Question

Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is $124$pm, compute the density (in g${\text{c}}{{\text{m}}^{ - 3}}$) of iron in bcc and fcc respectively.
A. $8.4,\,9.8$
B. $9.3,\,4.6$
C. $8.6,\,7.9$
D. $7.9,\,8.6$

Answer 1

Hint: To answer this question we should know the density formula of solids. Density of a solid depends upon the number of atoms, mass and length of a unit cell. First we will determine the edge length of each unit cell then by using the density formula we will determine the density of each unit cell.

Formula used:
\[{{\text{a}}_{{\text{bcc}}}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt {\text{3}} }}\], \[{{\text{a}}_{{\text{fcc}}}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt 2 }}\], ${\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{{\text{a}}^{\text{3}}}}}$

Complete step-by-step answer:
The formula to calculate the edge length of body-centered cubic unit cell is as follows:
\[{\text{a}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt {\text{3}} }}\]
Where,
${\text{r}}\,$ is the atomic radius.
${\text{a}}$ is the edge length of the unit cell.
First we will convert the atomic radius from Pico meter to centimetre as follows:
$\Rightarrow {\text{1pm}}\,{\text{ = }}\,{10^{ - 10}}{\text{cm}}$
$\Rightarrow {\text{124}}\,{\text{pm}}\,{\text{ = }}\,124\, \times {10^{ - 10}}{\text{cm}}$
On substituting $124\, \times {10^{ - 10}}{\text{cm}}$ for the atomic radius of the unit cell.
\[\Rightarrow {\text{a}}\, = \dfrac{{4 \times 1{\text{24}}\, \times 1}}{{\text{0}}^{ - 10}}{\text{cm}}{{\sqrt 3 }}\]
$\Rightarrow {{r}}\, = {{2.86\,\times 1}}{{\text{0}}^{ - 8}}{\text{cm}}$
So, the edge length of the body-centered cubic unit cell of the iron atom is $2.86\times 10^{-8}\,\, cm$
The formula to calculate the density of cubic lattice is as follows:
${\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{{\text{a}}^{\text{3}}}}}$
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.
The number of atoms in a bcc unit cell is $2$. The molar mass of iron is $55.85\,{\text{u}}$.
On substituting $2$ for number of atoms, $55.85\,{\text{u}}$ for molar mass of the metal, \[6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\]for Avogadro number, ${{2.86 \times 1}}{{\text{0}}^{ - 8}}{\text{cm}}$ for unit cell length.
$\Rightarrow {\text{d}}\, = \dfrac{{2 \times 55.85\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\, \times {{\left( {2.86\,{{ \times 1}}{{\text{0}}^{ - 8}}{\text{cm}}} \right)}^3}}}$
$\Rightarrow {\text{d}}\, = \dfrac{{111.7\,{\text{g}}}}{{14.08\,\,{\text{c}}{{\text{m}}^3}}}$
$\Rightarrow {\text{d}}\, = 7.9\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$
So, the density of iron in bcc is $7.9\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$.
The formula to calculate the edge length of face-centred cubic lattice is as follows:
$\Rightarrow {\text{a}}\,{\text{ = }}\dfrac{{{\text{4r}}}}{{\sqrt {\text{2}} }}$
On substituting $124\, \times {10^{ - 10}}{\text{cm}}$ for the atomic radius of the unit cell.
\[\Rightarrow {\text{a}}\, = \dfrac{{4 \times 1{\text{24}}\,\times 1}}{{\text{0}}^{ - 10}}{\text{cm}}{{\sqrt 2 }}\]
${\Rightarrow \text{r}}\, = {{3.51\, \times 1}}{{\text{0}}^{ - 8}}{\text{cm}}$
So, the edge length of face-centered cubic unit cell of the iron atom is ${{3.51\,\times 1}}{{\text{0}}^{ - 8}}{\text{cm}}$.
Now we will use the density formula to determine the density of FCC lattice as follows:
$\Rightarrow {\text{d}}\,{\text{ = }}\dfrac{{{\text{z}}\,{\text{m}}}}{{{{\text{N}}_{\text{a}}}{{\text{a}}^{\text{3}}}}}$
The number of atoms in an fcc unit cell is $4$. The molar mass of iron is $55.85\,{\text{u}}$.
On substituting $4$ for number of atoms, $55.85\,{\text{u}}$ for molar mass of the metal, \[6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}\]for Avogadro number, ${{3.51\, \times 1}}{{\text{0}}^{ - 8}}{\text{cm}}$ for unit cell length.
$\Rightarrow {\text{d}}\, = \dfrac{{4 \times 55.85\,{\text{g/mol}}}}{{6.02 \times {{10}^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}}\, \times {{\left( {3.51\,\times 1}{{\text{0}}^{ - 8}}{\text{cm}} \right)}^3}}$
$\Rightarrow {\text{d}}\, = \dfrac{{223.4\,{\text{g}}}}{{26.03\,\,{\text{c}}{{\text{m}}^3}}}$
$\Rightarrow {\text{d}}\, = 8.6\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$
So, the density of iron in fcc is$8.6\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$.
So, the density (in g${\text{c}}{{\text{m}}^{ - 3}}$) of iron in bcc and fcc are $7.9,\,8.6$ respectively.

Therefore, option (D) $7.9,\,8.6$ is correct.

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. The metal has more density in the fcc unit cell because in the fcc unit cell, the number of atoms is more than the bcc unit cell. Density is directly proportional to the number of atoms in the unit cell and indirectly proportional to the edge length.