Chapter 9, Problem 9.20P

Interpretation Introduction

(a)

Interpretation:

The critical radius of the nucleus in the liquid iron needs to be determined.

Concept Introduction:

Homogeneous nucleation is the process in which the nuclei that are formed randomly and spontaneously grows irreversibly and form into a new phase.

In this question, we need to calculate the critical radius of the nucleus.

The critical radius of the nuclei is the minimum size of the nucleus required for the formation of a new stable nucleus.

Answer to Problem 9.20P

From the equation of the critical radius for the homogeneous equation, the value of the critical radius is $10.128\times {10}^{-8}\text{cm}$.

Explanation of Solution

Given Information:

From the parameter of the iron, values of different properties are:

Latent heat of fusion, $\Delta H_f=173\text{7}{\text{J/cm}}^{\text{3}}$

Surface free energy of solid-liquid phase, ${\sigma }_{s-l}=204\times {10}^{-7}{\text{J/cm}}^{\text{3}}$

Formula used:

The critical radius of homogeneous nucleation can be calculated by,

  $\begin{array}{l}{r}^k=\frac{2{\sigma }_{sl}{T}_m}{\Delta H_f\Delta T_m}\\ \mathrm{Where},{\sigma }_{sl}=\text{Surface}\text{free}\text{energy}\text{of}\text{solid-liquid}\text{phase}\\ T_m=\text{Equillibrium}\text{solidification}\text{temperature}\\ \Delta H_f=\text{Latent}\text{heat}\text{of}\text{fusion}\\ \Delta T_m=\text{Temperature}\text{difference}\text{for}\text{undercooling}\end{array}$

Calculation:

Temperature difference for undercooling, $\Delta T=420C$

Equilibrium solidification temperature, $T_m=1538\text{C}=1538+273=1811\text{ K}$

The equation for the critical radius is,

  $r^k=\frac{2{\sigma }_{sl}{T}_m}{\Delta H_f\Delta T_m}$

Putting values in the equation,

  $\begin{array}{l}{r}^k=\frac{2\times 204\times {10}^{-7}\times 1811}{1737\times 420}\\ r^k=10.128\times {10}^{-8}cm\end{array}$

Interpretation Introduction

(b)

Interpretation:

The number of iron atoms in the nucleus needs to be determined.

Concept Introduction:

Homogeneous nucleation is the process in which the nuclei that are formed randomly and spontaneously grows irreversibly and form into a new phase.

In this section, we need to calculate a number of atoms in the iron nucleus.

The nucleus consists of a different number of neutrons, protons, and electrons.

Answer to Problem 9.20P

From the equation of volume of the nucleus with a critical radius for the homogeneous equation, we get the value of a number of iron atoms in the nucleus is 350 atoms.

Explanation of Solution

Given Information:

The lattice parameter of solid BCC iron, $a_0=2.92{\text{A}}^0=2.9\text{2}\text{nm}$

Formula Used:

The volume of the nucleus can be calculated by,

  $\begin{array}{l}$ V_{nucleus}=\frac{4}{3}\pi {(r^k)}^3$\text{where,}{\text{r}}^{\text{k}}\text{is}\text{the}\text{critical}\text{radius}\\ \text{Number}\text{of}\text{unit}\text{cell}=\frac{V_{\text{nucleus}}}{V}\\ where,V_{nucleus}isvolumeofnucleus\\ Visvolumeofunitcell\end{array}$

Calculation:

The volume of the unit cell,

  $V={(a_0)}^3={(2.92)}^3=24.897\times {10}^{-24}c{m}^3$

The radius of the nucleus,

  $r^k=10.128\times {10}^{-8}cm$

The volume of the nucleus,

  $\begin{array}{l}$ V_{\text{nucleus}}=\frac{4}{3}\pi {(r^k)}^3$=\frac{4}{3}\pi {(}^{10.128}\\ V_{\text{nucleus}}=4352\times {10}^{-24}{\text{cm}}^{\text{3}}\\ \text{Number}\text{of}\text{unit}\text{cell}=\frac{V_{\text{nucleus}}}{V}\\ =\frac{4352\times {10}^{-24}}{24.897\times {10}^{-24}}\\ =175\end{array}$

Atoms per nucleus = (Number of the unit cell) $\times$ (Number of atoms in the unit cell)

  $\begin{array}{l}=1752 \\ =350\text{ atoms}\end{array}$