 Everyone, welcome to this material characterization course. In the last class, we discussed about the x-ray diffraction intensity and the factors which are influencing this intensity and its consequences. And one of the factors which we talked about is about two kinds of intensity that is I max that maximum intensity as well as the integrated intensity. So most of the time we are interested in the integrated intensity and we try to interpret this integrated intensity to arrive at the possible crystal structure determination or the estimate the residual stress or the crystal orientation and so on. So in order to understand the factors which cause the peak broadening which is also very important aspect of understanding this intensity of the x-ray diffraction. So we will continue our discussion in this lecture also about the same thing and we will try to derive some of the expressions which will relate the peak broadening with the crystal structure and crystal size and so on. So I will continue this discussion from there and if you recall I just talked about two kind of theta effect. One is the x-rays exactly obeying the Bragg's law that means they the theta being Bragg angle that is they diffract exactly at the Bragg theta or a slightly away from this Bragg angle the diffraction takes place slightly away from Bragg angle. So these two things are going to give lot more effect on the final I mean expression or even the amount of intensity which we are going to get. So when we talk about exact Bragg angle and then slightly away from the Bragg angle so what is the meaning of this we have already discussed. So when you say that the crystal planes if you consider a polycrystalline material the crystal planes which are going to diffract slightly away from the theta B then their path difference also will be slightly different from the integral multiple of lambda. So what Bragg law states if the path differences in the range of integral multiple of lambda then the diffraction takes place and then we also talk about the destructive interference where you have this path differences exactly one half of the wavelength then they will cancel out each other. So this also has some influence I mean because of the slightly different from the theta B the diffraction will also have a significant influence on the amount of out of faceness and its relation to the crystal structure. So we will look at these concepts once again with a simple schematic and then we will get into this derivation of intensity with the effect of crystallite size and so on. So let me draw the schematic let us assume this spacing between the planes is d and this is the total crystal we are interested with the thickness t which is equal to md and this is mth plane. So this diagram we have already though we have already seen for the sake of completion let me start describing this. So now what we understand is that the scattered beam A prime will have the I mean will have the path difference from this d prime by one wavelength that is what we have seen. If you assume that lambda is equal to 2d sin theta and that is what we said. So that means what these two scattered beam will have fall in the same phase all of them will be in the same phase. So similarly when you have this ray m prime will have will be m wavelength out of phase because it is coming from the m plane so this is one wavelength out of plane this is m wavelength out of plane. Again m prime d prime and d prime will all have the same phase that means they all will contribute to the diffraction intensity so that is very clear that is Bragg glass states. So now the question is suppose if I have a beam which is slightly away from the Bragg angle like this let us call it as B, B prime and C, C prime and similarly here we can have here L, L prime, N, N prime this is incident beam and this is a scattered beam or a diffracted beam. So why we do this? This is for m case this is for a particular given case that is why we are comparing this these two. So now if you apply the similar rule suppose since these theta is slightly different from theta B it could be slightly excess or slightly lower. In this case it is slightly excess theta 1 and in this case it is slightly lower theta 2. Since it is slightly different from the theta B as I mentioned the path difference also will be very small fraction of integral multiple of lambda not exactly integral multiple of lambda the consequences is this is not going to cause the destructive interference. So the plane which is going to have the atoms which are scattered the scattered beam will have exactly half the wavelength out of phase will be somewhere inside the deep of the crystal which we are not seeing because this theta is only fractionally varying only the atoms which is scattered from the plane which is at least have half the wavelength only will cancel the scattered beam from this A prime. So that plane will somewhere lying in between we do not know but if you talk about this beam B and B prime it will have the phase difference between L and L prime m plus 1 wavelength and similarly your C and C prime will have a phase difference of m minus 1 wavelength with respect to n and n prime. So these are the wavelengths at which the I mean the intensity of the diffraction will be 0. So what are we trying to say here since you have the crystal planes which exist or which diffracts slightly away from the theta