 So in this lecture, I will continue with our discussion on small angle neutron scattering. I have introduced you to the subject and I told you that this is a small angle, basically experiments that we do at small q. And for small q, as I told you that q is equal to 4 pi by lambda sin theta, lambda sin theta. So we take recourse either to go to smaller q or to go to larger lambda. But in general, the technique is known as small angle neutron scattering. And I had just introduced you to the subject and I told you that when the q is small, when the q is small, the inherent resolution in an experiment is given by twice pi by q. So delta R becomes slightly larger. Actually, if q is equal to 0.1 angstrom inverse, in this expression, I can show you that it becomes immediately 6.3.1, 6.2 by 0.1. It becomes around 62 angstrom or around 6 nanometers. So now I have intentionally reduced the resolution of my experiment to look at my sample with a broader scale, which is called a nanoscale. And this is a very important technique used in soft condensed matter which we have used and also in precipitates and pores in rocks in this kind of fields. I will explain you with examples because that is important. And I also discussed with you the two instruments that we have in Dhruva. One is the slit-based sands and there is double crystal-based sands. In the slit-based sands, this is the most commonly used instrument in most of the sources because here we use basically primarily a long flight path. So you have got a large distance between two slits that dictates the resolution. Then you have a sample. Then you have a further large flight path before it goes to the detector. This large flight path because your sample scatters in a very small angular window in which you are interested. And to expand that angular window onto let us say a two dimensional detector or a linear detector, I need to move it outward. So these distances are long. So in this old module, we had a reasonably long flight path. You can see it in this diagram. And also then there is a large detector shielding at the back of which there is a one dimensional detector. There is a normal slit-based sands. Also we have something called a double crystal-based we call it M-sands, medium resolution sands. So basically what we have are two perfect silicon crystals in parallel geometry. Now please understand that if I have a perfect single crystal here, a perfect single crystal gives a beam width which is known as Darwin width. Intensity versus angle if I plot. For a Bragg angle, it will be a Darwin width. Typically looks like this, which is few arc seconds. Now this beam width falls on the second, a beam of this width falls on the second crystal. And then if I rock the second crystal, that means if I rock the second crystal, then it will erase the width of this beam on to my detector. So if it is a perfect single crystal, then the rocking the second crystal gives me the width of the beam in the first silicon single crystal or any perfect crystal. In this case it is silicon for this instrument. Now if I put a sample between the two crystals, then this sample causes a width, a widening of this incident very narrow beam. And when that happens, then the rocking of the second crystal actually catches the widening of this beam. So this is unlike other instruments. Here it is two single crystals cleverly manipulating a very low Q region. We can go to lower Q range and low Q range lower and resolution or studies of size at a larger. So this is here the here the in this instrument which I am discussing the incident wavelength is around 3 angstrom. So this is the double crystal monochromator based system. So you can see one crystal is here and this is the second crystal which is rocked which you can see here the second crystal and which is rocked with a very precise angular movement. And the sample comes right after the first crystal in the beam path. And so this rocking curve is arrested by theta to theta movement. This is the detector which is rotating in lock lock with the second crystal in a theta to theta mode. So there are two instruments which we have got. So briefly this is what we had earlier. So it was a beryllium oxide filter based monochromator. So it's not really monochromator it's more of a filter but it doesn't interesting purpose. So if I can just narrate to you basically it is a combination of various things. A guide has a transmission curve which is like this. This is transmission intensity versus lambda. A guide has a cutoff wavelength and this cutoff wavelength depends on for a bend guide how bent it is. I will stop here explaining why there is a cutoff but this is because of the radius of curvature of the guide and in the guide on which this instrument is put the cutoff is at 2.2 angstrom. So beyond that I have a flat path. Now that means what the guide transmits but the fact is that in my beam which is coming through the guide if it is a Maxwellian certainly the Maxwellian looks somewhat like this and because there is a cutoff at 2.2 angstrom so it will look somewhat like this or 2.2 angstrom maybe is a yarn. So there will be a cutoff below which it will not transmit and above which it will transmit. But now there is a beryllium oxide filter one more thing but does a beryllium oxide filter does it's the polycrystalline material. A polycrystalline block there are crystallites at all possible orientations inside the block. So if I have a lamp if I have a block satisfying 2D sin theta equal to lambda that will be scattered out. Now because it's a polycrystalline block there are many many many crystallographic planes and till you know sin theta sin theta equal to lambda by 2D so till lambda is less than 2D I will get sin theta less than 90 degree and some block or other will reflect out neutrons from the beam path and when lambda is greater than 2D so there is one more cutoff wavelength the polycrystalline allows their beam to pass through. There is a filter which has got a cutoff at lambda and lambda below this lambda minimum are not allowed to pass