Question

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 \( (111) \), respectively, as shown in the figure below. Dislocations AB and CD, have Burgers vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \), 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? 

• A. Slip plane is \( (001) \) and Burgers vector is \( \frac{a}{6} [110] \)
• B. Slip plane is \( (111) \) and Burgers vector is \( \frac{a}{6} [110] \)
• C. Slip plane is \( (001) \) and Burgers vector is \( \frac{a}{2} [110] \)
• D. Slip plane is \( (111) \) and Burgers vector is \( \frac{a}{2} [110] \)

Hint:

When dealing with dislocation reactions in FCC crystals, always consider the combination of partial dislocations' Burgers vectors and the changes in the slip planes resulting from their interaction. The resulting Burgers vector often depends on the specific orientation of the dislocations involved.

Answer 1

The Correct Option is A

In an FCC crystal, partial dislocations form as a result of the interaction of the primary dislocations. The problem involves determining the slip plane and the resulting Burgers vector after the interaction of dislocations AB and CD. The key steps are as follows:
1. Dislocations and Burgers Vectors:
- The Burgers vector for dislocation AB is \( \mathbf{b_1} = \frac{a}{6} [121] \), and for dislocation CD, it is \( \mathbf{b_2} = \frac{a}{6} [211] \). These are partial dislocations in an FCC crystal, where the dislocations are typically represented by vectors in the directions of the slip systems.
2. Resulting Burgers Vector:
- The resulting Burgers vector from the interaction of two dislocations in FCC crystals is the sum (or difference) of the Burgers vectors of the interacting dislocations. To find the resulting Burgers vector, we need to calculate the vector sum of \( \mathbf{b_1} \) and \( \mathbf{b_2} \), which involves the linear combination of the two dislocation vectors. This results in a Burgers vector of magnitude \( \frac{a}{2} [110] \), representing the magnitude and direction of the resultant dislocation.
3. Slip Plane:
- In the interaction of dislocations, the resulting slip plane is \( (001) \), as this is a common plane involved in slip systems for FCC crystals. The interaction of partial dislocations often involves a change in the slip system, with the slip plane of the resulting dislocation being \( (001) \). This is characteristic of the behavior of dislocations in FCC crystals, where dislocation reactions lead to a shift in the slip plane.
Thus, the correct answer is (A): Slip plane is \( (001) \) and Burgers vector is \( \frac{a}{6} [110] \).