Question #163094

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A cylindrical pill-like cluster of radius r nucleates on a dislocation that emerges from the substrate. The free-energy change per unit thickness is given by

DG = pr2Gv + 2pyr + A - B ln r,

where A — B In r represents the dislocation energy within the cluster.

(a) Sketch DG vs r (note at r = 0, DG = µ

(b) Determine the value of r*.

(c) Show that when AGvJB/7iy2 > i, AG monotonically decreases with r, but when AGvB/7ry2 < i, there is a turnaround in the curve. (The latter case corresponds to a metastable state and associated energy barrier.)


1
Expert's answer
2021-02-16T07:28:50-0500

a) dΔGdr=0\frac{d\Delta G}{dr} =0

b) r=πr+(π2r2+2BπΔGv)2πΔGyr^* = \frac{-\pi r +\sqrt (\pi^2 r^2+2B\pi \Delta Gv)}{2\pi \Delta Gy}

c) ΔGvBπr2<12\frac{\Delta Gv B}{\pi r^2} < \frac{1}{2}

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