B there exist at two limiting angles because we are talking about 2 theta here because that is what we measure 2 theta 1 and 2 theta 2. So we are talking about a range of angles which define this that means at these two within the range only your intensity will be varying and then these are two limiting theta where the diffraction intensity will become 0 that we will see. So that is the idea of doing all this. So let us now try to write few points. So the width of the diffraction curve increases as the thickness of the crystal decreases because the angular range 2 theta 1 minus 2 theta 2 increases as m decreases. As the m decreases and your range is going to increase that means it is going to cause peak broadening. So one more point you have to recall we have already discussed in the last class the as the crystal size becomes smaller and smaller and if you consider this theta 1 and theta 2 that is they are slightly away from the theta B then there is that the plane which is going to diffract half a length of phase difference may not exist. So that is another reason why we will see the peak broadening. So that is the bottom line that is the bottom line. So there is a connection between the amount of out of phaseness and the crystals size exist. So this you have to keep in mind before I mean when we talk about the peak broadening and one of the primary reason the physical basis for the peak broadening is this. There may not be a plane which will completely make the path difference out of phase. So now let us write the rough measure of angular width. So which is nothing but the B is equal to half 2 theta 1 minus 2 theta 2. So your B is measured in the angular range in the diffracted peak. So now you write the path difference of these two extreme angles with respect to the schematic what we have drawn. So the path difference for the two limiting angles related to the entire thickness of the crystal is 2t sin theta 1 equal to m plus 1 lambda. So that means we are talking about this ray B B prime with respect to L L prime they will differ in their path by m plus 1 lambda wavelength or m plus 1 lambda and in the second ray which is a less than theta B C C prime will have the path difference with respect to n n prime 2t sin theta 2 is equal to m minus 1 lambda. So now we will manipulate this by subtraction. So we can use some trigonometric relations to replace this. So we can write 2t cos theta 1 plus theta 2 by 2 which is equal to lambda. So what we have done is we are subtracting these two equations and we are replacing this with this expression trigonometric relation and now we say that some assumption theta 1 and theta 2 are very small or very nearly equal to theta B then we can assume this theta 1 plus theta 2 is equal to 2 theta B. This is an approximation to arrive at some relations and we can also assume one more thing that is sin theta 1 by theta 2 is nothing but theta 1 minus theta 2 by 2. This is also an assumption from this therefore so we can substitute all this assumption here. So what you get is 2t theta 1 minus theta 2 by 2 cos theta B is equal to lambda t is equal to lambda by B cos theta B and more exact treatment, treatment gives this as t is equal to 0.9 lambda divided by B cos theta B. So which is known as a sharers formula we will write it here which is known as popularly known as sharers formula and most of our scholars use this especially when we are interested in the fine grained material to find out the crystallite size they use this relation quite often. So we will see what all the precautions we have to take before we explicitly use this but this is a basic relationship between a crystallite size and the peak broadening effect in the x-ray diffraction. This is one of the fundamental aspects of x-ray diffraction. So now we will just illustrate this relation with some numerical example. Of course you have lambda is equal to 1.5 angstrom is equal to 1 angstrom theta equal to 49 degree for a crystal 1 mm in diameter. So what I am trying to say is what is that crystal size effect which will be having some visible effect on this peak broadening. Suppose if you assume that lambda is equal to 1.5 angstrom d is that is 1 angstrom and theta is 49 degree then for a crystal 1 mm in diameter the B due to the small crystal effect alone would be about 2 into 10 to the power minus 7 radians. It is going to be extremely small. So what is the size we will be able to see appreciably. There is some example for example we can say that suppose if the crystal size is about 500 angstrom thick then it would contain it would contain. So if you have a crystal size in the order of 500 angstrom thick then it would contain only 500 planes and the diffraction curve would be relatively broad namely about 4 into 10 to the power minus 3 radians which is easily measurable. So that is one simple example how to realize the effect of this relation and now we will move on to the strain effect whatever we have now seen is a size effect. Now we will talk about a strain effect on the peak broadening. So before we talk about the effect of strain on the peak broadening you have to bring certain things in mind especially when we talk about crystal and crystal size and grain size and so on. Suppose if you are suppose if you take any polycrystalline material which will have I mean which will contain a small units called grains multiple grains of single crystals. So each