 this material characterization course. In last class, we just looked at the X-ray scattering in terms of coherent and incoherent scattering and we found that a coherent scattering was explained by Thompson's equation and the incoherent scattering was explained by Compton effect and then we just try to understand the scattering of electron by X-rays and then we just moved on to the scattering by a unit cell and then we started looking at the structure factors and simple expressions, basic expressions for the structure factors how we can calculate. We started looking at a simple unit cell where you have the one atom per unit cell and then we started looking at the structure factors for a base centered unit cell. Today we will continue this exercise and we look at some of the basic very important crystal systems like a body centered cubic lattice as well as a face centered cubic lattice and then we also look at some of the important crystal systems or ordered crystal systems like a CCM chloride and sodium chloride and how this structure factor calculations are done. So the first slide which I am trying to show here is the structure factor calculation for a body centered cubic structure and what you are seeing in the schematic here is not a body centered cubic structure but rather it is a reciprocal lattice cell for the simple BCC structure. So we will now try to understand this what we are seeing is a reciprocal lattice cell. In fact we will try to understand this cell by looking into the structure factors in a usual manner. In a BCC cubic I mean the BCC structure the structure factor can be written like this F hkl is equal to F into e to the power 0 plus e to the power pi i into h plus k plus l that means you have two atoms per unit cell one at origin one at the body center and then we have the complex exponential function will have the relation like this and if you for the BCC crystal structure the structure factor rules are F hkl is equal to 2f if h plus k plus l are equal to 2n if structure factor F hkl is equal to 0 if h plus k plus l is equal to 2n plus 1. So that means we have to see whether this is true in this case what is reciprocal lattice here we will see the intensity only which are allowed reflections are seen here. So what you are seeing here is it is the cell of double the size if you see that this is not 1 0 0 but 2 0 0 2 2 0 and 2 2 2 and 2 0 2 and so on. So what you have to appreciate is here is for a BCC simple BCC structure your reciprocal lattice is an FCC unit cell in a reciprocal space and if you whatever the hkl sum is odd then you will see that they are all 0 for example you can see this all this other positions belong to this condition wherever you have this odd number. So like that you can verify the selection rule and then the structure factor from where you are going to get the intensity by looking at the reciprocal lattice. We will see this the reciprocal lattice in an actual case when we look at the electron diffraction we will also appreciate what kind of direction and orientation we will get the spot and we will also see the forbidden positions in the reciprocal lattice plane. So like that we can look at some other system and before we look at the one more system let us also recall this problem we have already discussed but then just to give you a perspective of how the forbidden reflections are actually visualized in terms of some of the cubic system here in this case you have this FCC unit cell and then this is D110 planes and then you also have D200 planes and we have already discussed this how this the phase relations really cancel the some of the intensities out of this reciprocal lattice and with respect to this diagram 2 diagrams that is A and B we will just now solve some simple problem to appreciate how the selection rules and the phase relations which whether it is add to the intensity or annual or and cancel the intensity to 0. So I would like to go to the blackboard and then solve this simple problem regarding this simple BCC unit cell. So now we will look at the extinction of peaks for some interplanar spacings in this case we will take a BCC unit cell. Consider a simple BCC unit cell with a cell parameter a0 equal to 3 angstrom let x raise with wavelength lambda which is equal to 1.5 angstrom be incident on the 1 0 0 phase. So now we will calculate the structure factor. So this is the structure factor equation because of the 2 atoms per unit cell. So when hkl are all even such as for 2 0 0 planes for example the terms for the corner atoms that is e power 0 and the body centered atom e power pi i into h plus k plus l atoms are plus 1 that means what does that mean that means all atoms scatter in phase and a 2 0 0 peak is present. So according to this rule all atom scatters in phase so these things you have to remember and 2 0 0 peak is also present. However when h h plus k plus l are all odd such as 1 0 0 plane the terms are plus 1 and minus 1 respectively and the 2 sides exactly out of phase and there is 1 0 0 peak. So in fact what I have written is what I have shown in the slide in the previous slide where I showed the reciprocal lattice where you could see only these kind of planes not the this kind of plane because of this extinction rules. Now we can prove this concept geometrically we first solve Bragg's law for scattering from Bragg's law we can write for scattering 1 1 planes lambda is equal to 2d. Let us understand what we have done so far from the Bragg law for scattering from 1 0 0 planes you can simply write lambda is equal to 2d 1 0 0 plane sin theta 1 0 0 from which we can find out the sin theta value sin theta is equal to 1.5 divided by 2 into 3 which is equal to 0.25 correspond to the theta 14.48. The difference in the path length between rays 1 and 2 the of