Q12 / 150 · MHT CET 2026 (13 April Shift 2) · Chemistry
A metal crystallizes to form simple cubic unit cell. Calculate the void volume of unit cell. [$a = 3.36 \times 10^{-8}$ cm]
- Correct answerA$1.807 \times 10^{-23}$ cm$^3$.
- B$1.214 \times 10^{-23}$ cm$^3$.
- C$3.662 \times 10^{-23}$ cm$^3$.
- D$1.986 \times 10^{-23}$ cm$^3$.
Solution
For a simple cubic unit cell, the relation between edge length $a$ and atomic radius $r$ is $a = 2r$.
The number of atoms per unit cell is $Z = 1$.
Volume of the unit cell, $V = a^3 = (3.36 \times 10^{-8})^3 = 37.933 \times 10^{-24} \text{ cm}^3 = 3.7933 \times 10^{-23} \text{ cm}^3$.
Volume occupied by the atom = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi \left(\dfrac{a}{2}\right)^3 = \dfrac{\pi}{6}a^3$.
Packing fraction = $\dfrac{\text{Volume occupied}}{\text{Total volume}} = \dfrac{\pi}{6} \approx 0.5236$.
Fraction of void volume = $1 - 0.5236 = 0.4764$.
Void volume = $0.4764 \times V = 0.4764 \times 3.7933 \times 10^{-23} \text{ cm}^3 \approx 1.807 \times 10^{-23} \text{ cm}^3$.
Answer: $1.807 \times 10^{-23}$ cm$^3$.
The number of atoms per unit cell is $Z = 1$.
Volume of the unit cell, $V = a^3 = (3.36 \times 10^{-8})^3 = 37.933 \times 10^{-24} \text{ cm}^3 = 3.7933 \times 10^{-23} \text{ cm}^3$.
Volume occupied by the atom = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi \left(\dfrac{a}{2}\right)^3 = \dfrac{\pi}{6}a^3$.
Packing fraction = $\dfrac{\text{Volume occupied}}{\text{Total volume}} = \dfrac{\pi}{6} \approx 0.5236$.
Fraction of void volume = $1 - 0.5236 = 0.4764$.
Void volume = $0.4764 \times V = 0.4764 \times 3.7933 \times 10^{-23} \text{ cm}^3 \approx 1.807 \times 10^{-23} \text{ cm}^3$.
Answer: $1.807 \times 10^{-23}$ cm$^3$.
Hence, the correct answer is option: 'A'.