So, today I this particular lecture will talk a little bit about crystallography, of course this particular thing which I am presenting here normally we do not teach in first year. So, all this particular portion which I am giving is generally a part of the basic solid state physics course which is a part of MSc course not really the first year course. So, till whatever I covered last time you know that was more or less what you used to do in first year with some modifications. So, when we start about the crystal structure the first idea that we introduce is what we call as a Bravais lattice. So, this is something which is inherent to any crystallography thing that we talk is a Bravais lattice. Now, a few statement which I want to make say people always get confused. So, many times we use a language which can further confuse people when we are talking about Bravais lattice it is not a crystal, crystal is something different about which I will talk later. It is just a geometrical arrangement of points, just a geometrical arrangement of points. A Bravais lattice is always in finite, we do not talk about finite let us say of course a crystal can be finite and crystal is always finite that thing is as we said we will discuss later. Now, this arrangement can be one dimensional, two dimensional, three dimensional depending on situation. So, as I said as a Bravais lattice is concerned this is just a set of points. It is a pattern of points which fill the entire space which obeys certain conditions. That is what basically is the idea that any infinite arrangement of points even any infinite regular arrangement of points may not form a Bravais lattice. So, that is what I am going to give some examples what we actually mean by a Bravais lattice and this is something very very important because the entire solid state physics is based on these concepts. In fact, if you start with a Ziemann's book the first statement that Ziemann makes that the study of physics of solids would have been impossible had the ideal structure for every material would not have been a perfectly crystalline lattice. So, we start from a I mean as I said last time also the we always want to start with the simplest problem and unfortunately or fortunately infinite lattice happens to be the simplest problem. So, let us accept that the Bravais lattice is a set of points which is infinite and obeys certain conditions. Now, let us look at what are those conditions it is an infinite array of discrete points with an arrangement that appears exactly identical from whichever point the array is viewed. So, it might be one dimensional, two dimensional, three dimensional basically the idea is that if you are sitting on a particular point and of course you are aware of the environment you know about your x direction, y direction and z direction so you know where you are located. And if I close your eyes and put you to some other point of the Bravais lattice you should not be able to find out whether you are at the original point or you have moved okay. If it means essentially the environment has to be exactly identical each point should be exactly identically placed. So, if the regular arrangement infinite arrangement obeys this particular condition then only we will call it a Bravais lattice. Now, this is for example a two dimensional simple square lattice this is the arrangement of points of course I cannot draw infinite set of points on this particular thing. So, we have to sort of accept that I have drawn only a few of the points and these points keep on continuing like this infinity in this particular direction, infinity in this particular direction, infinity in this particular direction, infinity in this particular direction. This is a purely a square lattice. If for example if I move from this particular point if suppose I am in this particular point and let us suppose this is x direction and this is y direction then I will find that around x direction there is a point and let us suppose this distance is a we find another point at a distance of a. If I move along the y direction again I find a distance of a, another point. If I move along i plus j direction I find at a point at a distance of root to a, all right? Now if I move from this particular point to this particular point exactly identical situation will happen. If you move on the right hand side you will find a point at a, if you move upwards you will find a point at a if you move along this particular direction you will find a point at distance with root to a, so this environment is exactly identical for these points, That is why we call this set of arrangement as a bravillatus. Now there is a 2D centered bravillatus, for example, here in addition to these two points there is also a point at the center, originally it may appear as if this point is differently located from this particular point, because this particular points are at the corner of the square, while this particular point is at the center of the square, but that is not really so, because if you draw another square, this point which was at the center would now appear to be the corner of the square and this point which was originally appeared to be at the corner of the square, now will appear at the center of the square. So my condition is still obeyed that if you are moving from this particular point to this particular point, environment is exactly identical. You cannot say looking on a point whether you are sitting at the center of the square or you are sitting at the corner of the square, it depends on how you want to look at it. You can always call that I am sitting at the center or you can always call that I am sitting at the corner. So this also is an arrangement of points which we can call as a two-dimensional bravillatus. Now this is an example of a non-bravillatus. Here we have two points which are located like this, this is also infinite arrangement. This is also a regular arrangement, symmetrical arrangement, however this does not form a bravillatus. The reason is that if I move from this particular point to this particular point, the environment becomes different. For example, if I am sitting on this particular point and if I move along this particular direction which is let us say minus i minus j, then I will find, let us suppose this small distance is x, I will find a point at a distance of x. But if I am sitting on this particular point, then at a distance of x, I will not find a point. Therefore, this particular point and this particular point are not identically located. Though this point and this point and this point, all these are identically located. These points are also identically located. But when I am looking at the lattice in its entirety, then I do not find that each point is identically located. This point and this point have different environments. Therefore, this is an example where you have infinite lattice, but this is not a bravillatus. This is another example of what we call as a honeycomb structure which is basically hexagons which are fitting with each other. This is also a non-bravillatus because if you are sitting on this particular point, here at a distance of a, for example, let us suppose this is a, you will find a point here. But if you move along this particular direction, you will not find another point at a distance of a. So, this is also a non-bravillatus. In fact, you will find out that as I move from this particular point to this particular point, the environment keeps on getting rotated. Nevertheless, because the environment is different for each particular point, this does not form a bravillatus. So, what is the key thing is that for a bravillatus, all the points must be identically located. I should not be able to tell if I move from one point to any other point, whether I am at the same point or I have moved to a different point. And remember, this has to be infinite. There is an alternate definition which can be shown that they are exactly identical definition. I will call this a more useful definition, more quantitative definition. And this definition I have given for a three-dimension. It can always be accordingly reduced to one-dimension or two-dimension. I will not show that they are equivalent, normally it is not shown, but you can always visualize and see that these two definitions are equivalent. So, this is using vectors. So, let us suppose we have three different vectors, a, b, c, which are not in the same plane. Now, I make a translation vector, which I am calling as t, which is n1 times a plus n2 times b plus n3 times c, where n1, n2, n3 are integers, positive, negative or 0, ok. I keep on varying the value of n1, n2, n3 starting from let us say 0, 0, 0, which gives me leads me to the origin, ok. Then take 1, 0, 0, 2, 0, 0 or whatever it is, take all possible values of that and then you will generate at a Bravais lattice, ok. These three vectors, a, b, c vectors are generally called primitive vectors of a lattice. So, each lattice is different, each Bravais lattice, ok, I mean, let us be clear, when I am saying many times in lattice, probably I will mean a Bravais lattice, only in the beginning I had to give you an example of a non-Bravais lattice, but most of the time when we say lattice, generally we mean Bravais lattice. So, you know, I mean, that is what I say, you know, while speaking many times we use things which are not 100 percent correct, but you know, so long we understand it is all right, ok. So, a 3D Bravais lattice can also be defined as a set of all points, the position vectors of which are given by this particular equation. Here n1, n2, n3 are integers, positive, negative or 0. If we are talking of 2 dimension, then instead of 3 vectors I will have only 2 vectors. If I am talking of 1 dimension, then I will have only 1 vector. The vectors a, b, c are not in the same plane and are called primitive vectors of the lattice. They are used to generate infinite set of points forming the Bravais lattice, all right.  