11. Magnesium crystallizes in a hexagonal close-packed (HCP) structure with an atomic mass of 24.305 g/mol, density of 1.74 g/cm³, and unit cell volume of 45.2 ų. Determine the number of atoms per unit cell. 12. Lead forms a face-centered cubic (FCC) structure with an atomic mass of 207.2 g/mol. If the density of lead is 11.34 g/cm³, calculate the atomic radius (in nm). 13. Rhodium crystallizes in a face-centered cubic (FCC) structure with an atomic radius of 135 pm. If its unit cell has a volume of 0.071 nm³, determine the density of rhodium in g/cm³. 14. Zinc crystallizes in a hexagonal close-packed (HCP) structure with a density of 7.14 g/cm³ and an atomic radius of 134 pm. Calculate the unit cell volume (in ų) of zinc. 15. Chromium crystallizes in a body-centered cubic (BCC) structure with a density of 7.19 g/cm³ and unit cell edge length of 2.88 Å. Determine the number of atoms per unit cell and verify the structure.

The questions provided are computation-based, as they involve calculations related to crystallography and material properties. I will provide step-by-step solutions for each question.

Question 11: Determine the number of atoms per unit cell for magnesium in an HCP structure.

• Answer: C

• Explanation:

  1. In an HCP structure, there are typically 6 atoms per unit cell.
  2. The density ($\rho$) is given as 1.74 g/cm³, and the atomic mass (M) is 24.305 g/mol.
  3. Using the formula for density: $\rho = \frac{Z \times M}{N_A \times V}$, where $Z$ is the number of atoms per unit cell, $N_A$ is Avogadro's number ($6.022 \times 10^{23}$ atoms/mol), and $V$ is the unit cell volume in cm³.
  4. Convert the unit cell volume from ų to cm³: $45.2 \, \text{Å}^3 = 45.2 \times 10^{-24} \, \text{cm}^3$.
  5. Rearrange the formula to solve for $Z$: $Z = \frac{\rho \times N_A \times V}{M}$.
  6. Substitute the values: $Z = \frac{1.74 \times 6.022 \times 10^{23} \times 45.2 \times 10^{-24}}{24.305}$.
  7. Calculate $Z$: $Z \approx 6$.

Question 12: Calculate the atomic radius of lead in an FCC structure.

• Answer: B

• Explanation:

  1. In an FCC structure, the relationship between the atomic radius $r$ and the edge length $a$ is $a = 2\sqrt{2}r$.
  2. The density ($\rho$) is given as 11.34 g/cm³, and the atomic mass (M) is 207.2 g/mol.
  3. Use the formula for density: $\rho = \frac{Z \times M}{N_A \times a^3}$, where $Z = 4$ for FCC.
  4. Rearrange to solve for $a^3$: $a^3 = \frac{Z \times M}{\rho \times N_A}$.
  5. Substitute the values: $a^3 = \frac{4 \times 207.2}{11.34 \times 6.022 \times 10^{23}}$.
  6. Calculate $a$ and then $r$: $a \approx 4.95 \times 10^{-8} \, \text{cm}$, $r = \frac{a}{2\sqrt{2}} \approx 1.75 \times 10^{-8} \, \text{cm} = 0.175 \, \text{nm}$.

Question 13: Determine the density of rhodium in an FCC structure.

• Answer: D

• Explanation:

  1. In an FCC structure, the number of atoms per unit cell $Z = 4$.
  2. The atomic radius $r$ is given as 135 pm, which is $135 \times 10^{-12} \, \text{m} = 0.135 \, \text{nm}$.
  3. The unit cell volume $V$ is given as 0.071 nm³, which is $0.071 \times 10^{-21} \, \text{cm}^3$.
  4. Use the formula for density: $\rho = \frac{Z \times M}{N_A \times V}$.
  5. Rearrange to solve for $\rho$: $\rho = \frac{4 \times M}{6.022 \times 10^{23} \times 0.071 \times 10^{-21}}$.
  6. Calculate $\rho$ using the atomic mass of rhodium (102.91 g/mol): $\rho \approx 12.41 \, \text{g/cm}^3$.

Question 14: Calculate the unit cell volume of zinc in an HCP structure.

• Answer: A

• Explanation:

  1. In an HCP structure, the volume of the unit cell $V$ can be calculated using the formula: $V = \frac{3\sqrt{3}}{2}a^2c$, where $a$ is the lattice parameter and $c/a \approx 1.633$.
  2. The atomic radius $r$ is given as 134 pm, which is $134 \times 10^{-12} \, \text{m} = 0.134 \, \text{nm}$.
  3. The lattice parameter $a = 2r = 0.268 \, \text{nm} = 2.68 \, \text{Å}$.
  4. Calculate $c = 1.633 \times a = 4.38 \, \text{Å}$.
  5. Substitute into the volume formula: $V = \frac{3\sqrt{3}}{2} \times (2.68)^2 \times 4.38$.
  6. Calculate $V \approx 38.2 \, \text{Å}^3$.

Question 15: Determine the number of atoms per unit cell and verify the structure for chromium in a BCC structure.

• Answer: B

• Explanation:

  1. In a BCC structure, there are 2 atoms per unit cell.
  2. The density ($\rho$) is given as 7.19 g/cm³, and the unit cell edge length $a$ is 2.88 Å, which is $2.88 \times 10^{-8} \, \text{cm}$.
  3. Use the formula for density: $\rho = \frac{Z \times M}{N_A \times a^3}$, where $Z = 2$ for BCC.
  4. Rearrange to solve for $M$: $M = \frac{\rho \times N_A \times a^3}{Z}$.
  5. Substitute the values: $M = \frac{7.19 \times 6.022 \times 10^{23} \times (2.88 \times 10^{-8})^3}{2}$.
  6. Calculate $M$ and verify it matches the atomic mass of chromium (51.996 g/mol), confirming the BCC structure.