Question

Calculate the radius of an atom of an element in pm if it forms bcc unit cell structure having edge length $4.3\times10^{-8}$ cm.

• A. 186.2
• B. 215.0
• C. 152.3
• D. 282.8

Hint:

BCC: $4r = \sqrt{3}a$; FCC: $4r = \sqrt{2}a$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the atomic radius (\( r \)) for a body-centered cubic (bcc) lattice.

Step 2: Key Formula or Approach:
For bcc: \( \sqrt{3}a = 4r \)
\[ r = \frac{\sqrt{3}a}{4} \]

Step 3: Detailed Explanation:
Given: \( a = 4.3 \times 10^{-8} \text{ cm} = 430 \text{ pm} \).
\[ r = \frac{1.732 \times 430}{4} = \frac{744.76}{4} \]
\[ r = 186.19 \approx 186.2 \text{ pm.} \]

Step 4: Final Answer:
The atomic radius is 186.2 pm.