you hear the following sound. Enjoy the show. Lecture 2 is a continuation of the discussion about crystals and periodic lattices. As you remember that I mentioned that the materials of interest in the semiconductor industry do not only include crystalline material. There are amorphous material, polysilicon material, of course, liquid crystals that may or may not have the periodic symmetry. But when they do have periodic symmetry, in that case, then a solution of a whole set of problems becomes particularly easy. And those conclusions that we get from crystalline silicon or crystalline material, we can translate back to amorphous material, polycrystalline material and others. So we are solving an easy problem in some way so that we can have a broader understanding of how all the materials interest in, that is of interest in semiconductor, how they work. Now let me start by a discussion about the volume and surface issues about body-centered cubic and face-centered cubic and cubic lattices. That's where we'll start up the discussion today. Now one thing I want to point out that in the bottom of the slide, you will see the reference, of course, for the book, the page number I'm approximately covering. But at the same time, there are a set of software that you can find in nanohub.org. Presumably you have had an account there already. You should be using this tool in order to get a better feeling about the various discussions here. Many of the pictures are three-dimensional and I can only take a snapshot of from a particular angle. Things may not be as clear. But in the nanohub, you may be able to rotate the crystals around so that you get an exact idea where the various atoms are and that will clarify the concepts. Now if you remember our discussion so far, that we are interested in calculating current through a semiconductor. And in order to calculate the current, we are interested in the resistivity that depends on the material as well as the arrangement of the atoms. Now in that case, we really want to know two quantities. One is this electron density and which is how many carriers do you have, free carriers you have available for conduction. Different material will have different number. And the other one is velocity. That how fast they move in response to an electric field, for example, or a density gradient. And in order to do that, we want to understand the periodicity of the atoms and the structure of the atoms. And that's why we are trying to understand the symmetry, the number of atoms per centimeter cube for a particular material and all sorts of basic information. So that if I tell you that in one centimeter cube of silicon, how many electrons do you have? You'll be able to say, okay, let me see first how many atoms do I have to begin with. And per atom, let's see how many electrons I have and thereby you might be able to estimate how many electrons you have. So that is what we are after today. And presumably after this discussion, you'll be able to make that count. Now if you remember that in the last class, I told you about Bravais lattices. In 1D, there is one type. In 2D, there are five types of Bravais lattices. And in 3D, there are 14 types. And 14 types I mentioned the other day are, of course, there are the seven symmetry based on the symmetry group. There are seven types and also depending on where the atoms sit, whether in the corner, whether in the body center, or whether in the face center, or on one side or the other of the face, which is the bottom line. Then we can sort of categorize these various crystal lattices. And that discussion we had. Now one very interesting thing is that although there are practical examples of all these crystals, if you go to a natural history museum, you'll be able to find actually examples of all these crystals of various sort. But 70 to 75% of all material actually occur in these three groups. One is this cubic, body centered cubic and face centered cubic, and as well as the last one is the hexagonal. And I'll explain these three primarily because the semiconductors of interest, silicon, germanium, silicon carbide, for example, these mostly occur in this cubic, body centered cubic and face centered cubic, and as well as hexagonal structures. So we'll spend some time. Now one example I want to very quickly tell you about that the cubic, the simplest one on the right hand side on the top, that one for many years, there is only one known example, and I think it still is, and that is only one, and that is polonium 84. It's a very strange material and the way other materials sort of work, this material is so strange, and that only very recently people are understanding how the crystal lattices work for a cubic lattice. It's a very exceptional material, but most other, most other will simply be face centered cubic, body centered cubic and hexagonal. So let's start with cubic lattices. Remember there's nothing practically here except polonium 84, but let's nonetheless discuss the certain number of concepts. For example, in terms of volume issues, if you wanted to know how many atoms do you have per centimeter cube, then what you could do, you could say that well let me take a unit cell, the unit cell of size A, and I construct a unit cell, how many atoms do I have inside the unit cell? Well one thing you could say 8, there are 8 corners, of course there you could say 8, but of course that will not be a correct answer, because you realize that every corner of that sphere, of that atom is being shared by 8 neighbors. It's a 3D, in 2D there have been 4 neighbors, but in 3D there are 8 neighbors, so what you really get inside your box is 1 eighth of that atom, and since you have 1 