Question

The ratio of the density of atoms between the (111) and (110) planes in a simple cubic (sc) lattice is ____. (up to two decimal places)

Hint:

To calculate planar density, it's crucial to correctly visualize the plane within the unit cell, determine its area, and count the effective number of atoms lying on that area. For atoms at vertices, edges, or faces, remember to divide by the number of cells sharing them (e.g., a corner atom on a plane area is shared by 4 cells if the area is a rectangle, or 6 cells if it's a triangle).

Answer 1

Correct Answer: 0.82

Step 1: Understanding the Concept:
The question asks for the ratio of the "density of atoms between" two different crystallographic planes. This is interpreted as the ratio of the planar atomic densities (\(\sigma\)) of these planes. Planar density is the number of atoms per unit area on a given plane.
Step 2: Key Formula or Approach:
The planar density \(\sigma_{hkl}\) is calculated as: \[ \sigma_{hkl} = \frac{\text{Number of atoms centered on the plane within a unit cell}}{\text{Area of the plane within the unit cell}} \] We will calculate this for the (111) and (110) planes of a simple cubic (sc) lattice with lattice constant \(a\). Then we will find the ratio \(\sigma_{111}/\sigma_{110}\).
Step 3: Detailed Explanation:
1. Planar Density of the (110) plane (\(\sigma_{110}\)): - In an sc lattice, the (110) plane cuts the x-axis at \(1a\) and the y-axis at \(1a\), and is parallel to the z-axis. - The area of this plane within the unit cell is a rectangle with side lengths \(a\) and \(a\sqrt{2}\). So, Area\(_{110} = a \times a\sqrt{2} = a^2\sqrt{2}\). - In an sc lattice, atoms are only at the corners of the unit cube. The (110) plane passes through four corner atoms. Each corner atom on the plane is shared by four adjacent unit cells that share that plane area. - Number of atoms on the plane within the unit cell = \(4 \times \frac{1}{4} = 1\). - Planar density of (110): \(\sigma_{110} = \frac{1}{a^2\sqrt{2}}\). 2. Planar Density of the (111) plane (\(\sigma_{111}\)): - In an sc lattice, the (111) plane cuts the x, y, and z axes at \(1a\). - The area of this plane within the unit cell is an equilateral triangle with vertices at (a,0,0), (0,a,0), and (0,0,a). - The side length of this triangle is the distance between any two of these points, e.g., \(\sqrt{(a-0)^2 + (0-a)^2 + (0-0)^2} = \sqrt{2a^2} = a\sqrt{2}\). - Area of the equilateral triangle Area\(_{111} = \frac{\sqrt{3}}{4}(\text{side})^2 = \frac{\sqrt{3}}{4}(a\sqrt{2})^2 = \frac{\sqrt{3}}{4}(2a^2) = \frac{a^2\sqrt{3}}{2}\). - The plane passes through three corner atoms. Each corner atom on this triangular area is shared by six unit cells that meet at that vertex. - Number of atoms on the plane within the unit cell = \(3 \times \frac{1}{6} = \frac{1}{2}\). - Planar density of (111): \(\sigma_{111} = \frac{1/2}{a^2\sqrt{3}/2} = \frac{1}{a^2\sqrt{3}}\). 3. Ratio \(\sigma_{111} / \sigma_{110}\): \[ \frac{\sigma_{111}}{\sigma_{110}} = \frac{1/(a^2\sqrt{3})}{1/(a^2\sqrt{2})} = \frac{a^2\sqrt{2}}{a^2\sqrt{3}} = \frac{\sqrt{2}}{\sqrt{3}} \] Now, calculate the numerical value: \[ \frac{\sqrt{2}}{\sqrt{3}} \approx \frac{1.4142}{1.7320} \approx 0.8165 \] Step 4: Final Answer:
The ratio of the density of atoms is approximately 0.82.