Homework%201_Solution - Problem I Figure (a): The equation...

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Problem I Figure (a): The equation of the plane is 1 3 (111) 333 xyz  Figure (b): The equation of the plane is 1 2 3 3 6 (233) 322     Problem II There are twelve equivalent directions: [110],[110],[110],[110],[101],[101],[101],[101],[011],[011],[011],[011] Problem III As shown in Figure 1-9, there are 11 224 24 4  atoms if the cube is cleaved along a (110) plane. (Four on the corners, two on the edges, and two inside the rectangle.) It is easy to calculate the area of the rectangle: 2 2 Sa Therefore, the area density is:  14 2 2 2 8 44 9.59 10 2 25 . 4 3 1 0 nc a cm m   Problem IV As shown in Figure 1-9. (a) On (100) plane, there are 4 atoms on the corners of the square, and one on the center. The area density on the (100) surface: 14 2 2 2 8 1 14 2 4 6.78 10 5.43 10 a cm m  (b) The crystal structure of InP is zincblende. Therefore, given the lattice constant a, the nearest atoms of the same kind have the distance: 8 8 . 8 71 0 2/ 2 4 . 1 51 0 2 In In cm da c m P.S: For nearest neighboring P, the answer should be the same.
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This note was uploaded on 02/09/2010 for the course EE 339 taught by Professor Banjeree during the Fall '08 term at University of Texas at Austin.

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Homework%201_Solution - Problem I Figure (a): The equation...

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