Chapter 9, Problem 9.19P

Interpretation Introduction

(a)

Interpretation:

The critical radius of the nucleus in the liquid nickel needs to be calculated.

Concept Introduction:

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

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

Answer to Problem 9.19P

The value of the critical radius is $\text{6}{\text{.65×10}}^{\text{-8}}\text{ cm (6}\text{.65 }\stackrel{\text{o}}{\text{A}}\text{)}$.

Explanation of Solution

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

Latent heat of fusion, $\Delta H_f=2756J/c{m}^3$

Surface free energy of solid-liquid phase, ${\sigma }_{sl}=255\times {10}^{-7}J/c{m}^3$

The critical radius of homogeneous nucleation can be calculated by,

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

Where ${\sigma }_{sl}$ is surface free energy of solid-liquid phase

  $T_m$ is equilibrium solidification temperature

  $\Delta H_f$ is Latent heat of fusion

  $\Delta T_m$ is Temperature difference for undercooling

Temperature difference for undercooling, $\Delta T=480C$

Equilibrium solidification temperature, $T_m=1453C=1453+273=1726K$

The equation for critical radius for homogeneous nucleation is:

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

Substituting the values in the equation:

  $\begin{array}{l}{\text{r}}^{\text{k}}\text{=}\frac{{\text{2×255×10}}^{\text{-7}}\text{×1726}}{\text{2756×480}}\\ {\text{r}}^{\text{k}}\text{= 6}{\text{.65×10}}^{\text{-8}}\text{ cm}\\ {\text{r}}^{\text{k}}\text{= 6}\text{.65 }\stackrel{\text{o}}{\text{A}}\end{array}$

Critical radius, ${\text{r}}^{\text{k}}\text{= 6}\text{.65 }\stackrel{\text{o}}{\text{A}}$

Interpretation Introduction

(b)

Interpretation:

The number of nickel atoms in the nucleus needs to be calculated.

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.

Nucleus consists of different number of neutrons, protons and electrons.

Answer to Problem 9.19P

The value of a number of iron atoms in the nucleus is 109 atoms.

Explanation of Solution

Given:

The lattice parameter of solid FCC nickel, ${\text{a}}_{\text{0}}\text{= 0}\text{.356 nm}$

The volume of the nucleus can be calculated by,

  $V_{nucleus}=\frac{4}{3}\pi {(r^k)}^3$

Where $r^k$ is the critical radius

Number of unit cell = $\frac{V_{nucleus}}{V}$

where $V_{nucleus}$ is volume of nucleus

  $V$ is volume of unit cell

The volume of the unit cell,

  $\begin{array}{l}{\text{V= (a}}_{\text{0}}{\text{)}}^{\text{3}}\\ \text{V= (0}{\text{.356×10}}^{\text{-7}}{\text{)}}^{\text{3}}\\ \text{V= 0}{\text{.04511×10}}^{\text{-21}}{\text{ cm}}^{\text{3}}\\ \text{V= 45}{\text{.118×10}}^{\text{-24}}{\text{ cm}}^{\text{3}}\end{array}$

The radius of the nucleus,

  ${\text{r}}^{\text{k}}\text{= 6}{\text{.65×10}}^{\text{-8}}\text{ cm}$

The volume of the nucleus,

  $V_{nucleus}=\frac{4}{3}\pi {(r^k)}^3$

  $\begin{array}{l}$ \frac{4}{3}\pi {(6.65\times {10}^{-8})}^3$V_{nucleus}=1232\times {10}^{-24}c{m}^3\end{array}$

  $\begin{array}{l}\text{Number}\text{of}\text{unit}\text{cell=}\frac{{\text{V}}_{\text{nucleus}}}{\text{V}}\\ \text{=}\frac{{\text{1232×10}}^{\text{-24}}}{\text{45}{\text{.118×10}}^{\text{-24}}}\\ \text{=27}\text{.3}\end{array}$

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

= 27.3 $\times$ 4

=109 atoms