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 \((\bar{1}11)\), respectively. 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?
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When two dislocations interact, always add Burgers vectors vectorially and check the crystallographic geometry of the intersecting planes. Simplify the resulting vector to the smallest lattice translation to identify the perfect dislocation.
Slip plane is (001) and Burgers vector is \(\tfrac{a}{6}[110]\)
Slip plane is (\(\bar{1}11\)) and Burgers vector is \(\tfrac{a}{6}[110]\)
Slip plane is (001) and Burgers vector is \(\tfrac{a}{2}[110]\)
Slip plane is (\(\bar{1}11\)) and Burgers vector is \(\tfrac{a}{2}[110]\)
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The Correct Option isA
Solution and Explanation
In the context of face-centered cubic (FCC) crystal systems, dislocations play a significant role in the deformation process. When considering the reaction of two leading partial dislocations, we are interested in finding both the slip plane and the Burgers vector of the resulting dislocation.
Given: Two partial dislocations AB and CD with Burgers vectors \(\vec{b}_1\) and \(\vec{b}_2\), on slip planes \((111)\) and \((\bar{1}11)\) respectively.
We need to find the correct slip plane and resulting Burgers vector for the newly formed dislocation.
Steps to solve:
1. The FCC structure has {111} planes as close-packed planes where slip typically occurs.
2. When two partial dislocations react, the resulting dislocation's Burgers vector is the vector sum of the two initial Burgers vectors: \(\vec{b}_{\text{result}}=\vec{b}_1+\vec{b}_2\).
3. Examine Burgers vector addition following FCC crystallographic rules.
4. For FCC, when dislocations dissociate or react, new planes like the (001) are often involved as they are less densely packed, allowing resolutions of dislocations.
5. For partial dislocations with Burgers vectors derived from \( \vec{b}=\frac{a}{6}[110]\), vector addition confirms \(\vec{b}_{\text{result}}=\tfrac{a}{6}[110]\).
After vector analysis and considering possible plane reactions typical in FCC crystals, the correct slip plane for the resulting dislocation is (001), since it involves secondary planes or auxiliary slip planes after primary Burgers vector reactions.
Thus, the solution is: Slip plane is (001) and Burgers vector is \(\tfrac{a}{6}[110]\).
