Chapter 15, Problem 15.81P

Interpretation Introduction

Interpretation:

Whether carbon atoms can be held between iron atoms in its unit cell has to be determined.

Concept Introduction:

A unit cell is smallest repeating unit of a crystal lattice. Unit cells are classified as primitive and non-primitive. Unit cells that have atoms at only corners are primitive and unit cells that have atoms at positions other than corners are non-primitive.

Unit cells that are non- primitive are of various types:

Face centered unit cell (fcc), body centered unit cell (bcc), edge centered unit cell, end centered unit cell.

Contribution of an atom at a position is as follows:

  $\begin{array}{cc}\text{position in unit}\text{cell}& \text{contributions}\\ \text{center of body}& \text{1}\\ \text{center of face}& \text{1/2}\\ \text{center of edge}& \text{1/4}\\ \text{corners}& \text{1/8}\end{array}$

Explanation of Solution

The formula to calculate the density of $\text{Fe}$ crystal is as follows:

  $\text{density}=\frac{Z\cdot M}{N_a\cdot a^3}$        (1)

Here,

$Z$ is the number of atoms of $\text{Fe}$ per unit cell.

$M$ is the atomic mass of the $\text{Fe}$.

$N_a$ is Avogadro's number.

$a$ is length of cube.

Rearrange equation (3), to determine of $a$.

  $a=\sqrt[3]{\frac{Z\cdot M}{\text{density}\cdot N_{\text{a}}}}$        (2)

Substitute 2for $Z$, $55.845\text{g/mol}$ for $M$, $6.023\times {10}^{23}{\text{mol}}^{-1}$ for $N_{\text{a}}$ and $7.86\text{g}\cdot {\text{cm}}^{-3}$ for density in equation (2) to determine $a$.

  $\begin{array}{c}a=\sqrt[3]{\frac{\left(2\right)\left(55.845\text{g/mol}\right)}{\left(7.86\text{g}\cdot {\text{cm}}^{-3}\right)\left(6.023\times {10}^{23}{\text{mol}}^{-1}\right)}}\\ =\sqrt[3]{2.3592\times {10}^{-23}{\text{cm}}^3}\\ =2.8680\times {10}^{-08}\text{cm}\end{array}$

The formula to relate radius of iron atom and length of unit cell is as follows:

  $\text{radius}\text{of}\text{Fe}=\frac{\left(\sqrt{3}\right)\left(\text{length}\text{of}\text{unit}\text{cell}\right)}{4}$        (3)

Substitute $2.8680\times {10}^{-08}\text{cm}$ for length of unit cell in equation (3) to determine radius of $\text{Fe}$.

  $\begin{array}{c}\text{radius}\text{of}\text{Fe}=\frac{\left(\sqrt{3}\right)\left(2.8680\times {10}^{-08}\text{cm}\right)}{4}\\ =1.2418\times {10}^{-08}\text{cm}\end{array}$

The formula to determine volume of sphere is as follows:

  $\text{volume}=\frac{4}{3}\text{π}{\left(\text{radius}\right)}^3$        (4)

Substitute $1.2418\times {10}^{-08}\text{cm}$ for radius in equation (4) to determine volume of iron.

  $\begin{array}{c}\text{volume}=\frac{4}{3}\left(\text{3}\text{.14}\right){\left(1.2418\times {10}^{-08}\text{cm}\right)}^3\\ =8.0146\times {10}^{-24}{\text{cm}}^3\end{array}$

So the volume of 2 iron atoms present in unit cell is determined as follows:

  $\begin{array}{c}\text{volume}\text{of}\text{2}\text{Fe}\text{atom}=\left(\text{2}\right)\left(\text{volume of Fe}\right)\\ =\left(\text{2}\right)\left(8.0146\times {10}^{-24}{\text{cm}}^3\right)\\ =16.0292\times {10}^{-24}{\text{cm}}^3\end{array}$

Substitute $77\text{pm}$ for radius in equation (4) to determine volume of carbon.

  $\begin{array}{c}\text{volume}=\frac{4}{3}\left(\text{3}\text{.14}\right){\left(77\text{pm}\right)}^3{\left(\frac{{10}^{-10}\text{cm}}{1\text{pm}}\right)}^3\\ =1.9113\times {10}^{-24}{\text{cm}}^3\end{array}$

The formula to determine volume of cube is as follows:

  $\text{volume of cube}={\left(\text{length}\right)}^3$        (5)

Substitute $2.8680\times {10}^{-08}\text{cm}$ for length in equation to determine volume of cube.

  $\begin{array}{c}\text{volume of cube}={\left(2.8680\times {10}^{-08}\text{cm}\right)}^3\\ =2.3590\times {10}^{-23}{\text{cm}}^3\end{array}$

The vacant space present in the cube is calculated as follows:

  $\begin{array}{c}\text{vacant space}=\text{volume of cube}-\text{volume of 2}\text{Fe}\\ =2.3590\times {10}^{-23}{\text{cm}}^3-16.0292\times {10}^{-24}{\text{cm}}^3\\ =7.5613\times {10}^{-24}{\text{cm}}^3\end{array}$

The vacant space is large so carbon atom can be held between iron atoms in its unit cell.