Question

Lead crystallizes in a face-centered cubic unit cell with an edge length of \(495 \mathrm{pm}\). What is the radius of a lead atom in picometers? What is the density of lead in \(\mathrm{g} / \mathrm{cm}^{3}\) ?

Short answer

The atomic radius is approximately 175.4 pm, and the density is about 11.4 g/cm³.

Step-by-step solution

Understand the Geometry of FCC

In a face-centered cubic (FCC) lattice, each corner atom is shared by 8 unit cells and each face-centered atom is shared by 2 unit cells. The atoms along the face diagonal touch each other, so the relationship between the edge length (a) and the atomic radius (r) is given by \(4r = \sqrt{2}a\).

Calculate the Atomic Radius

Given the edge length \(a = 495 \text{ pm}\), use the equation from Step 1: \[4r = \sqrt{2} \cdot 495 \text{. Hence, } r = \frac{\sqrt{2} \cdot 495}{4} \approx 175.4 \text{ pm}\].So, the radius of a lead atom is approximately 175.4 pm.

Calculate the Volume of the Unit Cell

The volume of a cube is given by \(a^3\). With \(a = 495 \text{ pm} = 495 \times 10^{-10} \text{ cm}\), the unit cell volume is:\[\text{Volume} = (495 \times 10^{-10})^3 = 1.21 \times 10^{-22} \text{ cm}^3\].

Determine Number of Atoms per Unit Cell

In a face-centered cubic structure, there are 4 atoms per unit cell. This is calculated as: 8 corner atoms \( \times \frac{1}{8} + 6 \text{ face atoms} \times \frac{1}{2} = 4 \) atoms per unit cell.

Calculate the Mass of a Unit Cell

The atomic mass of lead is approximately 207.2 g/mol. Use Avogadro's number \(6.022 \times 10^{23}\) atoms/mol to find the mass of one atom, and multiply by the number of atoms:\[\text{Mass per unit cell} = \frac{207.2 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \times 4 = 1.38 \times 10^{-22} \text{ g}\].

Calculate the Density

Density is mass per unit volume:\[\text{Density} = \frac{1.38 \times 10^{-22}\, \text{g}}{1.21 \times 10^{-22}\, \text{cm}^3} \approx 11.4 \text{ g/cm}^3\]. Hence, the density of lead is approximately 11.4 g/cm³.

Key concepts

Atomic Radius

In a face-centered cubic (FCC) unit cell, understanding the concept of atomic radius is essential. The atomic radius is the distance from the center of an atom to the outer boundary of its electron cloud, which determines the size of the atom. For FCC structures, atoms on a face diagonal are in direct contact with each other. This specific arrangement means the face diagonal equals four radii of the atoms.

The relationship can be mathematically expressed as follows:

• For the edge length (\( a \)) of 495 pm, the equation is given by \(4r = \sqrt{2}a\).
• Solving for \( r \) gives the atomic radius \( r = \frac{\sqrt{2} \cdot 495}{4} \approx 175.4 \text{ pm}\).

Thus, the atomic radius of lead in this configuration is approximately 175.4 pm.

Unit Cell Volume

Calculating the volume of the unit cell is a straightforward step that involves considering the geometric nature of the cube. In a cubic unit cell, the volume is simply calculated using the formula for the volume of a cube: \(a^3\), where \(a\) is the edge length of the cell. For a lead FCC unit cell:

• The edge length \(a\) is 495 pm, equivalent to \(495 \times 10^{-10} \text{ cm}\), converting it to centimeters to match the unit system of density.
• Thus, the volume becomes \[(495 \times 10^{-10})^3 = 1.21 \times 10^{-22} \text{ cm}^3\].

This cube represents the space that the atoms in the unit cell occupy in a crystalline structure.

Density Calculation

Density is a critical material property that describes how much mass is contained in a given volume. In the context of a crystal lattice like the face-centered cubic unit cell of lead, we use a few steps to calculate it:

• Begin with the mass of one unit cell. We calculated this from atomic data:
• For lead, its atomic mass of 207.2 g/mol, with Avogadro’s number (\(6.022 \times 10^{23} \) atoms/mol), leads us to find the mass of one atom and then of the unit cell containing 4 atoms:
  \(\text{Mass} = \frac{207.2}{6.022 \times 10^{23}} \times 4 = 1.38 \times 10^{-22} \text{ g}\).
• Finally, to find the density, divide the mass by the volume of the unit cell:
  \(\text{Density} = \frac{1.38 \times 10^{-22} \text{ g}}{1.21 \times 10^{-22} \text{ cm}^3} \approx 11.4 \text{ g/cm}^3\).

So, the density of the crystalline lead is about 11.4 g/cm³, indicative of how tightly packed the atoms are within the structure.

Avogadro's Number

Avogadro's number is a fundamental constant in chemistry and physics, representing the number of atoms, molecules, or other particles found in one mole of a substance. It plays a crucial role when determining the mass of atoms in a given quantity of material:

• This number is approximately \(6.022 \times 10^{23} \), which is huge and allows chemists to convert between the number of atoms and the mass of a sample efficiently.
• In our calculation of lead's unit cell, Avogadro’s number helps us understand how the atomic mass (207.2 g/mol) translates into the mass of individual atoms and subsequently the entire unit cell of FCC lead.

By using Avogadro's number, we can accurately ascertain how macroscopic measurements correlate to atomic scale insights, which is fundamental for material science and chemistry.