Question 15.6: Dislocations in MgO A sketch of a dislocation in magnesium o......
Dislocations in MgO
A sketch of a dislocation in magnesium oxide (MgO), which has the sodium chloride crystal structure and a lattice parameter of 0.396 nm, is shown in Figure 15-4. Note that the slip system for MgO is {{\overline{1}10}}<110>. Determine the length of the Burgers vector.
In Figure 15-4, we begin a clockwise loop around the dislocation at point x and then move equal atom spacings in the two horizontal directions and equal atom spacings in the two vertical directions to finish at point y. Note that it is necessary that the lengths of the two horizontal segments of the loop be equal and the lengths of the vertical segments be equal, but it is not necessary that the horizontal and vertical segments be equal in length to each other. The chosen loop must close in a perfect crystal. The vector b is the Burgers vector. Because b is parallel to a [110] direction, it must be perpendicular to (110) planes. The length of b is the distance between two adjacent (110) planes. From Equation 3-7,
sin θ = \frac{λ}{2d_{hkl}} (3-7)
d_{110} = \frac{a_{0}}{\sqrt{h^2 \ + \ k^2 \ + \ l^2}} = \frac{0.396 \ nm}{\sqrt{1^2 \ + \ 1^2 \ + \ 0^2}}=0.280 \ nm
The Burgers vector is a (110) direction that is 0.280 nm in length. Note, however, that two extra half planes of atoms make up the dislocation—one composed of magnesium ions and one of oxygen ions (Figure 15-4). This formula for calculating the magnitude of the Burgers vector becomes more complicated for non-cubic systems. It is better to consider the magnitude of the Burgers vector as equal to the repeat distance in the slip direction.