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

Step 1: Understanding the Concept:
The density of a crystal is related to the mass and volume of its unit cells. The total mass of a sample is the mass of one unit cell multiplied by the total number of unit cells in that sample.
Step 2: Key Formula or Approach:
1. Mass of one unit cell = \(\rho \times a^3\).
2. Number of unit cells (\(N_{uc}\)) = \(\frac{\text{Total Mass}}{\text{Mass of 1 unit cell}}\).
3. Total atoms = \(N_{uc} \times z\), where $z$ is atoms per unit cell.
Step 3: Detailed Explanation:
Given:
Total mass = \(1.58\text{ g}\)
Mass of one unit cell (\(\rho \times a^3\)) = \(1.58 \times 10^{-22}\text{ g}\)
For a BCC structure, \(z = 2\).
First, find the number of unit cells: \[ N_{uc} = \frac{1.58}{1.58 \times 10^{-22}} = 10^{22} \] Then, find the total number of atoms: \[ \text{Total atoms} = 10^{22} \times 2 = 2.0 \times 10^{22} \] Step 4: Final Answer:
There are \(2.0 \times 10^{22}\) atoms.