Discussion
In an FCC crystal with lattice parameter \(a\), consider the reaction of two leading partial dislocations, AB and CD, at the line of intersection of their slip planes \((111)\) and \((11\bar{1})\), respectively, as shown in the figure below. Dislocations, AB and CD, have Burgers vectors \(\vec{b}_{1}\) and \(\vec{b}_{2}\), respectively, as given in the figure. Which one of the following options for the slip plane and the Burgers vector of the resulting dislocation is correct?
[The figure shows the two intersecting slip planes, with \(\vec{b}_{1} = \frac{a}{6}[\bar{1}2\bar{1}]\) for AB on \((111)\) and \(\vec{b}_{2} = \frac{a}{6}[2\bar{1}1]\) for CD on \((11\bar{1})\).]
Answer: A
The resulting dislocation lies on \((001)\) with Burgers vector \(\frac{a}{6}[110]\).
Add the two Burgers vectors:
\[ \begin{aligned} \vec{b} &= \frac{a}{6}[\bar{1}2\bar{1}] + \frac{a}{6}[2\bar{1}1] \\ &= \frac{a}{6}[110] \end{aligned} \]
The product lies along the intersection of \((111)\) and \((11\bar{1})\), whose direction is \((1,1,1) \times (1,1,-1) = (-2, 2, 0)\), that is \([\bar{1}10]\). Its plane contains both this line and \(\vec{b}\), so its normal is \([\bar{1}10] \times [110] \propto [001]\): the plane is \((001)\). This is the Lomer-Cottrell stair-rod dislocation; \((001)\) is not a slip plane in FCC, so the dislocation is sessile and forms a barrier.
The energy drops from \(\frac{a^{2}}{6} + \frac{a^{2}}{6}\) to \(\frac{a^{2}}{18}\), so the reaction is favourable. \(\frac{a}{2}[110]\) would be the sum of two full dislocations, not two partials, and the planes \((1\bar{1}1)\) and \((\bar{1}11)\) do not contain both the line and \([110]\).
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