 Now, let us consider another condition which is called Lawyer's condition of diffraction. For x-ray crystallographer, everyone uses the Bragg's condition. Bragg's condition is taught even in high school. This is considered as a very, very important condition. But for solid state physics, Lawyer's condition turns out to be slightly more interesting and slightly more important. So, let us discuss Lawyer's condition. Of course, both are equivalent. It can be shown that in the case of Lawyer's condition can be derived from Bragg's law and vice versa. So, there is no difference. It is only the way we are going to put it. I generally call that Lawyer's condition is, you know, bottom sub-approach while Bragg's condition is a top-down approach. Because what in Bragg's condition you do? You first consider the lattice and then you divide into the planes and then you start studying it. Here in Lawyer's condition, we first consider two points and then you replace these two points eventually by the lattice. So, we start from the bottom just considering the diffraction condition from two points and then slowly decide that these two points will be actually a common transition vector of a lattice. So, it is a different approach. The final result that you get is a very, very interesting result which is very useful result for solid state physics and this can be shown to be exactly equivalent to the Bragg's condition. So, I have said consider a lattice, but you need not consider a lattice. You can consider any two arbitrary points. So, but this is the way normally we say that consider a lattice, consider any two arbitrary points of lattice. So, I start by two arbitrary points on a lattice. Now, these are only the definition. Whatever is the direction of incident x-ray beam, I call this as k. So, k consists, its magnitude is 2 pi by lambda and it is in the same direction as the incident x-ray. So, this is represented in the wave vector of the x-ray is considered as x and a unit vector in this particular direction, remember unit vector does not have a dimension. A unit vector in this particular direction k is written as n hat or n cap. Now, similarly if we are getting a diffracted beam in a particular direction that particular thing is going to be written in k prime as vector k prime. Remember, because in x-ray diffraction we always talk of what we call as elastic scattering of x-rays. So, we assume that lambdas are not changing the energy of the photon or wavelength of the x-ray does not change. So, the magnitude of k and k prime is same only its direction would become different. The x-ray beam would now move go in a different direction and let that particular direction be specified by a unit vector n prime. So, this is only a definition. So, I consider only 2 points here, just 2 points and let us suppose this is my incident beam coming which is parallel beam coming here and my d it is not very clear here, the d is pointing this way, this is the d. So, this is the arrow of the d, this is the end of the arrow. So, my vector is going this way, t vector was going this way. So, the thing is that if this is these are the 2 vectors, 2 points which are separated by a vector d. Let us suppose an x-ray is incident, the wave vector of which is given as k, find out what is the condition that a Bragg reflected beam comes or a diffracted beam comes with k prime in this particular direction. Remember again magnitude of k is same as magnitude of k prime. So, I am just looking for the condition that what happens if I get a diffracted beam in k prime direction when these 2 points are just separated by a vector t. So, this is the condition that I am going to write. You have to be careful in this particular thing because we are mixing 2 ideas, d is in the direct lattice space and k is in the reciprocal lattice space. But as I said you can always draw something by taking an appropriate unit. So, these dimensions I am not particularly looking at, it is basically looking specifically specifying the directions. Let us just define certain angles. So, let this particular angle be phi and let this particular angle be theta. So, let us not confuse between the Bragg's angle. These are just 2 angles at this particular moment. Let us not consider them in any different fashion. So, let us suppose this is the angle theta which is this is the vector d, this is the vector k, this is the vector k prime, this is the vector k prime and this is the angle phi, this theta and phi need not be same because I am looking for an arbitrary direction k prime and find out the condition that you will get actually a constructively diffracted beam in this particular direction. So, I have just taken theta and phi. Let us try to find out the path difference between these two. If I try to find the path difference, what I will do from this particular point, I will drop a perpendicular here, I will drop a perpendicular here. So, this particular beam which comes here travels an additional distance of this and this particular distance. So, this particular beam travels a larger distance and that path difference is going to be given by this expression. And as you know that if this particular path difference turns out to be equal to a constant, an integer times lambda then only I will get a constructively diffracted difference. So, let us calculate this path difference because this is theta, this is 90 degree. So, this particular distance which is the additional distance is d which is the magnitude of d into cos theta. While this particular thing because this is phi, this is equal to d, this angle is 90 degree. So, this is equal to d cos phi. So, total path difference will be d cos theta plus d cos phi, I hope this is clear. Remember, I mean we similarly calculated path difference when we are also calculating Bragg's law, but then we are calculating the path difference between two places. Here I am just taking two points. So, I am starting from two points. Now, let us try to express this in terms of unit vectors. So, as I said n is the unit vector in the same direction as k and n prime is a unit vector in the same direction as k prime. So, this should be the direction of n prime. So, this should be the direction of minus n prime because n prime is in the same direction as k prime. So, this is minus n prime. So, this d cos theta this distance you can see can be written as d dot n because d dot n will be d multiplied by n multiplied by cos theta. So, this is the direction of d, this was the direction of d, this is the direction of n. So, angle between d and n is this theta which is same as this theta. So, magnitude of n is of course unity. So, d dot n will be actually equal to d cos theta. Similarly, if I write this particular thing d cos phi this particular term actually the angle phi is angle between n prime and d because this is the d vector and this is n prime vector. So, this is the angle which is phi. So, therefore, I should take d dot minus n prime which is what I have written here. So, this will be the path difference. So, let us use this particular path difference. So, overall path difference