Question

Find the type of lattice for a cube having edge length $400$pm, atomic weight = $60$ and density = $6.25$g/cc.

Answer 1

Hint:We will use the density of lattice formula to determine the number of atoms. Every type of unit cell lattice has a fixed number of atoms. So, by determining the number of atoms we can determine the type of lattice. Density depends upon the number of atoms, mass and length of a unit cell.

Formula used: $d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$

Complete step-by-step solution:The formula to calculate the density of cubic lattice is as follows:
$d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$
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.
We will convert the edge length from Picometer to centimetre as follows:
${\text{1}}\,{\text{pm}}\, = \,{10^{ - 10}}{\text{cm}}$
$400\,{\text{pm}}\, = \,4\, \times {10^{ - 8}}{\text{cm}}$

On substituting $6.25$g/cc for density of cube lattice, $60$ for molar mass, \[6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\] for Avogadro number, $4\, \times {10^{ - 8}}{\text{cm}}$for unit cell length.

\[6.25\,{\text{g/c}}{{\text{m}}^3}\, = \dfrac{{{\text{n}}\,\, \times 60\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {{\left( {4 \times {{10}^{ - 8}}\,{\text{cm}}} \right)}^3}}}\]
$\Rightarrow {\text{n}}\, = \dfrac{{6.25 \times 6.02 \times {{10}^{23}} \times {{\left( {4 \times {{10}^{ - 8}}\,{\text{cm}}} \right)}^3}\,}}{{60}}$
$\Rightarrow {\text{n}}\, = \dfrac{{241\,}}{{60}}$
$\therefore {\text{n}}\, = 4.0$

So, the number of atoms in a cube lattice is $4$.

The face centered cube has $4$atoms in the lattice so the type of lattice for the cube is FCC.

Therefore, the type of lattice is face-centred cubic lattice (FCC).

Note: 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,
\[ = \left( {\dfrac{1}{8} \times 8} \right)\, + \left( {\dfrac{1}{2} \times 6} \right)\]
\[ = 4\]
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.