Question

Which one of the following options is correct?
In a face-centered cubic metal, Shockley partial is:

• A. Perfect and mobile dislocation
• B. Perfect and immobile dislocation
• C. Imperfect and immobile dislocation
• D. Imperfect and mobile dislocation

Hint:

Compare the Shockley partial's Burgers vector length to a full lattice translation vector, and check if it lies inside the glide plane.

Answer 1

The Correct Option is D

Step 1: Recall how a perfect dislocation is defined.
In a face-centered cubic (FCC) metal, a perfect dislocation has a Burgers vector equal to a full lattice translation vector, of the type \(\frac{a}{2}\langle110\rangle\). Moving a perfect dislocation across a plane restores the crystal to an identical, translationally equivalent structure, with no change in stacking order.

Step 2: Understand why the perfect dislocation splits.
The elastic strain energy of a dislocation is proportional to the square of its Burgers vector, \(E \propto b^2\) (Frank's rule). A perfect dislocation \(\frac{a}{2}[0\bar{1}1]\) can lower its total energy by splitting into two smaller partial dislocations that together still connect the same two lattice points:
\[ \frac{a}{2}[0\bar{1}1] \rightarrow \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2] \]
These two partials are called Shockley partials.

Step 3: Check why the Shockley partial is imperfect.
Each Shockley partial has Burgers vector \(\frac{a}{6}\langle211\rangle\), which is not a lattice translation vector of the FCC structure. Moving only one partial across the slip plane shifts the atoms into a wrong stacking position, so a thin ribbon of stacking fault is left between the two partials. Because the passage of a Shockley partial alone does not restore the perfect crystal, it is classified as an imperfect (partial) dislocation, not a perfect one. This rules out options (A) and (B).

Step 4: Check whether the Shockley partial can move by slip.
The Shockley partial's Burgers vector \(\frac{a}{6}\langle211\rangle\) lies entirely within the \(\{111\}\) glide plane on which it was created, since the vector has no component perpendicular to that plane. A dislocation whose Burgers vector lies in its own glide plane is glissile, meaning it can move by ordinary slip under an applied shear stress, just like a perfect dislocation. This is different from a Frank partial, formed by inserting or removing a plane of atoms, whose Burgers vector is perpendicular to the fault plane; a Frank partial cannot glide and can only move by climb, so it is sessile (immobile).

Step 5: Eliminate the remaining option.
Option (C) calls the Shockley partial imperfect and immobile. It is correctly imperfect, but it is not immobile, since it glides freely on the \(\{111\}\) plane. That description fits a Frank partial, not a Shockley partial, so option (C) is incorrect.

Step 6: Final Answer.
The Shockley partial in an FCC metal is an imperfect dislocation, because its Burgers vector is not a full lattice translation, but it is mobile, because that Burgers vector lies in the glide plane and allows slip.
\[ \boxed{\text{Imperfect and mobile dislocation}} \]