Chapter 18, Problem 18.121QA

Interpretation Introduction

To find:

The packing efficiency of pure silicon.

Answer to Problem 18.121QA

Solution:

The packing fraction of $Si$ is $33.5\%$.

Explanation of Solution

1) Concept:

We are asked to calculate the packing efficiency of pure silicon. Silicon crystallizes in a diamond lattice form, which contains eight atoms in the unit cell. The radius of the unit cell has been provided, which is used to find the volume of atom, and the density is given, which is used to find the density of the unit cell.

The packing efficiency is defined as the percentage of the total volume of the unit cell occupied by the spheres. The volume of the spheres is calculated by $V_{atom}= \frac{4}{3}\pi r^3.$ The density of the $Si$ is provided, and we know the number of atoms in the unit cell, which is used to calculate the volume of $Si$, and then, dividing the volume by Avogadro’s number gives us the $V_{unit cell}$. Then the packing efficiency is determined by the formula

$packing efficiency \%= \frac{V atom}{V_{unit cell}} \times 100$

2) Formulae:

i) $V_{atom}= \frac{4}{3}\pi r^3 c{m}^3$

ii) $density of Si \left(g/c{m}^3\right)= \frac{mass of Si}{volume of Si}$

iii) $V_{unit cell}= \raisebox{1ex}{$volume of Si$}\!\left/ \!\raisebox{-1ex}{$Avogadr{o}^{\text{'}}s number$}\right.$

iv) $packing efficiency \%= \frac{V atom}{V_{unit cell}} \times 100$

3) Given:

i) Radius of $Si$ atom $=117 pm=117 \times {10}^{-10} cm$

ii) Density of silicon $=2.33 g/c{m}^3$

iii) Number of atoms per unit cell $=8$

iv) Molar mass of $Si=28.08 g$

v) Avogadro’s number $=6.023 \times {10}^{23}$

4) Calculations:

Mass of silicon atoms in unit cell is

$m=number of atoms in unit cell \times molar mass g$

$m=8 \times \frac{28.08 g}{6.022 \times {10}^{23} }=3.730\times {10}^{-22} g$

From the given density equation, the volume can be calculated as

$density of Si \left(g/c{m}^3\right)=\frac{mass of Si \left(g\right)}{volume of unit cell \left(c{m}^3\right)}$

$volume of unit cell \left(c{m}^3\right)= \frac{mass of Si \left(g\right)}{density of Si \left(g/c{m}^3\right)}$

$volume of unit cell \left(c{m}^3\right)=\frac{3.730\times {10}^{-22} g}{2.33 g/c{m}^3}$

$volume of unit cell \left(c{m}^3\right)=1.601\times {10}^{-22} g c{m}^3$

The volume occupied by $Si$ atoms is

$V_{atom}= \frac{4}{3}\pi r^3= \frac{4}{3} \times 3.14 \times \left(117 \times {10}^{-10}\right) c{m}^3=6.7 \times {10}^{-24} c{m}^3$

The packing efficiency is

$packing efficiency \%= \frac{V atom}{V_{unit cell}} \times 100= \frac{8 \times 6.7 \times {10}^{-24} c{m}^3}{1.601 \times {10}^{-22} c{m}^3} \times 100=33.5 \%$

The packing fraction of the $Si$ is $33.5\%$.

Conclusion:

The packing efficiency of pure silicon is calculated, which shows that there are lots of empty spaces in the unit cell.