the rays shown in the slide you can see that we can obtain it from sin theta 1 0 0 equal to ac divided by d 1 0 0 which is equal to dc divided by d 1 0 0 which is equal to 1.5 angstrom which is 1 x ray wavelength. So that is what we have started with so it is 1 x ray wavelength. Now the question is what is what about the x ray scattering from 2 0 0 planes at the angle theta 1 0 0. So we can find out in a similar way the path length so the path length for 2 0 0 planes is 2 times a dash c dash which is equal to 2d sin theta 1 0 0 we can substitute this you will get 0.75 angstrom as a path length which is half of the so what is the consequence if you have the path length from 2 0 0 planes is exactly half the wavelength as we have discussed previously it will completely cancel out the intensity we can write that which will so when you have the path length for 2 0 0 planes is as half a wavelength then the wave from these planes completely cancel with those from 1 0 0 planes that is why you do not see the peaks like 1 0 0 3 0 0 and so on in that reciprocal lattice. Now let us consider the scattering from so let us consider the scattering from 2 0 0 planes at the angle theta 2 0 0 if you look at the slide the figure B shows this schematic excuse me and then we will now proceed with the calculations the Bragg law for this case gives sin theta 2 0 0 is equal to lambda by 2d 2 0 0 which is nothing but 1.5 angstrom divided by 2 times 1.5 angstrom which is worked out to theta 2 0 0 is 30 degree. Now at this angle the total path difference the total path difference is 2 times ac which is 2d 0 0 sin theta 2 0 0 which is nothing but 1.5 angstrom which is nothing but 1 wavelength now similar question what we have asked there what about scattering from similar question here what about the scattering from 1 0 0 planes at the angle of theta 2 0 0 now then we can work it out this also total path difference. So, the total path difference in the case of scattering from 1 0 0 plane at the angle of theta 2 0 0 is 2 times a prime c prime which is equal to 2d 1 0 0 sin theta 2 0 0 which is equal to 2 into 3 into 0.5 which is nothing but 3. 3 is 2 lambda this value is nothing but 2 lambda this is which is nothing but an integral multiple of a wavelength lambda and supports the constructive interference process. So, from this example we have seen how this the 2 0 0 type of planes exist and what is the meaning of extinction of certain peaks in for some interplanar spacing in crystal in a cubic crystal like a BCC. So, these 2 examples gave you some kind of an appreciation for this concept and now we will move on to the next example what I would like to show. Come back to this slides if you look at this slide this is a structure factor calculation for a phase centered cubic structure and if you write the expression you have 4 atoms per unit cell. So, you get the 4 terms here. So, 1 plus e pi i h plus k plus e pi i e to the power pi i h plus l plus e to the power pi i into k plus l. So, the rule is f h k l is equal to 4 f where h k l all even or odd and the intensity will become our structure factor will become 0 when h k l are mixed. So, now you can see that this is again a reciprocal lattice cell. So, do not confuse this with your real cell this is a reciprocal unit cell that means you will see the intensity spot actually in a double cell size and wherever you have the allowed reflection then only you will see the dark circles. So, here you can see that all h k l all even or all odd will have the allowed reflections. So, other positions are forbidden. So, what you can now appreciate is the reciprocal unit cell for an FCC lattice actually a BCC lattice and it is the same thing what we have seen before it is the other way around for an a real BCC lattice the reciprocal lattice is an FCC unit cell. So, that is clearly appreciated here. Now, we will see some interesting examples a a CCM chloride structure where you have a CCM chloride type of structure you can see that it is kind of body centered unit cell has got two atoms that is chlorine and CCM atoms per unit cell which is one in corner another is in the body centered position, but you can write the expression in terms of the structure factor equation like this and the selection rule is you see F h k l is equal to F cl plus F c s where h plus k plus l is equal to 2 n and if it is F cl minus F c s where h plus k plus l is equal to 2 n plus 1. See what you have to appreciate in this particular example is it is though it is a BCC crystal it is the since it is chlorine and the CCM are two different atoms it is to be considered as two BCC crystals are inter-percentrating with each other it is a kind of an ordered system and you have you should also appreciate that since these two atoms are very I mean different chlorine and a CCM their scattering factor will be will be very different induced as an individual atom. So, you will see instead of in some some reflections instead of a sum of the peak intensities you will see the difference in the scattering power that is why you see in some terms it is plus and some term it is minus here and the corresponding effect you will see in the bigger circle here that means higher intensity that is sum of intensity and then you have some of this it is the difference peak this is a difference peak this is a sum peak that is what it is written the peaks for which h k h plus k plus l is even or some peaks that means when h plus k plus l is equal to 2 n means a sum peaks and are bigger than these those with h plus k plus l is odd something like the red spot here which are all difference peak it is not a sum peaks it is called a difference peak here and you can this is again a reciprocal lattice cell and you will be able to appreciate this when you look at the electron