Consider the dislocation dissociation reaction (a/2) [110] rightarrow (a/6) [211

Question

Consider the dislocation dissociation reaction (a/2) [110] rightarrow (a/6) [211] + (a/6) [121] in an fcc crystal. Assume that the energy/length of a dislocation is given by E_L = Gb^2 and neglect any dependence of the energy on the edge versus screw nature of the dislocation. Assume that this reaction occurs and the partial dislocations move very far apart. Express the total decrease in energy/length of the original (a/2)[110] dislocation in terms of a and G. On which {111} plane must these dislocations lie?

Explanation / Answer

a. Energy of the perfect dislocation=Gb2=G (a/2[110])2=G*a2/4(12+12+02)=G*(a2/4)*2=Ga2/2

Energy of the two partial dislocations =2Gb2=2G(a/6[112])2=2G*a2/36(12+12+22)=2G*(a2/36)*6=Ga2/3

The total decrease in energy=Energy of the perfect dislocation - Energy of the two partial dislocations

=Ga2/2 - Ga2/3 = Ga2/6

b. The dislocations lie on [112] plane.

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