Q. If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius in fcc crystal is
A) 1.27 Å
B) 1.81 Å
C) 2.10 Å
D) 2.92 Å
Explanation
[Solution from source] In fcc, atoms touch along the face diagonal: 4r = √2a, so r = a/(2√2) = 3.6/2.828 ≈ 1.27 Å. 1.81 Å is just a/2, which applies to simple cubic. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.