All Exams Test series for 1 year @ ₹349 only
Question

Which one of the following dislocation dissociation reactions is feasible in face-centered cubic metals?

The correct answer is
a/2 $[0\bar{1}1] \rightarrow$ a/6 $[1\bar{2}1]$ + a/6 $[\bar{1}\bar{1}2]$

FCC Dislocation Dissociation Feasibility

In face-centered cubic (FCC) metals, dislocation dissociation reactions are governed by the principle of conservation of the Burgers vector. A perfect dislocation can split into two or more partial dislocations if the vector sum of the partials equals the original perfect dislocation vector. This dissociation is often driven by the reduction in strain energy, especially when it leads to the formation of a stacking fault.

Verifying Burgers Vector Conservation

The initial dislocation has a Burgers vector denoted as $ \vec{B}_{\text{initial}} = a/2 [0\bar{1}1] $. We need to check which of the proposed dissociation reactions conserves this vector through the sum of the Burgers vectors of the resulting partial dislocations.

  • Option 1: $ \vec{B}_1 = a/6 [1\bar{2}1] $ and $ \vec{B}_2 = a/6 [\bar{1}\bar{1}2] $

    Sum: $ \vec{B}_1 + \vec{B}_2 = a/6 [1\bar{2}1] + a/6 [\bar{1}\bar{1}2] = a/6 ([1-1], [-2-1], [1+2]) = a/6 [0, -3, 3] $

    Simplifying: $ a/6 [0, -3, 3] = a/2 [0, -1, 1] $. This matches the initial Burgers vector $ a/2 [0\bar{1}1] $. This reaction involves Shockley partials, which are common in FCC metals and dissociate on {111} planes.

  • Option 2: $ \vec{B}_1 = a/6 [112] $ and $ \vec{B}_2 = a/6 [21\bar{1}] $

    Sum: $ \vec{B}_1 + \vec{B}_2 = a/6 [112] + a/6 [21\bar{1}] = a/6 ([1+2], [1+1], [2-1]) = a/6 [3, 2, 1] $. This does not match $ a/2 [0\bar{1}1] $.

  • Option 3: $ \vec{B}_1 = a/6 [1\bar{1}2] $ and $ \vec{B}_2 = a/6 [\bar{1}\bar{2}\bar{1}] $

    Sum: $ \vec{B}_1 + \vec{B}_2 = a/6 [1\bar{1}2] + a/6 [\bar{1}\bar{2}\bar{1}] = a/6 ([1-1], [-1-2], [2-1]) = a/6 [0, -3, 1] $. This does not match $ a/2 [0\bar{1}1] $.

  • Option 4: $ \vec{B}_1 = a/6 [1\bar{2}1] $ and $ \vec{B}_2 = a/6 [2\bar{1}\bar{1}] $

    Sum: $ \vec{B}_1 + \vec{B}_2 = a/6 [1\bar{2}1] + a/6 [2\bar{1}\bar{1}] = a/6 ([1+2], [-2-1], [1-1]) = a/6 [3, -3, 0] = a/2 [1, -1, 0] $. This does not match $ a/2 [0\bar{1}1] $.

Conclusion on Feasibility

Only Option 1 satisfies the conservation of the Burgers vector, meaning the initial dislocation $ a/2 [0\bar{1}1] $ can indeed dissociate into the partial dislocations $ a/6 [1\bar{2}1] $ and $ a/6 [\bar{1}\bar{1}2] $. This type of dissociation into Shockley partials is a characteristic behaviour in FCC metals, leading to the formation of an intrinsic stacking fault.

Was this answer helpful?

Important Questions from Defects Dislocation Stress Field Burgers Vector

  1. With reference to edge and screw dislocations, which of the following statements is/are CORRECT?
  2. 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 _________

  3. A plastically deformed metal crystal at low temperature exhibits wavy slip line pattern due to
  4. 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.
  5. Determine the correctness (or otherwise) of the following Assertion [A] and the Reason [R]
    Assertion [A]: Refractory BCC metals like W and Mo are less ductile than FCC metals like Ni and Pt at room temperature
    Reason [R]: BCC metals have fewer independent slip systems than FCC metals
Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App