Question

The packing fraction for a two-dimensional hexagonal lattice having sides 2r with atoms of radii r placed at each vertex and at the center is _____. (up to two decimal places)

Hint:

The 2D hexagonal lattice is the most densely packed arrangement of circles on a plane. Its packing fraction, \(\frac{\pi}{2\sqrt{3}}\), is a fundamental result worth remembering. The question's description of a hexagonal unit cell is a non-primitive (conventional) cell for this lattice, but it correctly yields the same packing fraction as the primitive (rhombus-shaped) unit cell.

Answer 1

Correct Answer: 0.91

Step 1: Understanding the Concept:
The packing fraction (or packing efficiency) is the fraction of the area in a two-dimensional crystal structure that is occupied by atoms. It is a measure of how tightly the atoms are packed. The question describes a specific non-primitive unit cell (a hexagon) for a 2D lattice.
Step 2: Key Formula or Approach:
The packing fraction (PF) is calculated as: \[ \text{PF} = \frac{(\text{Number of atoms per unit cell}) \times (\text{Area of one atom})}{(\text{Area of the unit cell})} \] We need to find each of these quantities for the described hexagonal cell.
Step 3: Detailed Explanation:
The problem defines the unit cell as a regular hexagon with side length \(a = 2r\). Atoms of radius \(r\) are placed at each of the 6 vertices and at the center. 1. Area of the Unit Cell (\(A_{cell}\)): The area of a regular hexagon with side length \(a\) is \(A_{cell} = \frac{3\sqrt{3}}{2}a^2\). Substituting \(a=2r\): \[ A_{cell} = \frac{3\sqrt{3}}{2}(2r)^2 = \frac{3\sqrt{3}}{2}(4r^2) = 6\sqrt{3}r^2 \] 2. Number of Atoms per Unit Cell (\(N\)): - There is one atom at the center, which belongs entirely to this cell. - There are six atoms at the vertices. In a 2D lattice made of space-filling hexagons, each vertex is shared by three adjacent hexagons. Therefore, each vertex atom contributes \(1/3\) of its area to the cell. - Total number of atoms per cell: \(N = 1_{\text{center}} + 6_{\text{vertices}} \times \frac{1}{3} = 1 + 2 = 3\). 3. Area of Atoms (\(A_{atoms}\)): The atoms are treated as 2D circles of radius \(r\). The area of a single atom is \(\pi r^2\). Total area occupied by atoms within the unit cell is: \[ A_{atoms} = N \times (\pi r^2) = 3 \pi r^2 \] 4. Packing Fraction (PF): \[ \text{PF} = \frac{A_{atoms}}{A_{cell}} = \frac{3 \pi r^2}{6\sqrt{3}r^2} = \frac{3\pi}{6\sqrt{3}} = \frac{\pi}{2\sqrt{3}} \] Now, calculate the numerical value: \[ \text{PF} = \frac{\pi}{2\sqrt{3}} \approx \frac{3.14159}{2 \times 1.73205} = \frac{3.14159}{3.4641} \approx 0.9069 \] Step 4: Final Answer:
The packing fraction, rounded to two decimal places, is 0.91.