Question

Calculate the number of atoms present in \(1 \text{g}\) metal that forms simple unit cell structure if product of density and volume of unit cell is \(66.5\times 10^{-24} \text{g}\)

• A. \(1.504\times 10^{22}\)
• B. \(3.419\times 10^{22}\)
• C. \(2.516\times 10^{22}\)
• D. \(2.018\times 10^{22}\)

Hint:

Density times volume is the mass of the unit cell, and a simple cubic cell holds one atom.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Mass of a unit cell \(= \rho \times a^3 = \frac{Z \times M}{N_A}\). A simple cubic unit cell has \(Z = 1\) atom.

Step 2: Key Formula or Approach:
Here the product of density and unit cell volume, which is the mass of the unit cell, is \(66.5 \times 10^{-24}\) g. With \(Z = 1\) this is the mass of one atom.

Step 3: Detailed Explanation:
Mass of one atom \(= 66.5 \times 10^{-24}\) g.
\[ \text{Atoms in 1 g} = \frac{1}{66.5 \times 10^{-24}} = 1.504 \times 10^{22} \]
This is option A. The other options would result from using \(Z = 2\) or \(Z = 4\) (body or face centred cells), which do not apply to a simple unit cell.

Final Answer:
About \(1.504 \times 10^{22}\) atoms are present in 1 g, option (A). \[ \boxed{1.504 \times 10^{22}} \]