Let's break down the problem: **Given:** - BCC structure (Body-Centered Cubic) - Atomic weight = **167.5 g/mol** - Density = **12.5 g/cm³** **Asked:** - Calculate the **radius** of the atom. --- ### **Step 1: Formula for Density** For a BCC structure: - Number of atoms per unit cell, \( n = 2 \) - Molar mass, \( M = 167.5 \) g/mol - Density, \( \rho = 12.5 \) g/cm³ - Avogadro's number, \( N_A = 6.022 \times 10^{23} \) atoms/mol - Let \( a \) be the unit cell edge length (in cm) \[ \rho = \frac{n \times M}{a^3 \times N_A} \] --- ### **Step 2: Solve for \( a \)** \[ a^3 = \frac{n \times M}{\rho \times N_A} \] Substitute the values: \[ a^3 = \frac{2 \times 167.5}{12.5 \times 6.022 \times 10^{23}} \] \[ a^3 = \frac{335}{75.275 \times 10^{23}} \] \[ a^3 = \frac{335}{7.5275 \times 10^{24}} \] \[ a^3 = 4.45 \times 10^{-23} \text{ cm}^3 \] \[ a = (4.45 \times 10^{-23})^{1/3} \] \[ a \approx 3.56 \times 10^{-8} \text{ cm} \] Convert to nm (1 nm = \(10^{-7}\) cm): \[ a \approx 0.356 \text{ nm} \] --- ### **Step 3: Relation Between Radius and Edge Length (BCC)** For BCC: \[ \sqrt{3}a = 4r \] \[ r = \frac{\sqrt{3}a}{4} \] Substitute \( a = 0.356 \) nm: \[ r = \frac{1.732 \times 0.356}{4} \] \[ r = \frac{0.616}{4} \] \[ r \approx 0.154 \text{ nm} \] --- ### **Step 4: Compare with Options** Closest option: **0.153 nm** --- **Final Answer:** \[ \boxed{0.153 \text{ nm}} \]