 So far I have described how to find out macroscopic strains in a crystallographic sample, in a powder sample, and how to obtain the strain values from such measurements. Please note that when I say powder, it is not a powder in a sense that we commonly call something a powder. The powder means it is a solid object for all practical purposes, but it has got all possible orientation of the crystallographic planes, and this is the crystallographic definition of a powder sample and not the powder that we know of, like a talcum powder. So please note that when I say powder, it means that there are all possible orientations. So far I have discussed macroscopic strain. That means the entire crystal lattice is either compressed or elongated, and how to study various cases of interest, like I took two examples, one the welded pipes and another one was a railway track. There is also micro strain in such samples. Now what is the difference between micro strain and macro strain? The micro strain, especially for nanocrystals, they vary from part to part. So here, actually it causes a broadening. So we often use in extra diffraction what is known as a Sherer formula. So the finite size of a crystallite broadens the diffraction peak. So finite size means the crystallite is small. This is because when we say that the bright diffraction peak is a delta function, actually it is not a delta function. All of us know that a perfect crystal bright peak has something called a Darwin width which is in arc seconds. But for all practical purposes, they are delta function. But actually if there is a finite size of the crystal, then there is a broadening of the diffraction peak. So there are two kinds of broadening now which are coming together. One is the particle size and broadening due to that and the other is micro strain in the sample and broadening due to that. Now finite size of a crystallite, that is the crystallite size we obtain using what is known as the Sherer formula. Sherer formula tells us that delta is the broadening of the peak. This is often used in x-rays. It is given by K lambda by beta cos theta where K is a shape factor close to 1. Lambda is the wavelength of the x-rays. It can be neutrons also and lambda is for x as often it is 1.54 angstrom, we know for copper and neutrons it depends on the instrument you are using. Beta is the peak width for a Bragg angle at theta, so beta cos theta. This is often used in x-rays. Now I am taking you back to your master's days. So this we can also see that if we consider a one dimensional chain, actually for a one dimensional chain of n crystallite size, the broadening of the Bragg peak is given by sine square n qd by 2 upon sine square qd by 2 and as n goes to infinity, this expression goes to delta function. I have shown you here that using this formula, we could obtain the peak shapes for a highly oriented YBCO crystal and from there we could actually fit the n value that gave in this case it was a thin field. So for a one dimensional chain it is easy to evaluate, but even for a three dimensional chain the phenomenological formula of K lambda by beta cos theta is used often to find out the particle size broadening. So here this particle size broadening is due to the finite size of the crystal, but the fact is that the x-ray has got or the neutron has got a covariance length and the particle size should be smaller than the covariance length for this approach to be effective. Also there is strain related broadening. Here the strain related broadening is not a macroscopic broadening, but here there are defects, vacancies around the site and also there can be sometimes stresses due to a grain boundary and something migrating through it. So here again the black peak keeps shifting because there is a micro strain that means somewhere it is elongation and somewhere it is compressional and this black peak gets broadened due to this micro strain. So now the broadening of the peak one is due to particle size and one is due to micro strain. This is because this is different from what I did in the previous part of my talk. So these two cause a broadening of the peak and from these two we can find out the value of the particle size and also the strain. This is given by something known as Williams and Hull plot. Now the variation of the lattice spacing in micro crystallites coming from micro strain that is coming as micro crystallites and this is different from macro strain which I discussed earlier. Here is some constant C epsilon is the strain value and tan theta but theta is the Bragg angle and there is also the broadening due to particle size which is K lambda by beta cos theta. So we have two deltas so I have now delta total I consider this as sum of the two delta due to strain plus delta due to particle size. So this comes to C epsilon tan theta plus this comes K lambda upon beta cos theta. Now please note that I have written this to a summation. The summation means what I assume in this actually that the broadening due to strain and the broadening due to particle size they are independent of each other and I can add them up independent. So this is an assumption. If it is not true then I cannot use this formula but I go with the assumption that the micro strain and the particle size they don't interfere and the total broadening is given by the broadening due to strain and the broadening due to particle size. So then the delta total if I multiply it by cos theta it comes to C epsilon sin theta plus K lambda by beta. So now with this formula which is very handy but highly macroscopic and phenomenological I won't justify it all the time but often it is true for engineering samples. With this I can have a plot where I can plot this is the width because theta and if I plot it against sin theta you can see the epsilon is obtained from the slope of this because delta the b cos theta or delta cos theta is taken from a side so they are the same values and plotted against sin theta the slope gives me C epsilon and the intercept gives me K lambda by beta, beta is L here. So that means I can obtain from this Williams and Hull plot the value of the micro strain epsilon from this plot and also the particle size using Williams and Hull plot and you can see here actually various intensity of various black peaks have been plotted against sin theta. So this plot gives me strain as well as particle size in one go but this is not microscopic and there are other more elaborate theories which I will refrain from discussing now but Williams and Hull plot used for X-rays can also be used for neutrons. Not too many examples but in recent days this is the Williams and Hull plot has been used for nanocrystalline and micron size nickel powder. Nanocrystalline means this micron size powder also are having nanocrystallides which are 12 to 30 nanometer or 120 to 130 angstrom large and here these are diffraction peak time of flight diffraction peak I must point out to you that this is a time of flight peak that means this means this is just a mirror image of the kind of diffraction peak that you will get using neutron and monochromatic sources. So because this D spacing is increasing means time of flight is increasing. So the small angle peaks go to large D spacing peaks here but I just show you one the inset shows that how when you anneal this sample you can see there is a broad peak which is shifting to larger theta that means the lattice is relaxing to a lower value and also becoming narrower that means to annealing at 1000 degree centigrade the particles are growing nicely shows this plot. So this plot and with respect to the data obtained from here is the Williams and Hull plot it's only I add them linearly this Williams and Hull plot is actually square summation but in principle they are the same and you can see that the particle size and the microstrain both can be obtained from the Williams and Hull plot today using neutrons. So this completes the part that I want to discuss with you using neutrons microstrains microstrains particle size and these are these two techniques that I discussed now are more of interest for industrial problems and many many times these are done to find out strain in test welding joints or even samples which have been stressed for long time like the railway track I showed and here it is like nickel powder. So next part of my talk is the most interesting for the people who are in magnetism because I will be discussing the neutron diffraction from magnetic lattice and magnetic like magnetic lattice like iron, nickel, cobalt now these are the 3D elements there are also rare earths like gadolinium, samarium there are unfilled 4F orbitals. So these unfilled shells they give rise to magnetism and we can consider this magnetism as a spin sitting at a lattice side. So now there is a chemical structure that means there is long range order for a crystallographic lattice like iron, nickel, cobalt these are all crystallographic materials and along with that there is also magnetic long range order depending on the temperature of study. So extra diffraction is good for the physical long range order or crystallographic structure because X-ray is extremely powerful tool and in synchrotron you can have X-ray beams which are micron size and in intensity possibly at least 10 to the power 8 orders or 10 to the power 6 orders more strong than higher than a neutron source. So X-ray diffraction is very good for finding out long range physical order or crystallographic structure in powder samples except lozid materials because X-rays cannot see till we can say up to oxygen that means the Z values of 8, 8, 0, 16 so up to the mass number of 16. So hydrogen to oxygen they are not seen by X-rays and again for that you have to come back to neutrons and neutrons are tiny magnets. So neutron diffraction will be used not only for chemical structure but actually almost all the time when we do neutron diffraction we are doing it with an intention of finding out chemical as well as magnetic long range order. So my next lecture I will discuss in details I will introduce you to some extent with the elementary magnetic interactions in solid state and then how we can decipher them using neutron diffraction. Thank you.