Sketch the [111] direction and calculate the planar density in the (111) plane for BCC and FCC.
The Correct Answer and Explanation is:
[111] Direction and Planar Density in the (111) Plane for BCC and FCC
Sketch of the [111] Direction:
The [111] direction in a cubic crystal passes through the body diagonal of the unit cell. It connects one corner of the cube to the opposite corner, passing through the center.
In both Body-Centered Cubic (BCC) and Face-Centered Cubic (FCC) structures, the [111] direction passes through atoms at fractional coordinates:
- From (0,0,0) to (1,1,1)
This direction intersects several atoms depending on the crystal structure.
Planar Density in the (111) Plane
1. FCC Structure:
- Atoms per unit area:
- The (111) plane in FCC is densely packed.
- One (111) plane contains 6 atoms:
- 3 atoms at the corners contribute 1/6 each (0.5 atom total)
- 3 atoms at the face centers contribute 1/2 each (1.5 atoms total)
- 3 shared at edges in neighboring planes contribute 1/3 each (1 atom total)
- Effective atoms = 3 atoms per (111) plane
- Area of the (111) plane:
- The (111) plane is an equilateral triangle with side √2 * a
- Area = (√3 / 4) * (√2 * a)² = (√3 / 4) * 2a² = (√3 / 2) * a²
- Planar density = Number of atoms / Area
= 3 / ((√3 / 2) * a²)
= (6 / √3) * (1 / a²)
= (2√3 / a²) atoms per square angstrom
2. BCC Structure:
- Atoms per unit area:
- The (111) plane in BCC is not as densely packed as in FCC.
- Atoms at 3 corners contribute 1/6 each (0.5 atom)
- One atom at the body center contributes to the plane (if intersected)
- Effective atoms ≈ 1.5 atoms
- Area of the (111) plane = same as FCC
= (√3 / 2) * a² - Planar density = 1.5 / ((√3 / 2) * a²)
= (3 / √3) * (1 / a²)
= (√3 / a²) atoms per square angstrom
Summary:
- FCC (111) planar density = (2√3 / a²) atoms / Ų
- BCC (111) planar density = (√3 / a²) atoms / Ų
The FCC (111) plane is more densely packed due to the atomic arrangement, resulting in a higher planar density compared to the BCC (111) plane.
