Question

Calcium crystallizes with a body-centered cubic structure. (a) How many Ca atoms are contained in each unit cell? (b) How many nearest neighbors does each Ca atom possess? (c) Estimate the length of the unit cell edge, \(a\), from the atomic radius of calcium \((1.97 \AA) .\) (d) Estimate the density of Ca metal.

Short answer

(a) There are 2 Ca atoms in each unit cell in a body-centered cubic structure. (b) Each Ca atom in the BCC structure has 8 nearest neighbors. (c) The estimated length of the unit cell edge for Ca is approximately 5.58 Angstroms. (d) The estimated density of Ca metal is approximately 7.63 g/cm³.

Step-by-step solution

(a) Number of Ca atoms in a BCC unit cell

In a body-centered cubic (BCC) structure, there is one atom at each of the eight corners of the unit cell and one atom in the center. Since each corner atom is shared by 8 unit cells, only \(\frac{1}{8}\) of each corner atom belongs to a particular unit cell. So, the total number of Ca atoms in each unit cell is: \(1 \times (\frac{1}{8} \times 8) + 1 = 1 + 1 = 2\) Hence, there are 2 Ca atoms in each unit cell.

(b) Number of nearest neighbors for each Ca atom

In a BCC structure, each corner atom has 4 nearest neighbors in the same plane (one at each face center of the cube) and one in the center, for a total of 5 nearest neighbors. The center atom is a nearest neighbor to all 8 corner atoms. Since there is only one center atom for each unit cell, each Ca atom has 8 nearest neighbors.

(c) Estimate the length of the unit cell edge

We are given the atomic radius of Calcium, \(r = 1.97 \AA\). To find the length of the unit cell edge, \(a\), we can use the relation between the atomic radius and the edge length in a BCC structure: \[\frac{a}{2} = \sqrt{r^{2} + r^{2}}\] From this, we can find the value of \(a\): \[a = 2\sqrt{2r^{2}} = 2\sqrt{2(1.97)^{2}} \approx 5.58 \AA\] So, the estimated length of the unit cell edge for Ca is approximately 5.58 Angstroms.

(d) Estimate the density of Ca metal

To estimate the density of Ca metal, we use the expression for the density of a crystal lattice: \[\rho = \frac{n \times m}{V}\] where \(n\) is the number of atoms per unit cell, \(m\) is the molar mass of Ca divided by Avogadro's number \(N_A\), and \(V\) is the volume of the unit cell. We know that in Calcium BCC structure, \(n=2\). The molar mass of Ca is 40.08 g/mol. Using Avogadro's number (\(N_A \approx 6.022\times10^{23}\, \text{mol}^{-1}\)), we can determine the mass of a Ca atom: \[m = \frac{40.08 \,\text{g/mol}}{6.022 \times 10^{23}\, \text{atoms/mol}} \approx 6.65 \times 10^{-23} \, \text{g/atom}\] Now let's calculate the volume of the unit cell (\(V = a^{3}\)): \[V = (5.58\, \AA)^3 \times \left(\frac{10^{-8}\,\text{cm}}{\AA}\right)^3 \approx 1.74 \times 10^{-22}\,\text{cm}^{3}\] Finally, we can estimate the density of Ca metal: \[\rho = \frac{2 \times (6.65 \times 10^{-23}\, \text{g/atom})}{1.74 \times 10^{-22}\,\text{cm}^3} \approx 7.63\, \text{g/cm}^3\] So, the estimated density of Ca metal is approximately 7.63 g/cm³.

Key concepts

Understanding the Body-Centered Cubic (BCC) Structure

Metals like calcium can adopt various crystalline structures, one of which is the body-centered cubic (BCC) structure. In the BCC structure, each cubic unit cell has one atom at each of its eight corners and one atom at the center of the cube. However, the corner atoms are shared among eight adjoining unit cells, significantly impacting both the structure and properties of the metal.

The BCC structure is particularly important because it helps determine numerous physical properties, such as density and atomic packing. This is why understanding BCC is crucial for materials science and engineering. For instance, when considering the atomic arrangement in calcium's BCC, we see that the atoms are not tightly packed—compared to other structures like face-centered cubic (FCC)—resulting in different material characteristics such as lower density and greater ductility.

Each atom in a BCC structure effectively has 8 nearest neighbors. The geometry of the BCC arrangements allows each corner atom to be immediately adjacent to four atoms in the same layer and one in the center, while the one in the center touches all eight corner atoms. This pattern influences the strength and malleability of metals with BCC structures.

Calculating Atomic Packing in Unit Cells

The concept of atomic packing is key to understanding the density and stability of a crystalline structure. Atomic packing factor (APF) is the fraction of volume in a crystal structure that is occupied by atoms. To calculate the APF for the BCC structure, we must consider the volume occupied by the atoms versus the total volume of the unit cell.

For a BCC unit cell, where the atomic radius, r, is known, the volume of an atom can be approximated as that of a sphere, \(\frac{4}{3}\pi r^3\). Since there are two atoms per BCC unit cell, we double the volume of one atom and divide by the volume of the unit cell (\(a^3\), where \(a\) is the unit cell edge length) to get the APF.

By understanding the APF, students can better visualize the efficiency of space usage in a crystal and how it affects the material's properties. For instance, a higher APF generally implies a metal is more dense and has less space for dislocations to move, making it harder and less ductile.

Estimating Crystal Lattice Density

The density of a crystal lattice, such as for calcium with a BCC structure, is a critical property that reflects the mass of atoms within a given volume. Density can be estimated using an elegant formula that takes into account the number of atoms per unit cell, the mass of each atom, and the cube's volume. In mathematical terms, the density (\(\rho\)) is expressed by \(\rho = \frac{n \times m}{V}\), where \(n\) is the number of atoms per unit cell, \(m\) is the mass of an individual atom, and \(V\) is the unit cell's volume.

Through this calculation, one can understand how the atomic mass and the arrangement of atoms in space contribute to the overall mass of the crystal structure. This information is particularly useful for applications that depend on the material's strength, durability, and overall mass—for example, in aerospace, automotive, and construction materials. The ability to estimate the density of a crystal lattice based on its structure is, therefore, a valuable skill for professionals in various scientific and engineering fields.