The relationship between **edge length $a$** and **atomic radius $r$** in a **face-centred cubic (fcc)** unit cell is:
A$a = 2r$
B$a = 4r/\sqrt 2 = 2\sqrt 2\, r$
C$a = 4r/\sqrt 3$
D$a = 2r\sqrt 3$
Answer & Solution
Correct answer: B. $a = 4r/\sqrt 2 = 2\sqrt 2\, r$
In fcc, the face diagonal = $a\sqrt 2$ and equals 4r (4 atomic radii along the diagonal: corner + face + face + corner). So $a\sqrt 2 = 4r$ ⇒ $a = 2\sqrt 2 \, r$.
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