Of course, we have assumed, we have to assume that these lines which have been drawn are not there. Actually, what are there are only points, ok. It is only for our convenience, it is only for our drawing convenience that I have drawn these particular lines, ok. And these particular points which I have shown here are moving infinitely, they have infinite set of points in all the 3 directions or rather all the 6 directions plus x minus x plus y minus y plus z minus z. All right, you can verify yourself that if I use the primitive vectors a is equal to a times i, b is equal to a times j and c is equal to a times k, this will generate this particular lattice, ok. So, you take just linear combination of these vectors like I have given you in the previous example, ok. So, you will be able to generate the entire lattice or for that matter you look at any particular lattice point, ok. You will always be finding that that can be represent the position vector of that can always be represented as a linear combination of these 3 vectors n1a plus n2b plus n3c. So, primitive vectors of a simple cubic lattice is what I have given below. Now, this is what is called a body centered cubic lattice in which I mean I generally always take the example of a room while teaching because it is completely easier to visualize 3 dimensional structures become a little more difficult and see let us assume that we are sitting in a room which is a cubical room. So, all the corners of this particular room are centered and in addition to that there is a point at the center also. Now, looking like what we have said for square lattice this particular point which is at the center can also be looked as a corner of another cube. Therefore, this particular point is identically located. So, it should not appear that this point has a different environment from this particular point because this is at corner and this is at the center because using this particular center and the center of the neighboring cube you can form another cube for which this particular point will be at the corner. Now, what I have given here is let me also tell you there are multiple ways of having primitive vectors there number of ways in which you can define the primitive vectors for the same lattice. Normally we choose a particular convention now when we take for BCC lattice conventionally we take these as the primitive vectors of course you can take slightly different things also you can take a is equal to a i, b is equal to a j and c is equal to a by 2 i plus j plus k. But you know normally we try to take a more symmetrical set. So, this is the way we take the primitive vectors which is a is equal to a by 2 minus i plus j plus k. So, when you are talking of the first vector the first one is negative when you are talking of the second vector the second one is negative when I am talking of third vector it is the third one which is negative. So, if this is your origin this particular corner you are not looking at this particular point you are looking at one particular vector which is joining the center of the neighboring cube. There is another which is joining at the center of the cube which is bottom below. These are generally taken as the primitive vectors of a body centered cubic lattice. Of course, you can always I mean I mean I always give us an exercise to the students that you can show that any point for example, this particular point or this particular point or this particular point can always be taken as a linear combination of these vectors. So, there is absolutely no ambiguity about that. The next lattice that we talk I mean I am generally talk again example of cubical lattices because they are much more useful and are much more simpler also. So, this is a face centered cubic lattice in which you do not have a point at the center of the cube, but you have points at the centers of all the faces.  So, this face there is a one particular point in the center this face there is one particular point at the center this face there is one particular point at the center this particular face there is a point at the center.  So, if you go half the distance here then half the distance is above.  So, you will have point which is this face center. So, the vectors which are joining these things these are the primitive vectors of face centered cubic lattice and again you can see that every point of this particular bravillat is of this particular you know what we call as a unit cell can be represented as a linear combination of these vectors. Now, after defining a bravillat is the next important thing that we discuss is the concept of a basis that is the way always solid state faces look and sometimes you know you know there are some other people from different area have slightly different way of looking it, but physics people always want to define a crystal structure on the basis of a lattice and a basis. Now, basis is something what you actually put on these particular points. Now, there are three ways three reasons why we use to introduce the concept of a basis the first is that like the examples which I have given you there could be many set of points which are regular which are infinite, but they do not form a bravillat is. Now, we can always take more than one particular point and convert a non bravillat is into a bravillat is. So, many times we do that we convert a non bravillat is to a bravillat is by using a basis. So, I start with the bravillat is then at this particular point of this particular lattice I introduce let us say two points or three points or n number of points. So, my basis becomes a set of points and you have to place exactly identically on each point. What is most important is the symmetry and what we call it is the translation symmetry here. So, if I have chosen one particular basis on all the lattice points I have to put the basis exactly in an identical fashion. So, we use it to convert a non bravillat is into bravillat is that is the first use of this particular thing. Second use which we normally do in crystallography especially for cubic lattices that we convert one bravillat is to another bravillat is originally also we have to as a bravillat is, but we convert into a simpler bravillat is by using a basis and that is what is generally done in crystallography when we are talking of BCC lattice generally it is regarded as simple cubic lattice with two point spaces when we are doing FCC generally it is regarded as simple cubic lattice with four point spaces alright. So, this generally as I said is done crystallography. So, many times a bravillat is converted into a simpler bravillat is by using the concept of basis and third the most important thing is that you can put an atom or a set of atoms on each point of the lattice to generate what we call as an ideal infinite crystal an ideal crystal is always infinite ok. A practical crystal for example is never perfect ok it can never be infinite, but as we always say that in physics we always start with the simplest problem. So, we start assuming that my lattice is infinite my crystal is infinite and the effect of surface and effect of the finite mess is taken into consideration by putting appropriate boundary conditions that is what we do later alright or taking this as a perturbation depending on what type of problem you are talking. So, in principle I mean if you look at the Kittes book which is one of the most reputed books on solid state physics that is what he writes a very nice equation that bravillat is plus basis is equal to crystal. So, in order to generate a crystal you have to have a bravillat is you have to define what is bravillat is you have to define what is your basis ok. Once these two are defined then you generate what we call as crystal. So, for example I had taken this as a non bravillat is now what I could do I can imagine this particular set of points as an infinite set of points ok where you ignore the second point ok and say that this is my bravillat is and then I say then on each point of this particular bravillat is I put two points like this on each point. So, I generate another pattern which is also infinite pattern which does not form a bravillat is ok. So, starting from a bravillat is I could generate this particular regular arrangement of atoms ok by putting two points basis remember on all the points I am putting exactly the points exactly in identical fashion that is what is necessary that the points must be put exactly in identical fashion ok. So, this forms now this non bravillat is can be thought of as a bravillat is with two points basis. Similarly, if you look at that honey comb structure which I had discussed earlier then you can see that if I take these two points as one set of point these two points is one set of point these two points is one set of point ok this actually forms what we call as a simple hexagonal lattice. So, remember even in the honey comb structure if I had another point here that would actually have formed lattice only because that particular point was absent this was not forming a bravillat is. But if I take two points basis then you can see that this particular structure gets converted into hexagon with center point and this actually turns out to be bravillat is. So, I can take a two dimensional hexagonal bravillat is and put two points which are oriented like this on each of these particular points then you will be able to generate a honey comb structure which is not a bravillat is. So, this is one of the example where we convert a non bravillat is to a bravillat is by taking different number of points in the basis. Then as I mentioned just now a BCC lattice can be thought of as a simple cubic lattice with two points basis you start with a particular simple cubic lattice assume that at each end point you are having one point here and one point at the center of the room ok. Then you will essentially be able to generate a BCC lattice BCC lattice on its own was a bravillat is ok simple cubic lattice on its own is a bravillat is. But many times especially because I mean I am not talking about unit cell concept because it is much more simpler to talk in terms of drawing a cube. So, it is much simpler to talk of unit vectors i, j, k and a, b, c which are just simple mutually perpendicular to each other. While in these cases you know they may these vectors may not be mutually perpendicular to each other. So, it is much more simpler to talk in terms of simple cubic lattice rather than in talking in terms of body centered cubic or face centered cubic. So, body centered cubic lattice can always be thought that at each point of start of the simple cubic lattice at each point you put two points now one point at the origin another point at the body center which is a by 2 i plus j plus k ok. You will generate a body centered cubic lattice. Similarly, a face centered cubic lattice can be generated from a simple cubic by putting four points in the basis ok.  Keep on doing exactly for all the points you will generate a face centered cubic lattice. So, a face centered cubic lattice can be thought as a simple cubic lattice with four points basis. Now, in order to create a crystal for example, many times we say iron at room temperature is a BCC material ok. I mean this is again a sort of a short way of saying when we say that iron has a BCC structure it means it has a BCC lattice if you take a BCC lattice and on each point you put one atom of iron then you will generate a BCC structure then you will generate the crystal structure of iron ok. So, in order to say that much thing we expect to be to understand to say iron crystallizes and BCC structure. Effectively it means that I start with a BCC bravais lattice and on each point I put one atom of iron then I will get the iron structure. Sodium chloride I mean this is I mean I remember once I had an argument with some of the school teachers when you see sodium chloride is always I actually should have brought that particular thing also is considered as FCC lattice. When if sodium and chlorine would have been identical then this would have been simple cubic lattice, but you know sodium and chlorine are not identical and remember in basis in our bravais lattice that is the first condition that all points have to be identical. Now, if you look only at the sodium points ok forget about chlorine points then you will find that sodium form a FCC lattice or you look only at the chlorine points ok chlorine will form only in FCC lattice. So, sodium chloride actually has a FCC lattice in which you put two atoms basis one of sodium, sodium iron you can say or chlorine iron ok in exactly identical fashion and you will generate a sodium chloride structure. Similarly, cesium chloride has a simple cubic lattice I mean I am not sure whether you are familiar with cesium chloride structure you have all corners are let us suppose cesium then chlorine is at the body center. Now, if