will be d dot n minus n prime and it must be equal to n d j times lambda. I have written m here because I have used n here. So, makes no difference. Now, what I do I multiply this particular thing is by 2 pi by lambda. If I multiply right hand side by 2 pi by lambda lambda will go away and this will become 2 pi multiplied by m. And if I multiply this by 2 pi by lambda this will have the magnitude of the wave vector which remember lambda is same for incoming x-ray as well as outgoing x-ray. So, you will have 2 pi by lambda which will give you k. This will give you k prime. So, this becomes the d dot k minus k prime is equal to 2 pi m which essentially means that if I take an exponential of these two factors then e raise power i d dot k minus k prime exponential of this factor must be equal to exponential of 2 pi m. We have discussed earlier in the lecture today that e raise power 2 pi m is actually equal to 1. So, this particular factor will actually turn out to be equal to 1. It means Bragg diffraction condition or rather Lois condition says that diffraction from two points will be constructive and you will get an emerging beam or diffracted beam in k prime direction provided e raise power i d dot k minus k prime is equal to 1. Now, at this point we said that I have taken only two points now consider the entire lattice. Now, lattice consist of a large number of points rather infinite number of points and all these things points are separated you can draw similar type of pairs by taking any two arbitrary points of a lattice. So, let these two points be any two arbitrary points of the lattice. In case you are going to get Bragg diffraction in this particular direction k prime then with respect to all these pairs not only these I have just considered one pair. Consider all these pairs for all these pairs I must get this condition equal to 1. It means instead of this particular d if I replace it by any translation vector of direct lattice. So, remember two points any two points of a direct lattice are separated by a translation vector t this is what we have started yesterday. Now, take any two points difference between them will be represented by a translation vector t. So, if this d is replaced by any translation vector t and still this condition is satisfied then only you will get a Bragg diffraction from the entire lattice. It means this condition must get satisfied even when d is replaced by t where t is any direct translation vector of direct lattice. This condition must be satisfied for any general translation vector of the direct lattice. Hence, we must have e raise power i t dot k minus k prime equal to 1. Now, if you remember when we talked about the properties reciprocal lattice we said that it so happens that e raise power i g dot t turns out to be equal to 1. So, if I want this particular condition to be satisfied then all that you want that k minus k prime should be equal to any translation vector of reciprocal lattice that is what is important. In fact, in the Bragg's condition in the Lois condition you talk in terms of reciprocal lattice vectors while in the Bragg's language you talk in terms of planes that is what links these two Lois condition and the Bragg condition. So, you must have k minus k prime must be equal to gg. What it says that if the wave vector difference between the incident and the diffracted beam turns out to be equal to any reciprocal lattice any translation vectors in the reciprocal lattice in that particular case you will find out that Bragg reflection condition or the diffraction condition for constructive diffraction will get satisfied. As I said Lois condition can be shown to be equivalent to Bragg's condition both can be derived from each other. Important thing is that here we are talking of reciprocal lattice vectors not lattice planes. So, what was being talked in terms of lattice planes in the Bragg condition is now being talked in terms of the reciprocal lattice vectors. This one thing which I just will casually mention because we are not really having a course on solid state physics. See if you remember when we are talking about the quantum mechanics course at that time we said that h cross k represents a momentum of a photon of course it also represents the momentum of a free particle. Now, this condition can also be written as the condition for the momentum. So, what happens if you multiply this by h cross k? So, it becomes change in the momentum of the photon. Momentum of the photon will change by the amount h cross k minus h cross k prime. So, what essentially says that the change of momentum of the photon must be equal to h cross g where g is a reciprocal lattice vector. Of course, in this particular case these moment are real momentum and you will find out that this particular momentum because momentum has to be conserved. So, whatever this momentum you know where whatever your crystal is mount that will experience this particular recoil of course this is very small. So, and this particular crystal is tightly bound or rigidly bound to the machine you will never see this particular recoil but if you are talking from the physics point of E of course this system will get recoil. What I want to say that this is actually precursor to a what we call as a general conservation of crystal momentum equation. In solid state physics you get many type of elementary excitation. For example, excitation like phonons, excitation like magnons. Each one of them can be represented by wave vector for that matter even electrons inside a lattice what we call as a block electron. I am not sure probably maybe Professor Suresh will talk a little bit about when he talks about solid state physics. So, all these things are represented by wave vector and many times h cross k does not represent their actual physical momentum. We define a quantity which is called crystal momentum which is called h cross k irrespective of what type of excitation that we are talking because that excitation will always have a wave vector. And what essentially is a generalized equation of conservation of what we call as crystal momentum. In fact in solid state physics what we use is the conservation of crystal momentum which says that summation of all incident crystal momentum and summation of all final crystal momentum they can differ only by h cross g. g could be zero also which essentially means that in that particular case the crystal momentum is exactly conserved. But this is anyway just a sort of precursor to that thing. Again if you are not interested in solid state physics or you know you need not bother about this particular thing but it is important to specify to the student that why solid state physicists talk about Lawyer's condition and take this Lawyer's condition as very very rigorous and important condition because this turns out to be a precursor of a much more fundamental conservation law which happens within the crystals. Now there is a corollary to this particular thing which let me just try to take just expressing this particular term into a slightly different form. Lawyer's condition we have to explicitly assume