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Diamond Structure

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Publicada em

Engineering Physics,
CRYSTALLOGRAPHY,
Simple cubic, Body-centered cubic, Face-centered cubic,
DIAMOND STRUCTURE,
Atomic Packing Factor of Diamond Structure,
Projection of diamond lattice points on the base

Publicada em: Educação

Diamond Structure

  1. 1. Dr. Virendra Kumar Verma Madanapalle Institute of Technology and Science (MITS)
  2. 2. http://butane.chem.uiuc.edu/pshapley/Environmental/L26/1.html
  3. 3. http://www.metafysica.nl/turing/preparation_3dim_2.html http://www.chemgapedia.de/vsengine/vlu/vsc/de/ch/16/ac/elemente/vlu/38_56_88.vlu.html A face centered cubic unit cell has one atom at each corner and one atom at each face center.
  4. 4.  Diamond cubic structure is obtained when two FCC sublattices interpenetrates along the body diagonal by 1/4th cube edge. http://www.materialsdesign.com/appnote )ˆˆˆ( 4 zyx a  a
  5. 5. http://www.materialsdesign.com/appnote )ˆˆˆ( 4 zyx a  a  The space group of diamond is .  There are eight carbon atoms at the corner, creating a cube.  The six carbon atoms in the faces create an octahedron.  The four internal carbon atoms (black balls in figure) lie at ¼ of the distance along body diagonal forming a tetrahedran. mdF 3 C C F C C C C F C C F F F F I I I I The letters on the ball mean: C – Corner atom F – Face atom I – Internal atom
  6. 6. Square on the right hand side indicates the base of the cube. 1/4 1/4 a
  7. 7. a 0 0 0 0 0 1/4 1/4 The number ‘0’ denotes the height above the base. 0 0 0 0 0
  8. 8. a 0 0 0 0 0 0 0 0 0 01/2 1/2 1/2 1/2 1/4 1/4 The number ‘0’ and fraction ‘½’ denote the height above the base. 1/2 1/2 1/2 1/2
  9. 9. a 0 0 0 0 0 0 0 0 0 01/2 1/2 1/2 1/2 1/4 1/4 1/4 1/4 The number ‘0’ and fractions ‘½’ and ‘¼’ denote the height above the base. 1/2 1/2 1/2 1/2 1/4 1/4
  10. 10. a 0 0 0 0 0 0 0 0 0 01/2 1/2 1/2 1/2 1/4 1/43/4 3/4 1/4 1/4 The number ‘0’ and fractions ‘½’ , ‘¼’ and ‘¾’ denote the height above the base. 1/2 1/2 1/2 1/2 1/4 1/4 3/43/4
  11. 11. Atomic Packing Factor n = ? and r = ? 3 3 3 4 a rn cellunittheofVolume atomonetheofVolumecellunitainpresentatomsofNumber cellunittheofVolume cellunitainatomsbyoccupiedvolumeTotal APF      To solve APF of diamond structure, we need ‘n’ and ‘r’.
  12. 12. Atomic Packing Factor nfcc = (1/8 x 8 corner atoms) + (1/2 x 6 face atoms) = 1+3 = 4 atoms. n = 4+4 = 8 r = ?  There are atoms at all eight corners and all six faces.  In addition to that there are 4 full atoms inside the unit cell. Total number of atoms present in a unit cell.
  13. 13. Atomic Packing Factor 8 3 16 3 4 4)2( 2 2 222 a r a r rrXYBut    x r z a/4 a/4 a/4 y 16 3 48 844 222 222 222 2 aaa ZYXZXY aaa XZ                     16 3 4 3 444 )ˆˆˆ( 4 2 2 222 a XY aaaa YX zyx a YXVector                      OR )ˆˆˆ( 4 zyx a  a X Y
  14. 14. Atomic Packing Factor 34.0 16 3 8 3 3 4 8 3 4 3 3 3 3                    a a a rn cellunittheofVolume atomonetheofVolumecellunitainpresentatomsofNumber cellunittheofVolume cellunitainatomsbyoccupiedvolumeTotal APF %34 structurediamondoffactorpackingAtomic
  15. 15. please contact me via email for any further suggestions/comments. Email: virendrave@gmail.com Visit the below website for other topics http://sciencebank.wordpress.com/

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