Question

Calculate the number of atoms present in 1.625 g metal if it forms bcc unit cell structure.
\([ρ\times a^3 = 3.25\times 10^{-22} \text{g}]\)

• A. \(2.0\times 10^{22}\)
• B. \(1.5\times 10^{22}\)
• C. \(2.5\times 10^{22}\)
• D. \(1.0\times 10^{22}\)

Hint:

Mass of one unit cell = rho x a^3, and a bcc cell holds 2 atoms.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Density is mass per volume, so for one unit cell, mass = \(\rho \times a^3\). A bcc unit cell contains 2 atoms (8 corners at 1/8 each plus 1 body centre).

Step 2: Detailed Explanation
Mass of one unit cell = \(3.25 \times 10^{-22}\) g.
Number of unit cells in 1.625 g:
\[ \frac{1.625}{3.25 \times 10^{-22}} = 0.5 \times 10^{22} = 5 \times 10^{21} \]
Atoms = cells x 2:
\[ 5 \times 10^{21} \times 2 = 1.0 \times 10^{22} \]
Option (C), \(2.5 \times 10^{22}\), comes from assuming 5 atoms per cell, and (A) from 4 atoms (fcc) and a slip.

Final Answer:
The metal sample contains \(1.0 \times 10^{22}\) atoms, option (D). \[ \boxed{1.0 \times 10^{22}} \]