2材料科学基础英文版11PPT课件

1、 Unit cells Metallic crystal structures Ceramic crystal structures Crystal systems Crystallographic directions and planes第1页/共15页Unit cells Small repeat atom groups in a crystal structure Representation of a unit cell: (a) hard-sphere model; (b) reduced hard-sphere model; (c) an aggregate of many at

2、oms第2页/共15页Metallic crystal structures (1) Three major types: 1) Face-centered cubic (FCC) crystal structures 2) Body-centered cubic (BCC) crystal structures 3) Hexagonal close-packed (HCP) crystal structures第3页/共15页Metallic crystal structures (2)1) Face-centered cubic (FCC) crystal structure2Ra24aR

3、 42aR Each unit cell contains 4 atoms第4页/共15页Metallic crystal structures (3) Two other features: coordination number (CN) and atomic packing factor (APF) CN: the number of nearest-neighbour atoms 12 for FCC APF: the fraction of atom-occupied volume in a unit cell APF = (atom-occupied volume in a uni

4、t cell)/(unit cell volume) For FCC crystals: 4 atoms in a unit cell atom-occupied volume = 4 (4/3)( R3) = 4 (4/3) (3.14 (a 2/4)3 = 0.74a3 unit cell volume = a3 APF = atom-occupied volume/unit cell volume = 0.74Some common metals: Al, Ni, Au, Cu, and Mn第5页/共15页Metallic crystal structures (4)Interstic

5、es in FCC structures:octahedral interstices and tetrahedral intersticesRo = (a 2R)/2 = (22R - 2R)/2 = 0.414R(1/2,1/2,1/2)Rt = 3a/4 R = (3/4) 22R) R = 0.224R(1/4,1/4,1/4)Diagonal length: 3a第6页/共15页aMetallic crystal structures (5)2) Body-centered cubic (BCC) crystal structureaR3443aR Each unit cell co

6、ntains 2 atomsCN = 8APF = 2 (4/3)( R3)/a3 = 2 (4/3) (3.14 (a 3/4)3 /a3 = 0.68第7页/共15页Metallic crystal structures (6) Interstices in BCC structures:octahedral interstices and tetrahedral intersticesRoz = (a 2R)/2 = (4R/3 - 2R)/2 = 0.155RRoxy = (a 2 2R)/2 = (4R/3) 2 - 2R)/2 = 0.633RRt = a5/4 - R = (4R

7、/3)5/4 R = (5/3 1)R = 0.291Ra =4R/3第8页/共15页Metallic crystal structures (7)3) Hexagonal close-packed (HCP) crystal structureEach unit cell contains 6 atoms (12 (1/6) + 2 (1/2) + 3 = 6)2R = a R = a/2CN = 12c/a = ?APF = ?第9页/共15页Metallic crystal structures (8)c/2 xfgex=a3/3c/2aegfa2 = (c/2)2 + (a 3/3)2

8、c/a = (8/3)1/2 = 1.633(ideal ratio, actually 1.58 1.89)Equilateral triangleah = asin60o = a3/2x=2h/3fe第10页/共15页Metallic crystal structures (9)APF = atom-occupied volume/unit cell volumeunit cell volume =base height = (6 (1/2)a2sin60o) c = 3 a2( 3/2) 1.633a = 4.24a3atom-occupied volume = 6 (4/3)( R3)

9、 = 6 (4/3) 3.14 (a/2)3 = 3.14a3 APF = 3.14/4.24 = 0.74 (ideal packing)第11页/共15页Metallic crystal structures (10)Interstices in HCP structures:octahedral interstices and tetrahedral interstices(2/3,1/3,3/4)defgedgfaaRo = a 2/2 R = 2R R = 0.414R (a=2R)(2/3,1/3,7/8)Rt = (3/8)c R = (3/8)x1.633a R = 0.225

10、R (a=2R)7c/8第12页/共15页Metallic crystal structures (11)Density calculations for metals = (weight of atoms in each unit cell)/unit cell voulme = (nA/NA)/Vc n=number of atoms in each unit cell A=atomic weight NA=Avogadros number (6.0231023) Vc=unit cell volume(A/NA=weight of each atom)第13页/共15页Metallic

11、crystal structures (12)Density for CuFCC structure: n=4, A=63.54, a= 0.36148 nm , NA= 6.0231023weight of atoms in each unit cell = nA/NA = 463.54/(6.0231023) = 4.2198 10-22 g Unit cell volume = a3 = (0.36148)3 = 0.04723 nm3 = 4.7234 10-23 cm3 = (weight of atoms in each unit cell)/unit cell voulme = 4.2198 10-22 / 4.7234 10-23 = 8.93 g/cm3 The literature value is 8.94 g/cm3 Density for AlFCC structure: n=4, A=26.982, a= 0.40497 nm , NA= 6.0231023weight of atoms in each unit cell = nA/NA = 426.982/(6.0231023) = 1.7919 10-22 g Unit cel