Question

The density of Na is 0.613 g cm\(^{-3}\). If the edge length of the unit cell of Na is 5 Å, the effective number of atoms of Na per unit cell is (Atomic weight of Na = 23 u):

• A. 8
• B. 1
• C. 2
• D. 4

Hint:

Use the formula \( n = \frac{\rho \, a^3 N_A}{M} \) to calculate the effective number of atoms in a unit cell.

Answer 1

The Correct Option is C

Step 1: Recall formula for number of atoms per unit cell.
\[ n = \frac{\text{Density} \times V_\text{cell} \times N_A}{\text{Atomic mass}} \]

Step 2: Convert edge length to cm.
\[ a = 5 \times 10^{-8} \text{ cm} \implies V_\text{cell} = a^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

Step 3: Calculate total mass in unit cell.
\[ m_\text{cell} = \rho \times V_\text{cell} = 0.613 \times 1.25 \times 10^{-22} \approx 7.66 \times 10^{-23} \text{ g} \]

Step 4: Convert to number of atoms.
\[ n = \frac{7.66 \times 10^{-23} \times 6.022 \times 10^{23}}{23} \approx 2 \text{ atoms} \]

Step 5: Conclusion.
\[ \boxed{2 \text{ atoms per unit cell}} \]