 Hello everyone welcome to this material characterization course. In the last class we just discussed about the properties of x-rays in terms of phase relations and then how this phase relation influence the diffraction and then we just looked at little bit elaborately how we understand the Bragg law Bragg's law and then how it explains the diffraction intensity and so on. And then we have let us recall that what we have discussed in the Bragg law it is we we just said that it is not that the diffraction intensity is coming from the first layer of the atomic plane and also the the subsequent planes which is underneath the the surface also contribute to these diffraction intensities through constructive interference. And then we have seen that how this contribution from each atoms in the planes below the the surface and and how the overall diffraction intensity is envisaged. So and subsequently we looked at writing this Bragg law in a different form and then how it can be used to analyze the crystal system. And then we will continue in that line today and before that I would like you to make a brief a description on the the reciprocal lattice formally we have not introduced this concept so far but in the fundamentals of this course I mean the lectures we discussed little bit about this reciprocal lattice. However we will be using this concept throughout this course and not only this x-ray diffraction and also the now after this we will discuss about electron diffraction and transmission electron microscopy there also we will use this extensively and and what I request you to do is you should go through for all the mathematical treatment of this concept which is really out of the scope of this course you should refer a physics of materials which is also there in this NPTEL portal where you have a detailed mathematical treatment is given and also specifically there is a 10 hour course is being offered by professor Pratap Haridas where exclusively on the reciprocal lattice. What I request all of you to do is go through that physics of materials lecture notes or videos as well as the the 10 hour course exclusively on the reciprocal lattice then if you follow this it will be very easy in order to save time I am avoiding all this basic mathematical relations but then I will briefly talk about it. In in fact we will be dealing with more practical aspects of this reciprocal lattice concept and we will be actually seeing in in practice how we can visualize reciprocal lattice in reality. So in our course we will look at more the application part of it and not the basic mathematical part of it but nevertheless I will keep on referring this concept and then I will try to make you understand as much as possible in terms of physical phenomena. So if you look at the suppose if you want to talk about electron scattering from atoms in an a crystal lattice first we have to talk about the crystal lattice and this is how we define an unit cell a a convention is where you have this a b c all or vectors and then the angle between them or alpha beta gamma. So this is a called unit cell convention. So let us consider a lattice to be located by a reference atom at O defining the origin of unit cell and the position the jth atom of the unit cell by a vector r which we define normally r is equal to u j a plus v j b plus w j c where a b and c are unit vectors defining the unit cell. This is I am saying because we will be referring this a vector this is a vector in the real lattice which we will be using it extensively in the x ray diffraction intensity expressions as well as when we talk about reciprocal lattice. So this is a convention and it is a vector in a real lattice you have to remember that that is very important and excuse me what is reciprocal lattice it is a special lattice construction to a does in interpretation of the diffraction phenomena. So this is the fundamental aspect of it and how this lattice parameters or reciprocal unit vectors are defined which is a star is equal to 1 by v c into b cross c, b star is equal to 1 by v c into c cross a and c star is equal to 1 by v c into a cross b where a star b star and c star they are all reciprocal unit cell vectors where a b c they are all real crystal unit cell parameters and v c the crystal unit cell volume that is v c is equal to a dot b cross c is equal to b dot c cross a which is equal to c dot a cross b. So these are all very fundamental expressions which you might have already studied or must have come across much before that. Let us now go through some of the unique features of a reciprocal lattice a star in the reciprocal lattice is normal to the plane in the real lattice described by b and c. The dimensions in the reciprocal lattice are fractions of those in the real crystal where hkl describes a plane in real crystal it now describes a vector in the reciprocal space consequently the diffraction from a plane in a real crystal can be treated as common vector in the reciprocal lattice feeding its intensity into a point. Thus a diffraction intensity point in a reciprocal space corresponds to a plane hkl in the real crystal. You see these are all the some of the basic features of the reciprocal lattice. In fact we will be talking about this the reciprocal point in a real system in a practical situation as a electron diffraction pattern where we will try to interpret for its complete accountability what it means that we will do it in detail study. And what is said here is in fact it is not just one plane here hkl not necessarily one plane here it could be a set of plane it is always referred as a set of hkl plane which I mean for which correspond to each reciprocal lattice