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The density of lead is \(11.350 \mathrm{g} / \mathrm{cm}^{3},\) and the metal crystallizes in a face-centered cubic unit cell. Estimate the radius of a lead atom.

Short Answer

Expert verified
The radius of a lead atom is approximately 0.175 nm.

Step by step solution

01

Calculate the Volume of the Unit Cell

Since lead forms a face-centered cubic (fcc) crystal structure, each unit cell contains 4 atoms of lead. With the given density \( \rho = 11.350 \mathrm{g/cm^3} \), we can determine the volume \( V \) of the unit cell using the formula: \[ \rho = \frac{Z \times M}{N_A \times V_{cell}} \]where \( Z \) is the number of atoms per unit cell (4 for fcc), \( M \) is the molar mass of lead (207.2 g/mol), and \( N_A \) is Avogadro's number (approximately \( 6.022 \times 10^{23} \) mol\(^{-1}\)).Solve for \( V_{cell} \): \[ V_{cell} = \frac{4 \times 207.2 \mathrm{g/mol}}{6.022 \times 10^{23} \mathrm{mol^{-1}} \times 11.350 \mathrm{g/cm^3}} \approx 3.547 \times 10^{-23} \mathrm{cm^3} \]
02

Calculate the Edge Length of the Unit Cell

The volume of the cubic unit cell \( V_{cell} \) is equal to the cube of the edge length \( a \). Solving for \( a \) gives: \[ a = \left( V_{cell} \right)^{1/3} = \left( 3.547 \times 10^{-23} \mathrm{cm^3} \right)^{1/3} \approx 4.95 \times 10^{-8} \mathrm{cm} \]
03

Calculate the Atomic Radius

In a face-centered cubic lattice, the diagonal across the face of the cube consists of 4 atomic radii. The diagonal is also equal to \( \sqrt{2} \times a \). Therefore, the atomic radius \( r \) is: \[ 4r = \sqrt{2} \times a \rightarrow r = \frac{\sqrt{2} \times a}{4} \]Substitute the edge length \( a \) calculated in the previous step:\[ r = \frac{\sqrt{2} \times 4.95 \times 10^{-8} \mathrm{cm}}{4} \approx 1.75 \times 10^{-8} \mathrm{cm} \] Convert this value into nanometers, knowing that 1 cm = 10 nm:\[ r \approx 0.175 \mathrm{nm} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Face-Centered Cubic
A face-centered cubic (FCC) structure is a type of crystal lattice commonly found in metals. In this arrangement, atoms are positioned at each of the cube's corners and the centers of each face. This layout ensures a high packing density, with about 74% of the crystalline structure actually being occupied by the atoms. This tight packing is why metals with FCC structures, like lead, tend to be dense.
If you're visualizing this, imagine a cube. Each face of the cube has an atom in the center, and there's an atom at every corner. Because of symmetry and efficient packing, the FCC is often seen in ductile metals which can easily deform without breaking.
Importantly, the face-centered cubic unit cell contains four atoms, efficiently arranged to maximize density and minimize wasted space. This not only makes the structure stable but also plays a role in its physical properties such as melting point, ductility, and thermal expansion.
Atomic Radius
The atomic radius is an important concept when you're studying elements and their crystal structures. In simple terms, it describes the size of an atom. It’s usually measured from the nucleus to the outermost boundary of the surrounding cloud of electrons.
In the context of the face-centered cubic structure, calculating the atomic radius can be a bit tricky because the spacing between atoms isn’t straightforward. In a face-centered cubic system, the atomic radius influences how closely the atoms pack together. The diagonal across a face of the cube equates to four atomic radii, which helps us calculate the radius directly.
For lead, whose atoms are tightly packed in the FCC structure, you can use the diagonal of the cube face and the edge length to determine the atomic radius. By understanding the geometric relationship in FCC, you can efficiently calculate the atomic radius using the diagonal, which equals \( \sqrt{2} \times a \), where \( a \) is the edge length of the unit cell.
Unit Cell Volume
In crystallography, the unit cell volume is the measure of the cubic space that a crystal lattice occupies. It’s essentially the smallest repeating unit in the structure, providing a larger picture of the whole lattice.
For a face-centered cubic structure like that of lead, the unit cell’s volume can be calculated if you know the density of the material, the number of atoms per unit cell, and the element's molar mass. This can be done using the formula:
  • \( V_{cell} = \frac{Z \times M}{N_A \times \rho} \)
where \( Z \) is the number of atoms in the unit cell, \( M \) is the molar mass, \( N_A \) is Avogadro’s number, and \( \rho \) is the density.
This formula helps understand how tightly packed the atoms are within the unit cell. The unit cell’s edge length is later derived from this volume, allowing for further calculations such as the atomic radius.
Crystal Structure
Crystal structure refers to the ordered arrangement of atoms, ions, or molecules in a crystalline material. The face-centered cubic structure is one such arrangement that falls under cubic crystal systems, which are often seen in metallic elements.
Each element that crystallizes has a specific crystal structure that impacts its physical properties. For example, the FCC structure of metals like lead affects their density, mechanical strength, and how they conduct electricity.
The symmetry and repeating pattern within a crystal structure make it easier to predict the material’s properties. Because each crystalline structure like FCC is constant and repeats through the entire crystal, scientists can anticipate behaviors such as how the crystal will deform under stress or its thermal expansion characteristics.
Understanding crystal structures is crucial for material science, chemistry, and physics as it leads to applications in developing new materials and improving existing ones with tailored properties.

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