that the magnitude of k and magnitude of k prime is same but it has not been used in the mathematical form. I mean it has been used in its derivation but in the mathematical form it is not very clear. So there is a different way of putting this particular Lawyer's condition where we take into care the fact that k prime and k their magnitude is same. So this is expressed only in terms of k and we sort of remove k prime okay eliminate k prime. So all that I do I write k prime is equal to k plus g here and then square it. Square in the terms of vector means that you take a dot product of the vector with its own self. So if you take the dot product you will get k prime square here you will get k square because g is on the other side you will get g square minus 2k dot g because there is a k prime I am writing here. So this one is k minus g so you will get minus 2k dot g. Now the magnitude of k prime and k prime k square has to be same because we know that the wavelengths do not change and a standard diagonal friction so this gets cancelled out. So you write this equation as k dot g okay and take 2 on this particular factor is half g all right. Now take this g as a magnitude of this thing so if you divide by the magnitude of this particular g this you can write as k dot g where g is a unit vector in the direction of g. So in other words the lowest condition can be written in the fact that if you have a wave vector k then if k dot g turns out to be equal to half g then the condition of Bragg diffraction would satisfy it. Now this particular equation can be represented in terms of some sort of a geometrical way of expressing these things which we call as Brillouin's zone. As I said Brillouin's zone after reciprocal lattice is the very, very important concept in solid state physics of course I will not be able to justify it by this particular thing I will only mention why this particular thing is essential okay. So let us see how do we use this particular thing. Now remember the difference in this lowest condition is that we are always talking reciprocal lattice space you know k has also dimension of inverse of length g also has a dimension of inverse length. See k has a dimension of 2 pi by lambda which is the dimension of inverse of length. So all these things all these figures that we are going to make is now in reciprocal lattice space where everything is of the dimension of inverse of length. So let us suppose this is my reciprocal lattice you know now this is not the lyrical lattice but this is the reciprocal lattice okay. Let us take one particular point as origin I think I have taken this particular point as origin and try to geometrically express this particular condition of Bragg reflection or rather lowest condition in a pictorial form. So let us take at this particular origin. So let us take this and take one of its nearest neighbor. So this is my origin so I have taken this as the nearest neighbor. So what I do I draw in principle because two dimensions so I draw a line which bisects the line joining these two perpendicularly bisects. I take I join these two particular point okay. Then draw a line which is perpendicular to this line alright and bisects I mean this is equal to this. In three dimension I will draw a plane which bisects this particular thing normally. It means this particular plane is this particular direction is normal to this particular plane. If you can assume third dimension looking on this side and behind the paper and front of the paper going ahead then this will be a plane and this particular plane will be this particular direction will be normal to this particular plane in such a way that this distance and this distance is same. So let us suppose I draw a plane like this okay or how does it help? Let us see. Now let us start from this particular origin point. Let us write a wave vector and let us suppose the wave vector is such that this wave vector if I draw from this particular point remember this has the same dimension. So G also has a reciprocal lattice also has a dimension of inverse of length and K also has a dimension of inverse of length. So I can draw this particular thing exactly without any further convention only convention which I have to use or which only units which I have to take or scale which I have to take is when I am drawing the reciprocal lattice. Once I have chosen that scale using the same scale I can draw K also on that particular thing. So I draw K in this particular direction. If it so happens that K the tip of the K ends just on anywhere on this particular plane I insist that condition will get satisfied condition which I have written earlier which is this condition K dot G is equal to half G. So because this particular has G so this the G has a unit vector G has in this particular direction. So if I take K dot G this particular thing is theta different angle between K and G. So K dot G will be equal to just this which will turn out to be equal to half G. So if it so happens that using this origin I draw the incident wave vector K starting from this particular point then if it so happens that this vector terminates exactly anywhere on this particular plane then K dot G will be just equal to this much and this will automatically turn out to be equal to half G and then Bragg condition will get satisfied. But this Bragg condition gets satisfied or rather lowest condition I am sort of repeatedly sometimes these are equivalent. Now this is satisfied only for this G but then I must take all other values of G to realize whether this condition is going to be satisfied or not. It means I need not take only this particular thing I will take another nearest neighbor other nearest neighbor then I should go to the next nearest neighbors and keep on doing it forever and if it so happens that if I take a value of K if I know my incident X-ray beam I know its wavelength I know its direction with respect to the crystal that I have mounted it alright then in that particular case in principle I will be any if it so happens that this K terminates exactly in any of these planes let us call them Bragg planes if it terminates exactly in any of these Bragg planes then only you will get a constructive X-ray diffraction otherwise you will not get it. So let us take the second line now for this now I have taken this as my second nearest neighbor I draw this particular line then take this there is a third nearest neighbor which has exactly the same distance I draw this line here then I take this one and draw another line I draw another line here by taking this now it has four nearest neighbors I am talking of only two-dimensional here three-dimensional you have other nearest neighbors also which you can draw a similar type of thing it does not matter. Let us look for two-dimension where things are easy these things can be extended for three-dimensions easily alright now let us start second nearest neighbor second nearest neighbor will be this because this is a sort of a square structure if I take perpendicular bisector it will go like this then I will take the second nearest neighbors I take the