Q25 / 150 · MHT CET 2026 (17 April Shift 1) · Chemistry
Find the volume occupied by a particle in a simple cubic unit cell if the radius of a particle in it is 190 pm.
- A$6.410 \times 10^{-23}\ \text{cm}^3$
- B$3.625 \times 10^{-23}\ \text{cm}^3$
- C$5.715 \times 10^{-23}\ \text{cm}^3$
- Correct answerD$2.873 \times 10^{-23}\ \text{cm}^3$
Solution
The volume occupied by a particle is the volume of the spherical particle itself.
Given the radius of the particle, $r = 190 \text{ pm} = 190 \times 10^{-10} \text{ cm} = 1.9 \times 10^{-8} \text{ cm}$.
Volume of one particle is given by:
$V = \dfrac{4}{3}\pi r^3$
Substituting the value of $r$:
$V = \dfrac{4}{3} \times 3.1416 \times (1.9 \times 10^{-8} \text{ cm})^3$
$V = 4.1888 \times 6.859 \times 10^{-24} \text{ cm}^3$
$V = 28.73 \times 10^{-24} \text{ cm}^3 = 2.873 \times 10^{-23} \text{ cm}^3$
Answer: $2.873 \times 10^{-23}\ \text{cm}^3$
Given the radius of the particle, $r = 190 \text{ pm} = 190 \times 10^{-10} \text{ cm} = 1.9 \times 10^{-8} \text{ cm}$.
Volume of one particle is given by:
$V = \dfrac{4}{3}\pi r^3$
Substituting the value of $r$:
$V = \dfrac{4}{3} \times 3.1416 \times (1.9 \times 10^{-8} \text{ cm})^3$
$V = 4.1888 \times 6.859 \times 10^{-24} \text{ cm}^3$
$V = 28.73 \times 10^{-24} \text{ cm}^3 = 2.873 \times 10^{-23} \text{ cm}^3$
Answer: $2.873 \times 10^{-23}\ \text{cm}^3$
Hence, the correct answer is option: 'D'.