points. So that we will see it in a appropriate time. We will also see some of the other relations that a dot b I mean a star dot b is equal to a star c equal to b star dot a which is equal to b star dot c is equal to c star dot a is equal to c star dot b is equal to 0. And a star dot a equal to b star dot b equal to c star dot c is equal to 1. These are all some basic relations and very importantly now we will see the uniquely characterize the reciprocal lattice as it relates to the real crystal lattice. We can define a vector in the reciprocal lattice by r star which is equal to g hkl again this is which is equal to ha star plus kb star plus lc star. So this is a reciprocal lattice vector. So you just compare this with the real do not confuse this with the real lattice vector this is a reciprocal lattice vector it is also called a g vector. We will see in appropriate time what is the meaning of this g and how do we interpret those notations. So right now you have you have to keep in mind how a vector in a real lattice is represented and how the vector in a reciprocal lattice is represented and then what is the relationship mathematically. And if you can see further the dimensions in the reciprocal space are reciprocals to those in the crystal and we observe a mod of g hkl equal to 1 by mod of r which is equal to 1 by d hkl. We can now express the diffraction condition in an a vectorial form k minus k naught is equal to lambda by r which is equal to lambda times g hkl. This is essentially a vector form of law a condition and is equivalent of the Bragg reflection. Just for the completion I have brought this relation this is a I would say it is a vectorial form of a Bragg condition. For example Bragg law describes the diffraction in terms of scalar equation. What you see n lambda is equal to 2 d sin theta is a scalar equation. We can also define the diffraction condition by a vector equation. So this is one of the form. We will now see it in much more detail in 2, 3 slides later what how this law a condition or vectorial form of representation of a diffraction equation. So in the last class we just looked at a Bragg law and how it can be used in analyzing the crystal system. There are few more points we will see about this Bragg law. So Bragg law contains a great deal of useful information. We can write sin theta is equal to lambda by 2 d. There are two things we have to keep in mind. As lambda increases the scattering angle for a constructive interference theta also increases for a fixed d. As d increases theta decreases for a fixed theta. So diffraction is a very sensitive measure of changes in the crystal structure parameters in the crystalline materials. Hence sin theta by lambda is equal to 1 by 2 d which shows that the angular function sin theta by lambda which is used to tabulate scattering factors and incoherent scattering intensities can be related to the interplanar spacing of diffracting planes. So the diffraction is very sensitive measure of changes in the crystal. So this that is why this particular parameter is used as an angular function to identify the changes in the crystal structure which we will see in the subsequent slides. Now as I just mentioned before what we have seen so far as a diffraction condition in terms of a Bragg law or I would say Bragg's law it is in a scalar form. There we just try to use a parallel planes one after the other. So the all that we have showed in the form of animation is only the rays which are subjected to constructive interference or a diffracting beams. But we have not seen a diffraction condition for an individual atom such as rows or a 2d or a 3d lattice. So now we will get into that I mean discussion where we will try to arrive at a diffraction condition in a one dimension, two dimension and then and see three dimension and see what is the difference between what we have seen in a Bragg's law condition and what we are going to see in this discussion. So look at this schematic I have just drawn here a scatterers I would say it is a repeat distance in a 1d lattice or rows of atoms. The repeat distance is described and the direction is described by the vector a 1 and then you have the incident x rays S naught which is falling on this row and S is the scattered rays. We will say that it is a diffracted ray as well provided it follows some condition and in order if you look at the in order to define S as a diffracted beam we need to look at the condition whether it obeys the a kind of a diffraction I mean a condition or whether it obeys a diffraction condition or whether it facilitates the diffraction condition in terms of its geometry. So now what we can see here is you see the atom a and b which scatters and then this is the theta and this is the theta from this side and this is a perpendicular I have drawn for this line and this is a perpendicular drawn for this line. Now in order to be in the same phase of S and S naught they should have the a path difference should be having some certain conditions what is that condition suppose if you look at this atom and then look at this ray after scattering this is the path difference and similarly if you look at this ray which hits on the atom a and then it scatters this way and this length this is the path length. So there is a difference these two rays are having two different path length. So mathematically if you see what is this length this length is nothing but S naught dot a 1 similarly here this length is S dot S naught dot a 1 this is S naught is an incident wave vector and this is your lattice repeat distance that is also in a vector and this distance is S naught dot a 1 similarly this distance is S naught