second nearest neighbor this I draw a line like this actually this should pass exactly through this particular point then I take another nearest neighbor which is here next nearest neighbor then I take this as nearest neighbor I will get this particular thing then I will get another nearest neighbor which is this thing you get this particular line then I take second nearest neighbor second nearest neighbor is this thing third nearest neighbor rather okay this will be this so I must draw a line like this this is perpendicular bisector to this particular this particular g I have drawn multiple of them together so now you are getting multiple lines now it is essentially a mess when you keep on going to higher and higher things okay so we define certain zones out of this okay which is I mean of course in a realistic case you know what will be the typical order of magnitude of k so after certain thing you do not have to go to the larger nearest neighbors but as I said this concept is of great use so let us define certain things which we call as Brillouin zones so this is what we define as Brillouin zone a volume in the reciprocal lattice space of course in two-dimension it will be area a volume in reciprocal lattice space is called nth Brillouin zone if all the points within it except boundary of course if joined to the origin cross exactly n minus 1 planes or in two-dimension n minus 1 lines no fewer no larger then what is significance of this particular thing Brillouin zones if the tip of incident wave vector drawn from origin lands exactly on the surface of any Brillouin zone it will suffer a diffraction otherwise no okay it has to exactly land on one of the surfaces of any of the Brillouin zones all right so let us draw a few Brillouin zones so this was my original picture which I have done if you take any point here okay if you join this to origin it does not cross any line okay so it is first Brillouin zone this area will be considered as first Brillouin zone okay because any point within it if you draw to the origin does not cross any of these lines while if you take this particular point this particular area if you join any point from here to this particular origin if you cross exactly one point only one line you will cross only this line so this area will be a part of what we call as second Brillouin zone okay if you take any particular point here okay if it crosses here from example here to here that will cross this particular line if you cross I mean let us take somewhere here okay it will cross this line if you cross this line it will cross two lines so this will be a point in third Brillouin zone okay of course if you look at any textbook you find multiple number of Brillouin zones drawn very very beautifully I have just tried to put it approximately this is my first Brillouin zone okay any point within this when drawn when joined to the origin will cross no lines or no planes no Bragg planes this will be my second Brillouin zone because once any point is joined to here it will cross exactly one point or rather one plane this will be my third Brillouin zone because any point when joining to the origin will cross exactly two points you can see that you can keep on drawing various type of Brillouin zones in fact if you look at the any standard textbook you will find many Brillouin zones being drawn corresponding to this now if you have any Brillouin zone for example you have this Brillouin zones okay you if you draw a wave vector k if it ends exactly here exactly here not anywhere in between it must end exactly at the surface of this then only the condition for k of Bragg diffraction or Lois diffraction condition will get satisfied and you will get a Bragg reflection now it is very clear that if you just take everything arbitrarily if you take a single crystal okay you have just arbitrary value of lambda you know these lambdas do not come into clean numbers okay there I mean for example you take a copper k alpha 1.54 angstrom okay k that is constants also come in some arbitrary numbers chances that they will match that 2 pi by lambda will exactly hit at one of the zone surfaces it is very very unlikely so that is the reason I was telling that you actually require you are required to vary something continuously in order that you have a diffraction condition satisfied. Now what is the importance of Brillouin zone as I said that normally we have various type of excitations solid state physics like phonons which are lattice vibration you have magnons which are excitation of now fossil magnetic excitation with all these excitations you can assign a wave vector and it so happens that wave vector only in the first Brillouin zone are of importance because of the translation symmetry okay this again I am not showing okay this only when we do a proper solid state physics course we talk about this particular things you will find only the wave vectors within the first Brillouin zone need be considered for example if you want to describe the band structure of a material like say semiconductor or for that matter any material the wave vector which are of importance wave vectors of the electrons which are of importance are only those which are lying in the first Brillouin zone okay all other values of k can always be translated back to the first Brillouin zone by choosing an appropriately simple lattice vector. So this particular thing turns out to be extremely important also I want to say that if you are taking a 3-dimensional arbitrary lattice okay for example let us even body centered cubic or face centered cubic these Brillouin zones in 3-dimension other than of course simple cubic which turns out to be simple other lattices it can be fairly complex and then you have to consider these Brillouin zones whenever you are considering any elementary excitation within a solid so that is why the concept of Brillouin zone and concept of reciprocal lattice is a sort of brilliant butter for any solid state of this is so that is the reason they are always introduced right in the beginning. Now let us consider the effect of basis we do not have too much of time but let me just start talking that you know both these lowest condition and black conditions talked only of the diffraction from points we have said that actually x-rays are not diffracted from points x-rays are diffracted from wherever the electrons are lying there and electrons where they lie there is a probability of finding electron at a given point from their let us say center of the atom okay and that particular probability would depend on the wave function okay and wave function to find out the wave function of a complicated element is not all that simple task okay but nevertheless I mean if you are really interested in finding out how the x-ray diffraction is taking place okay then you have to look you have to find out that wave function and specifically if you are looking at the density of the x-ray beam you have to find out how these electrons are distributed across these particular lattice points okay but what is interesting is that you can separate out two different effects