Q7 / 150 · MHT CET 2026 (17 April Shift 1) · Chemistry
Calculate the number of unit cells in $1\text{cm}^3$ of metal if it forms simple cubic structure with unit cell edge length 500 pm.
- A$4.0 \times 10^{21}$
- B$2.0 \times 10^{21}$
- Correct answerC$8.0 \times 10^{21}$
- D$3.0 \times 10^{21}$
Solution
Edge length of the unit cell, $a = 500 \text{ pm} = 500 \times 10^{-12} \text{ m} = 5 \times 10^{-8} \text{ cm}$
Volume of one unit cell, $V = a^3 = (5 \times 10^{-8})^3 \text{ cm}^3 = 125 \times 10^{-24} \text{ cm}^3$
Number of unit cells in $1 \text{ cm}^3 = \dfrac{\text{Total volume}}{\text{Volume of one unit cell}}$
Number of unit cells $= \dfrac{1}{125 \times 10^{-24}} = \dfrac{1000 \times 10^{21}}{125} = 8.0 \times 10^{21}$
Answer: $8.0 \times 10^{21}$
Volume of one unit cell, $V = a^3 = (5 \times 10^{-8})^3 \text{ cm}^3 = 125 \times 10^{-24} \text{ cm}^3$
Number of unit cells in $1 \text{ cm}^3 = \dfrac{\text{Total volume}}{\text{Volume of one unit cell}}$
Number of unit cells $= \dfrac{1}{125 \times 10^{-24}} = \dfrac{1000 \times 10^{21}}{125} = 8.0 \times 10^{21}$
Answer: $8.0 \times 10^{21}$
Hence, the correct answer is option: 'C'.