Question

Naturally occurring gold crystallises in FCC structure and has a density of 19.3 g/cm ^3 .Find atomic radius of gold (Au= 197 g)

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Answer 1

Answer:

• Density of Gold = 19.3 g/cm³
• Atomic Mass of Gold = 197 g
• Type of Crystal Structure = FCC
• Avogadro Number (N) = 6.022 × 10²³

The Number of atoms in the unit cell of a face centred cubic structure :

⇴ $(z)=\sf\dfrac{1}{8}\times8+6\times\dfrac{1}{2}=4$

$\rule{100}{1}$

$\underline{\bigstar\:\textbf{According to the Question :}}$

$:\implies{\sf Density} =\dfrac{z \sf M}{ \sf Na^3}\\\\\\:\implies{\sf a^3}=\dfrac{z \sf M}{ \sf Density \times N}\\\\\\:\implies\sf a^3=\dfrac{4 \times 197g}{19.3 g\:cm^{-3} \times 6.022 \times 10^{23}}\\\\\\:\implies\sf a^3 = 6.78 \times 10^{ - 23}\\\\\\:\implies\sf a^3 = 67.8 \times 10^{ - 24}\\\\\\:\implies\sf a = \sqrt[3]{67.8 \times 10^{ - 24}}\\\\\\:\implies\sf a = 4.08 \times 10^{ - 8}\:cm$

$\rule{180}{2}$

$\underline{\bigstar\:\textbf{For FCC :}}$

$\dashrightarrow\sf\:\:r=\dfrac{a}{\sqrt{8}}\\\\\\\dashrightarrow\sf\:\:r=\dfrac{4.08 \times 10^{ - 8}\:cm}{\sqrt{8}}\\\\\\\dashrightarrow\sf\:\:r = 1.44 \times 10^{ - 8}\:cm\\\\\\\dashrightarrow\sf\:\:r = 144 \times 10^{ - 10}\:cm\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf r = 144\:pm}}$

⠀

$\therefore\:\underline{\textsf{The atomic radius of gold is \textbf{144 pm}.}}$

Answer 2

GIVEN :-

✦ Atomic mass of Gold(M) = 197 g

✦ Density of Gold(d) = 19.3 g/cm³

✦ Unit cell structure of gold = FCC

TO FIND :-

The atomic radius of Gold(Au).

ACKNOWLEDGEMENT :-

✦ The value of Avogadro's constant(N) = 6.022 × 10²³

✦ FCC crystal structure 4 atoms/unit cell.

FORMULA TO BE APPLIED :-

$\bigstar\ \displaystyle{\sf{d=\frac{zM}{a^3NA} }}$

SOLUTION :-

Substituting the known values in the formula above :-

$\displaystyle{\sf{19.3=\frac{4\times197}{(a)^3\times6.022\times10^{23}} }}\\\\\\\displaystyle{\sf{\implies a^3=\frac{788}{7.82\times10^{23}} }}\\\\\\\displaystyle{\sf{\implies a^3=\frac{788\times10^{-23}}{7.82} }}\\\\\\\displaystyle{\sf{\implies a=4.07\times10^{-8}\ cm}}\\\\\displaystyle{\sf{\underline{\implies a=407.53\ pm}}}$

Now,

$\displaystyle{\sf{Radius=\frac{a}{2\sqrt{2}} }}\\\\\\\displaystyle{\sf{\implies r=\frac{4.08\times10^{-8}}{2\sqrt{2} } \ cm}}\\\\\\\displaystyle{\sf{\implies r=\frac{407.53}{2\sqrt{2}}\ pm }}\\\\\\\underline{\boxed{\boxed{\displaystyle{\sf{\implies r=144.08\ pm}}}}}$

✦ So, the atomic radius of Golf(Au) is 144.08 pm.