Problem 29 - Chemistry in 30 days

Question

Total volume of atoms present in a face-centerd cubic unit cell of a metal is ( \(r\) is atomic radius) (a) \((20 / 3) \pi \mathrm{r}^{3}\) (b) \((24 / 3) \pi r^{3}\) (c) \((12 / 3) \pi \mathrm{r}^{3}\) (d) \((16 / 3) \pi r^{3}\)

Step-by-Step Solution

Answer

Option (d) \(\frac{16}{3} \pi r^3 \).

Step 1: Understand the FCC Unit Cell Structure

A face-centered cubic (FCC) unit cell contains atoms at each of the corners and in the center of each face. It effectively has 4 atoms per unit cell.

Step 2: Calculate Individual Atom Volume

The volume of a single atom, assuming it's a perfect sphere, is given by the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the atomic radius.

Step 3: Determine Total Volume of Atoms in FCC

Since a face-centered cubic unit cell contains 4 atoms, the total volume of atoms within one unit cell is \( 4 \times \frac{4}{3} \pi r^3 \).

Step 4: Simplify the Expression

Multiply the volumes: \( 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3 \).

Step 5: Compare with Options and Choose the Correct Answer

The simplified expression \( \frac{16}{3} \pi r^3 \) matches option (d), so that is your correct choice.

Key Concepts

Atomic Radius

The atomic radius is an important concept in understanding atomic structures, particularly in solid state physics and chemistry. The atomic radius is essentially the distance from the center of an atom's nucleus to the outer edge of its electron cloud. This is often visualized as the radius of a sphere that represents the atom. However, since electrons exist in a probability cloud rather than a defined orbit, defining the atomic radius can vary depending on the type of measurement being made.
Atomic radii are typically measured in picometers (pm) or angstroms (Å), with 1 Å = 100 pm. The size of an atom's radius can influence how atoms bond and pack together in a solid.

• In metallic elements, the atomic radius can influence metallic bonding and the strength of the material.
• Smaller radii often result in stronger bonds, as atoms can be packed closer together.

This is particularly important for elements arranged in crystal lattices, such as the face-centered cubic (FCC) structures.

Atom Volume Calculation

When considering the atomic volume, envision each atom as a sphere. This simplification helps in calculating the space occupied by an atom in a crystal lattice. To calculate the volume of a single atom, the formula for the volume of a sphere is used: \[ V = \frac{4}{3} \pi r^3 \]where \( V \) is the volume and \( r \) is the atomic radius.
This formula considers the three-dimensional nature of an atom, accounting for its entire volume. It's important to note that this formula assumes the atom is a perfect sphere, which simplifies calculations and provides a reasonable approximation in most cases.
To understand how atoms fit within a crystal structure, knowing the volume of individual atoms is crucial. It helps in estimating the material's density and the packing efficiency of the atoms within the lattice.

FCC Lattice Structure

A face-centered cubic (FCC) lattice is one of the most efficient ways to pack atoms in a crystal. In an FCC unit cell, atoms are positioned at each corner of a cube, and there are additional atoms located in the center of each of the cube's six faces. Despite having 14 locations, the sharing setup means only a total of 4 atoms are effectively part of one FCC unit cell.
Here's how they add up:

• 8 corner atoms contribute \( 1/8 \) of their volume to one unit cell each, totaling 1 atom.
• 6 face-centered atoms contribute \( 1/2 \) of their volume, adding 3 more atoms.

This arrangement results in high packing efficiency, meaning that FCC structures are typically denser than other types.
Understanding the FCC lattice is pivotal for recognizing material characteristics like ductility and conductivity. Many metals crystallize in this pattern, such as copper, aluminum, and gold, allowing them to be excellent conductors and relatively easy to deform without breaking.