Question

The lattice constant (in Å) of copper, which has FCC structure, is _________ (rounded off to two decimal places).
Given: density of copper is 8.91 g cm^-3 and its atomic mass is 63.55 g mol^-1 ; Avogadro’s number = 6.023 × 10²³ mol⁻¹

Answer 1

Correct Answer: 3.6 - 3.65

Given:
Density of Cu: \( \rho = 8.91\ \text{g/cm}^3 \) 
Atomic mass: \( M = 63.55\ \text{g/mol} \)
Avogadro number: \( N_A = 6.023 \times 10^{23}\ \text{mol}^{-1} \)
FCC structure → 4 atoms per unit cell

Step 1: Mass of one atom \[ m_{\text{atom}} = \frac{M}{N_A} = \frac{63.55}{6.023\times10^{23}} = 1.055\times10^{-22}\ \text{g} \] Step 2: Mass of atoms in one FCC unit cell \[ m_{\text{cell}} = 4 \times m_{\text{atom}} = 4.22\times10^{-22}\ \text{g} \] Step 3: Volume of one unit cell \[ V_{\text{cell}} = \frac{m_{\text{cell}}}{\rho} = \frac{4.22\times10^{-22}}{8.91} = 4.73\times10^{-23}\ \text{cm}^3 \] Step 4: Lattice constant \[ a^3 = V_{\text{cell}} \] \[ a = (4.73\times10^{-23})^{1/3} \approx 3.615\times10^{-8}\ \text{cm} \] Convert to Å:
\(1\ \text{cm} = 10^{8}\ \text{Å}\) \[ a = 3.615\ \text{Å} \] Final Answer:
\[ \boxed{3.61\ \text{Å}} \]