Question:medium

The atomic radius of atom in FCC structure having a lattice parameter 'a' is

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FCC: $4r = \sqrt{2}a$; BCC: $4r = \sqrt{3}a$.
Updated On: Jul 4, 2026
  • $a\sqrt{2}/2$
  • $a/2\sqrt{2}$
  • $a\sqrt{3}/4$
  • $a/2$
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The Correct Option is B

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