dot a 1 and that is how the path length differs. So now we have already seen that in order for the S rays that is the scattered ray to be in the on the same phase of S naught the path length or path differences should be a integral multiple of a wavelength. So that is what we have written here. So you see that S dot a 1 that is this minus S naught dot a 1 that is the path difference between these two should be equal to an integral multiple of a wavelength h lambda and equation one is a vector equation of cones of scattering around the row. So what actually we are seeing is this there is a intensity cone which is spread like this. So this is an incident intensity and this is scattered intensity actually this kind of a cone of intensity is generated around the row and we will now see one example how this cone is related to the angle of incident theta and then how it changes with the angle and the wavelength and the inter atomic distance and so on. So now you remember that this particular equation accounts for a diffraction in atoms lying in a row. So now we are just said that if this path differences is in the integral multiple of a lambda then these two waves that is incident wave as well as diffracted wave will be in the same phase and then constructive interference takes place so the diffraction happens. So this is a condition for that. Now we will look at one example to look at the effect of the angle similar schematic where the incident angle is about 80 and then let us assume that the diffraction geometry for a row of scatterers this is a row of scattering row of scatterers is 4 angstrom that is irradiated with the 1.54 angstrom x-ray beam. So now you see that the value for lambda is equal to I mean theta is equal to 80, 56 and 19 corresponding to H is equal to 0.5 angstrom 0, H is equal to 1 and H is equal to 2 and so on and how this intensity cone going to be different. You see that when the H is equal to 0 you can see that incident beam is almost on the same direction of the diffracted beam and as the value of the H increases you can see that the angle decreases you can see that it is 56 and 19 and you can also see that at the H is equal to 0 the cone is completely opened up you can assume that it is completely opened up and as the H value increases you can see that how the theta is decreasing. So this is another one example to visualize the how this the intensity cone varies with the different values of theta. Now we will move on to the two dimension. So diffraction from a net that means two dimension which defined by rows A1 and A2 another condition is added to the previous condition that is the S minus S0 dot A2 that is a path difference is also should also be equivalent to multiple I mean multiple of integral multiple of a lambda K lambda where K is an integer. A second series of cones is created around the second row of atoms the intersections of the cones give lines of maximum intensity. So this is where you have to be a little bit careful we are now talking about intersections of a cones from the two different rows first we talked about a single row now we talk about a two dimension that means one more row is added and then where that repeat distance is A2 and this also is going to produce a cones of radiation for a given theta around each scatterers that is here lattice positions you can say are atoms whatever it may be each one is going to produce a cone around it and these two cones are going to intersect. So all the cones are going to intersect and that they are going to form a line of maximum intensity finally a lattice is created by addition of a third row that is one more dimension where the path difference necessarily to be S minus S S0 dot A3 is equal to I lambda which is another dimension so this is again has to be satisfied for the diffraction to be occurring. Now common intersections are points of reciprocal lattice. So now we talked about one point and from there we talked about a line of maximum intensity and then now the line has become a point of intersection where we will have a maximum intensity that is where the reciprocal lattice point comes. All these equations 1 to 3 are known as the law way equation. Please understand in the Bragg diffraction schematic we were talking about only a parallel planes and then we said that the diffraction intensity comes from each of the rows or in fact all the atoms which are in phase are in otherwise contributing to the constructive interference then account for the diffraction intensity but in a law way equation it is in a vectorial form of I mean concept mathematically where we talk about an individual atom where in the form of a row or a 2D lattice as well as 3D lattice and then how each one is contributing to the diffraction or I would say that the each of the equation derives a condition for the diffraction to take place. So there is a difference between the concept discussed in a Bragg's law or a condition for a diffraction through Bragg's equation as well as conditions for the diffraction to take place through a law way equation. So in fact law way equation is much more general and you can talk about an individual scatterer in the 2D or 3D lattice. Now we will try to relate this reciprocal space I mean as well as the Bragg law. Look at this schematic I have just drawn a cross section of a solid sphere that is why the inside of the circle is shaded that means a cross section of the sphere where inside you are seeing that a plane and then this is incident ray and this is a diffracted ray and then you have the 2 theta angle is shown like this and let us now look at the remarks relation of reciprocal space to Bragg's law. Now let us relate