Question

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

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

Answer 1

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

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.

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

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