Question

Calculate the number of atoms present in 1 g of an element if it forms fcc unit cell structure. [$\rho \times \text{a}^3 = 6.8 \times 10^{-22} \text{ g}$]}

• A. $7.125 \times 10^{21}$
• B. $4.548 \times 10^{21}$
• C. $6.815 \times 10^{21}$
• D. $5.882 \times 10^{21}$

Hint:

Number of atoms in $m$ grams $= \frac{Z \times m}{\rho \times a^3}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The mass of a unit cell is the product of its density (\(\rho\)) and volume (\(a^3\)). The total number of atoms depends on how many unit cells are in the given mass.
Step 2: Key Formula or Approach:
Mass of unit cell = \(\rho \times a^3\).
Number of unit cells in mass \(m\) = \(\frac{m}{\rho \times a^3}\).
Total atoms = Number of unit cells \(\times Z\) (where \(Z=4\) for fcc).
Step 3: Detailed Explanation:
1. Mass of one unit cell = \(6.8 \times 10^{-22} \text{ g}\).
2. Number of unit cells in 1 g = \(\frac{1}{6.8 \times 10^{-22}} \approx 1.4706 \times 10^{21}\).
3. Since it is an fcc structure, each unit cell has 4 atoms.
4. Total atoms = \(4 \times 1.4706 \times 10^{21} = 5.882 \times 10^{21}\).
Step 4: Final Answer:
There are \(5.882 \times 10^{21}\) atoms in 1 g.