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What is the lattice constant for FCC crystal having atomic radius 1.476 Å?(a) 1.476 Å(b) 4.1748 Å(c) 5.216 Å(d) 0 |
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Answer» Right choice is (b) 4.1748 Å To explain I would say: Radius=1.476 Å or 1.476 × 10^-10 m Lattice constant for FCC = 4r/√2 Lattice constant = 4.1748 Å. |
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