Question

Gold has a cubic close packed structure where $Au$ spheres occupying $0.74$ of the total volume. If the density of gold is $20\text{g/cc}$, calculate the apparent radius of a gold atom in the solid. $(\text{Mass of Au}=197\text{amu})$
Given: $1\text{amu}=1.66\times {10}^{-24}\text{g}$

• A. $0.214\times {10}^{-8}\text{cm}$
• B. $2.512\times {10}^{-8}\text{cm}$
• C. $1.414\times {10}^{-8}\text{cm}$
• D. $3.6\times {10}^{-8}\text{cm}$

Answer 1

The correct option is C $1.414\times {10}^{-8}\text{cm}$

Gold has a cubic close packed structure with a packing fraction value of $0.74$. Since, it has a face - centred cubic unit cell. The number of ions in a face-centred unit cell is 4.

$\text{Mass of unit cell}=4\times 197\times 1.66\times {10}^{-24}\text{g}$
Now,
$\text{Density}=\frac{\text{Mass of unit cell}}{\text{volume of unit cell}}$

$\Rightarrow 20\text{g/cc}=\frac{4\times \left(197\right)\times 1.66\times {10}^{-24}\text{g}}{a^3}$
$\Rightarrow a^3=65.4\times {10}^{-24}{\text{cm}}^3$
$\Rightarrow a^3\approx 65\times {10}^{-24}{\text{cm}}^3$
$\Rightarrow a\approx 4\times {10}^{-8}\text{cm}$
In a face- centred cubic cell, the atom at corners and face centre are in contact with each other along the face diagonals.
Thus,
$4r_{Au}=\sqrt{2}a$
$\Rightarrow r_{Au}=\frac{\sqrt{2}a}{4}=\frac{\sqrt{2}\times 4\times {10}^{-8}}{4}=1.414\times {10}^{-8}\text{cm}$