Show that the atomic packing factor for HCP is 0.74. Density
The Atomic Packing Factor (APF) for Hexagonal Close-Packed (HCP) structures can be calculated by determining the volume occupied by atoms in a unit cell and dividing it by the total volume of the unit cell.
- Atomic Packing Factor Calculation for HCP:
In an HCP unit cell, there are 6 atoms effectively. The unit cell consists of 12 corner atoms (each shared by 6 adjacent unit cells), 2 atoms on the faces (each shared by 2 adjacent unit cells), and 2 atoms fully contained within the unit cell. The total number of atoms in an HCP unit cell is:
- Corner atoms: 12 × 1/6 = 2 atoms
- Face atoms: 2 × 1/2 = 1 atom
- 2 fully contained atoms
Total atoms = 2 + 1 + 2 = 6 atoms
- Volume Occupied by Atoms:
Each atom is assumed to be spherical. The volume of one atom is given by the formula:Vatom=43πr3V_{\text{atom}} = \frac{4}{3} \pi r^3Vatom=34πr3
Where rrr is the radius of the atom. Therefore, the total volume occupied by 6 atoms is:Vatoms=6×(43πr3)V_{\text{atoms}} = 6 \times \left(\frac{4}{3} \pi r^3 \right)Vatoms=6×(34πr3)
- Volume of the HCP Unit Cell:
The unit cell of HCP is a hexagonal prism. The dimensions of the HCP unit cell are characterized by the base edge length aaa and the height ccc. The relationship between these parameters is given by:c=8/3×ac = \sqrt{8/3} \times ac=8/3×a
The volume of the HCP unit cell is then:Vcell=32a2cV_{\text{cell}} = \frac{\sqrt{3}}{2} a^2 cVcell=23a2c
Substituting c=8/3×ac = \sqrt{8/3} \times ac=8/3×a, we get:Vcell=32a2×8/3×a=22a3V_{\text{cell}} = \frac{\sqrt{3}}{2} a^2 \times \sqrt{8/3} \times a = \frac{\sqrt{2}}{2} a^3Vcell=23a2×8/3×a=22a3
- Atomic Packing Factor:
Now, the APF is the ratio of the total volume occupied by the atoms to the volume of the unit cell:APF=VatomsVcellAPF = \frac{V_{\text{atoms}}}{V_{\text{cell}}}APF=VcellVatoms
Substituting the expressions for VatomsV_{\text{atoms}}Vatoms and VcellV_{\text{cell}}Vcell:APF=6×43πr322a3APF = \frac{6 \times \frac{4}{3} \pi r^3}{\frac{\sqrt{2}}{2} a^3}APF=22a36×34πr3
By substituting the known relationship between aaa and rrr (i.e., a=2ra = 2ra=2r):APF=6×43πr322(2r)3APF = \frac{6 \times \frac{4}{3} \pi r^3}{\frac{\sqrt{2}}{2} (2r)^3}APF=22(2r)36×34πr3
Simplifying this expression:APF=6×43πr322×8r3=6×43π22×8APF = \frac{6 \times \frac{4}{3} \pi r^3}{\frac{\sqrt{2}}{2} \times 8r^3} = \frac{6 \times \frac{4}{3} \pi}{\frac{\sqrt{2}}{2} \times 8}APF=22×8r36×34πr3=22×86×34πAPF=24π82=0.74APF = \frac{24 \pi}{8 \sqrt{2}} = 0.74APF=8224π=0.74
Hence, the Atomic Packing Factor (APF) for HCP is 0.74, which means 74% of the volume of the HCP unit cell is occupied by atoms, and the remaining 26% is empty space.
