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Before we leave the module associated with crystal structures,

Â I want to briefly introduce the concept of X-ray diffraction and how we use

Â X-ray diffraction to gain the information about the crystal structure of materials.

Â And in order for us to begin to start out,

Â what we're going to do is we're going to look at a number of sieves, and

Â I've illustrated what I mean by a sieve here in the figure.

Â On the left, I have a sieve which is very coarse so that the largest

Â diameter of particle that actually will pass through this sieve is pretty coarse,

Â in comparison to the size of the particle that would pass through

Â the mesh that I have to the right.

Â So this is a way that ceramists, for example, or

Â geologists can separate out particle size distributions of materials.

Â And the characteristics that are important about these meshes are obviously

Â the opening.

Â And so the larger the opening,

Â the larger will be the biggest particle that can pass through.

Â And depending upon how coarse these meshes are,

Â we'll have wires of different diameters.

Â So meshes are then developed based upon how big the opening needs to be and

Â therefore an associated wire diameter to accompany it.

Â Now what we're going to do is to take two different meshes and

Â note that the mesh size is on the order of microns.

Â We have something that's called 125 Mesh.

Â And that's a particular size.

Â And it corresponds to an opening of 112 microns.

Â If we look at Mesh 625, that's a much finer mesh.

Â And what we will see there is that the opening is on the order of 20 microns.

Â One of the things that I want to draw your attention to is the fact that

Â we have a distribution of points, and

Â those points are coming about as a result of passing a laser through the mesh.

Â And what we have on the left is a coarser mesh,

Â and the one on the right is a fine mesh.

Â But they're made up of an array of woven in a square pattern wires.

Â 2:28

Now what becomes very important in addition to the fact that we develop this

Â symmetry, what we see is that if we look at different mesh sizes starting out with

Â a mesh opening of a 112, progressing all the way down to an opening of 20 microns,

Â what we see is that the size of the spots that differentiate,

Â one mesh size from another, winds up having a reciprocal

Â relationship with the actual physical object that we're working with.

Â So, the courser the mesh, or

Â the larger the mesh size, the smaller will be the spacing between those scattering

Â points associated with the laser beam passing through the mesh.

Â So we develop then this concept of a reciprocal

Â relationship between the real image and

Â the associated diffracted image.

Â 3:34

Now let's turn our attention to how we actually

Â study using X-rays to scatter using a crystal.

Â Now, the atomic scale structures that we're looking at and

Â we're trying to understand are typically between 0.5 and 50 angstroms.

Â And in that particular case, we can use X-rays.

Â 4:00

Now, in addition to the fact that we're looking at spacings that

Â are on the order of 0.5 to 50 angstroms, we're looking at periodic arrangements.

Â And those periodic arrangement leads to what we refer to as constructive and

Â destructive interference.

Â Much like the patterns that we saw in the previous demonstrations.

Â Now what also becomes important, not only can we describe

Â the symmetry that's associated with the overall structure,

Â we begin to understand not only the symmetry, but the spacing, and

Â also the locations of the atoms that are within the crystal structure.

Â 4:46

And we know that, for example, that the scattering of these individual atoms

Â will depend upon the particular electron density or the particular element.

Â So, as we increase the atomic number, we increase the electron density.

Â And so, by understanding not only the scattering

Â phenomenon from the point of developing symmetry, we also will get

Â an understanding of the locations of the atoms in the crystal.

Â Now, using the concept of the duality between particles and

Â waves, rather than using X-rays, we can often use electrons.

Â And there are applications for looking at neutrons for

Â a variety of different purposes.

Â So, not only can we look at, use X-rays,

Â we can look at, we can use electrons and we can use neutrons.

Â 5:45

Now, we know with respect to a looking at a wave, and here I've indicated two waves.

Â And what I want to do is to take those two waves and add them together.

Â And what will happen is I will get a wave pattern associated with the sum of

Â those two wave patterns which is given in the figure to the right.

Â Now what I can also do is I can describe a phenomenon that's referred to

Â as constructive interference.

Â So if I take two sine waves, and they are in phase with one another,

Â I develop a constructive interference pattern

Â which the amplitude is associated with the sum of each of those individual waves.

Â And that leads to a constructive interference.

