9. Diffraction. Lattice vectors A real-space (direct) lattice vector can be represented as

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1 1 9. Diffraction Direct lattice A crystal is a periodic structure in real space. An ideal crystal would have complete translational symmetry through all of space - infinite, that is. So, oviously, there are no ideal crystals, ecause they all have surfaces. Anyway, a crystal can e sudivided into identical unit cells. All unit cells are equivalent; if we see at the corner of any one of them, the neighoring environment should e the same. There is more than one way to select the unit cell. We usually pick one that highlights the symmetry of the crystal, such as cuic or hexagonal. Lattice vectors A real-space (direct) lattice vector can e represented as ruvw ua1 va2 wa where uvw,, are integers and, a 2, and a are the direct-lattice asis vectors. A reciprocal-space lattice vector can e represented as ghk h1 k2 where hk,, are integers and 1, 2, and are the reciprocal-lattice asis vectors. Scalar and vector products There are two ways of multiplying vectors. Say we wave two vectors, expressed in Cartesian coordinates: r1 x1xˆ y1yˆ z1zˆ r2 x2xˆ y2yˆ z2zˆ where xˆxˆ 1, xˆyˆ 0, xˆzˆ 0 yˆxˆ 0, yˆyˆ 1, yˆ zˆ 0 zˆxˆ 0, zˆyˆ 0, zˆzˆ 1 The first type of multiplication gives the scalar (dot) product: r1r 2 x1x2 y1y2 z1z2 The second type is the vector (cross) product: xˆ yˆ zˆ r1r2 x1 y1 z1 y1z2 z1y2xˆ x1z2 z1x2yˆ x1y2 y1x2z ˆ x y z 2 2 2

2 2 Determining reciprocal-lattice asis vectors We can write the unit-cell volume using dot and cross products: V a1a2 a We usually want to pick right-handed set of asis vectors, such that a1 a2a 0. In that case, the reciprocal-lattice asis vectors are related to the direct-lattice asis vectors: 2 1 a a 1, 2 a a, a a V V V These have some useful properties. For example: 1 2 i a i 1, and i a j 0, i j Matrix representation In matrix form, vectors appear as column matrices: a1 x a2 x ax a 1 a 1y, a 2 a 2y, a a a y 1z a 2 z a z We can comine this into a direct-lattice asis matrix: a1x a2x ax A a1y a2y a y a1z a2z a z This lets us write any direct-space vector as: u ruvw A v w We can do the same thing for the reciprocal-space asis: 1 x 1 1y, 1z Then 2 x 2 2y, 2z x y z B and 1x 2x x 1y 2y y 1z 2z z h ghkl B k l Notice that

3 a a a T A B a2 1 a2 2 a a 1 a2 a where the T superscript represents the matrix transpose, and 1 is the identity matrix. From this we can see that A and B are different versions of the same thing: T 1 1 T B A A This points us toward a computation method to find B if we know A, and vice versa. For example, B is the inverse of the transpose of A. Another property of RLVs Notice that the dot product of a reciprocal-lattice vector with a direct-lattice vector gives: ghk ruvw hu kv w n where n is an integer. So we can say that, for any hk and uvw 2 e ighk ruvw 1 We often come across a so called lattice sum, which is a sum over many unit cells. The sum gives N N 2ighk 2 e ruvw ighk n e r 1 N uvw,, n1 n1 where N is the numer of unit cells. We will use this result later. Unit-cell specification The dimensions of a unit cell can, in general, e specified y six parameters: the lengths of the three direct-lattice asis vectors, called the lattice parameters (or lattice constants), and three angles etween the vectors, where a a, a a, a a and a2 a a2 acos1, aa 1 aa1cos2, and a1 a 2 a1a2 cos Fewer of these quantities may e needed if we have information on the symmetry of the unit cell. For example, a cuic crystal only requires a single lattice parameter. Example As an example, consider an orthorhomic lattice, meaning that a1 a2 a a1 and We might as well orient the asis vectors along the Cartesian coordinate axes:

