GUIDED BY :- MR. DEEPAK SAKSENA

PREPARED BY :- ROHIT KAVA (07DME033) ROHIT KAVA (07DME033) KAPIL PRAJAPATI (08DME313)KAPIL PRAJAPATI (08DME313)

What is APF?What is APF?atomic packing factor is the fraction of atomic packing factor is the fraction of

volume in a volume in a crystal structure that is occupied by atoms.

It is dimensionless and always less It is dimensionless and always less than unity. For practical purposes, the than unity. For practical purposes, the APF of a crystal structure is APF of a crystal structure is determined by assuming that atoms determined by assuming that atoms are rigid spheres.are rigid spheres.

It is represented mathematically by :- It is represented mathematically by :- 

APF for BCC structure = 0.68

APF for FCC structure = 0.74

APF = Volume of atoms in unit cell

Volume of unit cell

B.C.C. structureB.C.C. structure

F.C.C. structureF.C.C. structure

APF calculation for BCC structureAPF calculation for BCC structure• Arrangement of atoms :- Arrangement of atoms :-

» Unit cell is cubicUnit cell is cubic» One atom at each corner & one atom at the body center One atom at each corner & one atom at the body center

of cube.of cube.» Co ordination no. = 8Co ordination no. = 8

• E.g. :- Alpha Iron, Cr, Mo, Na, etc…E.g. :- Alpha Iron, Cr, Mo, Na, etc…• No. of atoms in unit cell :- No. of atoms in unit cell :-

» Ni = 1Ni = 1» Nc = 8Nc = 8» Nf = 0 Nf = 0

• There fore N = Ni + Nf/2 + Nc/8There fore N = Ni + Nf/2 + Nc/8 = 1 + 0/2 + 8/8= 1 + 0/2 + 8/8 = 2 atoms in unit cell= 2 atoms in unit cell• Relation between atomic radius(r) & Lattice parameter(a)Relation between atomic radius(r) & Lattice parameter(a)

» R =R =√3a/4√3a/4

Example: Alpha Iron, Cr, Mo, P, Na, W, TA, V• General Characteristics of BCC Metals

» Generally harder than FCC metals» Will not deform as much before failure» Tend to get harder as they deform

Co-ordination = 8

• Close packed directions are cube diagonals.Note: All atoms are identical; the center atom is shaded

differently only for ease of viewing.--

BODY CENTERED CUBIC STRUCTURE (BCC)

aR

• APF for a body-centered cubic structure = 3/8 = 0.68

Close-packed directions: length = 4R

= 3 a

ATOMIC PACKING FACTOR: BCC

APF = a3

4

3( 3a/4)32

atoms

unit cell atomvolume

unit cell

volume

Unit cell contains = 2 atoms/unitcell

APF calculation for FCC structureAPF calculation for FCC structure

• Arrangement of atoms :- Arrangement of atoms :- » Unit cell is cubicUnit cell is cubic» One atom at each corner & one atom at center of One atom at each corner & one atom at center of

each face of cube.each face of cube.» Co ordination number = 12Co ordination number = 12

• E.g. :- Cu, Ag, Au, Gamma Fe, etc…E.g. :- Cu, Ag, Au, Gamma Fe, etc…• No. of atoms in unit cell :- No. of atoms in unit cell :-

» Ni = 0 Ni = 0 No. of interior atomsNo. of interior atoms» Nc = 8Nc = 8 No. of corner atomsNo. of corner atoms» Nf = 6Nf = 6 No. of face atomsNo. of face atoms

• There fore N = 8/8+6/2+0/2 = 4 atoms in unit cellThere fore N = 8/8+6/2+0/2 = 4 atoms in unit cell• Relation between atomic radius & lattice parameter (a).Relation between atomic radius & lattice parameter (a).

» R = R = √2a/4 = a/(2√2)√2a/4 = a/(2√2)

Co-ordination = 12

• Close packed directions are face diagonals.• Note: All atoms are identical; the face-centered atoms are shaded differently only for ease of viewing.

FACE CENTERED CUBIC STRUCTURE (FCC)

APF = a3

4

3( 2a/4)34

atoms

unit cell atomvolume

unit cell

volume

a

APF for a body-centered cubic structure = APF for a body-centered cubic structure = /(3/(32) = 0.742) = 0.74

Close-packed directions: length = 4R

= 2 a

ATOMIC PACKING FACTOR: FCC

Unit cell contains = 4 atoms/unitcell

• Example: Cu, Ag, Au, Gamma Fe • General Characteristics of FCC Metals• !Fairly soft• !Malleable (easily formed) and do not harden

much as they deform• !Will deform a great deal before failure

Hexagonal Closed Packed Structure (HCP)

• (i) Arrangements of Atoms:• Unit cell consists of a Hexagon.• One atom at each corner of Hexagon.• One atom each at the center of two Hexagon faces & one

atom at center of line connecting the perpendiculars in the three rhombuses.

• - Coordination number: 12• Example: Ca, Zu, Cd, Mg• No. of atoms in unit cell|:• N: Ni+ Nf/2+ Nc/6 (as unit cell is a hexagon)• Here Ni:3, Nf: 2, Nc:12

Ni:3, Nf: 2, Nc:12

General Characteristics of HCP Metals• Tend to be soft• Generally will harden rapidly as they deform• Cannot usually deform a lot without failure