BCC, HCP, APF

1. Material Science and Technology
Unit 1 Lecture 05
By:
Saumy Agarwal
Asst. Professor (MED)
BTKIT Dwarahat

2. Body Centered Cubic Structure (BCC)
• In BCC, there are atoms at all eight corners of
the cube and a single atom at the center of cube.
• Examples- α-Iron, δ-Iron, Cr, W, Mo

3. • The atoms at each corner are shared with 8 unit
cells and a single atom is located at the center of
the unit cell.
• Thus, Number of atoms in a unit cell = 8 ×
1
8
+ 1
n = 𝟐
• Co-ordination number= 8
(the central atom is surrounded
by eight nearest neighbors)

4. Hexagonal Close-packed (HCP)
Structure
• Unit cell is a hexagonal in shape.
Atomic arrangement in HCP unit cell
In most cases,
c/a= 1.633

5. • The top and bottom faces of the unit cell consist
of six atoms that form regular hexagons and they
surround a single atom at the center of the face.
• Between the top and bottom planes, a plane is
situated that provides three additional atoms to
the unit cell.
• Examples- Zn, Cd, Co, Mg, Ti

6. • The corner atoms are shared with 6 unit cells
and the atoms at the face center are shared by 2
unit cells.
• Number of atoms in a unit cell
n = 12 ×
1
6
+ 2 ×
1
2
+ 3
n = 𝟔

7. One atom at the center of the top face is
surrounded by:
1. six neighbouring atoms in the plane of the face
2. three central atoms of the lower unit cell, and
3. three central atoms of the upper unit cell
• Co-ordination Number= 12

8. Atomic Packing Factor (APF)
• It is a fraction of volume occupied by all the
atoms in a unit cell to the total volume of a unit
cell.
APF =
volume of atoms in a unit cell
total volume of a unit cell
• A simple cubic unit cell has APF= 0.52

9. • APF of a FCC unit cell:
a= edge length
r= radius of an atomic sphere
𝑎 = 2 2 𝑟
In a FCC unit cell, n= 4
𝐴𝑃𝐹 =
4 × 4
3 𝜋𝑟3
𝑎3
𝑨𝑷𝑭 = 𝟎. 𝟕𝟒

10. • APF of a BCC unit cell:
a= edge length
r= radius of an atomic sphere
Length of body diagonal= 4r
𝑎 =
4𝑟
3
In a BCC unit cell, n= 2
𝐴𝑃𝐹 =
2 × 4
3 𝜋𝑟3
𝑎3
𝑨𝑷𝑭 = 𝟎. 𝟔𝟖

11. • APF of a HCP unit cell:
a= edge length of hexagon
r= radius of an atomic sphere
𝑐
𝑎 = 1.633, r=0.5 a
In a HCP unit cell, n= 6
𝐴𝑃𝐹 =
6×4
3 𝜋𝑟3
3 3
2
𝑎2×𝑐
𝑨𝑷𝑭 = 𝟎. 𝟕𝟒

12. Density of crystal
The density of a crystal can be calculated as:
𝜌 =
𝑛𝐴
𝑉𝐶 𝑁𝐴
where, n is number of atoms in a unit cell, A is
atomic weight, VC is the volume of unit cell, NA is
Avogadro's number (6.023×1023 atoms/mol)