Question

A screw dislocation in a FCC crystal has Burgers vector of $\frac{a}{2}[110]$, where $a$ is the lattice constant. The possible slip plane(s) is/are:

• A. $(1\bar{1}1)$
• B. $(111)$
• C. $(\bar{1}11)$
• D. $(11\bar{1})$

Hint:

In FCC crystals, slip occurs on $\{111\}$ planes along $\langle 110 \rangle$ directions. Always check slip compatibility by verifying $\vec{b} . \vec{n} = 0$.

Answer 1

The Correct Option is C, D

Step 1: Recall slip systems in FCC.
- FCC crystals slip along $\{111\}$ planes in $\langle 110 \rangle$ directions. - This gives 12 slip systems in total.
Step 2: Burgers vector direction.
The Burgers vector is: \[ \vec{b} = \frac{a}{2}[110] \] This is a $\langle 110 \rangle$ direction, which is a valid slip direction in FCC.
Step 3: Identify compatible planes.
For slip to occur, $\vec{b}$ must lie in the plane. That means the Burgers vector direction should be perpendicular to the plane’s normal vector. - For $(111)$ plane: normal $= [111]$ \[ [110] . [111] = 1+1+0 = 2 \neq 0 \ \Rightarrow [110] \in (111) \ \text{plane} \] - For $(1\bar{1}1)$ plane: normal $= [1\ -1\ 1]$ \[ [110] . [1\ -1\ 1] = 1-1+0 = 0 \ \Rightarrow [110] \in (1\bar{1}1) \] - For $(\bar{1}11)$ plane: normal $= [-1\ 1\ 1]$ \[ [110] . [-1\ 1\ 1] = -1+1+0 = 0 \ \Rightarrow [110] \in (\bar{1}11) \] - For $(11\bar{1})$ plane: normal $= [1\ 1\ -1]$ \[ [110] . [1\ 1\ -1] = 1+1+0 = 2 \neq 0 \ \Rightarrow [110] \notin (11\bar{1}) \]
Step 4: Conclusion.
The Burgers vector $\frac{a}{2}[110]$ lies in $(111)$, $(1\bar{1}1)$, and $(\bar{1}11)$ planes, but not in $(11\bar{1})$. Final Answer: \[ \boxed{(A), (B), (C)} \]