Chapter 11, Problem 11.43QP

Crystalline silicon has a cubic structure. The unit cell edge length is 543 pm. The density of the solid is 2.33 g/cm³. Calculate the number of Si atoms in one unit cell.

Interpretation Introduction

Interpretation:

The number of Silicon atoms in unit cell of Silicon 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 11.43QP

The number of Silicon atoms in unit cell of Iron cubic lattice is $8.$

Explanation of Solution

Record the given data:

    $\begin{array}{l}\text{edge}\text{length}\text{of}\text{unit}\text{cell}\text{=}543\text{pm}\\ \text{density of Silicon}\text{=}2.\text{33}{\text{g/cm}}^{\text{3}}\end{array}$

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

Calculate the volume of one unit cell of cubic lattice of Silicon.

   $\begin{array}{l}\text{given}\text{data:}\text{edge}\text{length,}\text{a}\text{=}\text{543}\text{pm}\text{=}\text{543}\text{×}{\text{10}}^{\text{-12}}\text{m}\\ \text{volume,}{\text{a}}^{\text{3}}\text{=}{\text{(543}\text{×}{\text{10}}^{\text{-12}}\text{m)}}^{\text{3}}\\ \text{=}\text{1}\text{.60}\text{×}{\text{10}}^{\text{-28}}{\text{m}}^{\text{3}}\text{=}\text{1}\text{.60}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\end{array}$

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 Silicon.

$\begin{array}{l}\text{known}\text{data}\text{: density }\text{= 2}{\text{.33 g/cm}}^{\text{3}}\\ \text{volume}\text{=}1.60\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\\ \text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{mass}\text{=}\text{volume}\text{×}\text{density}\\ \text{=}1.60\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}\text{×}2.\text{33}{\text{g/cm}}^{\text{3}}\\ \text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}3.73\times {\text{10}}^{\text{-22}}\text{g}\end{array}$

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 Silicon lattice.

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

  $\begin{array}{l}\text{Average mass of one Silicon atom}\text{=}\frac{\text{atomic}\text{mass}\text{of}\text{Silicon}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{28.08}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}4.\text{66×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

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

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

$\begin{array}{l}\text{Average mass of one Silicon atom}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{no}\text{.}\text{of}\text{Silicon}\text{atoms}\text{per}\text{unit}\text{cell}}\\ \text{no}\text{.}\text{of}\text{Silicon}\text{atoms}\text{per}\text{unit}\text{cell}\text{=}\frac{\text{mass}\text{of}\text{unit}\text{cell}}{\text{Average mass of one Silicon atom}}\\ \text{=}\frac{\text{3}\text{.73}\text{×}{\text{10}}^{\text{-22}}\text{g}}{\text{4}{\text{.66×10}}^{\text{-23}}\text{g}}\text{=}\text{0}\text{.80}\text{×}{\text{10}}^{\text{1}}\text{=}8\end{array}$

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

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

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

Conclusion

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