to the Bragg law. The Bragg law is related to reciprocal space through evolved sphere. The another concept similar to reciprocal concept for a diffraction is an evolved sphere. So what you are now seeing on the schematic is a cross section of an evolved sphere. So that is what you are seeing. Let us see A is the physical situation where the incident waves in the S0 scatter of Hkl planes in the direction S. B is the situation in the reciprocal space. So this is a real space like you have S0 and S and what you are seeing is an actual condition for the diffraction through an evolved sphere. You see that let us assume this as A and this is B and this is O. From the geometry we will be able to derive some expression based on which the diffraction condition can be arrived. So suppose if you define this A O vector S0 by lambda and A B vector S by lambda and you can look at the difference between A and B. So this is the difference between A and B. Look at this. This is called a diffraction vector S minus S0 by lambda by this relation. You see that the evolved sphere radius is 1 by lambda. So that is something which you have to remember. The evolved sphere radius is 1 by lambda and then now we will see how this geometry can be related to a diffraction. So from the figure B A O equal to 1 by lambda and sin theta is equal to B O divided by 2 divided by A O. So let us see what is that. So this is the triangle we are talking about. So this is B O. So B O by 2 for sin theta we can write like this. B O by 2 divided by A O. So that B O can be written as 2 sin theta by lambda and then since B O is equal to mod of S minus S0 by lambda which is nothing but this distance which can be rewritten like this S minus S0 by lambda mod is equal to 2 sin theta by lambda. From Bragg law 2 sin theta by lambda can be written as 1 by D. So the relation to the reciprocal lattice is as follows. The vector O to B is a reciprocal lattice vector r star hkl which has the magnitude 1 by D. When a reciprocal lattice point lies on the surface of the evolved sphere Bragg's law is satisfied and a diffracted beam passes through the point. So now you write the reciprocal lattice vector r star hkl which is equal to S minus S0 by lambda which is equal to hB1 plus kB2 plus lB3. The vector S minus S0 by lambda is called a diffraction vector which is nothing but r star hkl in a reciprocal lattice space. The diffraction angle 2 theta hkl is the angle in the space between the beam incident on the crystal and the diffracted beam that passes through the hkl diffraction peak. So you have to remember this how this is related to a diffraction condition through this diagram. So what it says is this is the diffraction vector which is nothing but r star hkl in a reciprocal space. So whenever this point hits on the surface of this sphere then the diffraction condition is satisfied and that is what is stated here. And now in order to visualize this concept little more we will look at some more schematic. What we are seeing here is in a schematic A and B a relation of reciprocal space to the evolved sphere. A is the incident beam is in 100 direction. Suppose this is the S0. The S0 is in the 100 direction, 100 direction. So the evolved sphere intersects the red mark as an evolved sphere which intersects 210 and 21 bar 0 reciprocal lattice points that is here as well as here. That means these two peaks will diffract. These two reciprocal points are diffracting set of planes. And what we are seeing in the schematic B is now the crystal has been rotated to 45 degree. So the crystal has been rotated to 45 degree. Then what happens is the evolved sphere now intersects 200 and 12 bar 0 reciprocal lattice points. You can see that 12 bar 0 and 200 reciprocal points which are coming on the intersecting the evolved sphere. So between these two rotation no other points are possible for the diffraction. That is what it means. So if you look at the diffraction spots 21 bar 0 and 210 after rotating to 45 degree you have a new planes 200 or 12 bar 0. And between these two rotation there are no other possibilities for the diffraction. That is what it means. So this is one way of interpreting the evolved sphere concept. And what is this schematic shows? This is also a set of evolved spheres but it says the limiting sphere. What do you mean by limiting sphere? The limiting sphere is the locus of the farthest point of the evolved sphere when rotated in all orientation. So right now we have seen that rotating into two direction. And suppose if you rotate another 45 degree, another 45 degree, another 45 degree. What are all the points will intersect in the reciprocal lattice? That is this is the reciprocal lattice. Where all it will intersect that is a maximum possible planes which will contribute to the diffraction or the planes maximum number of planes that will intersect the evolved sphere that is called limiting sphere. So that is what is shown here in a plane of B1 and B2 lattice. And this defines the limiting sphere like this. And another very interesting example I would like to show where the evolved spheres for molybdenum copper and chromium K alpha radiations in analyzing the a simple phase centered cubic structure with the cell parameter 4 angstrom. Suppose if you use one of this radiation for example let us use a chromium radiation. So which is 1 by lambda of chromium which forms the evolved sphere around these lattice points I mean reciprocal lattice points which have only 1 1 0 3 1 0 2 0 0. Since your reciprocal I mean your evolved sphere radius is 1 by lambda which is a function of a wavelength of a given radiation it can explore the possibility of finding the reciprocal planes