NANO 703-Notes. Chapter 12-Reciprocal space

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1 1 Chapter 1-Reciprocal space Conical dark-field imain We primarily use DF imain to control imae contrast, thouh STEM-DF can also ive very hih resolution, in some cases. If we have sinle crystal, a -DF imae can formed by tiltin the beam in one particular direction. But for polycrystalline solids, powders, or nanomaterials, there may be many crystalloraphic orientations within the field of view. One trick to hihliht a lare fraction of these in dark-field mode is to rapidly precess the beam around the optic axis, with a small OA in the BFP of the OL. Sinusoidal sinals can be sent to the CL deflection coils to drive the beam tilt in a circular path. We don t want any shiftin of the beam, only tilt, so alinment is important. A particular reflection, which appears as a rin in the diffraction pattern, can be brouht onto the optic axis to form the dark-field imae. We have achieved centered dark-field conditions for an entire diffraction rin, not just a spot. Conical dark-field imain of nanoparticles If the precession rate is several Hz or hiher, the beam motion is not even noticeable. Switchin the tilt anle to zero restores the BF imae. With the beam tilted, stoppin the precession ives a static DF imae that hihlihts just a subset of the crystallites. Partially crystallized material One nice thin about DF imain is it can hihliht only features of interest, while maskin others. Conical DF imaes of partially crystallized, hydroenated amorphous silicon (a-si:h) show very clear contrast between amorphous and crystallized areas. In the DF imae, the crystals enerally appear either white or black, dependin on whether they are at or away from the Bra condition of the reflection used

2 (111, in this case). The amorphous reions are essentially featureless in the DF imae, makin them easy to distinuish. In the BF imae, these can show contamination or toporaphy on the surface. Direct and reciprocal lattice parameters We know how to find the reciprocal-lattice basis vectors, iven a particular set of direct-lattice basis vectors. We can identify lattice parameters and anles for these reciprocal-space vectors in an analoous fashion. and b b, b b, b b b b 3 b b3cos1, 3 1 b3b1cos Say we have an orthorhombic crystal. a a, a b, a c, b b, and b1b b1b cos3 We saw that the reciprocal lattice is also orthorhombic, i.e.,: b1, b, b3, a b c Determinin interplanar spacin The interplanar (d) spacin used in Bra s law is iven by d 1, where hk hb1 kb b3. The eneral method to find the lenth of a vector is to dot it with itself and take the square root, so: hk hk hk hk 0.5 hb kb lb klb b lhb b hkb b hb kb lb klbbcos lhbbcos hkbbcos This is fairly complicated for a low-symmetry crystal, but simplifies reatly with hiher symmetry. Cubic, for example, reduces to: hk 0.5 hk 0.5

3 3 hkl h k l, d a hkl a h k l The d-spacin for orthorhombic is not what you miht uess off the top of your head: hkl h k l, dhkl 1 h k l a b c a b c Ewald sphere An important concept in diffraction is the Ewald sphere. It is a reciprocal-space construction used to examine the diffraction eometry. The Ewald sphere has radius 1. Its center is called the excitation point, which we will label M. The reciprocal-space oriin 0 is not at the center of the sphere, but rather on the surface of the sphere. The incident wave vector k (which has lenth 1 ) points from M to 0. We are mainly interested in elastic scatterin for now, so we know any scattered wave vector k will also have lenth 1. Thus, k can also be drawn from M to the surface of the Ewald sphere. So the sphere tells us all possible diffracted wave vectors. If we superpose the sphere on a crystal s reciprocal lattice, we see the relationship between k, k, and any RLV for this crystal. The Bra condition requires k k. So the reciprocal-space distance from the reciprocal-lattice point to the sphere is a measure of how far the correspondin lattice planes are from the Bra condition. Hiher-order Laue zones A plane of reciprocal lattice points is called a Laue zone. When the plane contains 0, and extends nearly perpendicular to the beam, it is called the zero-order Laue zone, or ZOLZ. But this is just one in a series of parallel planes. Each of the others is called a hiher-order Laue zone, or HOLZ. The HOLZs only et close to the Ewald sphere at hih anles, so we may have to look more carefully to see evidence of them in diffraction patterns. As the beam enery ets hiher, the Ewald sphere curvature ets smaller, so the intersection with each HOLZ moves out to a larer radius in reciprocal space. (Surprisinly, thouh, the scatterin anle for a HOLZ rin ets smaller as the beam enery increases.)

4 4 The HOLZ rins contain useful information. The reflections contributin to the rin arise from reciprocallattice points with RLVs that are not perpendicular to the beam direction. So the HOLZ rin radius can be used to determine the lattice parameter alon the beam direction. Also, notice that as we tilt away from the axis of the Laue zone, even the intersection of the ZOLZ with the Ewald sphere becomes a circle. Si <111> One of the most interestin examples of a HOLZ diffraction rin arises from the silicon crystals oriented in the <111> orientation. This shows up best in converent-beam diffraction, which we will discuss later. Excitation error (deviation parameter) A measure of how far a particular set of lattice planes is from the Bra condition is the distance of the correspondin reciprocal-lattice point from the Ewald sphere. The vector, which we call the excitation error s, connectin a point to the sphere, satisfies k k++s But this does not uniquely define. In different contexts, we may choose to measure it either: 1) perpendicular to ; ) parallel to k ; 3) normal to the surface of the thin TEM foil; or 4) alon the direction of shortest distance to the sphere. Reardless, when we are at the Bra condition, its lenth is zero (i.e., s 0 ). Notice that the lenth of s is unchaned if either the sample or the beam is rotated about.

5 5 Evaluatin excitation error We can pick a suitable orientation for s and find an expression for its lenth. Say the foil normal is vertical ( nˆ z ˆ ), The beam is directed downward with a tilt anle. k k sin xˆ cosz ˆ In TEM, we are almost always interested in reflections arisin from RLVs nearly perpendicular to ˆn. Let s assume x. ˆ For reasons we will explore later, a ood assumption is s sz. ˆ By convention, we have assumed s 0 if is outside the sphere. The lenths of the incident and scattered wave vectors must be equal: k k k++s Square the lenths k k k+s +s This ives s tan kcos1 1 k cos kcos There are two solutions: and. The solution represents the intersection with the top of the sphere, which is not a ood indication of the excitation. It is the solution we need. We are mainly interested in hih-enery diffraction, with k. In this case,

6 6 s tan k cos Sphere volume and ZOLZ intersection 3 The Ewald sphere has radius 1, so its reciprocal-space volume is 4 k 3. We expect the intersection with the ZOLZ to be a circle. All points on the circle have s 0, so the diameter can be found by: 0 tan ksin k cos If the tilt anle is small, the diameter of the circle becomes k