cesium and chlorine would have been identical this would have been VCC lattice, but they are not identical ok. So, actually this is a simple cubic lattice with two ions basis one with cesium another with chlorine. So, you put one cesium iron at the origin and one chlorine iron at A by 2 I plus A plus K or the other way it makes no difference. Similarly, diamond which is little more complicated structure is also an FCC lattice with two atoms basis you put two carbon atoms one at the origin and another at a distance of A by 4 not A by 2 A by 4 I plus A plus K. So, actually in diamond if you divide this particular room into 8 smaller cubes. So, you take A by 2 ok. So, this particular room will be divided into 8 cubes smaller cubes side of which will be A by 2 then you will find alternate of these are centered. If this is centered the neighboring will not be centered that will be centered this will not be centered. So, alternate of these cubes will be centered. So, that is what is a diamond structure. Now, next thing that I want to talk is about the planes. So, that is something which is very very important for us to go into the crystallography. When you are talking of the lattice remember we are not talking about the lattice ok. Then you can draw a very large number of planes inside this particular lattice. When I draw planes there are two conditions which has to be set a plane must contain a lattice point ok. I cannot draw a plane just here ok. There is no lattice point. So, whenever I am talking of planes in crystallography this particular plane must contain lattice points it must pass through lattice point. Otherwise this plane I will not regard this as plane. If no point is passing on their particular plane ok let us say between this wall and this wall these are parallel walls there could be infinite number of planes which are parallel to each other. But I will not accept those planes until a lattice point passes through that particular plane. So, whenever I am talking of distance between planes ok that is let us be clear that these planes are those in which there is a lattice point passing. Now, by symmetry which we have argued out because all points have to be identical in a gravel lattice. If a particular plane is passing through a particular lattice point a parallel plane must pass through all the lattice points. A parallel or the same plane either the same plane or a parallel plane must pass through any other lattice point. So, if there is a plane which is like this going like this which is passing through this point ok through this particular point also there is a plane which should pass through that which is parallel to it because all points are exactly identically located. There is no difference between this point and that point that perfectly symmetrically located. So, if there is a plane passing through this point a parallel plane must pass through the all the other lattice points. So, this is the example which I have given all my examples are two dimensional because they are rather easy to draw. For example, there is one set of planes which are red, another set of planes which are blue, third planes which are purple. So, these are various planes which have been drawn through the lattice. And remember see for example, this is one plane, this is another set of planes ok, this is another set of planes ok. They are all differently oriented with respect to each other ok, but they are always passing through a lattice point and a parallel plane must pass through the lattice point. For example, if you are taking here there must be a plane which must pass through this particular point which of course will pass through this particular point. If I am looking at this particular plane through this particular point also a parallel plane must pass. So, through every lattice point a parallel plane will pass. Now, next thing that we talk because it is very important for us to characterize these planes and for that we use what we call as a Miller indices.  So, that is what we call Miller indices. Miller indices are used to identify or characterize a particular set of parallel planes. Remember when I say a one particular plane let us say 1 0 0 plane or 1 1 1 plane it always represents a set of parallel planes ok. So, I am saying 1 0 0 plane it is not a particular plane, but all set of parallel planes will be 1 0 0 ok. So, it is used to characterize a set of parallel planes and this is the way we generate I will give you an example. This is the way we generate a particular Miller indices for a set of planes. First we take the intercept of this particular plane along the three primitive directions. Sometimes I do not take primitive, but I take conventional directions like for example cubic lattice I generally take along the x, y, z directions and not necessarily those primitive vectors which I have given for BCC and FCC which happen to be little more complex ok. So, we take intercepts of one of the planes, we take any plane of the set along a suitable set of axes and express in terms of lattice constants ok. So, we express what is the intercept of this particular thing ok in terms of lattice constant. Lattice constant A, B, C whatever are in general take reciprocal inverse of that numbers then multiply by a suitable constant so that there is no common factor nor there is any fraction. Then the set of planes that we would we put into a small bracket and that is called Miller indices of that particular plane. So, I am just giving one particular example and this a lattice is I think I probably missed that particular transparency. There are in general 14 Bravais lattices which are 7 crystal systems ok these in three dimension. So, this I am not listing that you know we look at any solid structures they will give different type of lattices. A lattice is depending upon the symmetry the more symmetrical the lattice becomes ok we classify that as a separate lattice. So, let us suppose we take some arbitrary lattice and let us suppose this is one plane which I am drawing I am drawing as a triangle, but this is supposed to be entire plane infinite plane. So, this plane cuts this particular let us call this is one axis I am not calling x, y, z axis because these in general for a lattice need not be even mutually perpendicular to each other mutually normal to each other ok. So, this is suppose one of the directions and this plane intercepts here ok and let us suppose this is the lattice constant which in general can be different for different lattice in different directions. So, this actually intercepts if we call this particular intercept a this intercepts at 2 a. So, the intercept here is 2 a because this intercept here this is supposed to be a this is supposed to be a. So, this intercept along a direction is 2 a then we look at the intercept along the b direction which is this particular direction I think I have I have called this as b does not matter ok. I have drawn my figure incorrectly, but let us just let us call this particular direction as b. So, intercept here is 3, but let us take this as b. So, if I take this as b the intercept is 3 times b. So, I write intercept as 3 times b if I take this as c this intercept is 4 times c. So, I take the intercept 4 c then I forget about a b c because I express this in terms of the constants a b c. So, I just take the numbers 2 3 4 and I take inverse of those numbers. So, this becomes 1 upon 2 1 upon 3 1 upon 4 ok. Now, these are fractions. So, I have to multiply by a suitable number. So, that eventually they become integers ok. Now, if you take 3 into 6 if you take 12 if I multiply by 12 ok. So, 12 divided by 2 is become 6 12 divided by 3 is become 4. So, this should be 4 ok and this is 4 3. So, it should be 6 4 3. So, I mean I will make a correction. So, it should be 6 4 3 plane alright. So, this is the way we define better indices. Now, in a simple cubic plane 1 0 0 plane refers to a set of planes which are parallel to y z axis. A 1 1 0 plane in a simple cubic lattice is something like this which divides this particular room into 2. A 1 1 1 plane is something like this ok. Here the intercepts are all a a a in all the 3 directions. Now, many times we have to give indices for directions. We have to talk of specific direction. A electron is moving in this particular direction in a crystal lattice for example, in the crystal and when we want to define the direction the notations are completely very simple. You just express their direction in terms of the a b c vectors which we have already defined. They can be primitive or they can be conventional ok. Then just write in square bracket and that is what is u v w.  Many times we have to refer to a set of parallel planes. For example, if you are talking of cubic lattice whatever is the property along this particular plane you expect the same property along this particular direction because what I am calling is a s x axis. You could have decided to call that as a y axis ok. You could have called this particular axis as z axis. So, this plane as for certain properties are concerned ok may be exactly identically located to this particular plane. Similarly, directions if I am talking of this particular direction let us say or let us say along x axis ok a particular property let us say band structure ok whatever is the band structure along this particular direction has to be identical along this particular direction because they are exactly symmetrically located. So, many times we are not looking at a specific plane or a specific direction, but looking at set of identically placed planes or identically placed directions ok. Then I use a different type of bracket. So, a set of equivalent planes I put into curly brackets. If I say Hkl plane for example, if I write for cubic lattices in curly bracket 1 0 0 plane it will mean 1 0 0 0 1 0 0 0 1. All the three sets of planes combined together can be represented by Hkl in a curly bracket. Similarly, if I am talking of equivalent directions then I put into angular bracket. If I say 1 0 0 directions equivalent put in curly bracket which it will include 1 0 0 it will include 0 1 0 it will include 0 0 1 because as far as the properties are concerned in all the three directions the properties are expected to be identical. So, therefore, we just put them into the angular brackets. So, it depends again on the type of lattice type of problem that you want to describe you may use these rotations to generalize that I am not talking with specifically 1 0 0 plane, but I am talking of 1 0 0 planes and all its equivalent planes which go by symmetry. Now, yeah here I have written that in three dimension you have total 14 gravel lattice types which are classified into seven crystal systems. And I mean of course, you can say the tri-clinic is the most general lattice, but on the other hand in the tri-clinic lattice as it becomes more and more symmetrical we classify into different lattice. In principle there are six lattice constants which are ABC and the angles alpha beta gamma. I will skip those things because that you can see into any standard solid state physics book. Now, let me talk about XA diffraction which is the Bragg's law one of the most famous laws which I think people study well in high school, but I always find that you know people do not really understand it so well. So, that is the reason I always put certain emphasis on this particular thing and take it from this particular point. So, here