Question #163099

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

∆G =πr^2∆G, + 2myr + A - B Inr, where A - B Inr represents the dislocation energy within the cluster. 0).


(a) Sketch ∆G vs r (note at r= 0, G


(b) Determine the value of r*.


(c) Show that when ∆G, B/ry >+, ∆G monotonically decreases with r,


but when ∆G, B/ay² < there is a turnaround in the curve. (The latter case corresponds to a metastable state and associated energy barrier.)


1
Expert's answer
2021-02-15T05:40:44-0500

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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