Discussion
A single crystal of copper contains a low angle tilt boundary on (001) plane with a tilt axis parallel to [010]. The tilt angle, if the spacing of the dislocation in the boundary is 3 × 10-6 m and their Burgers vector is 0.4 × 10-9 m, is :
Answer: D
The tilt angle is \(1.33 \times 10^{-4}\) rad.
A low-angle tilt boundary is a wall of edge dislocations. If their Burgers vector is b and their spacing D, the misorientation across the boundary is, for small angles,
\[ \begin{aligned} \theta &\approx \frac{b}{D} = \frac{0.4 \times 10^{-9}}{3 \times 10^{-6}} \\ &= 1.33 \times 10^{-4}\ \text{rad} \end{aligned} \]
The crystallographic planes given, (001) and [010], describe the geometry of the boundary but are not needed for the angle. The other values do not follow from \(b/D\); for example \(2.00 \times 10^{-4}\) would need a spacing of 2 \(\mu\)m.
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