Let's solve the problem step by step. ### Given: - **BCC structure** - **Atomic weight (M):** 167.5 g/mol - **Density (ρ):** 12.5 g/cm³ ### To Find: - **Radius (r) of the atom** --- ### Step 1: Formula for Density in a BCC Structure \[ \rho = \frac{n \cdot M}{N_A \cdot a^3} \] Where: - \( n \) = Number of atoms per unit cell (BCC: \( n = 2 \)) - \( M \) = Molar mass (g/mol) - \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mol) - \( a \) = Lattice parameter (cm) --- ### Step 2: Rearranging for \( a \) \[ a^3 = \frac{n \cdot M}{N_A \cdot \rho} \] --- ### Step 3: Substitute the values \[ a^3 = \frac{2 \times 167.5}{6.022 \times 10^{23} \times 12.5} \] \[ a^3 = \frac{335}{7.5275 \times 10^{24}} \] \[ a^3 = 4.45 \times 10^{-23} \text{ cm}^3 \] \[ a = (4.45 \times 10^{-23})^{1/3} \] \[ a \approx 3.56 \times 10^{-8} \text{ cm} = 0.356 \text{ nm} \] --- ### Step 4: Relation between **a** and **r** in BCC For BCC, \[ \sqrt{3}a = 4r \] \[ r = \frac{\sqrt{3}}{4}a \] --- ### Step 5: Calculate the radius \[ r = \frac{1.732}{4} \times 0.356 \text{ nm} \] \[ r = 0.433 \times 0.356 \text{ nm} \] \[ r \approx 0.154 \text{ nm} \] --- ### Step 6: Match with the options: - **0.183 nm** - **0.123 nm** - **0.153 nm** - **0.143 nm** The closest answer is **0.153 nm**. --- ## **Final Answer** **The radius of the atom is:** \[ \boxed{0.153 \text{ nm}} \]