Let's break down the problem step by step: --- ## Given Data - **Crystal Structure:** HCP (Hexagonal Close-Packed) - **Density (\(\rho\)):** \(6.51 \, \text{g/cm}^3\) - **Atomic Weight (M):** \(91.22 \, \text{g/mol}\) - **c/a ratio:** \(1.593\) (used for part b) - **Avogadro's Number (\(N_A\))**: \(6.022 \times 10^{23} \, \text{atoms/mol}\) --- ## (a) What is the volume of the unit cell in cubic meters? ### **Step 1: Find the Number of Atoms per Unit Cell (n)** For HCP, the number of atoms per unit cell: \[ n = 6 \] --- ### **Step 2: Write the Density Equation for Unit Cell** \[ \rho = \frac{n \cdot M}{V_{\text{cell}} \cdot N_A} \] Where: - \(\rho\) = density (\(\text{g/cm}^3\)) - \(n\) = number of atoms per unit cell - \(M\) = atomic mass (\(\text{g/mol}\)) - \(V_{\text{cell}}\) = volume of unit cell (\(\text{cm}^3\)) - \(N_A\) = Avogadro's number (\(\text{atoms/mol}\)) --- ### **Step 3: Rearrange for Unit Cell Volume** \[ V_{\text{cell}} = \frac{n \cdot M}{\rho \cdot N_A} \] --- ### **Step 4: Plug in the Values** \[ V_{\text{cell}} = \frac{6 \times 91.22}{6.51 \times 6.022 \times 10^{23}} \] Calculate numerator and denominator: - Numerator: \(6 \times 91.22 = 547.32\) - Denominator: \(6.51 \times 6.022 \times 10^{23} \approx 39.20 \times 10^{23}\) \[ V_{\text{cell}} = \frac{547.32}{39.20 \times 10^{23}} \quad \text{cm}^3 \] \[ V_{\text{cell}} = 13.96 \times 10^{-23} \quad \text{cm}^3 \] --- ### **Step 5: Convert to Cubic Meters** \[ 1 \text{ cm}^3 = 1 \times 10^{-6} \text{ m}^3 \] \[ V_{\text{cell}} = 13.96 \times 10^{-23} \text{ cm}^3 \times 10^{-6} \frac{\text{m}^3}{\text{cm}^3} \] \[ V_{\text{cell}} = 1.396 \times 10^{-22} \text{ cm}^3 = 1.396 \times 10^{-28} \text{ m}^3 \] **Final Answer (a):** \[ \boxed{1.40 \times 10^{-28} \ \text{m}^3} \] --- ## (b) If the c/a ratio is 1.593, compute the values of \(c\) and \(a\). ### **Step 1: Volume Formula for HCP Unit Cell** For HCP: \[ V_{\text{cell}} = \frac{3\sqrt{3}}{2} a^2 c \] --- ### **Step 2: Express \(c\) in Terms of \(a\)** Given \(c/a = 1.593\), so \(c = 1.593 a\). \[ V_{\text{cell}} = \frac{3\sqrt{3}}{2} a^2 (1.593 a) = \frac{3\sqrt{3}}{2} \times 1.593 \ a^3 \] \[ V_{\text{cell}} = K a^3 \] Where \(K = \frac{3\sqrt{3}}{2} \times 1.593\) Calculate \(K\): \[ \frac{3\sqrt{3}}{2} = \frac{3 \times 1.732}{2} = \frac{5.196}{2} = 2.598 \] \[ K = 2.598 \times 1.593 = 4.137 \] So, \[ V_{\text{cell}} = 4.137 a^3 \] --- ### **Step 3: Solve for \(a\) Using \(V_{\text{cell}}\) from Above** \[ a^3 = \frac{V_{\text{cell}}}{4.137} \] Recall: \(V_{\text{cell}} = 1.396 \times 10^{-23} \ \text{cm}^3\) \[ a^3 = \frac{1.396 \times 10^{-23}}{4.137} = 3.375 \times 10^{-24} \ \text{cm}^3 \] \[ a = (3.375 \times 10^{-24})^{1/3} \] Calculate cube root: - Cube root of 3.375 ≈ 1.50 - Cube root of \(10^{-24}\) = \(10^{-8}\) \[ a = 1.50 \times 10^{-8} \ \text{cm} \] Convert to nm (\(1 \text{ nm} = 10^{-7} \text{ cm}\)): \[ a = 0.150 \ \text{nm} \] --- ### **Step 4: Find \(c\)** \[ c = 1.593 \times a = 1.593 \times 1.50 \times 10^{-8} \ \text{cm} \] \[ c = 2.39 \times 10^{-8} \ \text{cm} = 0.239 \ \text{nm} \] --- ## **Final Answers** ### (a) **Volume of the unit cell:** \[ \boxed{1.40 \times 10^{-28} \ \text{m}^3} \] ### (b) **Values of a and c:** \[ \boxed{a = 1.50 \times 10^{-8} \ \text{cm} = 0.150 \ \text{nm}} \] \[ \boxed{c = 2.39 \times 10^{-8} \ \text{cm} = 0.239 \ \text{nm}} \] --- **Let me know if you need further clarification or a breakdown of any step!**