Answer to Question #126481 in Physics for BN KHATTAK

Question

calculate the spacing between dislocations in a tilt boundary in fcc copper crystal , when the angle of tilt is 10° (burgers vector = 2.6 a°)

Answer 1

As we know,

$$\text{sin}\frac{\theta}{2}=\frac{b}{2D},\\\space\\ D=\frac{b}{2\text{sin}\frac{\theta}{2}}=\frac{2.6\cdot10^{-10}}{2\text{sin}\frac{10°}{2}}=1.49\cdot10^{-9}\text{ m}.$$

The dislocation is 1.49 nm.