eighth, and you have 8 corners, so how many atoms? You have about 1 atom per unit box, and so the number of unit cells, 1 eighth points per corner, 8 corners, is 1 point per cell, remember these are geometrical points, inside it there may be many atoms, remember the basis idea that this is a geometrical point and the basis may contain multiple atoms inside it. Now from this, if you know the dimension A, you could easily say how many per centimeter cube, and from that you could say how many atoms per centimeter cube for cubic lattices. Now one thing you realize and you should convince yourself that although I have done it here in terms of primitive cells, even if my cell were bigger, you see that shouldn't matter because then it will contain more atoms, but of course the density cannot change depending on how you mentally partition it up in terms of blocks. And this would be important also for your homework, you will see that if you use this definition, the density is independent of the definition of the cell, then that will allow you to make some calculations very easily. Now in terms of cubic lattices, if you wanted to know what fraction of the volume is actually occupied, we just counted number, but what fraction of the volume is occupied, you will see that is approximately 52-53%. And this is how it comes about, you realize that if the two sides, if the lattice spacing is A, then the maximum size and spherical atom can be, the radius of it is A over 2, because A over 2 is at that point where the two spheres touch each other. It cannot be bigger than this because it is then the electron cloud from one has to penetrate the electron cloud from the other one, and that it cannot do, so it cannot go beyond A over 2, so the radius of the sphere is A over 2 and so if you wanted to know what fraction of the volume is occupied, then again you will start with, you will see the second line, 1 eighth, this is 1 eighth from each sphere, 4 by 3 pi r cube, you know this formula right, volume of a sphere multiplied by 8 corners, so that gives you this 8 corners from the cube divided by A cube, A cube is a volume, total volume, and the result is pi over 6 and pi over 6 is approximately 50%. So 50% of the volume is occupied and why did the other 50% go? You see there are empty spaces inside this, outside this sphere, that's why the other 50% has gone. One of the strange thing about materials that even if you didn't put things in terms of periodic structure, if you just packed it randomly, it's still the answer will be very close to 52%. It's a constant that you find over and over again, not only in periodic crystals, but also in all sorts of amorphous materials, about 54, 56% even for random packing. Okay, cubic lattice. What about surface issues? That if I wanted to know not only volume, but let's say I have a semiconductor that's terminated by a surface and I'm looking from the top and I want to know how many atoms per centimeter squared, not centimeter cube, not volume, but how many per centimeter squared, then what would you do? Then you would say, okay, first define me the surface. So for example, if I look at the top surface, then I realize that the two surfaces, each atom is now being shared by four neighboring squares because I'm looking from the top, not eight as before. And in that case, the density will be one by four atoms per corner. It's being shared by from the top, being shared by four neighbors and four corners, of course. And the area is a squared because it is a square on the top, looking from the top, area is a squared and the answer would be density will be a divided by a squared, one divided by a squared per centimeter squared. Now remember this is called an aerial density, not density, aerial density means surface density, number that you have per centimeter squared. Now this is, I will come back to this later on in the course. Surfaces in semiconductors are extremely important. You see, because that controls not only the reaction rates as you deposit semiconductor, but remember the example I gave you, crystalline material and an amorphous material on top, how they line up will depend on, when two materials essentially built on top of each other, that will depend on how many surface atoms you have. So it's a very important concept, we should be able to count it. Now what about if I had a different surface, because remember I can take a knife let's say, and cut surfaces at any angle, any angle I want. So if I wanted, if I cut it along diagonal, body diagonal angle, then should my surface density remain the same? Of course not. And the reason it will not remain the same is that because although in each corner, it's still one-fourth, you know these are rectangles, side-by-side rectangles, so there are four elements, so it's still one-fourth per corner, and there are still four corners, but you realize that since it's a body diagonal, so therefore the side length will be square root of two multiplied by A, right? It will not simply be A, and so therefore you can see on the bottom I have square root of two A squared, square root of two multiplied by A, and that gives me a little bit larger area, and so the density is actually a little lower compared to before. And density determines everything, reaction rates, the flow of electrons on the surfaces, so therefore this density on an inclined plane will be a little lower and therefore the properties will be different. Now let's talk about the other two lattices, body-centered cubic and face-centered cubic. Let's start with body-centered cubic. You can see that not only we have atoms on the corner, but we have one in the center as well in the body, and in that case it should be easy to