Question

Calculate the number of atoms present in 1.58 g metal if it forms bcc structure.
\([\rho \times \text{a}^3 = 1.58 \times 10^{-22}\text{ g}]\)

• A. \(1.0 \times 10^{22}\)
• B. \(2.0 \times 10^{22}\)
• C. \(3.0 \times 10^{22}\)
• D. \(4.0 \times 10^{22}\)

Hint:

BCC unit cell contains 2 atoms.

Answer 1

The Correct Option is B

Concept:
BCC → 2 atoms per unit cell. Step 1: Mass of one unit cell. \[ = 1.58 \times 10^{-22} \text{ g} \]
Step 2: Number of unit cells. \[ \frac{1.58}{1.58 \times 10^{-22}} = 10^{22} \]
Step 3: Total atoms. \[ = 2 \times 10^{22} \]
Step 4: Conclusion. Atoms = \(2.0 \times 10^{22}\)