Let's break down the problem: **Given:** - BCC structure (Body-Centered Cubic) - Atomic weight = 167.5 g/mol - Density \( \rho = 12.5 \) g/cm³ - Find the atomic radius (r). ### Step 1: Formula for Density \[ \rho = \frac{n \cdot M}{N_A \cdot a^3} \] Where: - \( n \) = number of atoms per unit cell (BCC has 2) - \( M \) = molar mass (167.5 g/mol) - \( N_A \) = Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) - \( a \) = edge length of unit cell (in cm) ### Step 2: Rearranging for \( a \) \[ a^3 = \frac{n \cdot M}{N_A \cdot \rho} \] \[ a = \sqrt[3]{\frac{n \cdot M}{N_A \cdot \rho}} \] ### Step 3: Plug in Values - \( n = 2 \) - \( M = 167.5 \) g/mol - \( N_A = 6.022 \times 10^{23} \) - \( \rho = 12.5 \) g/cm³ \[ a^3 = \frac{2 \times 167.5}{6.022 \times 10^{23} \times 12.5} \] \[ a^3 = \frac{335}{7.5275 \times 10^{24}} \] \[ a^3 = 4.45 \times 10^{-23} \text{ cm}^3 \] \[ a = (4.45 \times 10^{-23})^{1/3} \] Calculate the cube root: \[ a \approx 3.56 \times 10^{-8} \text{ cm} \] \[ a \approx 0.356 \text{ nm} \] ### Step 4: Relationship between \( a \) and \( r \) in BCC \[ a = \frac{4r}{\sqrt{3}} \] \[ r = \frac{a \sqrt{3}}{4} \] Plug in \( a = 0.356 \) nm: \[ r = \frac{0.356 \times 1.732}{4} \] \[ r = \frac{0.616}{4} \] \[ r \approx 0.154 \text{ nm} \] ### Step 5: Match with Options The closest answer is **0.153 nm**. --- ### **Final Answer:** **0.153 nm**