Let's solve the problem step by step: **Given:** - Body-Centered Cubic (BCC) structure - Atomic weight = 167.5 g/mol - Density (\(\rho\)) = 12.5 g/cm³ **Formula for density in BCC:** \[ \rho = \frac{n \cdot A}{N_A \cdot a^3} \] Where: - \(n\) = number of atoms per unit cell (BCC: \(n = 2\)) - \(A\) = atomic weight (g/mol) - \(N_A\) = Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) - \(a\) = lattice parameter (cm) **Step 1: Rearrange for \(a^3\):** \[ a^3 = \frac{n \cdot A}{N_A \cdot \rho} \] **Step 2: Substitute values:** \[ a^3 = \frac{2 \times 167.5}{6.022 \times 10^{23} \times 12.5} \] \[ a^3 = \frac{335}{7.5275 \times 10^{24}} \] \[ a^3 = 4.45 \times 10^{-23} \text{ cm}^3 \] \[ a = (4.45 \times 10^{-23})^{1/3} \] \[ a \approx 3.56 \times 10^{-8} \text{ cm} \] Convert to nm (\(1 \text{ cm} = 10^7 \text{ nm}\)): \[ a \approx 0.356 \text{ nm} \] **Step 3: Relation between \(a\) and atomic radius in BCC:** \[ a = \frac{4r}{\sqrt{3}} \implies r = \frac{\sqrt{3}}{4} a \] \[ r = \frac{1.732}{4} \times 0.356 \] \[ r \approx 0.154 \text{ nm} \] **Final Answer:** \[ \boxed{0.153 \text{ nm}} \] So, the correct answer is **0.153 nm**.