Question

Calculate the number of atoms in 1 g metal that forms bcc crystal structure \([ρ\times a^3 = 6\cdot 6\times 10^{-22}\text{ g}]\)

• A. \(4\cdot 05\times 10^{21}\)
• B. \(5\cdot 12\times 10^{21}\)
• C. \(3\cdot 03\times 10^{21}\)
• D. \(2\cdot 47\times 10^{21}\)

Hint:

Use density = Z M/(N_A a^3) with Z = 2 for bcc to get M, then divide N_A by M.

Answer 1

The Correct Option is C

Step 1: Mass of one unit cell:
Mass of a unit cell = density $\times$ volume = $\rho a^3 = 6.6\times 10^{-22}$ g.

Step 2: Count unit cells:
Number of unit cells in 1 g = $\frac{1}{6.6\times 10^{-22}} = 1.515\times 10^{21}$.
Each bcc cell has 2 atoms, so atoms = $2\times 1.515\times 10^{21} = 3.03\times 10^{21}$ (C).

Final Answer:
$3.03\times 10^{21}$ atoms. \[ \boxed{3.03\times 10^{21}} \]