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Nucleation on a substrate

  1. May 4, 2015 #1
    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
    [tex] \Delta G = \pi r^2 \Delta G_v + 2 \pi r \gamma + A - B lnr [/tex]
    Where [tex] A - Blnr [/tex] 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 [tex] r* = \frac{-2 \pi \gamma \pm \sqrt{4 \pi^2 \gamma^2 + 8 \pi \Delta G_v B}}{4 \pi \Delta G_v} [/tex]

    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. jcsd
  3. May 5, 2015 #2

    NascentOxygen

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    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:

    [tex]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} [/tex]
     
  4. May 5, 2015 #3
    To find the critical radius, you take [tex] \frac{ d\Delta G}{dr} = 0 [/tex] 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.
     
  5. May 5, 2015 #4

    NascentOxygen

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    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.
     
  6. May 5, 2015 #5
    Right, but I multiplied through by r to get rid of the 1/r that comes from differentiating the ln term.
     
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