Let's solve the problem step-by-step: Given: - Atomic radius (\( r \)) = **0.2 nm** = \( 2.0 \times 10^{-8} \) cm - Density (\( \rho \)) = **1.93 g/cm³** - Atomic weight (\( M \)) = **74.63 g/mol** We are to determine the **crystal structure** of the metal. The possible options are: - (a) BCC (Body Centered Cubic) - (b) SC (Simple Cubic) - (c) FCC (Face Centered Cubic) - (d) HCP (Hexagonal Close Packed) Let's proceed step by step: --- ### **Step 1: Relate Atomic Radius to Unit Cell Edge (a) for Different Structures** - **Simple Cubic (SC):** \( a = 2r \) - **Body Centered Cubic (BCC):** \( a = \frac{4r}{\sqrt{3}} \) - **Face Centered Cubic (FCC):** \( a = \frac{4r}{\sqrt{2}} = 2\sqrt{2} r \) --- ### **Step 2: Formula Relating Density, Atomic Weight, and Structure** \[ \rho = \frac{n \cdot M}{N_A \cdot a^3} \] Where: - \( n \) = number of atoms per unit cell - SC: \( n = 1 \) - BCC: \( n = 2 \) - FCC: \( n = 4 \) - \( M \) = atomic weight (g/mol) - \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \)) - \( a \) = edge length of unit cell (cm) - \( \rho \) = density (g/cm³) --- ### **Step 3: Calculate Edge Length (a) for Each Structure** \[ r = 2.0 \times 10^{-8} \text{ cm} \] #### **a) SC: \( a = 2r = 4.0 \times 10^{-8} \) cm** #### **b) BCC: \( a = \frac{4r}{\sqrt{3}} = \frac{4 \times 2.0 \times 10^{-8}}{1.732} \approx 4.62 \times 10^{-8} \) cm** #### **c) FCC: \( a = 2\sqrt{2} r = 2 \times 1.414 \times 2.0 \times 10^{-8} \approx 5.66 \times 10^{-8} \) cm** --- ### **Step 4: Calculate Density for Each Structure** Let's calculate for each: #### **a) SC (\( n = 1 \), \( a = 4.0 \times 10^{-8} \) cm):** \[ \rho = \frac{1 \cdot 74.63}{6.022 \times 10^{23} \times (4.0 \times 10^{-8})^3} \] \[ a^3 = (4.0 \times 10^{-8})^3 = 64 \times 10^{-24} = 6.4 \times 10^{-23} \text{ cm}^3 \] \[ \rho = \frac{74.63}{6.022 \times 10^{23} \times 6.4 \times 10^{-23}} \] \[ 6.022 \times 10^{23} \times 6.4 \times 10^{-23} = 6.022 \times 6.4 \times 10^{0} = 38.54 \] \[ \rho = \frac{74.63}{38.54} \approx 1.94 \text{ g/cm}^3 \] #### **b) BCC (\( n = 2 \), \( a = 4.62 \times 10^{-8} \) cm):** \[ a^3 = (4.62 \times 10^{-8})^3 \approx 98.7 \times 10^{-24} = 9.87 \times 10^{-23} \text{ cm}^3 \] \[ \rho = \frac{2 \times 74.63}{6.022 \times 10^{23} \times 9.87 \times 10^{-23}} \] \[ 6.022 \times 10^{23} \times 9.87 \times 10^{-23} = 6.022 \times 9.87 = 59.48 \] \[ \rho = \frac{149.26}{59.48} \approx 2.51 \text{ g/cm}^3 \] #### **c) FCC (\( n = 4 \), \( a = 5.66 \times 10^{-8} \) cm):** \[ a^3 = (5.66 \times 10^{-8})^3 \approx 181.3 \times 10^{-24} = 1.813 \times 10^{-22} \text{ cm}^3 \] \[ \rho = \frac{4 \times 74.63}{6.022 \times 10^{23} \times 1.813 \times 10^{-22}} \] \[ 6.022 \times 10^{23} \times 1.813 \times 10^{-22} = 6.022 \times 1.813 \times 10^{1} = 10.92 \times 10^{1} = 109.2 \] \[ \rho = \frac{298.52}{109.2} \approx 2.73 \text{ g/cm}^3 \] --- ### **Step 5: Match with Given Density** The calculated density for SC is **1.94 g/cm³**, which matches closely with the given value **1.93 g/cm³**. - SC: \( \approx 1.94 \) (**matches**) - BCC: \( \approx 2.51 \) - FCC: \( \approx 2.73 \) --- ### **Step 6: Final Answer** **The crystal structure is:** \[ \boxed{\text{Simple Cubic (SC)}} \] **Final Answer:** **(b) SC**

Let's analyze why the other options are incorrect based on our calculations: --- ### **1. Simple Cubic (SC): Correct Choice** - Our calculation for SC yields a density (~1.94 g/cm³) that closely matches the given density (1.93 g/cm³). - The atomic radius and unit cell dimension are consistent with the simple cubic structure. --- ### **2. Why BCC is incorrect** - Calculated density for BCC was approximately **2.51 g/cm³**, which is significantly higher than the given 1.93 g/cm³. - This discrepancy indicates that the metal's actual structure is unlikely BCC. ### **3. Why FCC is incorrect** - Calculated density for FCC was about **2.73 g/cm³**, even higher than BCC. - Again, this does not match the given density, ruling out FCC. ### **4. Why HCP is incorrect** - The problem provides no direct calculation for HCP because the atomic radius alone doesn't determine HCP, and the data does not match the calculated densities for other structures. - Typically, HCP structures have similar packing densities to FCC (~74%), but the data aligns best with the simple cubic arrangement. --- ### **Summary:** - The calculated density for the **Simple Cubic (SC)** structure matches the provided density **(1.93 g/cm³)**. - The other structures (BCC and FCC) give higher densities, inconsistent with the given data. - Therefore, **the metal has a Simple Cubic (SC) crystal structure**. --- **Final conclusion:** **Correct answer: (b) SC**