เทคโนโลยีวัสดุอุตสาหกรรม 1

Crystal St Academic Res

tructures source Center

Crystallinity: Repeating or atomic distances. 3-D patte bonded to its nearest neigh Crystal structure: the man molecules are spatially arra

r periodic array over large ern in which each atom is hbors nner in which atoms, ions, or anged.

Unit cell: small repeating e The basic building block of defines the entire crystal str positions within.

entity of the atomic structure. the crystal structure. It ructure with the atom

Lattice: 3D array of points positions (center of sphere

s coinciding with atom es)





Metallic Cryst FCC (face centered cubic) the corners and center the cell.

tal Structures ): Atoms are arranged at of each cube face of

FCC con Close packed Plane: On e face of the cube Atoms are assumed to to along face diagonals. 4 atoms in one unit cell. a = 2R 2

ntinued each ouch .

BCC: Body Ce • Atoms are arranged at t with another atom at th

entered Cubic the corners of the cube he cube center.

BCC cont • Close Packed Plane cuts the u diagonally • 2 atoms in one unit cell

tinued unit cube in half a = 4R 3

Hexagonal Close • Cell of an HCP lattice is visualiz as a top and bottom plane of 7 atoms, forming a regular hexag around a central atom. In between these planes is a half hexagon of 3 atoms. • There are two lattice paramete in HCP, a and c, representing th basal and height parameters respectively.

e Packed (HCP) zed 7 gon f- ers he Volume 6 atoms per unit cell

Coordination number – the number surrounding an For FCC and HCP systems, t BC

r of nearest neighbor atoms or ions n atom or ion. the coordination number is 12. For CC it’s 8.

Close Packed HCP

d Structures FCC

Atomic Pack • The ratio of atomic sph volume, assuming a har • FCC = HCP = 74% (26% • BCC = 68%

king Factor here volume to unit cell rd sphere model. void space in unit cell)

Problem • If the atomic radius for Pb= 0 of the unit cell. • Solution: Pb is fcc, hence R = 0.175nm a = 0.495 nm Volume for cubic = a^3 = 1.21e

m1 0.175nm, find the volume a = 2R 2 e-28 m^3

Proble Magnesium is hcp with c/a =1.624 the atomic radius of magnesium Solution: density = n*MW / (Vc*N For hcp n = 6 Vc = = 2.6*(1.624)a^3 = 4 c = 1.624a NA = 6.02*10^23 atoms/mol MW magnesium = 24.3 g/mol

em 2 4, density = 1.74g/cm^3. Find m. NA) 4.22a^3

Problem 2 4.22a^3 = (6*24.3g/mol)/(1.74g/c o l) a= 320 pm For hcp a = 2R R (atomic radius) magnes

continued cm^3*6.02e23atom/m sium = 160 pm

Crystallographic Poi Plan • Directions

ints, Directions and nes

Directions C

Continued

Directions C

Continued