Question

An element has a bcc structure with cell edge of 288 pm. The density of element is 7.2 g/cm\(^3\). What is the atomic mass of an element?

• A. 51.78
• B. 25.89
• C. 62.43
• D. 77.68

Hint:

For bcc crystals, the number of atoms per unit cell is 2. Use the unit cell volume and density to calculate the atomic mass.

Answer 1

The Correct Option is A

Step 1: Formula for atomic mass.
For a bcc structure, the relationship between the edge length \( a \), density \( \rho \), and atomic mass \( M \) is given by: \[ M = \frac{\rho \times a^3 \times N_A}{Z} \] Where: - \( a = 288 \, \text{pm} = 2.88 \times 10^{-8} \, \text{cm} \) is the edge length of the unit cell, - \( \rho = 7.2 \, \text{g/cm}^3 \) is the density, - \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \) is Avogadro’s number, - \( Z = 2 \) is the number of atoms per unit cell for bcc structure.

Step 2: Calculation.
Substitute the values into the formula: \[ M = \frac{7.2 \times (2.88 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2} \approx 51.78 \, \text{g/mol} \]
Step 3: Conclusion.
The atomic mass of the element is (A) 51.78.