4 4 a1 axˆ, a2 yˆ, a cz ˆ The unit-cell volume is easy to find: V a1a2 a ac Now find the reciprocal-lattice asis vectors: xˆ yˆ zˆ xˆ yˆ zˆ xˆ yˆ zˆ c a c c xˆ 1 1 a ˆ ac x, y ˆ, z ˆ a c a ac ac c Notice that the reciprocal lattice is also orthorhomic. The lengths of its asis vectors are the inverses of the lengths of the corresponding direct-lattice asis vectors. Direct- and reciprocal-lattice parameters We know how to find the reciprocal-lattice asis vectors, given a particular set of direct-lattice asis vectors. We can identify lattice parameters and angles for these reciprocal-space vectors in an analogous fashion.,, and cos1, 1 1cos2, and cos Say we have an orthorhomic crystal. a1 a, a2, a c, We saw that the reciprocal lattice is also orthorhomic, i.e.: , 2,, a c Determining interplanar spacing The interplanar (d) spacing used in Bragg s law is given y dhk 1 ghk, where ghk h k The general method to find the length of a vector is to dot it with itself and take the square root, so: 1 2. g g g g hk hk hk hk 0.5 h k l 2kl 2lh 2hk h k l 2klcos 2lhcos 2hkcos This is fairly complicated for a low-symmetry crystal, ut simplifies greatly with higher symmetry. Cuic, for example, reduces to: h k l, d a hkl g hkl a h k l The d-spacing for orthorhomic is not what you might guess off the top of your head: g hkl h k l, dhkl 1 h k l a c a c

5 5 Miller indices We can define a coordinate system with respect to the direct-lattice asis vectors using a1 a1a ˆ1, a2 a2a ˆ 2, and a aa ˆ Consider an aritrary direct-space vector, specified in terms of these asis vectors r xa1 ya2 za For any real numers x, yz,, r extends from the origin to a point on a plane normal to the RLV ghk if ghk r hx ky z A where A is a constant. A plane can generally e expressed in terms of its axis intercepts: x y z 1 x y z i i i Now we can find where the plane intersects any of these axes y setting the other coordinates to 0. A A A y z 0 xi, z x0 yi, and x y 0 zi h k Let s consider a plane passing through lattice points at p,0,0, 0, q,0, and 0,0,r, where p, q, and r are integers: x y z 1 p q r Multiply this y pqr : qrx pry pqz pqr So there is a plane with h qr, k pr, pq, and A pqr, where h, k,, and A are integers, that passes through lattice points on the coordinate axes. By allowing A to e any integer, we can generate a set of equally spaced, parallel planes, normal to ghk, with Miller indices hk. Interplanar spacing To find the distance from the origin to any of the hk planes, take the dot product of the unit vector normal to the planes with the a vector pointing from the origin to the plane of interest: ghk hx ky z A r g g g hk hk hk

6 6 For the hk planes, A is any integer. The distance etween adjacent planes, say from A 0 to A 1, is the lattice spacing: 1 dhk ghk So we have find that ghk is normal to the hk planes and has length 1 d hk. We also sometimes specify the direction of ghk y the Miller indices hk of the associated planes. Notation for planes, directions, and reflections For the purposes of clarity and revity, notations have een adopted to specify what is eing referred to y a particular set of indices, as follows: Direct Lattice Reciprocal Lattice Notation vector, direction, or point plane uvw family of vectors or directions family of planes uvw plane vector, direction or point hk family of planes family of vectors or directions hk diffracting plane reflection hk Visualizing direct-lattice planes: Cuic The planes are astractions, ut the orientation and spacing are not. We can pick any point as the origin, and we usually want the planes to intersect the origin. For a cuic crystal, with lattice parameter a, the (100) planes have spacing a. The (200) planes have the same orientation, ut a spacing of only a 2. The (110) planes are rotated y 45 and have spacing. a 2.. For cuic, any direction specified y its the direct-lattice indices, such as uvw, is the same as that specified y its reciprocal-lattice indices hk, whenever h u, k v, w.

