Question:medium

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

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Compare the Shockley partial's Burgers vector length to a full lattice translation vector, and check if it lies inside the glide plane.
Updated On: Jul 28, 2026
  • Perfect and mobile dislocation
  • Perfect and immobile dislocation
  • Imperfect and immobile dislocation
  • Imperfect and mobile dislocation
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept.
A dislocation in FCC metals is called perfect when its Burgers vector connects two atoms that occupy equivalent positions in the lattice, and partial (imperfect) when it does not. Whether a dislocation can glide or must climb depends on whether its Burgers vector lies inside the plane it sits on.

Step 2: Key Approach.
Two kinds of partial dislocations show up in FCC crystals. The Shockley partial comes from splitting a perfect glide dislocation, and the Frank partial comes from inserting or taking out a partial plane of atoms. Comparing these two tells us which box the Shockley partial fits into.

Step 3: Detailed Explanation.
A perfect FCC dislocation carries Burgers vector $\frac{a}{2}\langle110\rangle$. Splitting it, for instance
\[ \frac{a}{2}[0\bar{1}1] \rightarrow \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2], \]
gives two Shockley partials of type $\frac{a}{6}\langle211\rangle$. Since $\frac{a}{6}\langle211\rangle$ does not join two atoms sitting on equivalent lattice sites, each Shockley partial is a partial, or imperfect, dislocation, and a stacking fault ribbon sits between the pair.
Now check where this Burgers vector points. The $\{111\}$ plane on which the original dislocation glided has its normal along $\langle111\rangle$. Taking the dot product of $\frac{a}{6}[1\bar{2}1]$ with $[111]$ gives $1-2+1=0$, so the vector lies flat inside the $\{111\}$ plane, with no out of plane component. A Burgers vector confined to its own glide plane means the dislocation can move by slip, so it is glissile, that is, mobile.
This is unlike the Frank partial, whose Burgers vector points along $\langle111\rangle$, straight out of the fault plane. That out of plane vector cannot be swept through by shear, so a Frank partial only moves by climb (diffusion of atoms), making it sessile.

Step 4: Final Answer.
The Shockley partial is imperfect because its Burgers vector is a fraction of a lattice vector, and mobile because that vector lies inside the glide plane.
\[ \boxed{\text{Imperfect and mobile dislocation}} \]
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