Let's solve the problem step by step: **Question:** The volume of the BCC unit cell when the atomic radius is 0.662 nm is: **Step 1: Relationship in BCC** - For a Body-Centered Cubic (BCC) unit cell, the relation between the atomic radius \( r \) and the edge length \( a \) is: \[ a = \frac{4r}{\sqrt{3}} \] Given: \( r = 0.662\, \text{nm} \) **Step 2: Calculate Edge Length \( a \)** \[ a = \frac{4 \times 0.662}{\sqrt{3}} \] \[ a = \frac{2.648}{1.732} \] \[ a \approx 1.53\, \text{nm} \] **Step 3: Find Volume** \[ \text{Volume} = a^3 = (1.53\, \text{nm})^3 \] \[ \text{Volume} = 1.53 \times 1.53 \times 1.53 \] \[ \text{Volume} \approx 3.58\, \text{nm}^3 \] **Step 4: Match with Options** The closest answer to this calculation is **3.57 nm³**. **Final Answer:** > **3.57 nm³** Let me know if you need the explanation for any other concept!