only in a limited number that is 1 1 0 3 1 0 and 2 0 0 type. Suppose if you use a copper radiation then your radius increases then your evolved sphere become little bigger and now you see in addition to the 1 1 0 and 3 1 0 and 2 0 0 planes you have you are able to examine phi 1 0 4 2 0 2 2 0 etcetera. So as the lambda changes you are able to incorporate large number of reciprocal lattice points that means you are able to get into atomic planes more atomic planes which are satisfying the Bragg conditions. So the one last one is a molybdenum radiation you see that it covers quite a bit of reciprocal points that means it forms a huge sphere evolved sphere which will intersect which will intersect through many reciprocal point including large number of planes. So this is what the evolved sphere concept is readily visualized. So what you are now seeing is the black spot is as I said it is a reciprocal lattice points and then we will also look at this similar electron diffraction pattern and then again we will come back with the evolved sphere concepts to understand little more on the diffraction phenomena. But to start with this is very nice example and schematic to look at the different schematic to understand the relation between the Bragg's law and reciprocal lattice and evolved sphere. So the evolved sphere actually links the Bragg law and the reciprocal lattice that way we can consider it is a very nice concept to understand appreciate the diffraction phenomena. Now we will move on to intensities of diffracted beam. At the end of the day if you are interested in analyzing the crystal with X-ray diffraction we are interested in intensities and we will account for the intensity expressions and then we will see what are all the parameters which will influence the diffraction intensities when it happens with the amorphous material when it happens with the crystalline material or a single crystal or a poly crystal and so on. So the intensities are important and their quantification is important but we should now know what all the parameters which will control the diffraction intensities of X-rays. So we will begin our discussion with this. Let us look at the two I mean crystal structures simple crystal structures what you are seeing in the schematic A is a base centered unit cell and B is the body centered unit cell. The positions of the atoms in the unit cell affect the intensities but not the direction of the diffracted beams. So to prove this concept how the positions of the atoms in the unit cell is going to affect the diffracted beam we are going to illustrate through this two examples. One is a base centered unit cell another is a body centered unit cell and if you assume that X-rays are coming and then diffracting through both the unit cells and then you will get the corresponding ray diagram like this A and then B. I would like you to look at this ray diagram carefully the ray 1, 1 prime and 2, 2 prime and the distance is C and this two diagrams we have already seen on the in the previous classes you will recall when I talked about the importance of two zero zero planes. For example if you see that what is the path difference between these two 1, 1 prime and 2, 2 prime they are out of phase by a one wavelength or I would say that the path difference A, B and plus B, C is equal to some H lambda something like that. So but if you come to this unit cell the situation is slightly different suppose if you assume that these two planes diffract the X-ray and then their phase difference is is about one lambda here it is exactly half of that phase difference for example if because we have the other plane which is inserted in between so the 3 E 3 prime prime ray will have the phase difference exactly half of the the previous one so that means this phase difference is going to completely annul the intensities of the X-rays diffracted by this ray 1, 1 prime as well as 2, B, 2 prime so that means your phase difference is not going to get the intensities from one zero zero plane at all. So I hope you you get this idea this has been already I had told you and the phase difference I have also separately discussed how you have to visualize the phase difference and how they annul each other or they contribute to the constructive interference. So in this type of a crystal where you have this the path difference or the phase difference exactly half of this unit cell they are going to lose the intensity from one zero zero so that is what we are concluding this example shows how a simple rearrangement of atoms within the unit cell can eliminate a reflection completely. The intensity of a diffracted beam is changed not necessarily to zero but any change in the atomic positions and conversely the atomic positions can be determined only by the observation of diffracted intensities. So the aim is to establish a exact relation between atomic position and the intensity. The problem is complex because of the many variables involved and the relationship must be developed step by step by considering how X-rays are scattered by an electron then by an atom and finally by all the atoms in the unit cell. So we would we would look at the how the diffracted intensities are changing by step by step first by scattering by the electron then by an atom and then by unit cell then we will look at the the whole expression for the X-ray intensities diffracted by a crystal. So we will continue to look at this diffraction phenomenon and then we will in the next class we will start with the X-rays which when they are scattered by an electron what are all the physical phenomenon we will go through. Those things we will look at in the next class. Thank you.