what we are talking we are dividing the lattice into various planes which I have already said we can divide. Miller analysis of which I can always put it. So, a lattice is divided into a set of various sets of planes. Now, let us suppose XA is incident on a set of planes and its glancing angle is theta. So, let us again be clear that this glancing angle is different from angle of incidence. Angle of incidence is 90 minus theta. The angle of incidence when we calculate in the optics is always from the normal while here in XA rays the glancing angle is always calculated from the plane. So, this is the theta is the glancing angle then I will get Bragg reflection I will get a constructive interference provided two conditions are satisfied for that particular set of planes the angle of incidence must be equal to angle of reflection. So, this must also take off at the same angle from which the which is same as the glancing angle this one condition this is what we call condition of a specular reflection ok which essentially like a mirror like reflection. So, if I am putting a plane like this here a set of planes like this a light beam comes here it gets reflected in this particular direction. Similarly, the X ray comes here if at all it is going to get diffracted it has to get diffracted only in this particular direction. But the difference between normal reflection and the X ray reflection we call it sometime reflection but it actually means diffraction is that if a light beam is incident it will always get reflected provided this mirror is properly coated. But here if the X ray beam is incident here it may or may not get diffracted in this particular direction because it has to satisfy another condition that 2T sin theta is equal to L lambda should also be satisfied. If only both the conditions are satisfied then only I will get X ray beam diffracted in that particular direction. So, if I have just put a set of planes and I put I let the X ray beam incident on that it is not definite that I will get a diffracted beam making an angle theta with this like this ok it may or it may not happen it will happen only if the second condition 2T sin theta is equal to L lambda also get satisfied that is what is very very important both the conditions have to be simultaneously satisfied. Also just satisfaction of 2T sin theta is equal to L lambda is not enough it does not mean that you will get any any any direction if 2T sin theta is equal to L lambda is satisfied. If it is satisfied you will always get beam only in such a direction which essentially involves a normal reflection a speculative reflection. So, these 2 conditions have to be satisfied together. Now this is what I wanted to say this is again as I said you know normally I have found that the students do not understand the Bragg's law so well. If you use a single crystal and a monochromatic beam of X rays ok crystal means a normal crystal many times I have polycrystals you know in which you have multiple small crystallites which are all oriented with respect to each other normally most of the material that you see are polycrystalline. Then in this particular crystal I mean whatever is the lattice type ok let me also mention this particular thing Bragg reflection talks only of the reflection from lattice points it does not talk about the basis the effect of basis always to be accounted separately. If this gives sometimes a wrong picture to the students because they see because I always show the FCC. So, they think that atom is so small tiny here and there is another atom tiny here and you know connected by a bar that is not a correct picture. I mean a much better picture is that this atom is spherical and that another atom is spherical and they touch each other that is a much better picture. But in Bragg reflection every time you talk of only the point so people generally are confused. So, I always clarify Bragg reflection talks only of the condition of diffraction from lattice points what happens because of the basis that has to be accounted separately ok that we do later. This condition is only for the diffractions to take place from the lattice points I assume that diffraction is taking place only from that particular point though atom much may be much bigger there alright. Now as I said if X-ray beam is incident I can draw different set of planes. If I have put my crystal like that and X-ray beam is incident then angle theta on all the set of planes are defined ok. For this set of planes this is a theta for another set there is another theta in fact I have given a picture like this. See if for example X-ray beam is incident like this ok on this set of planes this is my theta on this set of planes this is my theta ok on this set of planes this is my theta. So, you can see that three different theta this set of planes this is my theta for this set of planes this is my theta for this set of planes this is my theta all three theta's are different. So, once I have mounted my crystal and allowed X-ray to be incident then on all different planes theta's are not fixed ok and they are all different alright. Now in case I get a Bragg diffraction then for at least one of the plane 2D sin theta is equal to L m dash to be satisfied then only I will get Bragg deflection. Let us suppose for this set of planes 2D sin theta is equal to L m dash satisfied then the transmitted beam or rather diffracted beam will go like this making same angle as this then it may not satisfy from any other plane may or may not ok. If it satisfy it gets satisfied for for example this set of planes then diffracted beam will go in this particular direction making this angle equal to that angle ok. So, depending upon from which plane it gets satisfied with that set of planes angle of incidence must be equal to angle of reflection theta's are all fixed once I have frozen my single crystals alright. So, this is what I said if everything is fixed chances that 2D sin theta is equal to lambda will get satisfied for any set of planes are very little. So, if you just take a single crystal and put X-ray on that particular thing chances are that you may not get any Bragg deflection because theta's for all the sets are fixed and at least for one set you must satisfy 2D sin theta is equal to lambda ok. These are all discrete values theta is not continuously varying depending upon what are your planes theta is discrete values is acquiring discrete values ok. Lambda is fixed I am using monochromatic beam ok there is no way that I mean there is good chance that you may not get any Bragg deflection. So, in any experiment that we perform generally one of the variables is continuous. So, it is standard what we used to call Lawyer's pattern ok in which we are not using monochromatic source, but we are using a continuous source of X-rays. So, lambda was not fixed. So, at least for one particular lambda Bragg deflection condition may get satisfied because there are multiple lambdas now there is a called lambda is continuously varying or you rotate the crystal you do not keep the crystal in a fixed position, but keep on rotating the crystal ok. Then at one particular angle for some set of planes you will get Bragg deflection condition satisfied. Is this the Bragg's law? Yeah. In that diffraction of the properties from the crystal that how we are understanding in terms of interference there is a some kind of there. No, no. So, basically I mean well it is generally called diffraction and not really interference, but it is essentially the same thing per skill. Let me put it like that you know I have not come to the derivation of 2D sin theta is called lambda. If you take a beam and which is coming only on one of the planes then you can show that the path difference must be equal to the n lambda then only this condition will be satisfied, but then only you will be getting a constructive interference alright. Now, for one set of planes it will is possible only when angle of incidence is equal to angle of reflection if there was only one plane. Now if you take one plane and another plane which is kept at a distance d apart then if this particular beam has to constructively interfere with the beam which is coming from the plane which is below it then 2D sin theta is equal to lambda should also be satisfied ok. So, what we are driving as basically the same interference condition which generally we call as a diffraction condition. They are essentially exactly identical, but this basically depends on how we derive our formula for the Bragg deflection. So, first we take into case of only one particular plane and say that for that to have a constructive interference I must have specular reflection condition satisfied. Then I take another set of parallel planes in that that if the beam coming from this plane and the plane which is below it must constructively interfere then you must have 2D sin theta is equal to n lambda also satisfied. My question is if a student asks how this thing which between particular deflection and diffraction. See the thing is that it is always a Bragg deflection, but no we use it it is a misnomer we call it reflection. I mean if you want to be I am exactly correct it is always x-ray deflection. Because as I say a normal deflection would mean that you will always get reflected light. I mean this is one of the very famous quiz questions which people used to ask what is the difference between a reflection and a Bragg reflection. No actually in crystal is 3D grating. Now in grating you can get. Let me put like that this is the way Bragg looked into there are other ways of looking into it also. But you are right 3D grating was the way Bragg looked into it. Yes and there is always a central maxima. So, central maxima is not due to diffraction. The thing yeah okay. So, how can I then because the problem is the incident beam. No the central maxima when you are talking you are talking about single center reflection. No, no, no. Here you have multiple points that are diffracting. Look the grating is acting as a reflecting grating. Yeah. The crystal is acting as reflection grating. So, in reflection grating also you always get a central maxima and then the diffracted maximas. Yeah. So, here the incident beam and the detector they are placed at the same angle. So, basically we are getting. No, no, see you can also get a second order of reflection because n becomes 2. See, you also get a second order of reflection and n can become 2. See, when you are saying 2 is there is a central maxima and there is another maxima. What is the difference between them? This is the central maxima okay. For that n value has become different. Here also you can have the same time because you can have n is equal to 2 you can have n is equal to 3. But we may have many n equal to 1. No, no, that's one thing which I would also like to mention. You are right, 100% right. The thing is that for example, okay. This is a good question. Let me try to answer it unfortunately. No, no, no. Let me come to that. Let me just explain this particular point a little bit. See, normally in Miller indices there is no common factor. There is nothing like 200 plane. There is nothing like 222 plane. We will always call it 111 plane or 100 plane. Now because of some convenience, which is little difficult. In fact, that becomes more obvious when I will talk about the lowest condition which I want to talk, okay. When you want to look at the parallelism between lowest