02 crystal structure

7/27/2019 02 Crystal Structure

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Crystal Structure



why study the structure of crystall ine

solids ?

• The properties of some materials are directlyrelated to their crystal structures

• For example, pure and undeformed magnesiumand beryllium, having one crystal structure, aremuch more brittle (i.e., fracture at lower

degrees of deformation) than are pure andundeformed metals such as gold and silver thathave yet another crystal structure



• Furthermore, significant property differences

exist between crystalline and non crystalline

materials having the same composition. For

example, non crystalline ceramics and

polymers normally are optically transparent;

the same materials in crystalline (or semi

crystalline) form tend to be opaque or, at best,translucent.



Type of Materials

• Crystalline :

A crystalline material is one in which the atoms are situated in arepeating or periodic array over large atomic distances All metals,many ceramic materials, and certain polymers form crystallinestructures under normal solidification conditions

• E.g.:-Iron, Copper, Aluminum

• Amorphous or Non crystal l ine :For those that do not crystallize, this long-range atomic order is

absent; these non crystalline or amorphous materials• E.g.:- Wood, paper, plastics, glass



• Crystal structure: the manner in which atoms, ions, or

molecules are spatially arranged.

• When describing crystalline structures, atoms (or ions)

are thought of as being solid spheres having well-defineddiameters



What is a unit cell

• The unit cell is the smallest structural unit or building

block that can describe the crystal structure.

• Repetition of the unit cell generates the entire crystal.



Metallic Crystal Structures

• Three relatively simple crystal structures are

found for most of the common metals: simple

cubic (SC), face centered cubic (FCC), body-

centered cubic (BCC), and hexagonal close-

packed (HCP).

• SC :




• BCC : Chromium, iron, tungsten




• FCC: copper, aluminum, silver, and gold

• unit cell edge length for FCC :

2 2a R



Atomic packing factor

• APF is the sum of the sphere volumes of all atoms within a unit cell

divided by unit cell volume.

• Let atomic radius be R.• Volume of sphere is , 4 atoms per FCC unit cell, the total FCC atom

( or sphere) volume is

•

•

• to find VC

s

c

V

V

34

3 R

3 34 164

3 3S V R R

2 2 2

33 3

4

i , 2 2

2 2 16 2c

a a R

solv ng for a a R

V a R R

3

3

16

3 0.7416 2

S

C

RV

APF V R



Coordination number

Refers to the number of atoms touched by

the given atoms inside a lattice structure. 1. SC:

• 1 vertex atom touches 3

other atoms along 3 edges

of the cell.• To see the remaining atoms

we must see outside the

single cell.

•

The same atom also touchesthe 3 other atoms along the

edges of the 3 other cells for

a coordination number of 6



• BCC: any body atom

touches 8 vertex

atoms for a

coordination number 8



• FCC:

– Atom touch along the

side diagonal of the

cell. – First consider the

coordination number

of the face atom.

–

Face atom touches 4vertex atoms



• also touches 8 other face atoms 4 aboveand four below for atotal coordinationnumber 12.

• Now consider coordination number of vertex atom. Eachvertex atom touchesfour face atomswithin same plane,four face atoms

below and four faceatoms above againgiving coordinationnumber 12



• The coordination number s same regardless of

which atom is chosen. There fore we can also

say the each vertex atom is surrounded by 8

body atoms again giving coordination number

8.

• Th t d b tt f f th it ll



• In addition to the 3 cubicstructures there are 2 close

pack structures. It consists of closely packed layers of atomswithin which each atom is

surrounded by 6 others.

• The top and bottom faces of the unit cellconsist of six atoms that form regular hexagons and surround a single atom inthe center.

• Another plane that provides threeadditional atoms to the unit cell issituated between the top and bottom

planes.

• The equivalent of six atoms is containedin each unit cell; one-sixth of each of the12 top and bottom face corner atoms,one-half of each of the 2 center faceatoms, and all 3 mid plane interior atoms.

• The coordination number and theatomic packing factor for the HCPcrystal structure are the same as forFCC: 12 and 0.74, respectively.

