Question

An element crystallizes in bcc type of unit cell, the density and edge length of unit cell is 4 g cm\(^{-3}\) and 500 pm respectively. What is the atomic mass of an element?

• A. 125.5
• B. 100.1
• C. 250.0
• D. 150.0

Hint:

For bcc unit cells, use the formula for density to calculate atomic mass when the edge length and density are known.

Answer 1

The Correct Option is D

Step 1: Using the formula for density of bcc unit cell.
For a bcc unit cell, the formula for density \( \rho \) is given by: \[ \rho = \frac{Z \times M}{a^3 \times N_A} \] Where \( Z \) is the number of atoms per unit cell (for bcc, \( Z = 2 \)), \( M \) is the atomic mass, \( a \) is the edge length, and \( N_A \) is Avogadro's number.

Step 2: Substituting the given values.
Given: \[ \rho = 4 \, \text{g/cm}^3, \, a = 500 \, \text{pm} = 500 \times 10^{-12} \, \text{m} = 5 \times 10^{-8} \, \text{cm} \] \[ N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \] Using the formula, we calculate the atomic mass \( M \) to be 150.0 g/mol.

Step 3: Conclusion.
The correct answer is (D) 150.0, as this is the atomic mass of the element.