Question

Nickel has a face-centered cubic unit cell. The density of nickel is \(6.84 \mathrm{g} / \mathrm{cm}^{3} .\) Calculate a value for the atomic radius of nickel.

Short answer

The atomic radius of nickel is approximately \(1.24 \times 10^{-8}\) cm.

Step-by-step solution

Determine the number of atoms per unit cell in a face-centered cubic structure

In a face-centered cubic (FCC) structure, there are 8 corner atoms contributing 1/8 of their volume each and 6 face atoms contributing 1/2 of their volume each. So the total number of atoms per unit cell is: Atoms per unit cell = 8 * 1/8 + 6 * 1/2 = 4

Calculate the mass of one unit cell

Given the density of nickel (ρ) is 6.84 g/cm³. We will use the formula to calculate the mass (m) of the unit cell. m = ρ * V Where m is the mass of the unit cell, ρ is the density, and V is the volume of the unit cell. We know that the number of atoms in a unit cell is 4, so the mass of one unit cell can be written as: m = 4 * (mass of one nickel atom) To find the mass of one nickel atom, we will use its molar mass (58.69 g/mol) and Avogadro's number (6.022 × 10^23 atoms/mol): mass of one nickel atom = (58.69 g/mol) / (6.022 × 10^23 atoms/mol) = 9.746 × 10^-23 g/atom Now, we can find the mass of one unit cell: m = 4 * (9.746 × 10^-23 g/atom) = 3.898 × 10^-22 g

Determine the volume of the unit cell

We can use the formula for the density (ρ) to find the volume (V) of the unit cell. ρ = m / V Rearranging to find the volume: V = m / ρ Now, plug the values of the mass and the density of nickel: V = (3.898 × 10^-22 g) / (6.84 g/cm³) = 5.7 × 10^-24 cm³

Calculate the edge length of the unit cell

The volume of a cube is given by: V = a^3 Where a is the edge length of the cube. Rearranging this equation to find the edge length: a = ∛(V) Substituting the value of the volume of the unit cell: a = ∛(5.7 × 10^-24 cm³) = 3.52 × 10^-8 cm

Determine the relationship between the atomic radius and the edge length in a face-centered cubic structure

In an FCC unit cell, the relationship between the atomic radius (r) and the edge length (a) can be derived from geometry and is given by: a = √8*r Rearranging the equation to find the atomic radius: r = a / √8

Calculate the atomic radius of nickel

Now, we will plug the value of the edge length (a) into the equation and calculate the atomic radius (r) of nickel: r = (3.52 × 10^-8 cm) / √8 = 1.24 × 10^-8 cm So, the atomic radius of nickel is approximately 1.24 × 10^-8 cm.

Key concepts

Face-Centered Cubic Structure

In the world of crystal structures, the face-centered cubic (FCC) structure is one of the most important and common arrangements of atoms in metals, including nickel as mentioned in our exercise. In an FCC structure, each unit cell consists of eight corner atoms and six face atoms. However, each corner atom is shared among eight adjacent unit cells and each face atom is shared between two, which means each corner atom contributes only 1/8 of its volume to one particular unit cell, and each face atom contributes 1/2.

This unique arrangement results in a dense, tightly packed structure with atoms at each corner of a cube, plus a single atom centered on each of the cube's faces. The FCC structure is particularly efficient in terms of space utilization within the cell—a feature that affects the material's properties, such as its density and ductility.

Unit Cell Volume

Understanding the volume of a unit cell is critical when delving into the microscopic properties of materials. The volume of a unit cell, particularly for cubic structures such as the FCC, is determined by cubing the edge length (\( a^3 \)).

This volume is a key piece in the puzzle when calculating properties like density and is used to establish the relationship between a substance's observable mass and its arrangement on the atomic level. In our exercise with nickel's FCC structure, we calculated the edge length based on the unit cell's mass and density—and from there, we derived the volume of the unit cell which plays a crucial role in further determining the atomic radius.

Density and Molar Mass Relationship

The connection between a substance's density and its molar mass is another fundamental concept in material science. Density reflects how tightly mass is packed within a given volume, while the molar mass denotes the mass of one mole of atoms of an element.

To find out more about the relationship between the two, we consider Avogadro's number, which provides the quantity of atoms in a mole, and this allows us to transition from discussing individual atoms to dealing with macroscopic amounts. During our problem-solving exercise, we put this relationship into action to calculate the mass of an individual nickel atom. Knowing the molar mass of nickel and using Avogadro's number, we calculated the mass of a nickel atom, and from there, the mass of the whole unit cell. These calculations link the atomic-level characteristics and the bulk properties of the material, providing a deep understanding of the matter's structure.