An element (atomic mass = ) having a face-centred cubic (FCC) crystal structure has a density of . W — Solid State Chemistry Question

💡 Solution & Explanation

Density $d = \frac{Z \times M}{N_A \times a^3}$. For an FCC unit cell, $Z = 4$. Substituting the given values: $6.23 = \frac{4 \times 60}{6.02 \times 10^{23} \times a^3}$. Solving for $a^3$, we get $a^3 \approx 64 \times 10^{-24}\text{ cm}^3$. Therefore, $a = 4.0 \times 10^{-8}\text{ cm} = 400\text{ pm}$.