Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10^-9m?(a) 0.115 nm(b) 0.144 nm(c) 0.235 nm(d) 0.156 nmI have been asked this question by my school principal while I was bunking the class.The doubt is from Solid State topic in chapter Solid State of Chemistry – Class 12

Gold (atomic mass 197 u) crystallises in a face-centred unit cell. Wha

1.	Gold (atomic mass 197 u) crystallises in a face-centred unit cell. What is its atomic radius if the edge length of the gold unit cell is 0.407 x 10^-9m?(a) 0.115 nm(b) 0.144 nm(c) 0.235 nm(d) 0.156 nmI have been asked this question by my school principal while I was bunking the class.The doubt is from Solid State topic in chapter Solid State of Chemistry – Class 12
Answer» The correct answer is (b) 0.144 nm The best explanation: Given, EDGE length of the Gold unit cell (a) = 0.407 x 10^-7m For FCC unit cell, the ATOMIC radius (R) = a/(2\(\SQRT{2}\)) = 0.407 x 10^-9/(2\(\sqrt{2}\)) = 0.144 nm.

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