7 7 Direct and reciprocal lattices: Non-cuic The Miller indices hk refer to planes in the direct lattice. But there are also planes in the reciprocal lattice, which are descried y the indices uvw. Take an aritrary reciprocal-space vector g x1 y2 z Consider some direct-lattice vector ruvw ua1 va2 wa The set of all points x, y, z in reciprocal space for which the dot product gr uvw is constant descries a plane. Let's call the constant B : gr uvw uxvywz B When B, the plane may contain some reciprocal-lattice points. So the direct-lattice vector r uvw, which we can also write as uvw, is perpendicular to the uvwreciprocal-lattice planes. In general, direct- and reciprocal-lattice directions with the same indices are not always parallel to one another. In other words, sometimes hk hk. We saw that the hk planes in the direct lattice are perpendicular to the reciprocal-lattice vector ghk, which can also e represented as hk. Similarly, the uvw planes in the reciprocal lattice are perpendicular to the direct-lattice vector r uvw, also represented as uvw. The distance from (000) (the reciprocal-space origin) to the reciprocal space point hk is 1 d hk, where dhk is the interplanar spacing in direct space. But the distance from 000 (the direct-space origin) to the point uvw in direct space is D uvw, where 1 D uvw is the interlayer spacing spacing of the perpendicular reciprocal-space planes. Confusing? Reciprocal lattice vectors and diffraction Diffraction spots, also called reflections, are indicated y their corresponding lattice planes hk (no parentheses). The vector ghk points from 0 to the reflection hk (if there is one).

8 8 An important lesson here is how to say an expression such as the one that follows: 2 4 This means the same as, 2, 4, ut is more compact. The - sign aove the 2 is equivalent to a - sign in front of the 2, as in -2, which is said negative two, or in this case three negative two four. But when we see 2, we say ar two, as in three ar two four. We do NOT say two ar, as in three two ar four, ecause that would imply 2 4, or, 2, 4. Whenever we might have said negative, we can just say ar. Bragg condition The Bragg condition implies that the scattered and incident wave vector differ y an RLV, i.e.: k k g The scattering is elastic, so 1 k k k We now know that 1 g g d Bragg s law must e hidden in there somewhere. Let s square the wave vectors: k kg k g 2kg 2

9 9 From the construction, we see that kgkgsin B So g 2ksin B Yes, this is Bragg s law: 2d sin B In vector form, it can e considered a statement aout momentum. The impulse (change in momentum) imparted to the electron in the scattering event is ph k -k hg Bragg s law: Oservations Bragg s law only mentions the spacing of the planes, not how they line up parallel to the planes. So crystals suject to stacking defects (such as graphite) that maintain a relatively constant interplanar spacing still show Bragg diffraction for those well-stacked planes. Consider the diagram elow: The constructive interference etween the adjacent planes, does not depend on how the atoms are lined up normal to the planes. We often see Bragg s law in the form 2dsin B n, where n is an integer. We don t need the n, ecause we already said hk (as in ghk 1 dhk ) can e any integers. Dividing y n 2d nsin B We can just interpret this to mean d1 hk, d2 hk, d hk, etc. Direct vs. reciprocal space A diagram of scattering in direct space shows trajectories. Bragg diffraction is similar to illiards, in that angles of incidence and reflection are equal, and the lattice planes play the role of umpers.

10 10 But we need the difference of the scattered and incident wave vectors to apply Bragg s law. To find the difference, we have to draw k and k, (or p and p ), with a common origin; the change in wave vector k (or momentum p ) closes the triangle. Selected-area diffraction error Selected-area diffraction is great ut not perfect. We previously assumed our ojective lens has no imperfections. Closer examination shows that, due to spherical aerration of the OL, the area selected is different for different reflections. Say we select our area in the usual way, with the SA in the first image plane of the OL. Paraxial (undiffracted) eams cross in the OL BFP to form the 0 spot. But diffracted eams make higher angles with respect to the optic axis, so they will e focused more strongly, slightly aove the BFP. The oject plane corresponding to this shorter focal length is closer to the lens than for the nominal focal length, and the size of the virtual aperture is smaller than the nominal size. When we trace rays diffracted at an angle passing through the edges of this virtual aperture, we find that they are sampling not only a smaller region, ut one that is laterally shifted from the direct eam. So, the larger the scattering angle for a reflection g, the farther off from center is the location of the area selected for that reflection. Influence of Cs on selected area In fact, we can use the SADP shift to measure the C s of the OL. At high magnification, the shift in the effective oject distance is aout equal to the shift in focal length: p f. With spherical aerration, 2 we saw that f C s. The lateral shift is then pc s.

11 11 We can measure the lateral shift etween images generated y 0 or g y defocusing the intermediate lens in diffraction mode. Now the diffraction spots ecome disks, and within each disk is an image of the sample area from which it originates. The lateral shift etween these images is. If we know the diameter of the virtual SA aperture, we can use that to find the length scale within these images. The scattering angle for the reflection is 2 B d. Then use Cs. However, typical estimates give much larger than expected values (e.g., Cs 19 mm ), so other contriutions appear to dominate the measurement.

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