Q90 · CSIR NET Chemistry, June 2012
Paper: CSIR NET June 2012 · Subject: Physical Chemistry · Chapter: Solid State Chemistry · Topic: Crystal Systems Lattices · Marks: 2 · Difficulty: Medium
The lattice parameter of an element stabilized in a fcc structure is $4.04 \AA$. The atomic radius of the element is:
(a)$2.86 \AA$
(b)$1.43 \AA$
(c)$4.29 \AA$
(d)$5.72 \AA$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
fcc: 4r = a√2 → r = 4.04/2.828 = 1.43 Å.
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