Question

Silver crystallizes in fcc structure. If edge length of unit cell is 316.5 pm, what is the radius of silver atom?

• A. 137.04 pm
• B. 111.91 pm
• C. 121.91 pm
• D. 158.25 pm

Hint:

In an fcc unit cell, the edge length is related to the atomic radius by the formula \( a = 2\sqrt{2}r \).

Answer 1

The Correct Option is B

Step 1: Formula for radius in fcc unit cell.
In the face-centered cubic (fcc) structure, the relationship between edge length \( a \) and atomic radius \( r \) is: \[ a = 2\sqrt{2}r \] Where \( a = 316.5 \, \text{pm} \). Solving for \( r \): \[ r = \frac{a}{2\sqrt{2}} = \frac{316.5}{2\sqrt{2}} \approx 111.91 \, \text{pm} \]
Step 2: Conclusion.
The radius of the silver atom is (B) 111.91 pm.