diffraction pattern much more clearly you will see this kind of an evidence actual practical evidence for the difference in the intensities and this structure factor calculation nicely shows that the difference in the intensities because of the factors like this. Now you will move on to another important crystal system called a sodium chloride structure and this is similar to C-C-M chloride you have a 2 interpret rating FCC crystal units and you can see that the 2 positions are mentioned here 1 and I mean you can this is a final expression what we have written and you can see that f h k l is equal to 4 times f cl plus f na where h k l are all even that is a sum peak you will see and there is a difference peak when h k l are all odd and if they are mixed there is the structure factor is 0. So, you can see this you can look at this selection rule and then corresponding the reciprocal lattice structure you can see that all a sum peak will have a higher intensity that is a larger circle dark circle and then there is a difference peak like 1 1 1 you will have a lesser intensity and then you have a 0 intensity in h k l are all mixed like this. So, it is just to give you an idea how the structure factor nicely gives you the idea of the intensity summing up or it is becoming 0 or it becoming at the difference in the scattering power of the atoms. So, these illustrations clearly shows the simply demonstrate the significance of the structure factor in appreciating the x-ray diffraction intensity. So, having said that we are only interested in appreciating the or understanding the intensity from the single crystal or it could be a poly crystal or it could be an amorphous material, but we have to now look at the intensity expressions what are all the term it contains. So, before we really get into that we have to appreciate what are all the factors which influence the intensity that we have to look at it. So, before I get into that discussion let me tell you the we have when you when you look at the x-ray diffraction spectrum in every peak there are two things you have I max and you have the the breadth of the peak or the integrated intensity there are two these three things are very important I max is always not important, but it is that integrated intensity which is important which is the area under the whole peak and this is influenced by so many factors and this we have to understand one by one and then how this the broadening of this intensity x-ray intensity peak is related to what we are looking at in the material that is also another important factor we will look at one by one. So, I will go back to the work board again. So, before we talk about the intensity of the x-ray peak we should know what is the theory behind it in when it in it being calculations are carried out and so you see that the kinematical theory is used for calculating the intensity of x-ray and it has got some assumption which is very important. So, I will write down one one three important assumptions. So, the kinetic theory of diffraction assumes these three points that is no interaction occurs between incident and the scattered waves. Secondly waves are scattered only once and thirdly scattered waves do not lose energy. So, based on these assumptions the the intensity of the x-ray is being calculated and then now I will just list out some of the factors which will influence the x-ray intensity. There are six factors. So, there are six factors that will affect the relative intensity of the diffraction lines in a polycrystalline material. One is a polarization factor which we have already seen it you know you have some idea about what is polarization factor and then structure factor which we have now seen. So, at least these two are familiar. The third one is multiplicity factor and fourth one is Lorentz factor, fifth one is absorption factor and sixth one is temperature factor. So, all these factors have to be taken into account when you write a complete intensity expression for a diffracted beam and before we get into all this you have to remember that of course, this intensity also what you obtained is depending upon the method by which you perform this experiments. There are primarily three methods that also I will write before we get into the intensity discussion. So, there are three methods. One is Laue method, another is rotating crystal method and third is a powder method. I will not get into the theory of all these things. I will take you to the lab laboratory and then actually show what are the typical methods we follow in the x-ray diffraction laboratory when we actually perform the experiment. Just for the sake of completion you should know the basic method of conducting this diffraction experiments. In a Laue condition or a Laue method where the lambda is variable and your theta is fixed. In a rotating crystal method lambda is fixed and your theta is variable partly and in a powder method you have a fixed lambda and then variable theta. These are all just about the methods by which you measure the x-ray diffraction intensity for a polycrystalline material. So, we will look at the importance of the variable lambda and fixed theta are fixed lambda and variable theta when we do the laboratory demonstration. So, now we will come back to the intensity of these x-rays. You have to have some idea about what all these factors mean. We will not get into the very detail of the calculation of this or the derivation of this which is out of the scope of this course, but you should have some idea what is this factor because all these factors are going to come in the final intensity equation and then we will briefly go through what is this multiplicity factor and Lorentz factor and absorption factor and temperature factor so on in the next class. Thank you.