condition which we feel from the solid state physics point of view is much more fundamental, okay. Rather than the black condition, though every crystallographer use blacks condition, okay. There in our when we draw parallelism what we say a second order reflection from 100, we call as a reflection from 200. No, replace reflection by deflection. Okay, second order defraction from 100 plane will be called defraction from 200 plane. So if you look at X-ray diagram, X-ray crystallization, I mean you will sign reflection or diffraction from 222 plane or 220 plane. Actually, these are actually diffraction from 111 plane or 110 plane. A second order diffraction from 111 plane in crystallography will be referred as a diffraction from 222 plane. A second order diffraction from 110 plane is referred as a diffraction from 222 plane. A third order diffraction from 111 plane will be referred as 330 diffraction. So it is very, very clear. If you look at, I mean in fact, I will give you some of the examples. In fact, I am planning to solve one particular example also. So if you look at that particular thing, things are very clear. You will always find diffraction coming from 220 plane. Now a student can always ask you 220, what you always said that there cannot be a common factor in real indices, okay. 220, there is nothing like 220. There has to be only 110. Now a second order diffraction from 111 0 will be called as 220. Because if you look at the de-expression, everything will turn out to be exactly identical, okay. So that's the reason there is a confusion about this particular thing. Otherwise, there is always a second order diffraction, if present, okay, if the space group does not reject it, okay. There always a possibility of third order diffraction and a third order diffraction. So, I mean, as you are talking, other diffraction, you know, things are present, okay. Of course, indices is different question because indices depend on so many other factors. Because remember here, we are talking only of diffraction from lattice points. Actual diffraction takes place from the atom which is there. And especially if you are talking of x-rays, okay, x-rays actually get diffracted from electrons. And depending upon the electron wave function, electrons could be anywhere in the unit cell, okay. And diffraction could take place from any of them. And principle, all of them have to be accounted before we can understand x-ray diffraction well. So, break diffraction only talks of diffraction from lattice points, all right, which takes care of only translational symmetry. It does not take care of bases. Okay, now there is a modern XRD machine. XRD machine is, I mean, let me put it like this. So, you mount crystal in a particular plane. Now, you always take this particular plane as a reference. You are always looking at the top frame. So, what you do, you have x-ray beam which is incident, which is like this. You rotate this plane by theta. And you rotate detector by 2 theta so that the break diffraction condition, specular reflection condition satisfied only for this particular plane.  And your detector keeps on moving by an angle of 2 theta. So, specular reflection condition always gets satisfied for this particular plane. Now, for some value of theta, whatever are the planes which are parallel to each, for that a break diffraction condition will get satisfied. Then at that particular value of theta, you will get a break diffraction. So, you will get a diffracted intensities. So, generally this is what is called theta 2 theta plot. Now, almost everyone does this thing. Now, we have X-ray machine in our department. We have multiple X-ray machines. So, if some of you are interested, you can always look into that. Okay, so this is what I was trying to tell you. Now, next we talk is the effect of bases. X-rays are generally scattered by electrons within the atom. Now, when we take the effect of bases, there are two things which we account. Okay, but this is a little more complicated calculation which I am not doing. If you want to look at this, any book like Ketel's book, they have derived all these things. Okay, so, for example, I may be putting multiple atom bases on a given lattice point. Okay, so there is something like geometry of these particular points. Okay, how these particular atoms are put on the lattice point? For example, let us say I have set for diamond. Diamond is FCC lattice with two atoms bases. One atom is being put at origin. Another atom is being put at a by 4 i plus j plus k, which is one-fourth the distance of the body diagonal here, which forms the small cubes center. Okay, so this is geometry. I have put one point here, one atom there, one atom here. Another is how if I put one atom here, how the electrons are distributed around the center of the atom? Okay, that depends on the wave functions. That is a much more complex problem because you have to know what are the wave functions, what are the probability of electron having at a particular point and then take diffraction from those particular points. Okay, so there are two factors which we separated out. One factor which we call as atomic four factor, which essentially tells that if I had just one atom, how the x-ray will get diffracted from this different points of the atom if I know the wave function, if I know the probability of electron being found there, for that you have to know quantum energy should be able to solve the wave function. It is a much more difficult problem. Another problem is purely the geometrical problem, how atoms are put around a particular lattice point? What are, how the centers of the atoms are put around a particular lattice point? So, first thing which is atomic four factor takes care of the electron distribution within an atom. So, take the center of the atom, how the electrons are distributed around that and what we call as a structure factor takes care of the geometrical distribution of the atoms within a basis, how the centers of the atoms are distributed geometrically around a lattice point. Now, if you take a cubic lattice, cubic structure and generally express this in terms of a simple cubic lattice, that is what we do. So, we consider a BCC lattice also as a simple cubic lattice with two atoms basis, a FCC lattice as also a simple cubic lattice with four atoms basis. Then the structure factor tells, this is a rather simple calculation, but you know I will take little longer time. So, I am not doing it, I can look into Kittel's book. You will find that if the lattice is simple cubic, all the reflections will be present, all the diffractions from all the planes will be allowed. But if it happens to be a body centered cubic, all those diffractions for which H plus K plus L is odd would be missing, ok. Because you will find that because the geometry of this particular thing, this particular atom here and this particular atom will constructively, I mean they will interfere destructively if H plus K plus L was odd. This is rather simple to show, ok. But basically the thing is that H plus K plus L which are odd will not be present in a BCC structure. So, it means a reflection from 1 0 0 plane because H is 1, K is 0, L is 0. So, H plus K plus L is 1 which is odd, ok, will not be present. A diffraction 1 1 0 would be present because H plus K plus L is 1 plus 1 plus 0 2. So, actually present 1 1 1 will not be present because 1 plus 1 plus 1 becomes odd which is 3 which will not be present. Now, if it is FCC lattice then the formula is somewhat different, ok. It means all individually H, K, L should be either even or all should be odd. 0 is always taken as even. So, if I take 1 0 0 reflection, 1 is odd, 0 is even, 0 is even, this will not be present, ok. 1 1 0, 1 is odd, 0 is even, it will not be present, ok. If you take 2 0 0, 2 is even, 0 is even, 0 is even, which is present. Remember 2 0 0 is second order reflection from 1 0 0. So, second order reflection from 1 1 0, 1 1 0 0 will be present, but original diffraction n is equal to 1 will not be present. For cubic lattice you can show that the distance between the planes is given by this A upon under root H square plus K square plus L square and you put this in 2 d sin theta is equal to n lambda, you find that sin square theta will be proportional to H square plus K square plus L square. I will just give one example how to index this particular photograph for a cubic lattice then we will go about it. So, this is what I have done for certain number of set of planes. So, for example, if you take 1 0 0, n is H square plus K square plus L square that is what I have written as n. So, this H square plus K square plus L square I have written as n. If you take n 1, which means this refers to 1 0 0 plane because 1 square plus 0 square plus 0 square is 1, ok. So, this will be allowed only in simple cubic, ok. If you take 1 1 0, this will be allowed in simple cubic and BCC. If it is FCC, only 1 1 1 will be allowed in simple cubic as well as an FCC. 2 0 0 will be allowed in all three of them. 5 will be allowed only in simple cubic. 7 cannot be represented as sum of 3 square. So, 7 will never be present. So, you can see in fact, the older time you know this is signature of FCC lattice, two very close lines separated by two close lines. In fact, in the olden time people will just look and say this is FCC, yes. So, hexagonal also you have exactly and I am not giving the expression, but I can give you the expression. This de-expression will become somewhat different, all right. And remember normal hexagonal, I mean the most common structure is what is called hexagonal closed packed structure, which is actually not forming a bravillatus, ok. Hexagonal closed packed is actually hexagonal lattice with two atom spaces, ok. There also you will find certain reflections present. But if you look any standard books, I mean any quality book on X-ray diffraction, there are many many standard books on X-ray diffraction. They will all talk about all these sort of particular planes, all right. So, this is just a table which I mean generally useful. Now, this is one particular value of, I mean just an example where I have given the theta values for various diffractions that you have seen through a particular crystal, that you have seen diffractions at these particular values of thetas. Then you take sine square thetas of these things and then you try to match. Then you realize that if it is FCC, then the ratio of the first two lines should be 3 by 4, ok. So, if you look here this ratio turns out to be actually 3 by 4 this particular ratio. So, you can identify that this is actually a FCC lattice. Then you just index that these particular thetas are coming from these set of planes. So, this is just a typical example of how we index the protocol. Now, utility of XRD, it is a powerful technique and we used to obtain many information. In presence of various phases and in fact, this is one of the things I mean I being essentially a condensed matter physicist and more of material scientist, but first thing that our students will do when they will prepare the material do in XRD to see whether you have really formed the material that you are interested in. You expect your material to have a particular crystal structure whether it is having that crystal structure or there is something impurity which has mixed in, some other crystal structure has mixed in. So, this is the most common thing that we do presence of various