one is what is called atomic form factor another is what we call as a structure factor see remember we have said that many times we put multiple number of atoms or multiple number of ions at each of these lattice points okay need not be I mean material simple material like you know elemental material like let us say sodium or iron or copper or whatever it is okay you are putting only one particular atom at one particular lattice point but we have said the concept of basis is used in three different contexts taking okay taking a non-bravil lattice trying to make it a bravil lattice the second thing is that make a more complicated bravil lattice into a simpler bravil lattice okay in all these cases the effect of basis has to be considered so what we do in this particular case we separate this problem fortunately it is possible to separate this particular problem I will not be able to derive in this thing because of the lack of the time but you can look at any standard book on solid state physics they will talk about the x-ray diffraction how does this particular intensity or what we call as structure factor how this particular overall in the amplitude of diffracted beam can be separated out in the form of atomic form factor and structure factor okay so you consider separate out as atomic form factor which essentially talks about the distribution of electrons or scattering particles or whatever they are okay across the center of the atom and second thing that we talk how the different atoms of basis are placed with respect to a given lattice point all right if you do that particular thing then in principle okay the things of which are totally related to the structure can be separated out from the point of atomic form factor now calculation of atomic form factor is a much more complicated thing but structure factor can be easily found out because you can know how these particular atoms are placed with respect to a particular point on the Bravais lattice okay now using just this concept of structure factor many times you can find out lot of information about the material and this is what I am going to describe briefly in another 4 to 5 transparencies. Now let us just take a specific example let us consider a cubic structure and express what is centered cubic lattice as simple cubic lattice with two atoms basis and a face centered cubic lattice as a simple cubic lattice with four atoms basis this is what I have said is normally done this also emphasizes that because you can take the unit cell in the form of cube which is much more simple and it emphasizes the symmetry of the particular lattice. So as I say generally express as simple cubic lattice with basis this is very helpful especially in accelerated diffraction so I take BCC as simple cubic lattice and FCC as simple cubic lattice as four atoms basis okay the advantage if I do that if I start with a unit cell which is simple cubic in that case okay the distance between planes which have Miller indices Hkl can be easily found out unfortunately I will not be able to derive these things okay just find it always express these things and your drag condition becomes this particular thing. Now what happens that you can show from the structure factor calculation that if you are having a simple cubic material then all reflections I mean I am using word reflection but it is actually diffraction as I said many times we will use you for all the planes are permitted then if it happens to be body centered cubic then H plus k plus l should be even only then they will be permitted if it happens to be face centered cubic Hkl all should be individually even or individually odd then only they will be permitted otherwise they will be absent. So by looking at the pattern you can find out whether your structure simple cubic body centered cubic or face centered cubic that is what is the interesting aspect about this particular structure factor calculation which is taking only the positions of the atoms okay starting from a braveritis position of the various atoms in the basis okay. Remember if I have to calculate the intensity properly then I have to also calculate the atomic form factor which is a much more difficult task I must know how the electrons are getting diffracted because there are different points within the atom where the electrons are could be present and the diffraction would amount okay a face difference will amount also of some sort of interference between the waves which are being transmitted or which are being diffracted from different parts of the atom so that also has to be a content which is much more difficult problem alright. So let us just take an example and let me first give you these values so let us call this n as H square plus k square plus l square alright and let us look at what are the possibilities of hkl so the lowest number n is hkl is equal to n is equal to 1 which corresponds to h is equal to 1 k is equal to 0 and l is equal to 0 this particular reflection from this particular plane will be allowed a is allowed not absent a is allowed is allowed in simple cubic but it is not allowed in BCC because 1 plus 0 plus 0 is odd remember 0 is always taken as even and 1 is odd and 0 and 0 is even so it is also not allowed for face centered cubic then 2 which is corresponding to 110 for simple cubic all are allowed for body centered cubic 1 plus 1 plus 0 is 2 which is even so this particular reflection will be present in BCC while in the FCC it will be absent because 1 and 1 both are odd and 0 is even so remember in FCC all 3 of them should either be all 3 of them are should be odd or all 3 of them should be even okay while in BCC some of them should be even if some of them is odd this will not be allowed 3 is 1 1 1 so for 1 1 1 you can see because all 3 of them are individually odd 1 is also odd 1 is also odd 1 is also odd so this will be allowed in FCC but 1 plus 1 plus 1 is 3 which is odd so this will not be allowed in FCC in fact you can see this particular table 7 of course cannot be expressed as some of 3 integers h square plus k square plus l square where hkl are all integers cannot be equal to 7 so this cannot be present okay anyway so you can look at various things okay and find out from this particular table you can draw this particular table to larger thing and find out which are the reflections which are present and which are the reflections which are not absent in a particular lattice. So I am just giving one particular example how to take when this a typical example which we have taken from you know some x-ray diffraction from given in some book. So let us suppose we are getting diffraction at various values of thetas for a particular material okay this is the values of thetas corresponding to which I have found out break diffraction first I calculate sin square theta why I calculate sin square theta because sin square theta is going to be proportional to lambda square which is the wavelength square divided by lattice constant square 4a square multiplied by h square plus k square plus l square I must know what are the values of hkl only then I can find out 