see that the number of atoms you have per unit box is two, right? You can see two because there is one eight multiplied by eight, that's coming from the eight corners that you have, and each corner is being shared by eight neighbors, so no problem. What about that one? Well the one is the one that is sitting in the body, and that is not being shared with anybody. So I have two and the number of points per cell is two for body-centered cubic. Now one thing you will realize in the body-centered cubic, let me step back a little, that if you look at the atoms and if you wanted to calculate the packing density, you know what fraction of the volume is occupied, then you should realize that along the diagonal the atoms essentially touch each other, the sphere essentially touch each other, and from that you could calculate what fraction of the total space is being occupied by these atoms. What about face-centered cubic? Face-centered cubic has not only atoms on eight corners, but there are six phases, and on each phase there is an atom sitting there, so in that case how many atoms do I have? Well I have eight corners, one eight multiplied by eight, that's one, just like the cubic one, but remember now each surface is being shared by two, one in the bottom, one in the top, one in the left, one in the right, so it's being shared by two, so each atom is being cut by half in the middle, and so you have half multiplied by six phases, and so from the phases you get three, from the corners you get one, so you have four points per cell, that's clear? And again in this case if you look, wanted to calculate the packing density, you should realize that look at the figure on the right, that the yellow, red, and the yellow one, that that series along the face diagonal, that essentially touches each other, and so that gives you an idea how to calculate the packing density very easily. Now these are very simple things, nothing complicated here, no reason to spend a lot of time. Okay, so remember body-centered cubic, two atoms, face-centered cubics, four atoms per unit cell. Now what about this hexagonal close pack? Now the hexagonal close pack, what should be the answer? The first thing is this is hexagonal, means that each face is in hexagonal shape, six atoms sitting on each face. On the top I have six, on the bottom say face I have six, a total of 12, right, 12 atoms, you can easily count. So points per cell, let's see whether we can do the counting. First is if you just think about the face, and the one that is sitting in the bottom face, and the top face, remember this is being shared by another corresponding column in the bottom, so each of these atoms will be cut in half inside the cylindrical hexagonal cylinder. So half per face, and there are two faces, so I have one. Now what about the other one? How many do I have, the ones that are being in the corner? Well let's think. Let's see how it looks. So you will have another column like this, and let's focus on any atom, let's say on the one where the red one is crossing, and so let's think about that atom. You see that if you look from the top, then that atom in the corner will be shared among three such columns, three hexagonal columns. You can see it here, the three hexagonal columns. So inside each hexagon, then we have one third of the atom sitting. But not really, because there is also one on the top and on the bottom. So you see that is one third, of course among the neighbors on the same plane, but between top and bottom, there's also another factor of half. And so that tells you that half between top and bottom, one third among the neighbors, and there are 12 corners. So what does it give you? Two, right? You multiply them, it gets two. So how many atoms in the hexagon? Then I have three atoms per volume. Whatever this volume is, you can calculate that will be the corresponding density of points per unit cell. So you should be able to calculate, for example, the packing density, the aerial density, surface density, and all other things from here. These are general concepts should be easy to calculate. Okay. Now remember, these are all Bravais lattices. We haven't said anything about practical crystals yet. These are just geometrical constructs. And we'll see how to map things back to these type of structures in a minute. So let's talk about a few material systems and see how they transfer back, translate back to the Bravais lattices. So one thing I wanted to point out, and that had been saying just a few minutes ago, that we are talking about Bravais lattice, and I told you about the two definitions, equivalent definitions of Bravais lattices in the last class. But most material do not occur exactly in the Bravais lattice. For example, the one that you see right with a yellow circle and green square, that of course is not Bravais lattice, it's because the environment looking from the yellow atom and the one looking from the green square is not the same. It looks about the same, but from the yellow one looking to the right, you have a very close green neighbor. But for the green one looking to the right, corresponding yellow neighbor, you have another one towards the left side, not on the right side. So it's not a Bravais lattice and the way to transform this to Bravais lattice is to assume the green and the yellow together is a unit and thereby transform it to one of the Bravais lattices. This we have done before. Now the point of this introducing this was introducing this was to come to this point actually. So one is this rock salt lattice structure and this occurs in a face centered cubic lattice as a face centered cubic lattice. Do you see one? First of all, this is not a type of lattice