Chapter 12, Problem 12.1P

Interpretation Introduction

Interpretation:

The critical nucleus size ${\text{r}}^{\ast }$ needs to be designed.

Concept introduction:

Before the formation of the new phase, the clustering of atoms or molecules is done together which results into critical radius. This is of the minimum size. It is stable and growing in nature.

Nucleation is the process of stable nuclei formation.

By nucleation process, sometimes this precipitation rate is limited. Formation of dendrite occurs when the change of phase forms the crystallization structure of solid into a matrix of liquid.

The growth of crystal is in the form of three-dimensional form and the atoms are attached through the preferred directions. Usually, dendrite is formed along the crystal axis. Dendrites are tree-like structures.

The temperature of the new phase is less than the temperature of transformation. Due to this, the nucleus volume energy is lower than the initial phase volume.

Answer to Problem 12.1P

The critical radius for homogeneous nucleation of $\text{β}$ phase in $\text{α}$ phase

  ${\text{r}}^{\text{*}}=-\frac{{\text{2πσ}}_{\text{αβ}}}{\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)}$

Explanation of Solution

Given Information:

Slope is zero when the total free energy change is the maximum

  $\frac{\partial \left(\text{ΔG}\right)}{\partial \text{r}}=0$

Calculation:

The total energy for the nucleation process −

  $\text{ΔG=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}{\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}{\text{+4πr}}^{\text{2}}{\text{σ}}_{\text{αβ}}\text{+}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\text{ε}$

Where,

Radius of spherical solid surface − $\text{r}$

Surface energy − ${\text{σ}}_{\text{αβ}}$

Interface − $\left(\text{α-β}\right)$

Strain energy − $\text{ε}$

Change in energy - ${\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}$

By rearranging the above equation,

  $\text{ΔG=}\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right){\text{+4πr}}^{\text{2}}{\text{σ}}_{\text{αβ}}$

Consider,

Critical radius for homogeneous nucleation − ${\text{r}}^{\ast }$

Slope is zero when the total free energy change is at the maximum.

So, the above equation gets partially differentiated with respect to radius $\text{r}$

  $\begin{array}{l}\frac{\partial \left(\text{ΔG}\right)}{\partial \text{r}}\text{=}\frac{\partial }{\partial \text{r}}\left(\frac{\text{4}}{\text{3}}{\text{πr}}^{\text{3}}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right){\text{+4πr}}^{\text{2}}{\text{σ}}_{\text{αβ}}\right)\\ \frac{\partial \left(\text{ΔG}\right)}{\partial \text{r}}\text{=}\frac{\text{4}}{\text{3}}\text{π}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)\left({\text{3r}}^{\text{2}}\right){\text{+ 4πσ}}_{\text{αβ}}\left(\text{2r}\right)\\ \frac{\partial \left(\text{ΔG}\right)}{\partial \text{r}}{\text{= 4πr}}^{\text{2}}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right){\text{+ 8πrσ}}_{\text{αβ}}\end{array}$

But,

  $\frac{\partial \left(\text{ΔG}\right)}{\partial \text{r}}=0$

Put

  ${\text{r = r}}^{\text{*}}$

Hence equation becomes -

  $\begin{array}{l}\text{4π}{\left({\text{r}}^{\text{*}}\right)}^{\text{2}}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right){\text{+ 8πr}}^{\text{*}}{\text{σ}}_{\text{αβ}}=0\\ \text{4π}{\left({\text{r}}^{\text{*}}\right)}^{\text{2}}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)=-{\text{8πr}}^{\text{*}}{\text{σ}}_{\text{αβ}}\\ {\left({\text{r}}^{\text{*}}\right)}^{\text{2}}=-\frac{{\text{8πr}}^{\text{*}}{\text{σ}}_{\text{αβ}}}{\text{4π}\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)}\\ {\text{r}}^{\text{*}}=-\frac{{\text{2πσ}}_{\text{αβ}}}{\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)}\end{array}$

Conclusion

Hence, by using free energy change expression, we get the critical radius for the nucleation of homogeneous mixture as −

  ${\text{r}}^{\text{*}}=-\frac{{\text{2πσ}}_{\text{αβ}}}{\left({\text{ΔG}}_{\text{v}\left(\text{α}\to \text{β}\right)}\text{+ ε}\right)}$