Question:
Pt has FCC structure with an edge length of unit cell \(392\) pm. What is the radius of Pt?
Pt has FCC structure with an edge length of unit cell \(392\) pm. What is the radius of Pt?
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In FCC the face diagonal is $4r$, so $r=\frac{a}{2\sqrt2}$.
Updated On: Oct 1, 2026
- \(138.6\) pm
- \(175.6\) pm
- \(185.6\) pm
- \(205.6\) pm
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The Correct Option is A
Solution and Explanation
Step 1: Use FCC geometry
In an FCC cell, atoms touch along the face diagonal. The face diagonal is \(\sqrt2a\) and holds \(4r\).
\[ 4r=\sqrt2\,a\;\Rightarrow\; r=\frac{a}{2\sqrt2} \]
Step 2: Substitute
\[ r=\frac{392}{2\times1.414}=\frac{392}{2.828}=138.6\text{ pm} \]
Step 3: Check
The value \(196\) pm would be \(a/2\), which applies to a simple cubic cell, not FCC. So option (A).
Final Answer:
\(r=138.6\) pm, option (A). \[ \boxed{\text{(A) } 138.6\text{ pm}} \]
In an FCC cell, atoms touch along the face diagonal. The face diagonal is \(\sqrt2a\) and holds \(4r\).
\[ 4r=\sqrt2\,a\;\Rightarrow\; r=\frac{a}{2\sqrt2} \]
Step 2: Substitute
\[ r=\frac{392}{2\times1.414}=\frac{392}{2.828}=138.6\text{ pm} \]
Step 3: Check
The value \(196\) pm would be \(a/2\), which applies to a simple cubic cell, not FCC. So option (A).
Final Answer:
\(r=138.6\) pm, option (A). \[ \boxed{\text{(A) } 138.6\text{ pm}} \]
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