Naturally occurring gold crystallises in face centred cubic structure and has a density of $19.3$ g cm $^{- 3}$ . Find atomic radius of gold.
(Au $= 197$ g mol $^{- 1}$ )

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Solution

density of Gold = $19.3 g / c m^{3}$
density = $\frac{z M}{9^{3} . N A}$
$19.3 = \frac{4 \times 197}{9^{3} . 6.022 \times 10^{23}}$
$9^{3} = \frac{4 \times 197}{6.022 \times 10^{23} \times 19.3}$ value of $z = 4$
$9^{3} = 6.78 \times 10^{- 23} = 4.077 \times 10^{- 8}$
$\sqrt{2} a = 4^{r}$
$r = \frac{\sqrt{2} 9}{4} = \frac{\sqrt{2} \times 4.077 \times 10^{- 8}}{4}$
$= 1.44 \times 10^{- 8} c m$

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Naturally occurring gold crystallises in face centred cubic structure and has a density of 19.3 g cm−3. Find atomic radius of gold. (Au=197g mol−1)

density of Gold = 19.3g/cm3density = zM93.NA19.3=4×19793.6.022×102393=4×1976.022×1023×19.3 value of z=493=6.78×10−23=4.077×10−8√2a=4rr=√294=√2×4.077×10−84=1.44×10−8cm

Naturally occurring gold crystallises in face centred cubic structure and has a density of $19.3$ g cm $^{- 3}$ . Find atomic radius of gold.
(Au $= 197$ g mol $^{- 1}$ )