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Crystal defects and diffusion

The Burgers vector of a perfect dislocation in an FCC metal is (a/2)<110>. Which value is closest to its magnitude in aluminum (lattice parameter 0.405 nm)?

  1. AAbout 0.405 nm
  2. BAbout 0.351 nm
  3. CAbout 0.286 nm
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Answer: C. About 0.286 nm

The magnitude of (a/2)<110> is (a/2) × √(1² + 1² + 0²) = a/√2, so 0.405/1.414 ≈ 0.286 nm. This equals the nearest-neighbor distance in FCC; dislocations take the shortest lattice translation vector along the close-packed direction as their Burgers vector. About 0.405 nm is the lattice parameter itself (a translation along <100>), which is unfavorable because the dislocation energy is proportional to b². About 0.351 nm applies the BCC (a/2)<111>, that is, √3a/2. The elastic energy per unit length of a dislocation is about Gb²/2, so dislocations with smaller b are more stable.

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