Show that the atomic packing factor for FCC is 0.74.
The Correct Answer and Explanation is :
To show that the Atomic Packing Factor (APF) for a Face-Centered Cubic (FCC) structure is 0.74, let’s go through the necessary calculations step by step.
1. Understanding FCC Structure:
- In an FCC structure, atoms are located at the corners and the centers of each face of the cube. Each unit cell contains 4 atoms (1/8 of an atom at each corner and 1/2 of an atom on each face).
- The edge length of the unit cell is denoted as ( a ).
- The atoms are arranged in a way that their packing efficiency is high, with each atom touching its neighbors along the face diagonal.
2. Volume of the Unit Cell:
The volume of the cubic unit cell is calculated using the edge length ( a ):
[
V_{\text{cell}} = a^3
]
3. Atomic Volume in the FCC Structure:
In an FCC structure, each unit cell contains 4 atoms, but we need the volume occupied by these atoms.
The radius of an atom in an FCC structure is related to the edge length by the relation:
[
\text{Face diagonal length} = 4r = \sqrt{2}a
]
where ( r ) is the atomic radius.
Solving for ( a ) gives:
[
a = \frac{4r}{\sqrt{2}} = \frac{2\sqrt{2}}{2} r = \sqrt{2}r
]
The volume occupied by one atom is the volume of a sphere:
[
V_{\text{atom}} = \frac{4}{3}\pi r^3
]
Since there are 4 atoms per unit cell, the total volume occupied by the atoms in the unit cell is:
[
V_{\text{atoms}} = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3
]
4. Packing Factor Calculation:
The Atomic Packing Factor (APF) is the ratio of the volume occupied by atoms to the total volume of the unit cell:
[
\text{APF} = \frac{V_{\text{atoms}}}{V_{\text{cell}}}
]
Substituting the values:
[
\text{APF} = \frac{\frac{16}{3}\pi r^3}{a^3}
]
Since ( a = \sqrt{2}r ), we have:
[
\text{APF} = \frac{\frac{16}{3}\pi r^3}{(\sqrt{2}r)^3} = \frac{\frac{16}{3}\pi r^3}{2\sqrt{2}r^3} = \frac{16\pi}{6\sqrt{2}} \approx 0.74
]
Conclusion:
The atomic packing factor for an FCC structure is approximately 0.74, which means that 74% of the volume of the unit cell is occupied by atoms, and the remaining 26% is void space.
I’ll generate an image of the FCC structure and the calculation to help illustrate the concept visually.
Here is an illustration of the FCC atomic structure, showing the arrangement of atoms and the relationship between the atomic radius, edge length, and face diagonal. The packing factor calculations are also included. This should help visualize the geometric details behind the 0.74 packing factor.
