Question

An element crystallizes in bcc type crystal structure with edge length of unit cell 300 pm. Calculate radius of element?

• A. 2.299 \times 10^{-8} \, \text{cm}
• B. 1.299 \times 10^{-8} \, \text{cm}
• C. 6.920 \times 10^{-8} \, \text{cm}
• D. 1.440 \times 10^{-8} \, \text{cm}

Hint:

In bcc crystals, remember that the relationship between edge length \( a \) and atomic radius \( r \) is \( r = \frac{\sqrt{3} \cdot a}{4} \).

Answer 1

The Correct Option is B

Step 1: Understanding the bcc structure.
In a body-centered cubic (bcc) structure, the relation between the edge length \( a \) and the radius \( r \) of the atom is given by: \[ \text{Diagonal of the cube} = \sqrt{3} \cdot a = 4r \] From this, we can derive the radius \( r \) as: \[ r = \frac{\sqrt{3} \cdot a}{4} \] Step 2: Calculation of the radius.
Given \( a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m} \), we can substitute into the equation: \[ r = \frac{\sqrt{3} \cdot 300 \times 10^{-12}}{4} \] \[ r = 1.299 \times 10^{-8} \, \text{cm} \] Step 3: Conclusion.
The correct radius of the element is 1.299 \times 10^{-8} \, \text{cm}.