[Solved] model answer written assignment group b

Sprightlier structure s composed of Sans tetrahedrons which are stacked in a layered stacking sequence of ABACA, while written structure has a hexagonal stacking sequence of ABA. The overall Sans tetrahedral structure of sprightlier viewed from [1 1 -1] direction is shown in Figure 1. The layer that is labeled with green for represents the top layer, follow by blue dot and red dot which represented the second and third layer respectively. Each dot represents the top off tetrahedral structure which differentiates one layer from another. Each layer of sprightlier can be seen from Figures I(b), 1 (c) and (1 (d)).

The boundary conditions used are (1 1 -1) and -1 -1 1) with central distances of 6. These boundary conditions are given so that only the three significant layers in Sans tetrahedral structure can be observed clearly. Figure 2 on the other hand, shows the overall Sans tetrahedral structure of written viewed from [001] direction. The two distinct layers and are identify with blue and red dot respectively. Unlike the boundary conditions used for written are (001) and (00-1), with central distances of 8. Similarly, each layer in written is shown in Figure 2(b) and figure 2(c).

Figure 1: Stacking arrangement in sprightlier. Figure 2: Stacking arrangement in written. Paragraph 2 The following describes the general principle of creating long-period polytypic. Firstly, all possible layer sequences is listed. Secondly, some special considerations will be taken into account to reduce the number of stacking sequences to be compared. Thirdly, the X-ray diffraction intensity of these sequences are generated. Finally, the experimental X-ray diffraction result of the sample is compared to the generated results to determine the stacking sequence of the long-period polytypic.

The Sans polytypic determined by Barman are as below: a. Sans OIL Since the stacking sequence is ABACA AC, 8 layers (ABACA) are in cyclic order and 2 layers (AC) are in antistatic order, the Cadenza notation will be (8 2). The polytypic nomenclature of this polytypic is OIL (8 2). B. Sans ALL Since the stacking sequence is ABACA CAB BC BACK, 5 layers (ABACA) are in cyclic order, 4 layers (CAB) are in antistatic order, 2 layers (BC) are in cyclic order and 3 layers (BACK) are in antistatic order, the Cadenza notation will be (5 4 2 3).

The polytypic nomenclature of this polytypic is ALL (5 4 2 3). C. Sans ALL Since the stacking sequence is ABACA CAB CAB CAB BACK BACK, 5 layers (ABACA) are in cyclic order, 3 layers (CAB) are in antistatic order, 5 layers (CAB) re in cyclic order, 3 layers (CAB) are in antistatic order, 5 layers (BACK) are in cyclic order and 3 layers (BACK) are in antistatic order, the Cadenza notation will be (5 3 53 5 3), in short, (5 3)3. The polytypic nomenclature of this polytypic is 241 (5 3)3. D.

Sans ALL Since the stacking sequence is BACCALAUREATE’S CAB BC BACK, 17 layers (BACCALAUREATE’S) are in cyclic order, 4 layers (CAB) are in antistatic order, 2 layers (BC) are in cyclic order and 3 layers (BACK) are in antistatic order, the Cadenza notation will be (17 4 2 3). The polytypic nomenclature of this polytypic is 261 (17 423). E. Sans 281 Since the stacking sequence is ABACAS ABACA BACK BACCARAT, 9 layers (ABACAS) are in cyclic order, 5 layers (ABACA) are in antistatic order, 5 layers (BACK) are in cyclic order and 9 layers (BACCARAT) are in antistatic order, the Cadenza notation will be (9 5 5 9).

The polytypic nomenclature of this polytypic is ALL (9 5 5 9). Figure 3: Sans polytypic stacking sequences. If Brahmas had not discover that birefringence can be used as the “special considerations” to limit the possible stacking sequences, it would have been impossible to determine these 5 long period polytypic of Sans since the working ill be extremely tedious. For example, the number of possible stacking sequences for merely for a OIL polytypic = 310 = 59049, and Abraham’s method successfully reduced the stacking sequences in question to 16. Nevertheless, the Sans polytypic is diverse.

As the number of layers (the exponent) increases, the working grows more complicated and will eventually be too tedious to solve even after applying Abraham’s method. Paragraph 3 CARD is used to characterize crystal structure of crystalline materials by collecting the X-ray diffraction beams. Crystalline materials is multifaceted in various erection and have well-defined peaks in specific directions. In Sans polytypic, single crystal CARD can be used to determine not only crystal structure, but also other properties such as lattice parameters, preferred orientation, phase composition, thickness, defects, strain, etc.

The test itself is done by scanning the Sans crystal across various angles to form a continuous spectrum from the collected peaks. Consequently, relationship between intensity (y-axes) against 20 angle (x-axes), which is a function of crystal structure, can be obtained. Using Brags Law, interplant spacing and Miller indices (hike) can be determined abstinently. Preferred crystal orientation can also be observed by the highest intensity count. From the resulting CARD pattern, crystal structure can be reconstructed using both qualitative and quantitative information.

The most convenient way to explain this by introducing the concept of real and reciprocal space. In real space, lattice points represent atomic arrangement inside crystal structure, while in reciprocal space, each lattice points refers to set of planes in a crystal, since they are inversely proportional with respect to each other. During CARD test, collected peaks will form diffraction spots on resulting reciprocal space ND it can vary depending on which axes to be observed, especially when it is not cubic structure.

For hexagonal Sans, it forms hexagonal diffraction spots from c-axes, but looks different from other axes, which is orthogonal with it. As mentioned in previous paragraphs, Sans has various polytypic with variable amount of Zen atoms along c-axes. Afterwards, this results in incident x-ray beams diffract in different angle and preferred orientation with rotation of the crystal, and the extra Zen atoms diffract beams in other angle or preferred orientation if we repeat each Sans polytypic over long-range order.

Hence, this results in distinct position and spacing of diffraction spots with respect to repeating unit of polytypic. Quantitatively, we should refer to intensities of OIL (8 2) polytypic in table 1. Barman et. Al. Also lists similar tables with different values for each ALL, ALL, ALL and ALL polytypic, but not 151 and ALL. However, we can safely assume that ALL is nearly similar towards ALL and ALL towards ALL, but each polytypic in both types should have distinct value of intensities. As c-axes getting significantly longer from ALL to ALL there will be more diffraction beams from the Zen and

S atoms, so some peaks will have higher multiplicity and intensity due to hexagonal structure. Since interplant spacing is related through this equation: haw/AAA +aka/ABA + IA/ca with duke as interplant spacing, h, k and I as miller indices and a, b and c as lattice parameters. With this relation, it is expected that interplant spacing increases as c increases. Finally, since in reciprocal space diffraction spots are formed, then the spacing is inversely proportional to duke (real), hence the spacing of diffraction spot shrinks as it goes on from OIL to ALL polytypic.

[opskill_register_order]

"Looking for a Similar Assignment? Order now and Get a Discount!

Hey, wait!You Don't want to miss this offer!

Before you go, let us offer you a 20% discount coupon for your next purchase.