4a now looking at various patterns I have to actually individually find out what are the values of hkl that is what is called indexing effort of a x-ray diffraction sorry I will take another 5 minutes just to explain this particular thing. So you take sin square theta these will be the values of sin square theta now you try to take various ratios of this particular thing if you take this divided by this you find out that this thing matches 4 by 3. So chances are that this will be 111 plane and this will be 200 plane okay let me just mention at one particular point that you know earlier when we are talking about Miller indices we have said that a common place is not allowed here so I mean 110 and 220 planes are exactly the same planes why we in x-ray diffraction we write 220 actually at that time I had also mentioned that in x-ray diffraction we do talk of planes which are the common planes let me without any more sort of you know sort of derivation about this particular thing let me just that is a convention of break diffraction is concerned second order diffraction from 110 plane will be termed as a 220 diffraction second order reflection from 111 plane will be reflected will be written as 222 when I say 400 plane it means it is a fourth order diffraction from 100 plane so that is the way we write it. So effectively you take the ratios of these things and try to find out which are the ratios which will fit these ratios of sin square theta value and then you can look at this particular thing okay and then here you can see that there is a 111 present and there is a 200 plane which is present okay this is typical reflection of FCC lattice. So this particular material is actually a phase centered cubic lattice okay so looking at these particular sin square theta you can find out what type of structure you have and of course once you have index this you know h square plus k square plus l square you can also find out what is the value of lattice constant. Just last few transparent some of the utilities I have said some of the utilities because x-ray is a very powerful technique used in for a varieties of things okay I will just I am very very briefly try to mention it is very very powerful technique can be used to obtain many information. In fact you are getting better and better x-ray machines which can give you much better information about so many things okay you can talk about stresses strains and so many things okay the texture texture is one of the very very important aspect that we discussed especially when you are talking of minerals or for example steel okay. So for example if you have presence of various phases if your material is a mixture of two different type of withstand structure then relative concentration can be found out their texture and orientation can be found out how do you align a single cell that can be done okay in fact obtaining lattice constant. But if you talk of the most common things which most of our students do most of our PhD students do two things which are done very very commonly I think very very large number of time we use x-rays only for one of these two things okay one is to check whether you have formed a phase if you have generated a material if you have created a material okay you would like to know whether this particular material is a single phase it means whether it has only one crystal structure or you have got some impurities okay it can be checked by x-rays to find out whether you are getting a single phase material or not okay for all standard materials okay there is something which is called jcpedias data these are cards which are now available in the form of you know softwares okay you can have these particular cards and you can match your theta values and find out whether your structure is really resembling one of the standard structure for any of the material okay then to determine the structure of unknown material that is also many times done that you have generated purely new material and you have no idea of the structure okay if you have no clue of the structure whether it is cubic or non cubic or whatever it is that is not a very simple thing okay many times people use softwares now to find out what is the structure. The last thing which I would just like to mention this is I mean this is one of the students which we have taken this particular data so from our group okay just to show how typical theta theta curve looks like remember here things are always plotted as a function of 2 theta these are the intensities which are arbitrary like things and you get various peaks here which shows reflections from or diffractions from various planes and eventually you have to index and find out what are the planes from which these dexers in excess reflections are coming. The important thing here is that you are getting two mixture of two different phases which we call a zinc ferrite and cobalt ferrite which are coming and you can see a small splitting on this particular x-ray diffraction beam which actually refers to this particular being two phases present. Just to give you idea of how these particular x-ray diffraction things look like so that is the end of this particular thing. So if you have I mean we have already exceeded time but you know two or three questions probably we can just quickly take if you have. S.T.M. College yes. Hello sir. Good morning sir. Good morning. Sir yeah the working where these lattices are in seven crystals. Yes. Yeah in some system we have all types of unit cells in simple cubic BCC and abscissure to base angle. Yeah that's right. But in some stuff like monoclinic and prime clinic we have only simple cubic. That's right. So we don't have other types of unit cells. That's right. What is the reason for that? See these as I say these if you are looking at the reasons this have to be found out from what we call as group theory. See basically thing is that you can show that if you look at that particular structure you can I mean see if you take for example triclinic I mean let's like put it like that in terms of general lattice constant there are six lattice constant which you call as abc and alpha beta gamma which are the angles. So in principle they are sort of six constants which define a particular lattice. Now it can happen that if they take a specific values then the symmetry of that particular lattice changes. If the symmetry of that particular lattice changes we classify this into a separate lattice. Okay that's one aspect. The second aspect is that many times for example if you say that why not take face centered let's say cubic. If you take face centered cubic this can also by by re-looking into different fashion can be classified as a different lattice and it does not change the symmetry. So these particular things as I call are sort of group theoretical way of looking into it where you look at all the symmetry operations and from that you find out what are the symmetry operations which are common. Okay then define crystal systems and in that crystal system what additional points that you can put it so that essentially symmetry is not altered. All