If the lattice parameter for a crystalline structure is 3.6, then the atomic radius in fcc crystal is (a) 2.92 (b) 1.27 (c) 1.81 (d) 2.10

Answer: (b). In FCC, atoms touch along the face diagonal: 4r = √(2)a, so r = (a)/(2√(2)). r = (3.6)/(2 × 1.414) = (3.6)/(2.828) r ≈ 1.27 So the answer is (b) 1.27.

If the lattice parameter for a crystalline structure is 3.6, then the… | NEET 2008 (AIPMT)

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Other / Mixed

If the lattice parameter for a crystalline structure is 3.6 $\text{\AA}$ , then the atomic radius in fcc crystal is

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Answer: (b)

In FCC, atoms touch along the face diagonal: $4r = \sqrt{2}a$ , so $r = \frac{a}{2\sqrt{2}}$ .
$r = \frac{3.6}{2 \times 1.414} = \frac{3.6}{2.828}$
$r \approx 1.27\ \text{\AA}$
So the answer is (b) 1.27 $\text{\AA}$ .

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