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: Use the density formula to find Z.
\(Z = \frac{\rho \times a^3 \times N_A}{M}\). With \(a = 5 \times 10^{-8}\) cm, \(a^3 = 1.25 \times 10^{-22}\) cm\(^3\).

Step 2: Calculate.
\(Z = \frac{0.613 \times 1.25 \times 10^{-22} \times 6.022 \times 10^{23}}{23} = \frac{46.2}{23} \approx 2\). Z = 2 corresponds to a BCC unit cell.
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