right on the other hand if for example if you take a triclinic and put some let's say base centered you may eventually generate some other lattice which is exactly the same time and you are not you are not trying to talk of any further symmetry enhancement. Sir actually there are 14 various lattices. Only that much is there but if in future or something was is it possible to develop some or are only depending upon group theory of mathematics. Yes so this can be mathematically shown it seems I am not a group theoretical person but group theory it can be shown that only possibility exists are of 14. You cannot have anything additional than other than 14. All right now out of these 14 it is a different question whether you have crystals which have all those 14 symmetries they may not have I am not very aware. Okay as I said most of the crystals generally have happened to have a cubic symmetry or hexagonal symmetry. Of course I know system which are you know for example triclinic no not triclinic but monoclinic tetragonal okay rhombohedric where there are certain type of structures there are certain materials which crystallize in that but I am not sure whether your materials which crystallize in all 14 of them I am not very sure. College of engineering Bhuvaneshwar. Why polycrystalline materials are more stable than single crystalline materials? That is not a question of being stable I think they are equally stable if you can make them. The question is that I mean that is what is my feeling in general. Thing is that to make a single crystal materials you require special efforts and what you try we should normally make is polycrystalline material okay. If you have to grow a single crystal specifically if you have to grow a high quality single crystal okay it requires lot of effort because you do not want the I mean whenever you are generating the material as we call it as a nucleation you have to have only single nucleation across that particular point you know you should have the entire structure built in. So you require very specialized technique to generate single crystals so which I mean the basic let me put it like that I mean generally you will find it is much more expensive to build good quality single crystals. So in fact many of the applications that people envisage they envisage on the polycrystalline materials and and because the applications are which which are which turn out to be cheaper and because the applications are envisaged for polycrystalline materials people on the network tend to work also more on polycrystalline material but if you want to study some of the fundamental properties you really require good quality single crystal materials and those people I mean growing a good quality single crystal is an art not everyone can do it but I do not think it is related to stability of the material. Can I give an example of a reciprocal lattice? Can I give an example of a reciprocal lattice? I had given you an example that you know for example if your body is centered cubic lattice the reciprocal is a face centered cubic what type of example are you looking see remember reciprocal lattice is something which is in a reciprocal lattice space okay this is what you have to imagine you know putting assume that there is three-dimensional graph paper okay so you assume that these are points spaced in a three-dimensional graph paper like for example a two-dimensional reciprocal lattice I have just drawn on a paper all right so only thing the length that I have taken is of dimension of inverse of length I am not sure whether I have explained it or not. So what is space group okay these are a space group and point groups these are again you know group theoretical things okay which essentially looks at the symmetrical aspects including the translation symmetry and this I am not very sure so I should not be answer something which is wrong but these are essentially group theoretical models to describe a particular type of symmetry operations including in this particular lattice okay so these are sort of group theoretical language which people use which tells you I mean I mean using a particular space group you can find out what are the symmetry aspects a particular space group is defined how many type of mirror symmetry you have how much of rotational symmetry you have all those things are described in that particular thing. Yeah so which one is more advantageous polygistral and material or single crystal? See the thing is that if you want to study basics physics then single crystal is much more useful because that actually shows a regular arrangement and you can really study whatever aspects that we want to study but as I said to grow a good quality single crystal material is much more expensive so I mean commonly if you are people you are using in a commercial application I mean of course I can imagine you know I mean for example many of the semiconductor devices you are using semiconductor single crystals okay but you know semiconductor at least the growth of semiconductor single crystals have been you know you know have been because of their applications have been very very well utilized but if you take any arbitrary material okay generally you will not find it very easy to grow a single crystal material at least in my area which is of magnetic materials okay you do not I mean there are hardly any applications which you envisage using single crystal material okay because they are so expensive to do other than magnetic bubbles you know when people are talking about single stand material and eventually they lost only because you know they were too expensive okay most of the time we use materials which are polycrystalline so unless you are looking for very very specific application people generally because they are cheaper you know eventually the word unfortunately or fortunately is ruled by economics it depends on whether the you know the application how cheap is the application you know the what you are going to industrialize okay MGM College of Engineering Nanded Sir yesterday you have shown that how non-bravious lattice becomes to be as an bravious lattice that's right that's right my question my question is is it possible that sorry basis basis is simply a group of atoms we can say then that basis can make a primitive unit cell to a non-primitive unit cell and also loosely placenta loosely packed a structure to a closely packed structure and in that way how the coordination number is going to be becoming to be like that no no see I am not very clear about the question that you are talking see the thing is that see when we are talking of close packed structures we have been talking of close packed structures in which we are talking of single atoms okay when you have multiple atoms when you have a complicated type of material for example the type of material which I have been working on let's say ferrite which have you know at least three different components okay how these atoms and ions are going to be fitted okay may not really be an exactly close packed forms okay so then still we describe in terms of basis okay for