Step 1: Understanding the Concept.
When a perfect dislocation in an FCC metal splits into two partials, the reaction has to obey two physical rules at the same time: the Burgers vectors must add up correctly (nothing can be created or destroyed at the split point), and the total elastic energy, which goes as $b^2$, must go down after the split, otherwise the crystal would never bother splitting the dislocation.
Step 2: Key Approach.
Instead of checking energy for all four options, it is faster to first check which option even adds up correctly as vectors. A reaction that fails vector addition is not physically possible regardless of energy, so it can be dropped right away. Only the surviving option needs the energy check.
Step 3: Detailed Explanation.
Write the target vector with a common denominator of $6$: $\frac{a}{2}[0\bar{1}1] = \frac{a}{6}[0\,\bar{3}\,3]$. Now add up each option's two partials component by component.
Option (A): $[1\bar{2}1]+[\bar{1}\bar{1}2] = [0\,\bar{3}\,3]$. This matches the target exactly.
Option (B): $[112]+[21\bar{1}] = [3\,2\,1]$. This does not match $[0\,\bar{3}\,3]$.
Option (C): $[1\bar{1}2]+[\bar{1}\bar{2}\bar{1}] = [0\,\bar{3}\,1]$. The last entry, $1$, should have been $3$, so this fails too.
Option (D): $[1\bar{2}1]+[2\bar{1}\bar{1}] = [3\,\bar{3}\,0]$. This does not match $[0\,\bar{3}\,3]$ either.
Only option (A) survives the vector addition check. As a final check, compare the squared magnitudes: the original vector gives $b_0^2 = \left(\frac{a}{2}\right)^2(2) = \frac{a^2}{2}$, while each partial in option (A) gives $\left(\frac{a}{6}\right)^2(6) = \frac{a^2}{6}$, so together $b_1^2+b_2^2 = \frac{a^2}{3}$, which is smaller than $\frac{a^2}{2}$. The split is energetically favorable, confirming option (A) is the real Shockley partial reaction.
Step 4: Final Answer.
$\frac{a}{2}[0\bar{1}1] \rightarrow \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2]$ is the only reaction where the Burgers vectors add up correctly and the energy drops after the split.
\[ \boxed{\frac{a}{2}[0\bar{1}1] \rightarrow \frac{a}{6}[1\bar{2}1] + \frac{a}{6}[\bar{1}\bar{1}2]} \]
