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Understanding Crystal Structures in Diamond and Zinc Blende Lattices

Explore the diamond crystal structure through covalent bonds and the zinc blende lattice with gallium and arsenic atoms. Calculate molecular densities and analyze hexagonal symmetry in CdS crystal lattice.

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Understanding Crystal Structures in Diamond and Zinc Blende Lattices

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  1. ECE 874:Physical Electronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu

  2. Lecture 02, 04 Sep 12 VM Ayres, ECE874, F12

  3. Silicon crystallizes in the diamond crystal structureFig. 1.5 (a): Notice real covalent bonds versus definition of imaginary cubic unit cell VM Ayres, ECE874, F12

  4. VM Ayres, ECE874, F12

  5. Corners: 8 x 1/8 = 1 Faces: 6 x ½ = 3 Inside: 4 x 1 = 4 Atoms in the Unit cell: 8 VM Ayres, ECE874, F12

  6. What do you need to know to answer (b)? Size of a cubic Unit cell is a3, where a is the lattice constant (Wurtzite crystal structure is not cubic) VM Ayres, ECE874, F12

  7. VM Ayres, ECE874, F12

  8. 1 Which pair are nearest neighbors? O-1 O-2 O-3 2 3 O VM Ayres, ECE874, F12

  9. 1 Which pair are nearest neighbors? O-1 O-2 O-3 a/2 2 a/2 3 z O y x VM Ayres, ECE874, F12

  10. (1/4, ¼, ¼) is given To get this from drawing would need scale marks on the axes VM Ayres, ECE874, F12

  11. 1 Which pair are nearest neighbors? O-1 O-2 O-3 a/2 2 a/2 3 z O y x VM Ayres, ECE874, F12

  12. Directions for fcc (like Pr. 1.1 (b)):8 atoms positioned one at each corner.6 atoms centered in the middle of each face. VM Ayres, ECE874, F12

  13. The inside atoms of the diamond structure are all the ( ¼, ¼, ¼) locations that stay inside the box: VM Ayres, ECE874, F12

  14. All the ( ¼, ¼, ¼) locations are: VM Ayres, ECE874, F12

  15. You can match all pink atoms with a blue fcc lattice, copied from slide 12: VM Ayres, ECE874, F12

  16. You can match all bright and light yellow atoms with a green fcc lattice copied from slide 12: VM Ayres, ECE874, F12

  17. Therefore: the diamond crystal structure is formed from two interpenetrating fcc lattices VM Ayres, ECE874, F12

  18. But only 4 of the (¼, ¼, ¼) atoms are inside the cubic Unit cell. VM Ayres, ECE874, F12

  19. Now let the pink atoms be gallium (Ga) and the yellow atoms be arsenic (As). This is a zinc blende lattice. Compare with Fig. 1.5 (b) VM Ayres, ECE874, F12

  20. Note that there are 4 complete molecules of GaAs in the cubic Unit cell. VM Ayres, ECE874, F12

  21. Example problem: find the molecular density of GaAs and verify that it is: 2.21 x 1022 molecules/cm3 = the value given on page 13. VM Ayres, ECE874, F12

  22. VM Ayres, ECE874, F12

  23. Increasing numbers of important compounds crystallize in a wurtzite crystal lattice: Hexagonal symmetry Fig. 1.3 forms a basic hexagonal Unit cell Cadmium sulfide (CdS) crystallizes in a wurtzite lattice Fig. 1.6 inside a hexagonal Unit cell Wurtzite: all sides = a VM Ayres, ECE874, F12

  24. Example problem: verify that the hexagonal Unit cell volume is: = the value given on page 13. VM Ayres, ECE874, F12

  25. Hexagonal volume = 2 triangular volumes plus 1 rectangular volume a a 120o a c VM Ayres, ECE874, F12

  26. VM Ayres, ECE874, F12

  27. VM Ayres, ECE874, F12

  28. Hexagonal volume = 2 triangular volumes plus 1 rectangular volume a a 120o a c VM Ayres, ECE874, F12

  29. Work out the parallel planes for CdS: 3 S 7 Cd 7 S 3 Cd 3 S 7 Cd Note: tetrahedral bonding inside VM Ayres, ECE874, F12

  30. How many equivalent S atoms are inside the hexagonal Unit cell?Define the c distance as between S-S (see Cd-Cd measure for c in Fig. 1.6 (a) ) 3 S 7 S 3 S VM Ayres, ECE874, F12

  31. How much of each S atom is inside: Hexagonal: In plane: 1/3 inside Hexagonal: Interior plane Top to bottom: all inside: 1 Therefore: vertex atoms = 1/3 X 1 = 1/3 inside 6 atoms x 1/3 each = 2 Also have one inside atom in the middle of the hexagonal layer: 1 Total atoms from the hexagonal arrangement = 3 VM Ayres, ECE874, F12

  32. How much of each atom is inside: Note: the atoms on the triangular arrangement never hit the walls of the hexagonal Unit cell so no 1/3 stuff. But: they are chopped by the top and bottom faces of the hexagonal Unit cell: = ½ atoms. Therefore have: 3 S atoms ½ inside on each 3-atom layer layers: = 3/2 each layer Total atoms from triangular arrangements = 2 x 3/2 = 3 VM Ayres, ECE874, F12

  33. How many equivalent S atoms are inside the hexagonal Unit cell?Total equivalent S atoms inside hexagonal Unit cell = 6 3 S 3/2 7 S 3 3 S 3/2 VM Ayres, ECE874, F12

  34. Pr. 1.3 plus an extra requirement will be assigned for HW: You are required to indentify N as the Cd atoms in Fig. 1.6 (a) VM Ayres, ECE874, F12

  35. So you’ll count the numbers of atoms for the red layers for HW, with c = the distance between Cd-Cd layers as shown. Cd  N: 7 Cd 3 Cd 7 Cd VM Ayres, ECE874, F12

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