Chapter 18, Problem 18.125QA

Interpretation Introduction

To find:

a)  The density of bcc iron, assuming the radius of an iron atom is $126 pm$.

b) The density of hexagonal iron, given a unit cell volume $5.414 \times {10}^{-23} c{m}^3$

c) The density of $Si$ substituted $Fe$ crystal.

Answer to Problem 18.125QA

Solution:

a) The density of bcc iron is $7.53 g/c{m}^3$.

b) The density of hexagonal iron is $3.42 g/c{m}^3$.

c) The density of silicon substituted iron crystal is $3.38 g/c{m}^3$.

Explanation of Solution

1) Concept:

The unit cell of the iron is given as bcc unit cell. The edge length can be calculated for the given unit cell as the radius of the $Fe$ is known, by the relation $4r= \sqrt{3}l$. From the edge length of the unit cell, we can calculate the unit cell volume. We will calculate the mass of the unit cell from the number of atoms present in unit cell. From volume and mass of the unit cell, we will calculate the density of the unit cell.

2) Formula:

i)  $4r= \sqrt{3}l$

i) $V_{unit cell}=l^3 c{m}^3$     

ii) $density= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

3) Given:

i) Atomic radii of $Fe=126 pm=126 pm \times \frac{1.0 \times {10}^{-10}cm}{1 pm}= 1.26 \times {10}^{-8}cm$

ii) Molar mass of $Fe=55.845 g/mol$

iii) Number of atoms in unit cell $=2$

iv) Avogadro’s number $1 mol =6.023 \times {10}^{23} atoms$

v) Volume of the hexagonal unit of $Fe=5.414 \times {10}^{-23} c{m}^3$

vi) Molar mass of $Si=28.08 g/mol$

4) Calculations:

a) Calculate the density of bcc iron, assuming the radius of an iron atom is $126 pm$.

Since the unit cell of $Fe$ in this case is bcc, there are two atoms of $Fe$ present in the unit cell.

$2 atoms \times \frac{1 mol}{6.022 \times {10}^{23}atom } \times \frac{55.845 g}{1 mol }=1.855 \times {10}^{-22}g$

Mass of the unit cell is $1.855 \times {10}^{-22}g$.

Calculate the edge length from atomic radius.

$l= \frac{4r}{\sqrt{3}}= \frac{4 \times 1.26 \times {10}^{-8}cm}{\sqrt{3}}=2.9098 \times {10}^{-8}cm$

Volume of unit cell

$V_{unit cell}=l^3 c{m}^3= {\left(2.9098 \times {10}^{-8}\right)}^3 c{m}^3=2.4638 \times {10}^{-23} c{m}^3$

Density of the bcc iron

$density, d= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

$d= \frac{1.855 \times {10}^{-22} g}{2.4638 \times {10}^{-23} c{m}^3}$

$\therefore d=7.53 g/c{m}^3$

Hence, the density of the $Fe$ in bcc unit cell is $7.53 g/c{m}^3$.

b) Calculate the density of hexagonal iron, given a unit cell volume $5.414 \times {10}^{-23} c{m}^3$.

Since the unit cell of $Fe$ in this case is bcc, there are two atoms of $Fe$ present in the unit cell.

$2 atoms \times \frac{1 mol}{6.022 \times {10}^{23}atom } \times \frac{55.845 g}{1 mol }=1.855 \times {10}^{-22}g$

Mass of the unit cell is $1.855 \times {10}^{-22}g$.

Density of the hexagonal iron

$density, d= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

$d= \frac{1.855 \times {10}^{-22} g}{5.414 \times {10}^{-23} c{m}^3}$

$d=3.42g/c{m}^3$

Hence, the density of $Fe$ in hexagonal form is $3.42 g/c{m}^3$.

c) Calculate the density of $Si$ substituted $Fe$ crystal.

It is given that $4\%$ by mass of $Si$ is substituted for $Fe$. This means that from the total mass of the $Fe$ unit cell, $4\%$ has been replaced by $Si$.

The mass of $Fe$ remaining in the unit cell is calculated as

$mass of Fe remaining=\frac{2 atoms Fe }{unit cell}\times 55.845\frac{g}{mol}- \left(\frac{4 \times 28.0855}{100}\right)\frac{g}{mol}$

$mass of Fe remaining=\left(111.69-1.12342\right) g/mol$

$\therefore mass of Fe remaining=110.56658 g/mol$

Mass of iron in unit cell is

$\frac{110.57 g}{1 mol} \times \frac{1 mol}{6.022 \times {10}^{23} atoms}=1.83 \times {10}^{-22} g$

Density of the system

$density= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

$density= \frac{1.83 \times {10}^{-22} g}{5.414 \times {10}^{-23} c{m}^3}=3.38 g/c{m}^3$

Hence the density of such a crystal is $3.38 g/c{m}^3$.

Conclusion:

The density of the $Fe$ atom in bcc unit cell is calculated from the edge length and volume of the unit cell.