Question

The density of Na is 0.613 g cm$^{-3}$. If the edge length of unit cell of Na is 5\AA, 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:

Density formula for crystals: $\rho = \frac{Z \times M}{N_A \times a^3}$.
Ensure all units are consistent. If density is in g/cm$^3$, then $a$ should be in cm, and $M$ in g/mol.
$1 \text{ \AA} = 10^{-10} \text{ m} = 10^{-8} \text{ cm}$.
$Z$ (effective number of atoms per unit cell) is typically:
1 for simple cubic (SC)
2 for body-centered cubic (BCC)
4 for face-centered cubic (FCC) or cubic close-packed (CCP)
The calculated value of $Z$ should be close to an integer.

Answer 1

The Correct Option is C

The density ($\rho$) of a crystal is related to its unit cell parameters by the formula: $\rho = \frac{Z \times M}{N_A \times a^3}$ where: $Z$ = effective number of atoms per unit cell $M$ = atomic weight (or molar mass in g/mol) $N_A$ = Avogadro's number ($6.022 \times 10^{23}$ mol$^{-1}$) $a$ = edge length of the unit cell We need to find $Z$. Rearranging the formula: $Z = \frac{\rho \times N_A \times a^3}{M}$. Given values: Density $\rho = 0.613 \text{ g cm}^{-3}$. Edge length $a = 5 \text{ \AA} = 5 \times 10^{-8} \text{ cm}$ (since $1 \text{ \AA} = 10^{-8} \text{ cm}$). Atomic weight of Na, $M = 23 \text{ u}$, which means molar mass is $23 \text{ g/mol}$. Avogadro's number $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$. First, calculate $a^3$: $a^3 = (5 \times 10^{-8} \text{ cm})^3 = 5^3 \times (10^{-8})^3 \text{ cm}^3 = 125 \times 10^{-24} \text{ cm}^3$. Now, substitute all values into the formula for $Z$: $Z = \frac{(0.613 \text{ g cm}^{-3}) \times (6.022 \times 10^{23} \text{ mol}^{-1}) \times (125 \times 10^{-24} \text{ cm}^3)}{23 \text{ g/mol}}$. The units g, cm$^{-3}$, mol$^{-1}$, cm$^3$, g/mol will cancel out, leaving $Z$ dimensionless, as expected. $Z = \frac{0.613 \times 6.022 \times 125 \times 10^{23} \times 10^{-24}}{23}$. $Z = \frac{0.613 \times 6.022 \times 125 \times 10^{-1}}{23}$. $Z = \frac{0.613 \times 6.022 \times 12.5}{23}$. Calculate the numerator: $0.613 \times 6.022 \approx 3.691$ $3.691 \times 12.5 \approx 46.1375$. So, $Z \approx \frac{46.1375}{23}$. $Z \approx 2.0059...$ Since $Z$ must be an integer (representing the number of atoms per unit cell for common crystal structures like BCC, FCC, simple cubic), the value $Z \approx 2.0059$ rounds to $Z=2$. A value of $Z=2$ corresponds to a body-centered cubic (BCC) structure for an element. Sodium (Na) crystallizes in a BCC structure at room temperature. This matches option (c). \[ \boxed{2} \]