phases and their relative concentration. We also look at texture and orientation because if you have polygocyan material, all the crystallites may be oriented in a particular fashion, alignment of a single crystal and of course, obtaining lattice constants. These are typical diffraction patterns that you get you know these are the what we call as theta to theta. Remember this is always given as a function of 2 theta because detector always moves at an angle of 2 theta and these are the different peaks that you get and here for example, it is a mixture of two materials. So, you can see that these peaks get split and you can see that this is mixture of I mean this is taken by one of my students zinc ferrite and cobalt ferrite they have small difference in lattice constant and you can clearly see that both the phases are present together. Now, let me define reciprocal lattice because that is what we had discussed about last time. See as we have just now discussed that a lattice can be defined using a set of primitive vectors. So, if I know primitive vectors of a direct lattice, I can define primitive vectors of a reciprocal lattice and as the name suggests reciprocal lattice the dimensions that I am talking as are of inverse of length. But we can always imagine thing I mean I always tell the students see in graph paper we do not always plot things which are only of dimension of position. If you take simple pendulum you plot a graph between t square and whatever it is you take an appropriate scale that you know 1 centimeter will be equal to that much value of t square. Similarly, we can always imagine a lattice which does not have a dimension of length but has a dimension of inverse of length by taking an appropriate scale. So, you can always say that 1 millimeter will be equal to 1 angstrom inverse and then image in this particular lattice in 3 dimension. So, we have ABC vectors of which are the vectors primitive vectors of a direct lattice then I can find out the primitive vectors of reciprocal lattice using these particular things. I would just like to mention that you know this factor of 2 pi is only recent reduction not recent it is quite old almost from the time when I was student. But if you look at very very old books like Azeroth then this 2 pi factor was not there earlier. So, but does not matter because the earlier things that they were deriving had a 2 pi factor but it is like that you know when MKS units started initially there was just a constant then we put started putting 1 upon 4 pi epsilon or not. So, this 4 pi was put so that all other equations become little more easier. So, that is the way it happened. So, this 2 pi factor now is generally believed. So, if you are talking of A you must have B cross C, if you are talking of B you must have C cross A, if you have C you must have A cross B. Of course, denominator is common on A dot B cross C. This is the way you define the reciprocal of A direct lattice. If you take a simple cubic lattice I have already given you what are the primitive vectors of simple cubic lattice you can show that just use those particular definitions you will find that capital A will turn out to be 2 pi by A, I, capital B will turn out to be 2 pi by A, B and capital C will turn out to be 2 pi by A, K. So, this also happens to be simple cubic lattice because all the three dimensions are same ok, but with a lattice constant of 2 pi by A because A is a of dimension of length. So, as I have mentioned to you reciprocal lattice is always of the dimension inverse of length.  So, a reciprocal lattice of a body centered cubic lattice turns out to be face centered cubic lattice of lattice constant 4 pi by A not 2 pi by A. It is simple cubic it is 2 pi by A in face centered cubic in body centered cubic it is 4 pi by A. Similarly, you can show that for a face centered cubic lattice the reciprocal lattice turns out to be a body centered cubic lattice of lattice constant 4 pi by A. Just use those particular definitions there is no other way of doing it. Now, there are certain properties of these particular lattices sorry I am going a little faster which shows that A dot A is equal to 2 pi by 2 pi which is very easy to show because if you go even to this particular definitions if you take A dot A then you have A dot B cross C this will cancel it. So, A dot A will be equal to 2 pi. But on the other hand if you take A dot B you remember B cross C numerator will give you a vector which is both perpendicular to B and C. So, if you take B dot of that particular vector okay the cos theta will be 90 it will be 0. So, these are the typical equations that A dot A is 2 pi B dot B is 2 pi C dot C is 2 pi all other dot products become 0. That is the reason it is called reciprocal because in fact if we did not have 2 pi vector A dot A will just be equal to 1. So, this appears to be almost perpendicular to each other. Now, one thing which is very interesting I am rather really running out of time. So, we can always define a translation vector in a direct lattice which we had just now defined when we are trying to talk of rival lattice T as n 1 A plus n 2 B plus m 3 C. Similarly, we can define a translation vector in reciprocal lattice which is given as G is equal to it is another constant m 1 A plus m 2 B plus m 3 C. Now, you can show that G dot T because of the property that I have just now mentioned if I take A dot A I will give 2 pi if I take B dot B I will get 2 pi if I get C dot C I will get 2 pi A dot B will be 0 A dot C will be 0 B dot A will be 0 B dot C dot C will be 0 ok. If you just take G dot T this will become this particular thing and because all those numbers in the bracket are integers. So, essentially it will be equal to 2 pi times sum integer therefore, E this power I G dot T will be E raise power 2 pi m which is cos 2 2 m pi plus sin 2 m pi which will give you 1. So, E raise power I G dot T turns out to be equal to 1. Now, this is one thing which I wanted to go through because this for solid state physics is much more important. See there is another condition of diffraction which was derived by Lawyer which is a totally different way of looking into it ok. In Bravais lattice I will generally call it in in brass condition it was a top down you had lattice and then you started dividing into planes while Lawyer's condition you start with 2 points then generate lattice. So, it is bottom up. So, you first find out diffraction condition from 2 points ok then replace these 2 points by lattice. So, we consider a lattice consider any 2 arbitrary points of a lattice. Let the incident wave vector be k in the direction n. Let the outgoing wave vector be k prime in the direction of n. I want to find a condition that a beam will really go in k prime direction. What is the condition for that I get a diffracted beam in k prime direction if I have only 2 points ok. So, this is what I have said that let us suppose this is my 2 points which are separated by a position vector d and this is the way my x-ray is coming. So, this is my wave vector k ok. I want to find out what is the condition that my x-ray diffraction goes out in this particular k prime direction. This is my incident beam this is my d these are my k primes I want to find out the condition only from 2 points all right. So, I put these angles. So, let us suppose this angle is theta this angle is phi ok I find what is the path difference that is what we do. So, the path difference this particular point this particular ray travels this distance and this distance extra all right. This distance as you can see is d if I take the magnitude this as d is d cos theta plus d cos phi theta and phi are generally arbitrary ok.  Now, I generate my unit vectors. So, this is my n vectors this is my n prime because n is a unit vector in the direction of k n prime is a unit direction in the direction k prime and so, this particular direction will be minus n prime. So, this path difference can be written as d dot n plus d dot minus n because n is a unit vector.  Now, condition for constructive interference is that this path difference should be equal to m lambda standard diffraction condition. So, this should be equal to m lambda which means that d dot k minus k prime should be equal to 2 pi by m into 2 pi m which implies that e raise power i d dot k minus k prime should be equal to 1. So, diffraction condition from these two points is that if this happens to be equal to 1 then I will get a constructive interference in k prime direction. Now, remember this is being talked only for two arbitrary points of a lattice. Now, if I really get a k prime in k prime direction a diffracted condition diffracted beam then not only from these two points, but you take any two pairs of the lattice this condition must get satisfied. Not only for these two points, but you take any two pairs, any two atoms, any two points, not atoms, any two points this condition must get satisfied. It means this condition should get satisfied if d is replaced by any translation vector of a direct lattice because two points of a lattice is always connected by a translation vector t which is m 1 a plus m 2 b plus m t c. So, in principle this condition should get satisfied if d is replaced by any translation vector t this condition must be satisfied for any general translation vector of the direct lattice. Hence, we must have e raise power i t dot minus dot k minus k prime is equal to 1. And just now we said that this condition is satisfied if we have e raise power i t dot g equal to 1 it always is equal to 1. It means you must have k minus k prime should be equal to g. So, this is what is the called the law is condition that change in the wave vector of the x-rays must be equal to a reciprocal lattice vector. This condition is actually a little more fundamental from solid state physicist. I will tell you the reasons why this is much more fundamental because this generally refers to what we call as the conservation of crystal momentum which is a very very important condition for solid state physicists. So, this is just a precursor of that particular condition. Of course, it can be shown that these two conditions law is condition and the brass condition are equivalent. Here we talk of reciprocal lattice vectors and not lattice planes. In law is condition always talking of lattice planes here the same are being replaced by g. So, a particular diffraction is coming what is the g involved what is the reciprocal lattice vector which is involved here. So, we always talk in terms of g and as I said this is a special form of conservation of crystal momentum. Of course, in this case H cross k actually turns out to be momentum, but last time I had talked a little bit about crystal momentum. There is a general law of conservation of crystal momentum. Remember conservation of crystal momentum may not get completely conserved there is always a possibility for addition of a reciprocal lattice vector. Anyway that is rather fast. I take a corollary of this particular thing. See in all the x-ray diffraction we assume that the wavelength of light does not change after a diffraction. So, in fact this is called elastic scattering there is also possibility that x-ray may get scattered inelastically where part of the energy the photon is used for some other processes. But in x-ray diffraction we are only looking at what we call as elastic scattering that magnitude of k and magnitude of k prime are same initial