they are reflected out by Bragg reflection and lambda greater than lambda M they will pass through it. So this is a transmission curve there is a transmission spectrum and there is this polycrystalline block which sort of in an idealistic case will give a transmission like this with lambda. But all these things three things are multiplied together and what to get actually a sharp rise and a fall because on the longer wavelength side anyway in the maxolian your intensity is falling and in the shorter wavelength side you have cutoff with this bradylium oxide and also the on the guide. This cutoff is not so sharp as we look at it ideally but it is slightly longer so it will be somewhat like this and now if I look at it now if I look at it one second this is how it looks like. So this is the bradylium oxide filter beam this is not a perfectly monochromatic beam actually delta y lambda is quite large 15 percent and the mean lambda is 5.2 angstrom this is what we are using earlier. So and when I impinge this beam on a sample I have got two parts one is the transmission data which you see here that means the beam the direct beam which has passed through the sample and put its footprint on the detector and you have scattering data on both the sides of the detector. So on a one-dimensional detector we have this is a 0.1 molar solution of C-tab and you can see this is the small angle data this is the small angle data that we collected. You can see also at the back we have also shown the data from D2 which does not have any anything like this C-tabs are basically large inhomogeneities in a solution which can be seen D2 has nothing like that it is just a solvent and you get a flat background with a direct beam profile. So that means in these experiments we could not go below these angles because at the smaller angle side this will cut the direct beam and we have got data up to a certain angle this is coming as channel number over here in a one-dimensional detector which will dictate the Q value and I have just taken the trouble of the whole thing to plot the resolution which changes you can see with the Q or the angle or the channel number in the one-dimensional detector and this is what the resolution function is. So this has been changed now with a velocity selector. What does the velocity selector do? Now I earlier I told you I had a beryllium oxide filter cool the beryllium oxide filter was cool because if I don't cool it if I don't cool the polycrystalline block which is acting as a filter it will cause thermal scattering of the neutron which is undesirable. I want a sharp cutoff I want neutrons of wavelength lambda below lambda cutoff cutoff to be completely stop from going so that's why we used to cool the beryllium oxide polycrystalline block. Now we have a velocity selector now in case of small angle neutron scattering or any we know that Q is equal to let me just quickly mention this to you 4 pi by lambda sin theta so this you can work out delta Q by Q will be equal to delta lambda by lambda square plus delta theta by theta square just single one differentiation delta Q will be 4 pi by lambda square delta lambda minus and one will be 4 pi by lambda cos theta delta theta cos theta and sin theta and dividing one with each other and assuming cos theta equal to one and sin theta equal to delta theta at low angle you will get this relation for the small angle scattering now when it is delta lambda by lambda square delta theta by theta square usually delta theta by theta will be in the of the order of 10 percent first small angle 10 percent I'm sorry not 10 angstrom 10 percent or 10 percent means delta theta by theta will be 0.1 better to write in number 0.1 0.1 which is 10 percent so if this is acceptable then you can see from this expression delta lambda by lambda equal to 0.1 should be acceptable to us so one take away from this expression is that if I am working in the small angle neutron scattering zone I can relax my wavelength resolution this I can do when I'm working at a larger angle in a crystallographic diffraction so I can accept delta by lambda by lambda at 10 percent and how to do it we can do it mechanically so I told you this earlier again I will quickly tell you there is a cylinder a velocity selector is nothing but a cylindrical block which has got helical slits cut on the body cut on the body and the neutron which is traveling a certain lambda traveling in a straight line path in this rotational medium when I convert this it becomes a helix and that helix which matches the helicity of these will pass through the velocity selector so it's called velocity selector basically it's the wavelength selector undeterred and will go through and in this mechanical assembly I can get 10 to 15 percent 10 to 15 percent velocity width which will add to the intensity and at the same time will not disturb the experiment because I know my delta theta by theta is of the order of 10 percent so here you can see that we can actually calculate out the source slit is 3 centimeter by 2 centimeter the sample slit is 1.5 centimeters by 1 centimeter so before the sample at 2 meter distance these have been kept so this will be for simplicity let me just say there is a 3 centimeter slit this is 2 centimeter slit so this is the resolution angular resolution theta and this distance is 2 meters so this is a let us say 2 centimeter this is 3 centimeters and this is 200 so I can do it trigonometrically very simple task so but without doing it if they were equal it will be approximately delta x by 200 you can work out and here so I have if I assume both of them are 3 centimeters or 2 centimeters here it is slightly different it will be delta x by 200 this delta theta assuming 10 theta equal to delta theta and that's around how much if I say 2 centimeter if I consider 2 by 200 so it is around 0.01 as you can see so in this case this is the angular width which I can tolerate here and distance from s2 and the detector is 1.5 meter and here the detector is a multi detector setup so I have given the expression for the