grain or a single crystal will orient with respect to the neighboring grain with some degree. It could be a low angle boundary or a high angle boundary. So whatever we discussed just previously about the Bragg theta here also it will come into the effect because if your neighboring grain is oriented with respect to a given single crystal the angle is if it is very extremely small. So that orientation also is going to contribute to the x-ray diffraction. So suppose if you say that an x grain or an x single crystal in a polycrystalline material diffracts with theta B if the neighboring grain is oriented for example theta 1 then the diffraction is going to be from theta B plus theta. So that effect we are going to consider similar to what we have seen before. So you have to visualize that we are now talking about a diffraction coming from a polycrystalline material where it contains units of single crystals or grains and then we are now considering a diffraction being coming from two three different grains together all together. So this is with this background we can discuss the effect of strain also. So first we will define type of stresses because what we measure is actually a stress and then we convert that into strain or vice versa. If you measure if you are able to measure strain and then we can convert that into stress we will see that. There are two types so there are two types of stresses which can be identified through x-ray diffraction. One is micro stress which will vary from one grain to another or from one part of the grain to another part on the macroscopic scale. On the other hand the macroscopic stress which may be quite uniform over a lot distances. So these are the two types of stresses which we are talking about here. So similarly we will talk about the strain also a uniform strain and as well as a non-uniform strain. To illustrate that again I have to draw one schematic then we will talk about it. So what I have drawn here is three schematic. Suppose this is a crystal lattice with the deep planes distance from D0 that is an equilibrium distance and then this is the incoming x-ray and this is the diffracted x-ray. So what we are now seeing is that D0 is equally spaced so we describe this as a no strain and then if you assume that this plane is being pulled in this direction and then we say that all the D is stretching into a distance away from D0 that is more than D0 then it is called uniform strain and in the third case what we do is we try to bend the crystal like this towards this direction. So in that case what happens is with respect to the equilibrium position the D0 is going to increase on the top and the which is going to get compressed in the bottom and somewhere in between it will be equal to D0. So there are three situations one is under tension where the D is more than D0 and at the bottom it is D is less than D0 and somewhere in between it will be equal to D0 then we say that the specimen is subjected to or the crystal lattice is subjected to non-uniform strain. So we will now see what is the corresponding peak diffraction peak line. So you see that a typical x-ray diffraction line this is for the node strain state you will have the peak like this and if it is a uniform stretching then your peak is shifted to the left hand side and when you have the non-uniform state you have a bit of broadening with little bit of complexity inside you will see why it is so. So the first point to note down here is so the shifting of the line is the basis of the x-ray method for the measurement of macro stress that is the fundamental information one you should keep in mind because of this x-ray methods are used for measuring the macro stress and then like I said this particular case where you have D is more than D0 and here the D is less than D0 and somewhere in between it is equal to D0 because of that you will have a region in a grain where the D spacing is constant for some region with respect to the neighboring region within the grain itself. So because of that kind of a situation each one of this peak will produce a small, small sharp peak inside and which eventually we observe or record as a broad intensity peak that is why you see this as a very broad intensity spectrum and we will now see that how do we measure this. So we can say that typefracting or I would say that let me write that equation directly if we can directly differentiating yields the relation between the broadening and non-uniformity. So how to differentiate the Bragg law you will get a relation like this. So you differentiate n lambda, I mean lambda is equal to 2D sin theta, uv dash is equal to uv dash plus vu dash kind of formula can apply and then you will be able to derive a relation between delta theta is equal to 2D, I mean 2 delta D by D tan theta but what we measure is 2 theta not theta so delta theta can be written like delta 2 theta is equal to minus 2 delta D by D tan theta. So this is the expression where V is extra broadening over and above the instrumental broadening. So now the question is we are talking about peak broadening because of crystal size. Now we talk about broadening because of the instrument error and also broadening due to the strain. So how to separate these two, how do we know that a particular amount of broadening is because of the strain or because of the crystal size. So that aspects we will discuss in the next class. Thank you.