Question

Silver metal crystallizes in face-centred cubic lattice. The length of the unit cell is found to be $4.077 \times {10^{ - 8}}\,{\text{cm}}$. Calculate atomic radius and density of silver. (Atomic mass of ${\text{Ag}}\,{\text{ = }}\,{\text{108u}}$, ${N_a}\, = \,6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}$)

Answer 1

In a face-centred cubic lattice, the radius is one-fourth of diagonal length that is $a\sqrt 2$. Density depends upon the number of atoms, mass and length of a unit cell.

Formula used:
$r\, = \dfrac{{a\sqrt 2 }}{4}$ and $d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$

Step by step answer: The formula to calculate the atomic radius of face-centred cubic lattice is as follows:
$r\, = \dfrac{{a\sqrt 2 }}{4}$
Where,
$r\,$ is the atomic radius.
$a$ is the length of the unit cell.
Substitute $4.077 \times {10^{ - 8}}\,{\text{cm}}$ for unit cell length.
$\Rightarrow r\, = \dfrac{{4.077 \times {{10}^{ - 8}}\,{\text{cm}} \times \sqrt 2 }}{4}$
$\Rightarrow r\, = 1.441 \times {10^{ - 8}}\,{\text{cm}}$
So, the atomic radius of the face-centred cubic lattice is $1.441 \times {10^{ - 8}}\,{\text{cm}}$.
The formula to calculate the density of cubic lattice is as follows:
$\Rightarrow d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$
Where,
$d$ is the density.
$z$ is the number of atoms in a unit cell.
$m$ is the molar mass of the metal.
${N_a}$ is the Avogadro number.
$a$ is the length of a unit cell.
Substitute $4$ for number of atoms, $108\,{\text{u}}$ for molar mass of the metal, $6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}$for Avogadro number, $4.077 \times {10^{ - 8}}\,{\text{cm}}$ for unit cell length.
$\Rightarrow d\, = \dfrac{{4 \times 108\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {{\left( {4.077 \times {{10}^{ - 8}}\,{\text{cm}}} \right)}^3}}}$
$\Rightarrow d\, = \dfrac{{432\,{\text{g}}}}{{40.796\,{\text{c}}{{\text{m}}^3}}}$
$\Rightarrow d\, = 10.589\,\,{\text{g/c}}{{\text{m}}^3}$
So, the density of silver is $10.6\,\,{\text{g/c}}{{\text{m}}^3}$.

Therefore, the atomic radius of face-centred cubic lattice is $1.441 \times {10^{ - 8}}\,{\text{cm}}$and density of silver is $10.6\,\,{\text{g/c}}{{\text{m}}^3}$.

Additional information: In face-centred cubic lattice, eight atoms are present at the corner and six atoms are present at each of the face-centre. Each atom of corner contribute $1/8$ to a unit cell and each atom of face contribute $1/2$ to a unit cell so, the total number of atoms is,
$= \left( {\dfrac{1}{8} \times 8} \right)\, + \left( {\dfrac{1}{2} \times 6} \right)$
$= 4$

Note: The value of the number of atoms depends upon the type of lattice. For face-centred cubic lattice, the number of atoms is four whereas two for body-centred and one for simple cubic lattice.