The Burgers vector of a mixed dislocation line is \(1 / 2[\overline{1} 10]\). The dislocation line lies along the [ \(\overline{1} 12\) ] direction. Find the slip plane on which this dislocation lies. Find also the screw and the edge components of the Burgers vector.

The concept of the Burgers vector is fundamental when studying dislocations in crystals. It measures the magnitude and direction of the lattice distortion resulting from a dislocation. This vector helps quantify how far and in which direction the atoms have moved to accommodate the dislocation. If you march around a dislocation line along a closed path within the crystal, the Burgers vector is the vector that closes the loop. Without it, it would not connect back to your starting point.

In the given exercise, our Burgers vector is \(\frac{1}{2}[\bar{1} 10]\). It tells us that the displacement in the crystal due to this dislocation is half the distance in the crystallographic direction [\(\bar{1} 10\)].

The slip plane is the plane within the crystal where dislocations occur, and slips can take place, thus maximizing the shear stress. In our particular example, the dislocation line lies along the [\(\bar{1} 12\)] direction. To find the slip plane, we need to determine the plane perpendicular to the dislocation line on which the dislocation can glide. This is achieved by taking the cross product of the Burgers vector \(\frac{1}{2}[\bar{1} 10]\) and the dislocation line direction [\(\bar{1} 12\)].

Performing the cross product, we get:

• \(\frac{1}{2}[\bar{1} 10] \times [\bar{1} 12] = \frac{1}{2}[010 - 2(-110)]\)

This simplifies to:

• \(\frac{1}{2}[-2, 0, -1]\)

The resulting slip plane is thus \(010\). This plane allows the atoms to slip past each other most efficiently.

Among the types of dislocations, screw dislocations are a bit tricky. In these dislocations, the atoms in the crystal move parallel to the dislocation line. One way to visualize this is to think of it as a spiral ramp around a cylinder. In the context of our exercise, the screw component of the Burgers vector is the projection of the Burgers vector onto the direction of the dislocation line.

Using the given Burgers vector \(\frac{1}{2}[\bar{1} 10]\) and dislocation line direction [\(\bar{1} 12\)], we calculate the projection to find the screw component. This part of the Burgers vector represents the displacement that is parallel to the dislocation line.

Edge dislocations represent a different scenario compared to screw dislocations. Here, the distortion of the lattice happens perpendicular to the dislocation line. Imagine the crystal structure has an extra half-plane of atoms, which causes the dislocation. In our exercise, the edge component of the Burgers vector is the portion orthogonal to the dislocation line. We isolate this by subtracting the screw component from the original Burgers vector.

For the Burgers vector \(\frac{1}{2}[\bar{1} 10]\) and the dislocation line direction [\(\bar{1} 12\)], any displacement not parallel to the direction [\(\bar{1} 12\)] gives us the edge component. This part of the vector quantifies the distortion perpendicular to the dislocation line.

Understanding crystallographic directions is essential when dealing with problems in materials science, especially those involving dislocations. These directions provide a way to describe orientations and planes within a crystal lattice using specific notations like Miller indices. For example, [\(\bar{1} 10\)] and [\(\bar{1} 12\)] are such crystallographic directions in a cubic system.

In the given exercise, these directions help us determine the Burgers vector and the dislocation line's orientation. They serve as coordinates to navigate the crystal lattice and are indispensable for calculating vectors like the Burgers vector and identifying slip planes. Understanding these directions makes it easier to solve complex crystal structure problems and to visualize atomic interactions within materials.