Question

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

• A. \(\dfrac{a}{2}[0\bar{1}1] \rightarrow \dfrac{a}{6}[1\bar{2}1] + \dfrac{a}{6}[\bar{1}\bar{1}2]\)
• B. \(\dfrac{a}{2}[0\bar{1}1] \rightarrow \dfrac{a}{6}[112] + \dfrac{a}{6}[21\bar{1}]\)
• C. \(\dfrac{a}{2}[0\bar{1}1] \rightarrow \dfrac{a}{6}[1\bar{1}2] + \dfrac{a}{6}[\bar{1}\bar{2}\bar{1}]\)
• D. \(\dfrac{a}{2}[0\bar{1}1] \rightarrow \dfrac{a}{6}[1\bar{2}1] + \dfrac{a}{6}[2\bar{1}\bar{1}]\)

Hint:

Check which option's two partial Burgers vectors add up, component by component, to the original vector before checking energy.

Answer 1

The Correct Option is A

Step 1: Recall why perfect dislocations dissociate in FCC metals.
In an FCC crystal, a perfect dislocation glides on \(\{111\}\) planes with a Burgers vector of type \(\frac{a}{2}\langle110\rangle\). Because the strain energy of a dislocation scales as \(E \propto b^2\) (Frank's rule), it is energetically favorable for this perfect dislocation to split into two Shockley partials, each of type \(\frac{a}{6}\langle211\rangle\), separated by a stacking fault:
\[ \frac{a}{2}\langle110\rangle \rightarrow \frac{a}{6}\langle211\rangle + \frac{a}{6}\langle211\rangle \]

Step 2: State the two conditions a valid reaction must satisfy.
Any proposed dissociation reaction \(b_0 \rightarrow b_1 + b_2\) must satisfy:
(i) Conservation of the Burgers vector: \(b_0 = b_1 + b_2\), added component by component, since the crystal displacement far from the dislocation core must stay unchanged.
(ii) Frank's energy criterion: \(b_1^2 + b_2^2 < b_0^2\), so that the split genuinely lowers the elastic energy and is favorable.

Step 3: Test option (A) for vector conservation.
\[ \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2] = \frac{a}{6}\big[(1-1),\,(-2-1),\,(1+2)\big] = \frac{a}{6}[0\,\bar{3}\,3] = \frac{a}{2}[0\bar{1}1] \]
This equals the left-hand side exactly, so option (A) conserves the Burgers vector.

Step 4: Test option (A) for the energy criterion.
\[ b_0^2 = \left(\frac{a}{2}\right)^2\left(0^2+1^2+1^2\right) = \frac{a^2}{2} \]
\[ b_1^2 = \left(\frac{a}{6}\right)^2\left(1^2+2^2+1^2\right) = \frac{a^2}{6}, \qquad b_2^2 = \left(\frac{a}{6}\right)^2\left(1^2+1^2+2^2\right) = \frac{a^2}{6} \]
\[ b_1^2+b_2^2 = \frac{a^2}{3} < \frac{a^2}{2} = b_0^2 \]
The split lowers the energy, so option (A) is both vector conserving and energetically favorable, meaning it is a feasible reaction.

Step 5: Test option (B).
\[ \frac{a}{6}[112] + \frac{a}{6}[21\bar{1}] = \frac{a}{6}[3\,2\,1] \]
This does not equal \(\frac{a}{2}[0\bar{1}1] = \frac{a}{6}[0\,\bar{3}\,3]\), since the components \((3,2,1)\) do not match \((0,-3,3)\). The Burgers vector is not conserved, so option (B) is not feasible.

Step 6: Test option (C).
\[ \frac{a}{6}[1\bar{1}2] + \frac{a}{6}[\bar{1}\bar{2}\bar{1}] = \frac{a}{6}[0\,\bar{3}\,1] \]
The third component, \(1\), does not match the required \(3\), so the vector sum does not equal \(\frac{a}{2}[0\bar{1}1]\). Option (C) is not feasible.

Step 7: Test option (D).
\[ \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[2\bar{1}\bar{1}] = \frac{a}{6}[3\,\bar{3}\,0] \]
This does not equal \(\frac{a}{6}[0\,\bar{3}\,3]\) either, since the first and third components do not match. Option (D) is not feasible.

Step 8: Final Answer.
Only option (A) conserves the Burgers vector and satisfies Frank's energy rule, so it is the physically feasible Shockley partial dissociation reaction.
\[ \boxed{\frac{a}{2}[0\bar{1}1] \rightarrow \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2]} \]