lambda and then final lambdas are same. So, if you just put this particular equation brass condition in a slightly different form just square these things it means take dot product of this particular thing with its own. Then this condition can be obtained this square this will become k I have taken this and put this on the other side taken k prime on the other side. So, you will become k prime square is equal to k square plus g square minus 2 square root g because the magnitude of this and this is same. So, this I can cancel it out. So, it becomes k dot g is equal to half times g which actually means if I take a unit vector along the direction of a reciprocal lattice this is k dot g is equal to half g. This is another way of looking into what we call as the law is condition. Now, this particular condition can be used to construct one of the very important things which is called bell war zones. So, that is what I will discuss now. So, let us suppose we have a two-dimensional lattice and of course you can imagine this two-dimensional lattice nothing will change. Okay, remember the condition was k dot g is equal to half g. All right. Now, let us take one particular point of this particular lattice as origin. Yes. Suppose if I have a nano material and then in a powdered form if I take the XRD. So, basically through that peak 2 theta by knowing that theta you can calculate capital D which is written as a crystallite size. So, just I want to know sir the basic difference between that crystallite size particle size and a grain size. So, these three are a bit. Okay. Okay, let me first explain this problem before I go. So, the thing is that as far as I am concerned crystallite size is same as grain size. There is something which is called particle size. Okay, which means different. See, you have powdered your sample. Okay, if you have powdered your sample if you take one grain when small part piece of that particular particle. Okay, this may not be a single crystal. It depends again on the size. Okay, the size is very big rather comparatively big. You may find that in this particular small particle also there may be multiple grains. When we talk of generally grains it means different type of single crystals. So, what is a polycrystalline material? A polycrystalline material is actually consist of very large number of crystalline materials and these crystals are rotated randomly with respect to each other. So, if you take this particular material, this particular end may be a small crystal. This may be another small crystal, but this crystal and this crystal are rotated with respect to each other in a random fashion. So, this is what we call as a polycrystal. Now, generally in material science we call this a single crystallite whatever is that we call as a grain. Okay. So, generally it is confused because we are talking about a powder sample because powder also we call it a grain. But this particular powder in one small piece of powder may still consist of multiple grains because within this there may be multiple single crystals which are oriented with respect to each other. Okay. So, this is what we call as grain. Now, in X-ray we do use it to find out the grain size because remember, okay, this is by a formula which is generally called a Scherer formula, almost all of our students use it. Okay. The thing is that if we have really an infinite crystal, if we really have infinite crystal in that particular case you will show, you can show that other than the natural line width of your X-rays, there will be no other width of the X-rays. But for example, if I am talking only two planes. Now, if I am talking only two planes you will not get completely destructive interference for all values of theta. So, you find that as the grain size keeps on going down, it means your crystal is no longer that infinite, you will find that the width that you are getting becomes broadened. So, this width which I have been showing here, if you are looking at this particular width, okay, you see there always finite width, there is always a finite width, okay. It is not very sharp. If you take a single, if you take a material with a large grain or a single crystal material, you will find this extremely sharp, okay. Now, Scherer has given a formula which correlates this width to the, what we call as the grain size. So, what people do, take this particular width and from that they calculate what is approximately the grain size, which gives you only an approximate value. Let us be very, very clear. In fact, we normally ask the students to do this particular experiment and say that, you know, you verify from the other method. There are certain other methods like, you know, electron microscopes from which you can get much clearer idea of the grain sizes, okay. The grain size is equal to the grain size. Yeah, yeah.  But when you are saying the particle size, that is generally different. See, if you are taking a powder sample and you are taking a particle, that particle size may be different. And also, it is valid for 100 nanometer. Big partner? 100 nanometer thickness. Generally, Scherer formula is not applicable beyond 100 nanometers. So, if your thickness, grain sizes are less than 100 nanometers, then only most of the time your Scherer formula will work. And I mean, generally, I say, I never like to bet on the grain size evaluated from Scherer formula. If you are looking at a trend, I will believe it. If you are measuring, preparing samples by different techniques, and then you are finding out that Scherer formula is giving you different values, okay. The trend, I will agree. But if somebody calculates and says that I have found 25 nanometers and if you want me to bet that it is only 25 and not 26, I will not agree. Because, you know, it is giving you a very, very average behavior of the grain size, okay. You are not really watching the grain size. I mean, the best method to calculate the grain size is you do a dark field imaging in the transmission electron microscope. That is the one which gives you the most correct. But it has also its own limitation. To make a sample for transmission electron microscope is the hell of a job. When you ask our PhD students, you know, they essentially cry when you have to make some samples for a while. Because the electron beam has to get transmitted. So, the material has to be so thin, especially if you are talking of thin films, where there is substrate behind it, okay. Then also, you are watching a very, very small area. Now, if your film or material is non-uniform, okay, just counting the some 30 or 40 grains may not actual representation of the grain size. So, each method has its own limitation. And then one has to adopt a proper method in the search environment, okay. So, we said, let us suppose we take one particular point in origin and let us take another and let us suppose this is a reciprocal lattice. Let me put it like that. See, let us suppose this is a reciprocal lattice, not the direct lattice. Remember, wave vector also has a dimension of 2 pi by lambda. So, this is also inverse of length. So, wave vector can be drawn without any other approximation in this particular reciprocal lattice space. So, let us suppose we draw a wave vector starting from this particular point, okay. What I insist? Let us suppose this is another reciprocal lattice pi vector, okay. So, if I join this particular point to this particular point, this is one of the G's, okay. Now, let us draw a plane which is bisecting this line, this vector G to be G is normal. So, this is what I am drawing, alright. Now, what I am insisting that if a wave vector ends anywhere on this particular point, then K dot G, G is this. This is one of the, at the moment I am looking at only one point of the lattice, okay. I will keep on looking at other points. So, I am looking, this is my origin, this is another point of the reciprocal lattice. This is one of the valid G. I took one particular plane in three-dimension, a line in two-dimension, which divides this particular G into half. If my K starts from here to here, then K dot G will be just this particular distance. If this happens to be just ending on this particular line, then K dot G will be equal to half G. It means if I draw a wave vector from here to any other point, this particular plane, if it ends, if the tip just ends at that particular point, then this particular K will satisfy a condition of Bragg diffraction, okay. So, this is my K and this is my G. If this just ends at the tip, K dot G, remember unit vector will be this particular distance. This will be just equal to half G. This, what is that particular condition? Then I will get Bragg diffraction. It means if I draw this particular plane in three-dimension and start my K wave vector from origin, if the tip of this particular vector ends any point on this particular plane, then this particular value of this particular K, this particular x-ray will get Bragg diffraction. In which direction? You can always find out from K minus K prime equal to G. You just take the vector difference, you will find out what is the K prime direction, all right. Now, I have taken only one of the directions. Now, I must take all the other points also, okay. So, now I move, this was only from one point. Let us suppose I go to this particular point. If I have to go to this particular point, then I must join this particular point and draw a perpendicular bisector or a normal bisector. So, this is what I have to do. Now, if a wave vector ends here, then this will suffer a Bragg diffraction from this particular G. If it ends here, it will suffer two Bragg diffraction, one from this particular G and another from this particular G, okay. Now, I keep on drawing. Now, I have taken this as nearest neighbor. I keep on further drawing. I have taken this particular nearest neighbor. I have taken, I mean, just in two dimensions, I have taken only nearest neighbor in two dimensions. Then I must talk of next nearest neighbor. So, I must look at these particular points, this particular point. This is the next nearest neighbor. So, you draw this particular line. Then you draw this particular line, okay. You draw this particular line. You draw this particular line. Keep on going to next to next nearest neighbor. You draw this particular point. You have gone to point which is much above. Now, if you keep on drawing, eventually you find the whole, you have a mesh of lines, you know, here, there, here, there. And if you start a value of K, okay, from the origin, if tip of this particular K ends on any of these particular lines, okay, then it will suffer a Bragg diffraction. Now, in order to simplify these issues, this particular complicated figure, which happens in two dimensions, fairly complicated, three dimensions becomes more complicated, okay, we divide into certain zones which are called Brellois zones. So, this is what is the definition of a Brellois zone. A volume in a reciprocal lattice space is called nth Brellois zone. If all the points within it, except, of course, boundaries, if you join to the origin cross exactly n minus 1 planes, no fewer, no longer. If the tip of the incident wave vector drawn from origin lands exactly on the surface of any Brellois zone, it will suffer a diffraction, all right. So, this was the picture which I have drawn. Now, I will draw first Brellois zone. So, you take these particular points, if you cross, if you join any point to the origin, this will not cross any of the planes, any of these lines. So, this is my first Brellois zone, all right. Now, if I talk of second Brellois zone, these will be these portions, when you draw to the origin, this will just cross one of the lines. This is my second Brellois zone, all right, because take any particular point you join to the origin, it just crosses line. If you take this particular point, it just crosses this only one line. If you take this particular point, it just crosses one line. This is my second Brellois zone. This is my third Brellois zone. If you take this particular point, joins to this particular point, it will cross this line, this line.  