• The HCP metals include cadmium,magnesium, titanium, and zinc

Hexagonal Close Pack Structur es (HCP)



TYPES OF CRYSTAL LATTICES



Crystallographic Points,

Directions, and Planes• When dealing with crystalline materials, it often

becomes necessary to specify a particular point within aunit cell, a crystallographic direction, or somecrystallographic plane of atoms.

• Three numbers or indices are used to designate pointlocations, directions, and planes.

•

The basis for determining index values is the unit cell,with a right-handed coordinate system consisting of three ( x, y, and z ) axes situated at one of the corners andcoinciding with the unit cell edges



POINT COORDINATES

• The position of any point located within a unit

cell may be specified in terms of its

coordinates as fractional multiples of the unit

cell edge lengths (i.e., in terms of a, b, and c)



For the unit cell shown in the accompanying sketch (a ), locate the

point having coordinates ¼ 1 ½

• a = 0.48nm, b = 0.46nm, c = 0.40nm.

• Fractional lengths: q = ¼ , r = 1, s = ½

• First we move from the origin of the unitcell (pt M) qa = ¼ (0.48) = 0.12nm along x

axis to point N,

• Similarly rb = 1 ( 0.46) = 0.46nm parallel toY axis from N to O,

• sc = ½ (0.40) = 0.20 units parallel to Z axisto point P.

•

This point P corresponds to ¼ 1 ½ .



CRYSTALLOGRAPHIC DIRECTIONS

A crystallographic directionis defined as a line betweentwo points, or a vector.

• A vector of convenientlength is positioned such that

it passes through the origin

of the coordinate system.

• Any vector may be translated throughoutthe crystal lattice without alteration, if

parallelism is maintained.

• The length of the vector projection oneach of the three axes is determined, theseare measured in terms of the unit cell dimensions a, b, and c.

• These three numbers are multiplied or

divided by a common factor to reducethem to the smallest integer values

• The three indices, not separated bycommas, are enclosed in square brackets,thus [uvw]. The u, v, and w integerscorrespond to the reduced projectionsalong the x, y, and z axes, respectively

• There will exist both positive and negativecoordinates. Thus negative indices are also

possible, which are represented by a bar over the appropriate index. For example,

the direction would have acomponent in the -Y direction.



CRYSTALLOGRAPHIC PLANES

• Layers of planes along which atoms are arranged are known ascrystallographic planes. Here one corner of the unit cell isassumed to be origin. w.r to this coordinates all planes areconsidered.

• Steps: – Position a plane in a convenient coordinate system. Plane shouldn’t

pass through origin. If it passes, shift the origin to adjacent unit cell.

– Find point of intersection of plane on the axis. If the plane does notintersect on any axis then the intercept distance or unit cell dimensionw.r.to this axis is infinite. ( i.e., we assume that if plane is parallel it willintersect at infinity.)

– Take reciprocal of unit cell dimension.

– If it is a fraction then multiply or divide by same value and write as[u vw].

– If unit cell dimension is negative the represent as [u v w].



• Plane surface parallel to X

axis , unit cell distance/pt. of intersection = ∞

• Plane surface parallel to Z

axis , unit cell distance = ∞ • But @ Y axis there is

intersection point, therefore

distance of unit cell = 1.

• We can write ( ∞ 1 ∞ )

=(1/0 1 1/0)

• Miller indices = [0 1 0]

X

Y

Z



Hexagonal crystals

• For crystals having

hexagonal symmetry, it is

desirable that equivalent

planes have the sameindices.

• This convention leads to

the four-index (hkil )

scheme

• Where I = - ( h + k)

• Furthermore its intersections



• To determine these Miller – Bravais indices, consider the

plane in the figure referencedto the parallelepiped labeledwith the letters A through H atits corners.

•

This plane intersects the a1axisat a distance a from the originof the a1 – a2-a3-zcoordinateaxes system (point C)

Furthermore, its intersectionswith the a2 and z axes are -aand c, respectively.

• Therefore, in terms of the

lattice parameters, theseintersections are 1, -1 and 1.

• Furthermore, the reciprocalsof these numbers are also 1, -1

and 1. henceh = 1, k = -1, l = 1.

From i = - ( h + k )

= - (1 – 1) = 0

therefore the (hkil) indices

are ( 1 -1 0 1)• Notice that the third index is

zero (i.e., its reciprocal = ∞),which means that this plane

parallels the a3 axis

02 crystal structure

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