Â On the other hand, if these waves are out of phase,

Â then what I'd get is a destructive interference pattern

Â where I get that flat line that's given on the right.

Â So we're going to use this idea of constructive interference

Â to describe how we can begin to understand the crystal structure of a material.

Â What I'd like to introduce now is the simple geometry

Â that's used in understanding the development of structure in a material.

Â And we're going to use X-rays to probe this.

Â And what's illustrated here is a diffraction set up where we have

Â an incident beam of X-rays and a scattered beam of X-rays that go to a detector.

Â And remember we want to have constructive interference,

Â which means that those waves that are going to the detector

Â must all be in phase with one another to have constructive interference.

Â 8:18

And then I'm going to look at the ray that goes from point C to the detector.

Â The reason that I'm looking at those particular lengths is that I want to

Â make sure that the conditions are such that the incident and

Â defracted rays, when I look at all of the rays, like ray one and ray two,

Â those are in phase with one another and will lead to constructive interference.

Â In order for that to occur, I must have a particular geometry that comes about.

Â Namely the distance between A to B has got to be

Â equal to the distance between B and C.

Â And more importantly, what has to be true is that that distance,

Â A plus B, or AB plus BC, has to be an integral number of

Â wave lengths in order to have the X-rays in phase.

Â And so I'm going to describe those path lengths.

Â The path length AB and the path length BC are going to be equal to one another and

Â they're going to be equal to that simple geometry that I have of d sin theta,

Â where d is the spacing between the parallel lines of the crystal.

Â And, as a result, I'm going to be able to add those two together so

Â that I get AB + BC, and that's going to be equal to 2 times d sin theta.

Â 10:23

wind up creating a distance

Â of AB + BC times the quantity 2.

Â So that winds up giving us the fact that that path length

Â must be now twice the number of wavelengths.

Â So that set of beams, to be in phase with the other set will be equal to an integer,

Â namely 2 times lambda.

Â As a result what we find is,

Â if we start looking at multiple layers below the surface, so

Â we're talking about going in further and further, deeper into the crystal.

Â And for all of these waves to be in phase, the requirement is that n lambda,

Â where n represents the integer and that's going to represent what we refer

Â to as the order of diffraction, is equal to 2 d sin theta.

Â If we're only interested in the spacing we can talk about

Â something we refer to as first order, where n's going to be equal to 1 and

Â we can solve the geometry lambda is equal to 2 d sin theta.

Â So what that means is, if I understand or I know what the value of lambda is,

Â and I know where the diffraction is occurring for

Â this particular distance of my grid, namely my d spacing,

Â I would then be able to determine what that value of d is.

Â I'd like to illustrate an example here where we

Â consider a series of parallel 1, 1, 1 planes.

Â And I've put up the figure where we describe the stacking sequence A, B, C.

Â So each one of those then is a 1, 1, 1 plane.

Â Now what I can do is I can come up with a relationship between

Â the spacing between these planes and the edges of the unit cell.

Â And what I'm going to do is to write that expression as

Â d for the inner planer spacings of 1, 1,

Â 1 is going to be equal to a0 onto the square root of 3.

Â Now that's the body diagonal that goes through the cube, and

Â because I'm only talking about the distances between one 1, 1,

Â 1 plane and the next parallel, I then have to divide that expression by 3.

Â So that would then give me the relationship between the d spacing, and

Â therefore that d spacing, given a certain wavelength,

Â that diffraction will occur at a specific angle theta.

Â 13:16

And then what I can do is I can describe the relationship,

Â Bragg's Law, for first order diffraction then as lambda equals 2 d sin theta.

Â So if I have lambda, I know what d is.

Â I can tell you where the angle should be for the defraction.

Â Or alternatively, if I actually measure the diffraction and

Â I see at what angle I get diffraction for this particular set of planes,

Â then I can determine the spacing, the despacing from this expression.

Â There is a simple general relationship that exists between the despacing for

Â any set of planes, and remember these planes are integers.

Â And that relationship simply reduces to a0 divided

Â by the sum of the squares of the indices, h squared,

Â k squared, and l squared, in the form that's on the slide.

Â So now what we've done is to talk a little bit about the concepts

Â of the rudimentary geometry associated with X-ray diffraction.

Â Thank you.

Â