CHAPTER 12 - LECTURE 2

CHAPTER 12 - LECTURE 2 - CHAPTER 12 LECTURE 2 For cubic...

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CHAPTER 12        - LECTURE 2:     For cubic  structures of simple atoms (such as metals),     there are 3 different possibilities :     a)   simple  cube:  cell edge  a =2r , because the atoms         touch each other along the edge .  Therefore, there         are 6  nearest neighbors.          r  = a/2     b)   fcc :  face diagonal =  2 1/2 a = 4r , because         3 atoms touch each other along the FACE diagonal .                   h = (a 2  + a 2 ) 1/2  = (2a 2 ) 1/2  =  2 1/2 a = 4r         There are  12  nearest neighbors.        r =   2 1/2 a/4       c)   bcc :  body diagonal =  3 1/2 a = 4r , because        3 atoms touch each other along the BODY diagonal .         Therefore, there are  6  nearest neighbors.                                r =   3 1/2 a/4   USEFUL EQUATION:     rho = Z*M / V*N,         where rho = the density of the material in g/cm 3                  Z  = number of atoms (particles) in the cell                  M = molecular (atomic) weight in g/mol or amu                  V = volume of the unit cell = a 3 , in cm 3                  N = Avogadro's number =                                 6.022 x 10 23    particles/mol   CONVERSIONS:  1 pm = 1 x 10 -12  m = 1 x 10 -10   cm                                          = 1 x 10 -2  Angstrom               1 Angstrom =  1 x 10 -8  cm = 1 x 10 -10  m = 100 pm
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Types of problems:     1)  Calculate Unit-Cell Dimension from Unit-Cell Type and           Density:       Pt crystallizes in a face-centered cubic ( fcc ) lattice with       all atoms at the lattice points.  It has a density of       21.45 g/cm 3  and an atomic weight of 195.08 amu.       Calculate the length  of a unit-cell edge.  Compare this       with the value of 392.4 pm obtained from x-ray       diffraction. STRATEGY :  We can calculate the mass of the unit cell  from             the atomic weight.  Knowing the density and the             mass of the unit cell, we can calculate the volume  of             a unit cell and then the edge length  of a unit cell.      195.08 g Pt   x                1 mol Pt                                 =3.239 x 10 -22 g Pt       1 mol Pt        6.022  x 10 23 Pt atoms         1 Pt atom    Since there are 4 atoms  of Pt per  fcc  unit cell, the mass         per unit cell =  4 x 3.239 x 10 -22   = 1.296 x 10 -21  g Pt
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