Let's solve the question step by step: ### Given Data: - **c/a ratio** = 1.5 - **Density (ρ)** = 1.25 g/cm³ - **Atomic weight (M)** = 24.31 g/mol - **Structure:** HCP ### HCP Structure Information: - **Number of atoms per unit cell (n):** 6 - **Volume of HCP unit cell:** \( V = \frac{3\sqrt{3}}{2} a^2 c \) - **c = 1.5a** (from c/a ratio) - **Avogadro's number (\(N_A\))** = \(6.022 \times 10^{23}\) atoms/mol ### Step 1: Write the density formula \[ \rho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \] \[ \rho = \frac{n \cdot M}{N_A \cdot V} \] ### Step 2: Rearranging for 'a' \[ V = \frac{3\sqrt{3}}{2} a^2 c = \frac{3\sqrt{3}}{2} a^2 (1.5a) = \frac{9\sqrt{3}}{4} a^3 \] \[ \rho = \frac{6 \cdot M}{N_A \cdot \frac{9\sqrt{3}}{4} a^3} \] \[ \rho = \frac{24 M}{9\sqrt{3} N_A a^3} \] \[ a^3 = \frac{24 M}{9\sqrt{3} N_A \rho} \] \[ a = \left(\frac{24 M}{9\sqrt{3} N_A \rho}\right)^{1/3} \] ### Step 3: Plug in the values - \( M = 24.31 \) g/mol - \( N_A = 6.022 \times 10^{23} \) atoms/mol - \( \rho = 1.25 \) g/cm³ \[ a^3 = \frac{24 \times 24.31}{9 \times 1.732 \times 6.022 \times 10^{23} \times 1.25} \] Calculate denominator: - \( 9 \times 1.732 = 15.588 \) - \( 15.588 \times 6.022 \times 10^{23} = 93.88 \times 10^{23} \) - \( 93.88 \times 1.25 = 117.35 \) - \( 117.35 \times 10^{23} = 1.1735 \times 10^{25} \) Calculate numerator: - \( 24 \times 24.31 = 583.44 \) So, \[ a^3 = \frac{583.44}{1.1735 \times 10^{25}} \] \[ a^3 = 4.973 \times 10^{-23} \text{ cm}^3 \] \[ a = (4.973 \times 10^{-23})^{1/3} \] \[ a \approx 3.68 \times 10^{-8} \text{ cm} \] \[ a \approx 0.368 \text{ nm} \] ### Step 4: Calculate Atomic Radius for HCP For HCP: \[ a = 2r \implies r = \frac{a}{2} \] \[ r = \frac{0.368}{2} = 0.184 \text{ nm} \] ### Closest Option: - **0.183 nm** **Final Answer:** \[ \boxed{0.183 \text{ nm}} \]