# Nucleation on a substrate

1. May 4, 2015

### muskie25

1. The problem statement, all variables and given/known data
A cylindrical pill-like solid cluster of radius r nucleates from the vapor phase on a dislocation that emerges from the substrate. The free-energy change per unit thickness is given by
$$\Delta G = \pi r^2 \Delta G_v + 2 \pi r \gamma + A - B lnr$$
Where $$A - Blnr$$ represents the dislocation energy within the cluster.

a) Sketch $\Delta G$ vs $r$
b) Determine the value of r*
c) Show that when $| \Delta G_v | B / \pi \gamma^2 > 1/2$, $\Delta G_v$ monotonically decreases with r, and when $| \Delta G_v | B / \pi \gamma^2 < 1/2$, there is a turnaround in the $\Delta G$ vs $r$ curve.

2. Relevant equations

3. The attempt at a solution
For r*, I got $$r* = \frac{-2 \pi \gamma \pm \sqrt{4 \pi^2 \gamma^2 + 8 \pi \Delta G_v B}}{4 \pi \Delta G_v}$$

I don't exactly know how to plot $\Delta G$ vs $r$ for this heterogenous nucleation with the natural log function in there. I also don't really understand part c, but I think that's because I have part b wrong. Any help/input would be greatly appreciated.

2. May 5, 2015

### Staff: Mentor

It is not clear how you arrived at this. What happened to the $lnr$ from your top equation? How does r* differ from r?

I don't know about the physics in this question, but looking at the maths you can take a factor outside the radical:

$$r*=\frac{ -2 \pi \gamma \pm 2\pi\gamma\sqrt{1 + \dfrac{2\Delta G_v B} {\pi\gamma^2}}} {4 \pi \Delta G_v}$$

3. May 5, 2015

### muskie25

To find the critical radius, you take $$\frac{ d\Delta G}{dr} = 0$$ and solve for r. The reason that I am concerned is due to the fact other solutions that I have found (homogeneous nucleation) are much more simple.

4. May 5, 2015

### Staff: Mentor

Your first equation shows ΔG in terms of r2 and r. If you differentiate this to find dΔG/dr the result won't contain an r2 term.

5. May 5, 2015

### muskie25

Right, but I multiplied through by r to get rid of the 1/r that comes from differentiating the ln term.