Question:
Gold crystallizes as fcc unit cell, the edge length of unit cell is \(408\) pm. What is the radius of gold atom?
Gold crystallizes as fcc unit cell, the edge length of unit cell is \(408\) pm. What is the radius of gold atom?
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In an fcc cell atoms touch along the face diagonal, so a = 2 root 2 times r.
Updated On: Oct 1, 2026
- \(86.6\) pm
- \(115.4\) pm
- \(144.2\) pm
- \(175.2\) pm
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
In a face-centred cubic cell, the atoms touch along the face diagonal. Its length is \(\sqrt{2}\,a\), and it equals four atomic radii.
Step 2: Key Formula:
\[ \sqrt{2}\,a = 4r \Rightarrow r = \frac{\sqrt{2}\,a}{4} = \frac{a}{2\sqrt{2}} \]
Step 3: Calculation:
\[ r = \frac{1.414 \times 408}{4} = \frac{576.9}{4} = 144.2\ \text{pm} \]
Step 4: Check the Other Options:
The other values do not come from the fcc relation. Using the bcc relation \(r = \sqrt{3}a/4\) would give 176.6 pm, which is close to (D) but belongs to a different lattice. (C) is correct.
Final Answer:
The radius of the gold atom is 144.2 pm, option (C). \[ \boxed{\text{(C) } 144.2\ \text{pm}} \]
In a face-centred cubic cell, the atoms touch along the face diagonal. Its length is \(\sqrt{2}\,a\), and it equals four atomic radii.
Step 2: Key Formula:
\[ \sqrt{2}\,a = 4r \Rightarrow r = \frac{\sqrt{2}\,a}{4} = \frac{a}{2\sqrt{2}} \]
Step 3: Calculation:
\[ r = \frac{1.414 \times 408}{4} = \frac{576.9}{4} = 144.2\ \text{pm} \]
Step 4: Check the Other Options:
The other values do not come from the fcc relation. Using the bcc relation \(r = \sqrt{3}a/4\) would give 176.6 pm, which is close to (D) but belongs to a different lattice. (C) is correct.
Final Answer:
The radius of the gold atom is 144.2 pm, option (C). \[ \boxed{\text{(C) } 144.2\ \text{pm}} \]
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