Question:
A metallic element crystallises in FCC type crystal lattice. What is the radius of atom if edge length of its unit cell is 405 pm?
A metallic element crystallises in FCC type crystal lattice. What is the radius of atom if edge length of its unit cell is 405 pm?
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In FCC lattice, atoms touch along the face diagonal, not along the edge.
Updated On: Feb 11, 2026
- 113.2 pm
- 175.3 pm
- 143.2 pm
- 202.5 pm
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The Correct Option is C
Solution and Explanation
Step 1: Recall FCC relation between edge length and radius.
For FCC lattice:
\[ a = 2\sqrt{2}\,r \]
Step 2: Substitute the given value.
\[ a = 405\ \text{pm} \]
\[ r = \frac{a}{2\sqrt{2}} = \frac{405}{2.828} \]
Step 3: Calculate radius.
\[ r \approx 143.2\ \text{pm} \]
Step 4: Conclusion.
The radius of the atom is 143.2 pm.
For FCC lattice:
\[ a = 2\sqrt{2}\,r \]
Step 2: Substitute the given value.
\[ a = 405\ \text{pm} \]
\[ r = \frac{a}{2\sqrt{2}} = \frac{405}{2.828} \]
Step 3: Calculate radius.
\[ r \approx 143.2\ \text{pm} \]
Step 4: Conclusion.
The radius of the atom is 143.2 pm.
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