Q113 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Chemistry
An element has bcc structure with atomic radius 200 pm. Find the volume of unit cell.
- A$5.625 \times 10^{-23}\ \text{cm}^3$
- B$7.513 \times 10^{-23}\ \text{cm}^3$
- C$8.41 \times 10^{-23}\ \text{cm}^3$
- Correct answerD$9.848 \times 10^{-23}\ \text{cm}^3$
Solution
For a bcc unit cell, the relation between edge length $a$ and atomic radius $r$ is given by:
$\sqrt{3}a = 4r$
$a = \dfrac{4r}{\sqrt{3}}$
Substituting $r = 200 \text{ pm} = 2 \times 10^{-8} \text{ cm}$:
$a = \dfrac{4 \times 2 \times 10^{-8}}{1.732} \approx 4.618 \times 10^{-8} \text{ cm}$
Volume of the unit cell, $V = a^3$
$V = (4.618 \times 10^{-8})^3 \approx 9.848 \times 10^{-23} \text{ cm}^3$
Answer: $9.848 \times 10^{-23}\ \text{cm}^3$
$\sqrt{3}a = 4r$
$a = \dfrac{4r}{\sqrt{3}}$
Substituting $r = 200 \text{ pm} = 2 \times 10^{-8} \text{ cm}$:
$a = \dfrac{4 \times 2 \times 10^{-8}}{1.732} \approx 4.618 \times 10^{-8} \text{ cm}$
Volume of the unit cell, $V = a^3$
$V = (4.618 \times 10^{-8})^3 \approx 9.848 \times 10^{-23} \text{ cm}^3$
Answer: $9.848 \times 10^{-23}\ \text{cm}^3$
Hence, the correct answer is option: 'D'.