we have seen before. Here we have sodium and chlorine. It looks like, if you just look at it, it looks like face centered, right? But then you see there's one or another atom sitting in the body, in the middle. So is it face centered or is it body centered? In fact, it will turn out that if you look at this way, there's no unique definition. But if you do it this way, then it will work. Assume you pair a sodium and chlorine together. Yes, like the blue, the yellow circle and the green square before in the last slide, you paired them together. When you pair them together and you replace them with one unit, geometrical point. And then you do it for everybody. So you push it up at one point and let's take everybody and their neighbor and pair them up. So you can see the second one is sort of the bottom chlorine and smaller sodium that have been pushed up. And you can see that you'll also pair up something from below because that's where the chlorine atom was for that corner and you do that for everyone, all the corners. Now do you realize that why this should be a face centered lattice? That's clear, right? But you also realize this is a face centered lattice with a basis and basis is sodium chloride, right? Sodium and chlorine atoms together. Now you don't have to do it this way. What about you could not only have to go, you can go in the vertical direction if you like, it's fine. You could go in the horizontal direction. So you could form basis in any way you wanted and at the end you will still get a face centered cubic lattice, okay? But this is of course not a semiconductor, right? This is sodium chloride is not a semiconductor but there are more important semiconductors which is a little bit more complicated that we'll discuss next. The first one that I will discuss is this zinc blend structure for gallium arsenide and you can see here the white atoms. Those are arsenic and the bigger black atoms are gallium and we want to know what to call this face centered, body centered, what should we call this? Again we'll see that only if we transform it into a basis first then look at the arrangement we'll come to an conclusion. Now I have given you the answer already. This is like an FCC lattice but let's see how it comes about. So first of all you can see there are eight arsenic atoms in eight corners. No problem, right? And you can also see that on six faces there are six arsenic atoms. So that's also no problem, this is white. But where is this black one sitting inside the volume that requires some thought and let's think about it. In order to know where that one sits this is how you should go about. You should first find a corner atom on one of the arsenic corner atoms and then go body diagonal on the other side. You see on the other side that the blue one rotating blue and if you join through it a body diagonal and one fourth down the road if you stop and place the black atoms there that will be the position of the first black atom. What about the remaining three black atoms? There are four inside the volume. Well you get the same idea. The first thing is identify the red atoms and the red atoms have been constructed from the first red one, the other three have been because these are all neighboring faces, face diagonal on neighboring faces. Do you see that? They are the first face front face. You can see one on the top red, the side right side face and you can see the top right, the red corresponding red one and the bottom surface. You can see face diagonal on the bottom side is the red corresponding red one and then from a given red for example look at the blue and the corresponding red again you draw a body diagonal one fourth down the road you put a black atom and you do the same for all corners. This is how you create a zinc blend structure. Now zinc blend and diamond structure that we will discover a little bit later these are exactly the same except that in zinc blend structure you have two types of atoms, gallium and arsenide. If you had been one type of atom then it would have been called a diamond structure. Now how many atoms do I have per unit cell? By the way none of these are primitive cells. In the homework you will learn how to define primitive cells for this structure. This is not a primitive cell but for this cell how many? Well you could easily see now that if you just look at the white ones then this is face centered cubic and if you have face centered cubic then we already know one eight multiplied by eight half multiplied by six because there are six faces half of it on each face and how much do you have the black atoms? You have four black atoms so a total of eight so that gives you a total of eight atoms and one thing people often talk about that these are tetrahedrally bonded structures and you can see why the black atoms has essentially four arsenic atoms on them as neighbors and that looks like a tetrahedra this is like a triangular faces this is a triangular pyramid and so the tetrahedral structure it defines a zinc blend structure is it clear more or less? Now this is zinc blend structure now what about for gallium arsenide it works indium phosphide remember the table we had gallium arsenide, indium phosphide and there are many other materials that will work in zinc blend structure and semiconductors now for diamond a lattice for silicon both silicon and germanium these elemental structures from group four also occurs in diamond structure or the tetrahedral structure and you can see essentially these two are exactly the same in terms of structure and the same rules but the ones that are sitting within the volume again the same rules one fourth down the body diagonal everything is the same so you