Importance of Brellois zone, this actually concept of Brellois zone is one of the most important fundamental concept of solid state. Concept of reciprocal lattice and concept of Brellois zone, these are what we call as the bread and butter of solid state physicists. That's the first thing that people have to learn. Generally, we assign a wave vector for excitations like phonons, etc. In fact, that's one of the most important uses of Brellois zone. In fact, you can show, and I had talked about it yesterday, that wave vector, there's not a single unique way of representing the wave vector of electron, of a block electron, okay. In fact, the block, the wave vector that you assign, you can always add to it a reciprocal lattice wave vector and that will be as good as a wave vector as the original one. In fact, we get into one condition that almost all the excitations, their wave vector can always be represented only in the first Brellois zone. So, that's why Brellois zone becomes very, very important because all the k values which are confined into one Brellois zone, they are generally are the most important. Even the band structures, they are just expressed in just one first Brellois zone. So, I think I have just ended. I'll just like like to just mention you before we leave that this is one model which I have brought which has got for a Brellois zone for a FCC lattice. So, remember a reciprocal of FCC lattice is a BCC lattice. So, if you look at these particular points, okay, there is a point at the center which you can probably see inside it, okay. Now, you take this particular point at the center, you go to the nearest neighbor which is this nearest neighbor, body centered cubic, these two are the nearest atoms. So, you draw and this is a plane which bisects it. So, for example, this particular plane which bisects it. Take at the a particular point, it has eight nearest neighbors in body centered cubic. Remember, this is FCC lattice, but the reciprocal is BCC. And I have to draw this Brellois zone in the reciprocal lattice space, not in the direct lattice space, okay. So, this is for FCC, reciprocal of which is BCC, okay. Now, you have eight corners, eight these things. So, you will have eight of these corners. So, this will be octahedral, okay. There is small mistake here in this particular thing for this side should be equal to this side. This has not been drawn very correctly, but let us just accept it, okay.  Second nearest neighbor is distance A here, okay. So, you draw a plane like this, okay, which bisects it. So, this gets truncated. So, this is called truncated octahedral. This is actually the first Brellois zone of a FCC lattice, okay. Now, you can see the shape is fairly complex. If you have to express your band structure for FCC solid, okay, if you express, let us say, along one zero zero direction, you have to go along this particular direction. Then the time when you are going to hit Brellois zone boundary is going to be different. If you are going along one one one direction, this is the way you are going to hit the zone boundary. So, it basically depends on which direction you are using, but important fact is that essentially is the first Brellois zone, all the k values which are confined into first Brellois zone, they are generally of the physical significance. All the k values can be translated back to the first Brellois zone. I think I will end here. I took a lot of your time. Thank you.


Hope everybody and welcome back. Let me show you guys inside my full dental L tutoring course. So if you are taking the dental hygiene board exam, either in Canada or in the States, I have a course for you. So I've been teaching now for 17 years and my tutoring program is all online. So I can and will help you pass the exam because unfortunately with the exam, the main thing is they're going to make things very hard for you. You're going to need to know which answer is the most correct. And two of them are always going to be correct. So I teach you guys that all throughout my tutoring course. So this is the main website here. If you scroll down, you will see the dental hygiene board exam prep Academy for Canada, but there's also the dental hygiene board exam prep Academy for the States also. So you just pick which one if you're taking the exam in in Canada or if you're taking the exam in the States, you would click on either one of those. Let's just say you're taking the exam in Canada. So you want to click on this one here and then it will sort of it will explain things for you. So pretty much I have everything that you need to pass the exam. There are mock exams, there are case studies, all kinds of stuff. So you will see all of that here for you. And I'm going to actually show you guys inside so you have an idea of what that looks like. So just switching my screen here. So this is what it looks like on the inside. So if you guys look sort of in the middle here and then you will also see on the left hand side, but look at all of the different modules. So you will see modules depending on the topic that you want to learn. And there is also a private group on Facebook for members only. So we are constantly keeping in touch and then you can also keep in touch with other students as well. So there's a lot of things here for you. But then let's say, actually, let's just click on something. Let's say you want to learn about, let's just pick one, eruption patterns. So let's just click on that. You, it's then going to ask you to download the PDF file. So once that's done, I'm just going to click on that to show you guys. So this keeps everything very, very simple for you. If you were to read the textbook, it would be very confusing and very stressful. So I have kind of summed it all up for you. And everything here is what you have to know for the board exam. Again, nothing more and nothing less. So let me just exit out of this. But let's say you don't want to do any like PowerPoints, none of that. You just want to listen to me teach you a topic. I do have that for you too. So you would click over here and just for fun, let's click on dental hygiene. And it's going to show you all of our session recordings for the past two to three years. The new ones are on the bottom, but I'm going to actually change that. So the new ones are at the top. So you can pick and choose whichever one you want to learn. So let's say you were going through medical emergencies and you didn't want to have to take the time to actually read something, but you want to hear me teach it to you. So then you would just click on that. It takes a moment. And then it should sort of show up for you.  But this one, for example, is 41 minutes. What I'm actually teaching you, medical emergencies. And this one happens to be medical emergencies and infection control. So I'm just going to speed it up for you. So you guys have an idea. So this is where we are actually going through mock exam questions. So you also don't have to think, well, what if I can't attend live? That is okay. The session recordings are usually posted by the next morning. So you can watch them anytime. And as you guys can see over to the right-hand side here, there's many more. So if you're after 41 minutes, you have done this one. Well, next you want to move over to, let's just say implants and perio. Then you would just click that. And then this will load up for you. This one looks like it's hour of, it looks like a lecture and questions as well. So I do everything here for you.   So let me know you guys if you have any questions and I will see you in the next one.


So why is it important that more women and people from a diversity of backgrounds are included in and contribute to Wikipedia? Well, those who contribute help to shape its content that we now all rely on. So it's important that it represents all of us. Let me explain. Wikipedia is open source, so its content is used in lots of different ways across the web, like in the info box that you see on the right when you do a Google search. You can see it's drawing that summary information straight from Wikipedia. And if you are Siri or Alexa, a question for example, there's a good chance that some of that information is also being drawn straight from Wikipedia. So it's important that the content is accurate and of a high quality, but also that it reflects the diversity of our society. But unfortunately, this is not always the case. Currently, only roughly one in 10 Wikipedia editors identifies female. The majority are male, white, straight and from the global north and mostly English speaking as well. This affects the kind of content that's being written on Wikipedia. It creates a kind of bias that then permeates out across the internet. And an example of that is that only 20% of all biographies on English Wikipedia are actually about women. And that's even worse for gender diverse folks or people of underrepresented backgrounds as well. Bias shows in the language that we use too. For example, the woman is often described as a socialite or a wife, as opposed to an entrepreneur or a business person, which is the language that we might find use for a man on Wikipedia. So it's critically important that we broaden who is contributing to Wikipedia and we don't just need them editing content about women, we need them editing all kinds of content because the more diversity of editors there are on Wikipedia, the better and more equitable the content will be.


Hi there, I'm Sandy Alnach and let me start out by saying welcome and I hope you brought your sketchbook. I call this little journey of mine an art venture. I've been on this path since I was just a little girl and with an art degree and decades as a graphic designer and illustrator before going out on my own to start online art classes. Here on YouTube I love to swap ideas to talk about art, try new things because all of us are in different stages but we share one great commonality and that's that we're all growing wherever we're at. I start and end each week on YouTube on a theme or with a medium an idea just something I feel like exploring and then I create more of the same the rest of the week. Five days gives me time to mull over ideas and sleep on them play in my sketchbooks and all of this at a slower pace that's more conducive to learning. I would love for you to join me by subscribing here on YouTube. Tapping that bell so you don't miss anything and joining my art venture community linked in the doobly-doo down below. I've included a few other links so you can get to know me a little better and maybe you can leave a comment on this video with a little bit about you and your journey so that I and others can get to know you too. Thank you so much for stopping by and go out and create something every day.