The planar density (PD) of a crystal plane in a crystal structure is defined as the number of atoms per unit area on that plane. For a simple cubic structure, where each atom is at the corners of the unit cell, we calculate the planar density for the $(1, -1, 1)$ plane as follows:

Step 1: Determine the atom positions on the (1, -1, 1) plane.

• A simple cubic (SC) unit cell has atoms at each corner of the cube, and the unit cell is characterized by its lattice parameter $a$, which is the distance between adjacent atoms.

• The $(1, -1, 1)$ plane intersects the corners of the cubic cell in a certain pattern. Atoms at the corners of a cube in a simple cubic structure are shared among eight unit cells, so each corner atom contributes $\frac{1}{8}$ of an atom to the unit cell.

Step 2: Calculate the number of atoms per unit area on the $(1, -1, 1)$ plane.

Number of atoms on the plane:

To find how many atoms are centered on the $(1, -1, 1)$ plane, we need to consider only the atoms that lie directly on the plane. For a simple cubic structure:

• In a single $(1, -1, 1)$ plane passing through the origin, there are atoms located at the corners that contribute partially.
• The plane passes through 3 atoms at the corners of the cubic unit cell.

Thus, for a simple cubic unit cell:

• Each corner atom contributes $\frac{1}{8}$ of an atom, and there are 3 such atoms, giving us: $\text{Number of atoms} = 3 \times \frac{1}{8} = \frac{3}{8} \text{ atoms}$

Area of the $(1, -1, 1)$ plane:

The area of the $(1, -1, 1)$ plane is calculated using the lattice parameter $a$. The Miller indices $(h, k, l)$ give the orientation of the plane, and the interplanar spacing $d_{hkl}$ for a simple cubic structure is given by: $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$ Substituting $h = 1$, $k = -1$, and $l = 1$, we get: $d_{1,-1,1} = \frac{a}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{a}{\sqrt{3}}$ The area of the plane can be approximated as a square with side length $a$, so: $\text{Area of the plane} = d_{1,-1,1}^2 = \left(\frac{a}{\sqrt{3}}\right)^2 = \frac{a^2}{3}$

Step 3: Calculate the planar density.

Now that we have the number of atoms on the plane and the area of the plane, we can calculate the planar density (PD) as: $\text{PD} = \frac{\text{Number of atoms on the plane}}{\text{Area of the plane}} = \frac{\frac{3}{8}}{\frac{a^2}{3}} = \frac{3}{8} \times \frac{3}{a^2} = \frac{9}{8a^2}$

Thus, the planar density of the $(1, -1, 1)$ plane in a simple cubic unit cell is: $\boxed{\frac{9}{8a^2}}$

Do you need further details or have any questions?

Here are 5 related questions for deeper exploration:

1. What is the difference between planar density and linear density?
2. How does the planar density of the (1, 1, 1) plane compare to that of the (1, -1, 1) plane in other crystal structures, such as FCC?
3. How is the atomic packing factor (APF) related to the planar density in a simple cubic structure?
4. What is the effect of changing lattice parameters on the planar density of a given plane?
5. How would you calculate the planar density for a BCC or FCC unit cell?

Tip: The Miller indices $(h, k, l)$ are crucial in determining the geometry of planes and directions in crystal structures, so mastering them is key to understanding crystallography.