example oxygen for example take take take a normal ferrite oxygen forms A what we call as a phase centered cubic structure there but all these oxygen atoms need not be touching each other the way we are describing okay so it depends on what type of inclusions you are having but let me put it like that you know I am not sure whether I am really answering your question see thing is that you cannot I mean a structure is a structure now a structure has to be described in terms of a Bravais lattice and a basis okay there are multiple ways in which you can do that particular thing okay that's all about that's all I would like to say I cannot really see you know how you can say that a close packed structure can be or a loosely packed structure can be put to a close packed structure because structure its own cell see packing is very very different from the geometrical aspect when I am talking of the lattice and the basis I am more or less looking at its geometric aspects all right so packing etc becomes a slightly different question I am not sure whether I have answered your question you know if I have not answered can you repeat the question what I have not answered yeah means that you have shown that non Bravais lattice how basis basis is going to be added it becomes to be as an Bravais lattice that particular part in that way in that way I am asking that non primitive primitive unit cell can be becoming to be as an non primitive unit cell in that sense okay basis is going to be added okay if it I do the other way I take a non okay what I do a non primitive cell I convert into what I call as conventional cell by using a basis okay so for example a conventional unit cell let us say when I draw a cube for a body centered cubic okay show eight points in the corner and one point at the center this is not a primitive unit cell all right because it contains on the average two points all right on the other hand a primitive unit cell should contain only one point now this non primitive unit cell I can convert into a primitive unit cell by converting take into a basis that is what precisely I have done I considered a basis a BCC lattice by taking one atom at the origin another atom at the body center and converted this unit cell into a simple cubic unit cell so I actually converted a non primitive which is a conventional unit cell to a primitive unit cell so in that sense you are right yeah second question is that reciprocal you have shown that lattice is to be there is exist then that vectors are having that lattice points are having the identical surroundings that's of course of course any lattice has to have that particular quality so reciprocal lattice is also a Bravais lattice let's be very clear okay so these translation vector will generate a Bravais lattice okay which is has exactly the same quantity same quality it has to have exactly the same environment otherwise no longer a Bravais lattice that's perfectly right good evening sir sir my question is sir we are having the direct lattices simple cubic as a reciprocal lattice has a simple cubic that's right for BCC we are having reciprocal BCC FCC for FCC we are having the BCC right sir yeah so why there is a difference for the BCC as a FCC reciprocal lattice for FCC as a BCC in simple cubic we are having the direct lattice same as reciprocal lattice see there why is a difficult question because the thing is that that's what we get it okay there is no okay I mean we have defined in a particular fashion and eventually we have got it so I mean it's very difficult to say why you know because we have obtained it so the why question is a little difficult question but you know of course one could get surprised why it happens like that way if you want to go into little intricate intricacies of that little more complex it requires a little bit of visualization in fact what you can do if you look at the simple cubic lattice in fact see we said that certain type of reflections are absent okay in x-ray diffraction see for example you start with a simple cubic lattice and start omitting those points take one origin and take the coordinates of all these points and start omitting those points for which x plus k plus l is odd you take only those points h plus k plus l is even you'll find that this lattice gets converted into one of the lattices so these things have little more if you want to look into some sort of thought process you can sort of visualize why it should happen like that but we cannot really answer why does it happen it happens because it happens because from our definition we get it sir one more question is that what is the meaning of packing fraction okay packing fraction essentially means that total volume which is occupied divided by the total volume available for example if I take for example let us say body centered cubic lattice in a body centered cubic lattice you know if I take a unit cell conventional unit cell okay you have points here you have point here okay now total volume of this particular unit cell is a cube okay now let us assume that these are we are having hard spheres which are making this particular lattice okay so at these points all these things are hard spheres so let us find it out how many number of what is the volume which is occupied now all these eight corner atoms will contribute on the average one atom in this unit cell because each one of them is being shared by eight cubes and there are eight such corners okay if you take one this there will be one two three four and four on the top there will be eight cubes which will be sharing this particular corner okay so and there are eight such so you have only one plus you have one at the center so you have total volume which will be occupied as two multiplied by four by three pi r cube okay so this will be what we will call as a packing fraction of course in order that you calculate the packing fraction you should know the relationship between r and a you should know which is the nearest neighbors okay if you take this particular atom and this particular atom okay the distance between them is a on the other hand if you take this particular atom which is the body center and this particular thing the distance is root 3a by 2 okay because this distance body diagonal is root 3a so distance between this atom and this atom or this point in this point is root 3a by 2 root 3 is 1.732 so this distance is smaller so if I really put spheres here this sphere will touch this sphere and this will touch this but this will not touch this there will be a small gap between these two okay so you can calculate what will be the r and a versus a relationship because root 3a by 2 okay this is the distance between them this must be equal to 2r because there will be one r here and there will be one r here so you find out the relationship between r and a substitute it here you will get the packing fraction this basically tells you how much of this space is occupied within a unit cell okay in comparison to the total space which is available I think that is close enough we already bought them one meet after lunch thank you thank you