Gold (atomic radius $= 0.144 n m)$ crystallises in a face-centred unit cell. What is the length of a side of the cell?

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Solution

For fcc unit cell, $a = 2 \sqrt{2} r$ .
Here, $a$ is the edge length and $r$ is the atomic radius ( $0.144$ nm).
$a = 2 \sqrt{2} \times 0.144 = 0.407$ nm
Hence, the length of a side of a cell is $0.407$ nm.

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Gold (atomic radius =0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

For fcc unit cell, a=2√2r.Here, a is the edge length and r is the atomic radius (0.144 nm).a=2√2×0.144=0.407 nmHence, the length of a side of a cell is 0.407 nm.

Gold (atomic radius $= 0.144 n m)$ crystallises in a face-centred unit cell. What is the length of a side of the cell?

For fcc unit cell, $a = 2 \sqrt{2} r$ .
Here, $a$ is the edge length and $r$ is the atomic radius ( $0.144$ nm).
$a = 2 \sqrt{2} \times 0.144 = 0.407$ nm
Hence, the length of a side of a cell is $0.407$ nm.