How Do You Determine Burger Vector Orientations in HCP Metals?

[somebodyelse5]

Homework Statement

sketch atomic arrangement, label coordinate axes, determine miller indices of the plane and the burger orientations in the slip plane of an HCP metal.

Homework Equations

no equations neeeded

The Attempt at a Solution

I know the atomic arrangement, a close packed hexagon on top and bottom and a triangle in the middle.

Coordinate axis are simple, a1, a2, a3, c

Miller indices of slip plane, (0 0 0 1) this is the close packed plane (it has the highest linear density) and it does not intersect any of the a axes

I am having trouble with the orientation of the burger vectors. My book gives that there are 3 "slip systems" , the slip plane is (0 0 0 1) and the slip direction is <1 1 -2 0>

My issue is, how do I find and draw the Burger vector orientations?

 

[KataKoniK]

To determine the Burger vector orientations, we can use the following formula:

B = n1a1 + n2a2 + n3a3

where n1, n2, and n3 are integers that represent the number of unit cell translations in each direction, and a1, a2, and a3 are the lattice vectors.

In the case of an HCP metal, the a1 and a2 lattice vectors are in the basal plane, and the a3 lattice vector is perpendicular to the basal plane. Therefore, the Burger vector orientations will also be in the basal plane.

To find the Burger vector orientations for the given slip direction <1 1 -2 0>, we can first find the magnitude of the slip direction:

|<1 1 -2 0>| = sqrt(1^2 + 1^2 + (-2)^2 + 0^2) = sqrt(6)

Next, we can use the formula for the Burger vector to find the different orientations that correspond to this slip direction:

B = n1a1 + n2a2 + n3a3

We can set n1 = n2 = 1 and n3 = -2 to represent the slip direction <1 1 -2 0>.

So, B = a1 + a2 - 2a3

This means that there are three possible Burger vector orientations for this slip direction, which are:

B1 = a1 + a2 - 2a3
B2 = -a1 + a2 - 2a3
B3 = a1 - a2 - 2a3

We can sketch these three orientations on the slip plane (0 0 0 1) to visualize them.

Furthermore, since there are three slip systems in an HCP metal, there will also be three different slip planes (0 0 0 1) for each system. This means that there will be a total of nine possible Burger vector orientations in an HCP metal.

In summary, the Burger vector orientations for the given slip direction <1 1 -2 0> in an HCP metal are:

B1 = a1 + a2 - 2a3
B2 = -a1 + a2 - 2a3
B3 = a1 - a2 - 2a3

We can label these