Let's solve the problem step by step: **Given:** - BCC structure - Atomic weight = 167.5 g/mol - Density, ρ = 12.5 g/cm³ **To Find:** Radius of the atom (r) --- ### Step 1: Formula for Density For a cubic unit cell: \[ \rho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \] #### For BCC: - Number of atoms per unit cell = 2 --- ### Step 2: Mass of Atoms in Unit Cell \[ \text{Mass of 1 atom} = \frac{\text{Atomic weight}}{N_A} \] \[ \text{Mass in 1 unit cell} = 2 \times \frac{\text{Atomic weight}}{N_A} \] where \( N_A = 6.022 \times 10^{23} \) atoms/mol (Avogadro's number). --- ### Step 3: Volume of Unit Cell Let the edge length be \( a \). \[ \text{Volume of unit cell} = a^3 \] --- ### Step 4: Plug into Density Formula \[ \rho = \frac{2 \times \frac{167.5}{6.022 \times 10^{23}}}{a^3} \] But remember to convert \( a^3 \) to cm³ since density is in g/cm³: - If \( a \) is in cm, \( a^3 \) is in cm³. --- ### Step 5: Relationship Between a and r in BCC For BCC: \[ \sqrt{3}a = 4r \implies a = \frac{4r}{\sqrt{3}} \] --- ### Step 6: Substitute \( a \) in Density Formula \[ \rho = \frac{2 \times \frac{167.5}{6.022 \times 10^{23}}}{\left(\frac{4r}{\sqrt{3}}\right)^3} \] \[ \rho = \frac{2 \times \frac{167.5}{6.022 \times 10^{23}}}{\frac{64r^3}{3\sqrt{3}}} \] \[ \rho = \frac{2 \times \frac{167.5}{6.022 \times 10^{23}} \times \frac{3\sqrt{3}}{64}}{r^3} \] \[ \rho = \frac{2 \times 167.5 \times 3\sqrt{3}}{64 \times 6.022 \times 10^{23} \times r^3} \] --- ### Step 7: Solve for \( r \) \[ r^3 = \frac{2 \times 167.5 \times 3\sqrt{3}}{64 \times 6.022 \times 10^{23} \times \rho} \] Plug in values (\( \rho = 12.5 \) g/cm³): \[ r^3 = \frac{2 \times 167.5 \times 3 \times 1.732}{64 \times 6.022 \times 10^{23} \times 12.5} \] Calculate numerator: \[ 2 \times 167.5 = 335 \] \[ 335 \times 3 = 1005 \] \[ 1005 \times 1.732 \approx 1739.16 \] Calculate denominator: \[ 64 \times 6.022 = 385.408 \] \[ 385.408 \times 10^{23} = 3.85408 \times 10^{25} \] \[ 3.85408 \times 10^{25} \times 12.5 = 4.8186 \times 10^{26} \] \[ r^3 = \frac{1739.16}{4.8186 \times 10^{26}} \] \[ r^3 \approx 3.611 \times 10^{-24} \text{ cm}^3 \] \[ r = (3.611 \times 10^{-24})^{1/3} \] \[ r \approx 1.53 \times 10^{-8} \text{ cm} \] Convert cm to nm (\(1 \text{ nm} = 10^{-7} \text{ cm}\)): \[ r = 0.153 \text{ nm} \] --- ### **Final Answer:** \[ \boxed{0.153 \text{ nm}} \] So, the correct option is: - **0.153 nm**