What is the planar density of the (1,-1,1) plane in a simple cubic unit cell with equal sized atoms?

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.

Explore how to calculate the planar density of the (1,-1,1) plane in a simple cubic unit cell with equal-sized atoms. This solution involves key crystallography concepts such as the Miller indices, interplanar spacing, and geometry of planes in cubic structures. Perfect for undergraduate students studying materials science or solid state physics.

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