Question

Gold occurs as face centred cube and it has a density of $19.30kg$ dm${}^{-3}$.
Calculate atomic radius of gold. (Molar mass of $Au=197$)

Answer 1

Unit cell of fcc type gold contain

$=\frac{1}{8}\times 8+6\times \frac{1}{2}=4atoms$

Mass of unit cell of fcc type,

$=4\times \frac{197}{6.022\times {10}^{23}}(Au=197)$

$=130.85\times {10}^{-23}g$

Density of gold $=19.3gc{m}^{-3}$

$Density=\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$

Volume of unit cell $V=\frac{130.85\times {10}^{-23}g}{19.3c{m}^{-3}}=6.78\times {10}^{-23}c{m}^3$

$\therefore V=a^3=6.78\times {10}^{-23}c{m}^3$
where, a is the edge of unit cell,

Hence, $a=\sqrt[3]{36.780+{10}^{-23}c{m}^3}=4.08\times {10}^{-8}cm$
For fcc type unit cell $a=\sqrt{8}r$

Hence, $r=\frac{a}{\sqrt{8}}=\frac{4.08\times {10}^{-8}cm}{\sqrt{8}}$