Question:medium

Gold crystallizes as fcc unit cell, the edge length of unit cell is \(408\) pm. What is the radius of gold atom?

Show Hint

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
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Geometry:
Look at one face of the cube. Three atoms lie in a line along the diagonal: corner, face centre and opposite corner, so the diagonal holds 4 radii.

Step 2: Work:
Diagonal $= 408\sqrt{2} = 577$ pm. Divide by 4: $r = 144.2$ pm.

Step 3: Check:
Using the wrong relation $r = a/4$ would give 102 pm, and the bcc relation gives 176.6 pm. Neither is an option except by confusion. Option (C) is right.

Final Answer:
Option (C), 144.2 pm. \[ \boxed{\text{(C) } 144.2\ \text{pm}} \]
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