Question

Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is $407$ pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold $=197$ amu.

• A. $19.4g/c{m}^3$, $143.9$ pm
• B. $38.8g/c{m}^3$, $143.9$ pm
• C. $19.4g/c{m}^3$, $287.8$ pm
• D. None of the above

Answer 1

The correct option is A $19.4g/c{m}^3$, $143.9$ pm

For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
$4r=\sqrt{2}a$, where $a$ is the edge length of unit cell and $r$ is the radius of atom.
Substituting values in the above expression, we get
$r=\frac{\sqrt{2}\times 407}{4}=143.9$ pm
Number of atoms per unit cell (FCC) $=4$
Mass of $1$ unit cell $=197\times 4=788amu=1.66\times {10}^{-27}\times 788kg=1.30808\times {10}^{-24}kg$
Volume of $1$ unit cell $=(407\times {10}^{-12}{)}^3{m}^3=6.741\times {10}^{-29}{m}^3$ $=2.824\times {10}^{-28}{m}^3$
$\text{Density}=\frac{1.30808\times {10}^{-24}}{6.741\times {10}^{-29}}=19404.83kg/m^3=19.4g/c{m}^3$