can do the calculation exactly the same way now one thing you have to realize that when things are within the volume in the body there are four neighbors sort of tugging you along for every atom and things are in equilibrium state and nice zinc blend structure occurs but when you think about things on the surface then of course things are a little bit more complicated because you can see if you are looking on the top surface now there are these three gallium atoms arsenic atoms on the top they are shown marked by red but the fourth one is down down in the volume so in one side you have air on the top and on the bottom side you have a bunch of atoms as a result what will happen and the top view is shown on the bottom left side the big red atoms are the corresponding ones mark on the top with a red square and what will happen because the arrangement of the atoms on the surface is no longer symmetric is no longer symmetric therefore what will happen that on the surface of gallium arsenide there will be this atomic rearrangements I am not going into the details of this but there will be a rearrangement of the atoms and as a result surface of gallium arsenide or silicon are significantly more complicated than the bulk of the volume but surface is very important I told you and we will go get back to that later in the course what about hexagonal close packed this is cadmium sulphide this requires some thought this looks like a hexagonal except all those complicated atoms so first let's think so I have just like a hexagonal on the six corners I have six atoms and also on the face on the top face I have one just like before right the hexagonal one and then I have the corresponding sulphur atoms in their plane so cadmium has their plane sulphur has their plane same sulphur has the same plane I have similar six in the corner and one in the face and also the same so this plane cadmium plane will repeat how many atoms here well let's think so if I am looking on the top and the black points I have shown here in green on this side you can see that although my drawing is not exactly very good but you can see the hexagonal looking from the top though you can see the hexagonal the ones that I have connected up and you can see the face centered one so that is on that plane for the cadmium atoms now the one the three cadmium atom that you see within the body just below the sulphur surface you can see those occurs in the gap those three marked by blue those occur in the gap of the original one so if you are looking from the top you will find the ones that are in the body those ones are occurring correspondingly here on the middle plane and correspondingly it will continue for the other surface and in that case how many atoms here do you have again we have the same one six multiplied by twelve those are the surfaces the corners that I have and then half multiplied by two where that half multiplied by two come from top surface and a bottom surface and what about that last three the three is sitting in the volume shown here marked here in blue so that is all six atoms I have in cadmium and I will leave for you as an exercise to show that how to convert it into an hexagonal lattice you have to do something to it in order to convert it to a pure hexagonal lattice form a basis and then convert it so I hope you will be able to work it out I will get started on the Miller indices and continue in the next class and so let's talk about so I have talked about atoms, volumes, area densities, surfaces and all sorts of complicated things but if you go home and read for an hour these things will become clear let's talk about surfaces a little bit more now in semiconductor as you saw the atoms sit in various places various corners in various places and it's very important many times to define the surfaces so for example and the surfaces have names and for example the blue one on the left they are giving three indices zero zero one and saying that that's the name of their surface the yellow one they are saying zero one zero so every surface you see they have given a name now how did the names come about this is what the Miller indices are and how the name come about is what I am trying to explain the naming convention of surfaces now this is how it works assume that you have three atoms any three you want take three from the crystal and you know from the three three points you can always pass a plane through it so I pass a plane through it what is this what is the name of this surface that's what I am trying to calculate so the first step would be to normalize the intercept so for example in the x axis you see I have the intercept of two you see that on the y axis how many three and z axis one so I have two three and one I am giving you the algorithm how the naming comes about so I have two three and one now you invert it so two becomes half three becomes one third and one becomes one and then you rationalize it so that the numerator becomes the same and in that case half becomes three six one third is two six and one is six by six take away the six from the numerator and that's your surface name three two and six that's your surface name this is a Miller index for this surface so this is as simple as this so let's take some more examples think about I have a surface where the plane actually the atom is on the z axis but in the negative z axis what do I do about that so I will do the same procedure two in the x axis three in the y axis but minus two in the z axis not plus two minus two in the z axis again invert half one third and minus half right rationalize what will be half will become three sixth right one third is two six and half is again minus half is minus three