Chapter 11, Problem 11.39QP

Metallic iron crystallizes in a cubic lattice. The unit cell edge length is 287 pm. The density of iron is 7.87 g/cm³. How many iron atoms are within a unit cell?

Interpretation Introduction

Interpretation:

The number of Iron atoms in unit cell of Iron cubic lattice has to be determined.

Concept introduction:

In crystalline solids, the components are packed in regular pattern and neatly stacked. The components are imagined as spheres and closely packed. This phenomenon is called “close packing” in crystals. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing.

Answer to Problem 11.39QP

The number of Iron atoms in unit cell of Iron cubic lattice is $2.$

Explanation of Solution

Record the given data:

    $\begin{array}{l}\text{edge}\text{length}\text{of}\text{unit}\text{cell}\text{=}\text{287}\text{pm}\\ \text{density of Iron}\text{=}\text{7}\text{.87}{\text{g/cm}}^{\text{3}}\end{array}$

The unit cell is assumed that of a cubic shape and the edge length of the unit cell is given. Density of Iron is given.

Calculate the volume of one unit cell of cubic lattice of metallic Iron.

   $\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}\text{287}\text{pm}\text{=}\text{287}\text{×}{\text{10}}^{\text{-12}}\text{m}\text{=}\text{2}\text{.87}\text{×}{\text{10}}^{\text{-10}}\text{m}\\ \text{volume,}{\text{a}}^{\text{3}}\text{=}{\text{(2}\text{.87}\text{×}{\text{10}}^{\text{-10}}\text{m)}}^{\text{3}}\\ & \text{=}\text{23}\text{.64}\text{×}{\text{10}}^{\text{-30}}{\text{m}}^{\text{3}}\\ \text{=}\text{23}\text{.64}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\end{array}$

Edge length of the cubic unit cell is given. The cubic value of edge length gives the volume of the unit cell.

Calculate the mass of one unit cell of cubic lattice of metallic Iron.

$\begin{array}{l}\text{known}\text{data}\text{: density = 7}{\text{.87 g/cm}}^{\text{3}}\\ \text{volume}\text{=}\text{23}\text{.64}\text{×}{\text{10}}^{\text{-30}}{\text{cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{mass}\text{=}\text{volume}\text{×}\text{density}\\ \text{=}\text{23}\text{.64}\text{×}{\text{10}}^{\text{-24}}{\text{cm}}^{\text{3}}\text{×}\text{7}\text{.87}{\text{g/cm}}^{\text{3}}\\ \text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{186}\text{.05}\text{×}{\text{10}}^{\text{-24}}\text{g}\text{=}\text{1}\text{.86}\text{×}{\text{10}}^{\text{-22}}\text{g}\end{array}$

Density of the unit cell is given. The mass of unit cell is calculated using the equation $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\text{.}$ the values are substituted in the equation and rearranged to obtain mass of the unit cell of Iron lattice.

Calculate the average mass of one Iron atom in unit cell.

  $\begin{array}{l}\text{Average mass of one metal atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{metal}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{55}\text{.85}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{9}\text{.27×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

The average mass of one Iron atom in its crystal lattice is calculated using the atomic mass value of Iron.

Calculate the number of Iron atoms present in a unit cell.

$\begin{array}{l}\text{Average mass of one Iron atom}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{no}\text{.}\text{of}\text{Iron}\text{atoms}\text{per}\text{unit}\text{cell}}\\ \text{no}\text{.}\text{of}\text{Iron}\text{atoms}\text{per}\text{unit}\text{cell}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{Average mass of one Iron atom}}\\ \text{=}\frac{\text{1}\text{.86}\text{×}{\text{10}}^{\text{-22}}\text{g}}{\text{9}{\text{.27×10}}^{\text{-23}}\text{g}}=0.20\times {10}^1=2\end{array}$

The average mass of one Iron atom in its crystal lattice is related to number of atoms per unit cell as follows –

$\begin{array}{l}\text{Average mass of one Iron atom}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{no}\text{.}\text{of}\text{Iron}\text{atoms}\text{per}\text{unit}\text{cell}}\\ \text{no}\text{.}\text{of}\text{Iron}\text{atoms}\text{per}\text{unit}\text{cell}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{Average mass of one Iron atom}}\end{array}$

Using this equation, the number of Iron atoms present per unit cell has been calculated.

Conclusion

The number of Iron atoms in unit cell of Iron cubic lattice has been determined.