Dissociation of a dislocation into two partials in an FCC metal is given by the following equation. $\frac{a}{2}[1 \bar 10] \rightarrow \frac{a}{6}[2 \bar 11] + \frac{a}{6}[1 \bar 2 \bar 1]$ On which plane do these two partial dislocations lie?
$(11\bar{1})$
The question involves the dissociation of dislocations in Face-Centered Cubic (FCC) metals. The primary equation provided is:
\(\frac{a}{2}[1 \bar{1} 0] \rightarrow \frac{a}{6}[2 \bar{1} 1] + \frac{a}{6}[1 \bar{2} \bar{1}]\)
We need to determine on which plane these two partial dislocations lie.
Step-by-Step Solution:
Understanding FCC and Dislocation Dissociation: In FCC metals, dislocations under stress can split into two partial dislocations. The dissociation involves movement on specific crystallographic planes, often the {111} family of planes, which are closed-packed planes in FCC structures.
Vector Representation: The Burgers vector involved in the question is converted into partial dislocations:
- Initial dislocation: \(\frac{a}{2}[1 \bar{1} 0]\)
- Partial dislocations: \(\frac{a}{6}[2 \bar{1} 1]\) and \(\frac{a}{6}[1 \bar{2} \bar{1}]\)
Plane Selection: The partial dislocations must lie on one of the {111} planes, which are the most densely packed planes in FCC structures. To find the specific plane, we add the indices of the partial dislocations:
- Adding the two partial Burgers vectors, \(\frac{a}{6}[2 \bar{1} 1] + \frac{a}{6}[1 \bar{2} \bar{1}] = \frac{a}{2}[1 \bar{1} 0]\), which confirms the original Burgers vector.
- The resultant indices \([1 \bar{1} 0]\) show the glide plane must be perpendicular to this vector. In FCC, this aligns with the {111} plane vectors.
Conclusion: The dislocation dissociation of the question fits best with the plane \((11\bar{1})\) based on geometric compatibility with the partials' resultant vector.
Thus, the correct answer is: \((11\bar{1})\).
Important Questions from Defects Dislocation Stress Field Burgers Vector
- Which one of the following dislocation dissociation reactions is feasible in face-centered cubic metals?
- With reference to edge and screw dislocations, which of the following statements is/are CORRECT?
The Burger's vector of a dislocation in a cubic crystal (with lattice parameter a) is $\frac{a}{2}[110]$ and dislocation line is along $[112]$ direction. The angle (in degrees) between the dislocation line and its Burger's vector is _________
- A plastically deformed metal crystal at low temperature exhibits wavy slip line pattern due to
- The c/a ratio of Zn (hcp) is 1.856. Slip at room temperature occurs most easily on which of the following slip systems in Zn:
Note: In hcp metals, the ideal c/a ratio is 1.633.