over six take away the six two three two and instead of writing a minus three will take that minus and pull it on the top and you can see that three has been pushed as a hat on top of three so the Miller index is three two three with a bar right for this surface what about this surface if it doesn't cut the z axis at all then what happens well then you say the same thing so the x axis is two y three and then z axis doesn't cut at all so probably it cuts at infinity so two three and infinity inverted half one third and zero and then rationalize then it becomes three six two six and zero over six take away the six three two and zero and that's the answer pretty simple right this is nothing in here now one other thing people often use for hexagonal lattice just for convenience is something called the Breve Miller index so Breve has inserted his name there but let's see how that works so if you have a surface which is shown here in that colored line in the transparent colored line and if you have three axis but instead of x y and z which are 90 degrees with each other you have four axis here three in the same plane which is given by a one a two and a three they are one twenty degree apart and the z axis is the same z axis before so then how would I describe this surface in this coordinate this is how I'll do it do you see whether it makes sense if you look at the a one axis the surface is such that if you extended it it will never cut a one in this particular case so therefore it cuts a one at infinity I'll say so the first index on the top is infinity what about a two well it cuts a two at one position one that's where it cuts it a three well a three is going in that direction but my surface is on the negative side cutting it is a negative side so I have minus one and what about the z axis again it's parallel to z axis so it will be cutting at infinity again you invert it zero one minus one and zero and you can very quickly check that villa Breve Miller index for this particular structure zero one one bar and zero and check that out but one thing about this you should notice if you haven't already that if you have done your math correct during exam this is a good way to check things if you have done your math correct then the first three must always sum to zero this is you can see zero one minus one so one and minus one gives you zero and no matter what surface it is it will always be always be this condition would be always be satisfied but remember for every Breve Miller indices there is a corresponding Miller index because you know it's a matter of putting axis and then getting the intercepts so if you do it in a more complicated way conceptually it's easier when you are drawing a few diagrams but you should always be able to convert it to Miller index if you needed to now I will show you you know I gave you an algorithm how did they get this algorithm I mean is it something if I you know dream up something on my myself that this is set of rules and give it my name will it stick well actually it will not stick because this is a very good reason why this algorithm works and this is how remember the first surface three two six how does that what does this number come from if you look at the vector r1 of the surface do you agree that since the z axis is along one intercept and x axis has two intercepts so therefore the vector connecting those two points is one c minus two a a being the unit unit right what about the other one r2 so r2 y is the final point there is three then I get a vector perpendicular from r1 and I take a cross product I hope that you remember how to take cross product but look at the answer three a plus two b plus six c do you see the correspondence between Miller index and these three numbers three two and six so Miller index essentially is the vector the integers that multiply the unit vectors for a vector perpendicular to the surface that's where it come from so it's a poor man's vector algebra that's what this is if you don't want to do vector algebra then you can go it by this particular way now I will stop here and I will maybe I can continue on there are five more minutes so let me let me continue on this two more slides then we'll be done if we wanted to calculate the angle between two planes right many times you will see that's very important quantity how will you do it now if you remember from high school or maybe from college how do you calculate the angle between two vectors you take dot product right not cross products dot products you take and if I have first of all a unit vector this h1 k1 and l1 are the Miller index for indices for plane one and the unit vector for plane two h2 k2 and l2 are the Miller index for indices for plane two so if I simply take the cross product dot product of this quantities then that will give me the cosine theta one thing people often forget is the normalize the vector in terms in the denominator you need to normalize it otherwise your cosine theta will not be between zero and one it will become a large number so you want to normalize it and thereby get the cosine theta the angle between it and one example is something like this for example if you wanted to calculate so this is a silicon silicon volume and people will generally cut small wafers out of this it's like cheese so you take small wafers out of this now there is something called a primary flat these are historical things you know the book is little bit old this is how it used to be these days of course you don't have any of this primary and secondary flat these days the robot come in and there is a small notch the robot knows it doesn't require this type of visual cue it knows which plane is which one with