Question

What is the void space per unit cell for metallic silver crystallizing in the FCC system, the edge length of the unit cell being \(4 \AA\) ? (a) \(47.36 \AA^{3}\) (b) \(30.72 \AA^{3}\) (c) \(20.48 \AA^{3}\) (d) \(16.64 \AA^{3}\)

Short answer

The void space per unit cell for metallic silver is not given by any of the provided options (a) 47.36 ų, (b) 30.72 ų, (c) 20.48 ų, or (d) 16.64 ų - the calculation for the correct void space must be done.

Step-by-step solution

Find the Volume of the Unit Cell

Calculate the volume of the FCC unit cell using the edge length. The unit cell is cubic, so the formula for the volume is: volume = edge length^3. With the edge length being 4 Å, the volume is (4 Å)^3.

Calculate the Volume Occupied by Atoms

In an FCC unit cell, there are 4 atoms per unit cell. The atomic volume is calculated using the volume of a sphere, which is: volume = (4/3)πr^3. Given that each silver atom touches its neighbors along the face diagonal, the atomic radius r can be determined using the face diagonal length, which is 4r = √2 times the edge length, hence r = edge length / (√2 * 2).

Compute the Void Space

Subtract the total volume occupied by atoms (calculated in Step 2) from the volume of the unit cell (calculated in Step 1) to find the void space per unit cell.

Key concepts

Crystal Structure

Understanding crystal structure is fundamental in materials science as it describes the arrangement of atoms in a crystalline solid. The crystal structure not only defines the shape of the solid but also has a profound impact on its physical properties, such as strength, electrical conductivity, and melting point. In layman's terms, you can think of it like the architecture of a building, which determines whether it's a skyscraper or a bungalow.

Different materials crystallize in different structures depending on their atomic interactions. These structures are defined by the type of unit cell they form, which is the basic repeating unit that makes up the crystal. Understanding the unit cell helps determine many important characteristics of the material, including the one addressed in our exercise: the void space.

Face-Centered Cubic

The face-centered cubic (FCC) structure is one such arrangement where atoms are located at each corner of a cube and at the center of each face, hence the name. This leads to a specific way in which the atoms pack together, known as close-packed structure.

Imagine eight oranges in a box, one at each corner, and an orange on each of the six faces of the box, partially inside the box. This is the 3D structure of an FCC unit cell. This structure is highly efficient in terms of space utilization and is found in many metals, including silver, as in our example. Each silver atom in the FCC cell is shared with adjacent cells, meaning we don't have a whole atom inside our specific cell but rather portions of it that add up to equivalent atoms.

Atomic Radius

The atomic radius is half the distance between the centers of two atoms that are just touching each other. It's a crucial measurement because it directly relates to the size of atoms and how close they can get to each other.

This is important in the context of the FCC crystal structure since the atoms touch along the face diagonal. For silver and other metals with the FCC arrangement, we can find the atomic radius by diving into some geometry. If we draw a line through the center of a cube diagonally across a face (the face diagonal), passing through two atoms, this line would be as long as four atomic radii, or to be precise, \(4r = \sqrt{2}\) times the edge length of the cube. Knowing the edge length gives us a direct way to calculate the atomic radius.

Void Volume Calculation

When it comes to void volume calculation in crystal structures, we're basically figuring out the unused space within the unit cell. In layman's terms, it's like finding out how much empty room you have left in a suitcase packed with spheres.

In the case of FCC, although the unit cell is tightly packed with atoms, there are still some interstitial spaces—tiny pockets between the atoms that remain unoccupied. By calculating the volume of the unit cell and subtracting the volume occupied by atoms (which themselves are tiny spheres), we can determine the void space. This void space is not wasted, though; it can affect properties like diffusion and the overall density of the material. In our specific task concerning silver, we can use the atomic radius that we have calculated from the edge length to work out the volume taken up by atoms and thus the remaining void within the cell.