resolution excuse me and I have also given you the detector setup which is 1.5 meters away so it's an improved version and this is being used till last few years so this is the double crystal based medium resolution sounds which I explained to you just now so these are actually both these crystals one here and one here the silicon 1 1 reflection from the incoming guide beam the guide actually traveling like this I can show you later that on this guide actually there are so this guide goes on then a beam comes out to this instrument guide goes on further then beam comes out to the reflector meter then this guide goes further then there's a billets detector and the sound machine so there are one two three instruments on the same guide here I am talking about the first gap in this the first outlet in this neutron guide and that's the advantage of neutron guides that on a single guide you can put number of instruments here we have put three so this is one of them so in this instrument we have silicon 1 1 1 as the first monochromator and also silicon 1 1 1 as the analyzer which rocks with the theta in the theta to theta mode to obtain the broadening caused by the sample the lambda average is 3.12 angstrom and here the delta lambda by lambda is much better with one person and the but the flux is low there it is around the previous sound machine that I showed the flux was around 10 to our 5 neutrons per centimeter square per second here the flux is much smaller much smaller 500 neutrons per centimeter square per second these numbers are actually quite small if you compare them with the standard extra machine but we can work with them there's a bf3 detector and the q-ring is actually this is what something which you need to see it is 0.003 to 0.17 nanometer inverse so if I consider this twice pi by 0.003 it can see much larger objects but there is a q minimum that you can reach and twice pi by 0.003 would be around 6 divided by 0.003 so 6 1 2 3 by 3 you can see things which are around 200 nanometers large the other instrument which I described to you I can tell you typically around 1 nanometer to 10 nanometer will be its range it can be slightly less slightly more approximately here I can go to much larger objects and the studies done include ceramics metallurgical alloys micro granules and fractals I can't give you example from all of these but I'll try to choose some of the examples to explain to you the utility of this particular imsaans machine so here it's given actually 40 to 1000 it's actually sorry I said 200 but 1000 nanometer they want to say okay so and another important tool in case of neutron is using the contrast matching this is interesting we are talking in terms of scattering length density and not in terms of atomic scattering strength scattering length density again and again it'll come to this term known as SLD so we are talking in terms of macroscopic things like density and not in terms of microscopic atomic structure because here I have inherently reduced the resolution reducing the resolution is making the resolution broader in terms of size broader so in that size scale I don't talk about atomic structure of the medium but we'll talk in terms of density so here suppose I have a sample what is the what the sample will be I'll complete later but suppose it has got a scattering length density as I told you earlier and it is in a medium mostly in a liquid medium so such samples are basically soft materials sometimes proteins sometimes surfactants and micelles and polymers these are the materials mostly studied by small angle neutron scattering then in a medium in a liquid medium you please see I worked it out for last time the scattering length density for D2O is 6.38 centipus centimeter square the way I did it if you remember I showed I worked out the scattering length for one hydrogen which is 2 into BH plus BO or 2 into BD plus BO that is H2O and D2O then I calculated in 18 grams of H2O grams you had 6.023 10 to the power 23 molecules so in 1 cc how much you have you can calculate out it is just 1 by I mean 1 cc weighs 1 gram so 1 by 18 grams similarly you can do the same thing for D2O increase of D2O not 18 because it is 1 H2O so it is 20 grams of D2O we will have the same number of molecules each molecule you can calculate out the scattering length so now it is how much is in 1 cc we know so that we can calculate out and finally we get the scattering length density which is per centimeter square because this comes in femtometer which is a length scale you can put it in centimeter you divide it by 1 cc because you are doing per cc so centimeter by centimeter q gives you centimeter to the power minus 2 and I have got scattering length density of water scattering length density of D2O one is positive one is negative that is the most interesting thing now as you can see I can enlighten up different parts of my scatterer so I can lighten up the if it is a core shell structure I can lighten the core I can make it the contrast strong and I can match the contrast of the shell with the surrounding solvent by judiciously choosing D2O H2O mixture and I can only see this part or this part either the core part or the shell part and you can see by color showing it that when I match the this thing the scattering length density of the solvent with that of the shell then what I see is this or when I match the scattering density of the core with the solvent what I see is the shell and I can play with this thing and it is very routinely done for neutron small angle scattering and it's a routinely used technique so this is what is just a plot of that you can here given some of the routine things which are used regularly in small angle like lipid RNA protein water is just scattering length densities and you can see that percentage of D2O H2O if it is 100% it goes to 6. something this is minus 0.57 and as you keep mixing D2O H2O you can go across the whole band of scattering densities where you can match one or other match the solvent scattering density with various parts of a scatterer as I showed you