simple marking and it can pick it up and do the processing but for older days this is how it used to be a silicon volume then you will have to cut it and various angles so for example if the surface is one zero zero and if you wanted to know that zero one one which direction is zero one one then you could easily calculate the angle between them so first of all one zero zero that's the Miller index for the top surface that means that's the vector direction of the vector coming out of that surface from the primary flat if you say that that's zero one one another vector so if I multiply take the dot product of these two vectors I should be able to calculate the angle between them and the angle between them is cosine theta do you see how it works one multiplied by zero so I'm taking the pair wise indices the x index the y index and z index multiply them together and what is the answer then the answer for cosine theta is zero you see that one multiplied by zero plus zero multiplied by one and zero multiplied by one so numerator is zero do you see where the denominator comes from so square root of one is coming from one zero zero this is h one squared k one squared and l one squared so that gives you one and for the other vector is square root of two do you see that because there are two ones there on the vector and from that you can calculate the angle 90 degree because cosine theta when it's zero the angle between them is 90 degree so one is perpendicular to the other one now one thing as mentioning was that if you know the Miller indices and the information about this two other vectors for example in this case one vector is coming perpendicular out of the surface of one zero zero that's one direction and the other is perpendicular to the primary cut which is zero one one if you know these two vectors then let's say your professor asks you to make a device along zero to one direction because he believes that it will have interesting transport properties so you can easily solve that problem and this is how you can see either on the left side which is the top view of the wafer or the right side which is the perspective view that I have drawn those two vectors one zero zero perpendicular to one zero zero surface and zero one one shown here in red perpendicular to the primary cut now if you wanted to know very very easy because you could easily calculate the first the angle theta one and the theta one is easily calculated by the dot product of the two vectors and you know the Miller indices so therefore you can calculate the dot product in this particular case the dot product turns out to be zero and you can see why because one of the Miller indices is one zero zero while the other one is zero to one so at the end when you multiply them out you get zero and therefore the angle is 90 degrees so the angle between one zero zero and zero to one is 90 degrees and zero to one will be lying on the surface of one zero zero one zero zero plane what about the second angle then the second angle you will easily calculate with respect to zero to zero one one and zero to one right the second vector you have and that you easily calculate again you get an angle of 18 degrees let's say for this case 18.5 degrees so you know the two angles with respect to the two vectors so you uniquely determine the position of the blue vector which is zero to one now in this case we were lucky that the vector turned out to be in the same plane as zero to one if the angle was let's say 85 degrees or 120 degrees then you would have to saw the wafer in angle to define that 120 degree and then rotate it along and so therefore at the end you will determine that zero to one is in 18 degree with respect to the primary flat and you will create your wafer or your device they are shown here in the blue patch and your advisor will be very happy now let me conclude this lecture two by saying that we have tried to understand the periodicity of the lattice in order to really calculate the total number of atoms total number of atoms per centimeter of the unit volume and if we can use this information about one unit cell you know you talked about face centered body centered and other types of structures from that information and from the information of the basis then we can easily calculate that how many atoms do I have per centimeter cube of the material so you may think we are done we can go home but that shouldn't be good in the beginning actually we are not done the reason is let's say I have just calculated the number of atoms per unit volume from the crystal structure which is shown here in row and from the periodic table let's say for silicon I want to calculate how many free electrons I have I multiply it with number 14 because silicon has 14 electrons so let's say so in that case if I multiply them out then the total number of electrons is not really the number of electrons that is available for conduction if you just multiply with the total number of electrons throughout then you can see that the conductivity you will get for different material has no resemblance to anything that you measure experimentally so what it means that whatever number of electron material has well while all those electrons are the same physically but they do not all participate in the conduction process equally a fraction of them do and the whole purpose of the next lecture and in fact next three lectures would be to find what fraction what fraction of them is actually available for conduction if you know the fraction not the whole only then we have an understanding about the transport property and that requires an understanding of the quantum mechanics and that's the purpose of next three lectures