Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius R.
The Correct Answer and Explanation is :
Planar Density Derivation for BCC (100) and (110) Planes
Planar Density Definition
The planar density (PD) is defined as the number of atoms per unit area in a specific crystallographic plane. Mathematically,
[
PD = \frac{\text{Number of atoms centered on the plane}}{\text{Area of the plane}}
]
where the number of atoms is determined by the atomic positions in the given plane, and the area is calculated based on the lattice parameters.
1. Planar Density for BCC (100) Plane
Step 1: Number of Atoms in the (100) Plane
In the (100) plane of a BCC (Body-Centered Cubic) structure, we observe:
- Four corner atoms, each contributing ( \frac{1}{4} ) of an atom (since each corner atom is shared among four adjacent unit cells).
- No atoms at the center of the (100) face in a BCC structure.
Thus, the total number of atoms in the (100) plane:
[
\text{Total atoms} = 4 \times \frac{1}{4} = 1
]
Step 2: Area of the (100) Plane
The unit cell edge length a in terms of the atomic radius ( R ) for BCC is:
[
a = \frac{4R}{\sqrt{3}}
]
Since the (100) plane is a square with side length a, its area is:
[
A_{100} = a^2 = \left(\frac{4R}{\sqrt{3}}\right)^2 = \frac{16R^2}{3}
]
Step 3: Planar Density Calculation
[
PD_{(100)} = \frac{1}{A_{100}} = \frac{1}{\frac{16R^2}{3}} = \frac{3}{16R^2}
]
2. Planar Density for BCC (110) Plane
Step 1: Number of Atoms in the (110) Plane
In the (110) plane of a BCC structure, we observe:
- Four corner atoms, each contributing ( \frac{1}{4} ) of an atom.
- One full body-centered atom lies entirely within the plane.
Thus, the total number of atoms in the (110) plane:
[
\text{Total atoms} = (4 \times \frac{1}{4}) + 1 = 2
]
Step 2: Area of the (110) Plane
The (110) plane forms a rectangular surface with dimensions a (one unit cell edge) and ( \sqrt{2}a ) (diagonal of the square face). The area is:
[
A_{110} = a \times \sqrt{2}a = a^2\sqrt{2}
]
Substituting ( a = \frac{4R}{\sqrt{3}} ):
[
A_{110} = \left(\frac{4R}{\sqrt{3}}\right)^2 \times \sqrt{2} = \frac{16R^2}{3} \times \sqrt{2} = \frac{16\sqrt{2} R^2}{3}
]
Step 3: Planar Density Calculation
[
PD_{(110)} = \frac{2}{A_{110}} = \frac{2}{\frac{16\sqrt{2} R^2}{3}} = \frac{3}{8\sqrt{2} R^2}
]
Final Expressions
[
PD_{(100)} = \frac{3}{16R^2}
]
[
PD_{(110)} = \frac{3}{8\sqrt{2} R^2}
]
Explanation
Planar density provides insight into how closely packed atoms are within a specific crystallographic plane. It is important for understanding surface energy, slip systems, and material properties such as plastic deformation and diffusion rates.
For BCC (100) vs. (110):
- The (110) plane has a higher planar density than (100), which means atoms are more closely packed in (110).
- The (110) plane is a common slip plane in BCC metals because of this higher atomic density, allowing for easier dislocation movement.
These expressions are fundamental in materials science, particularly for mechanical properties and failure mechanisms in BCC metals like iron, tungsten, and molybdenum.