# 2.9: Worked Examples

- Page ID
- 35351

## Example A

The figure below is a scanning electron micrograph of a niobium carbide dendrite in a Fe-34wt%Cr-5wt%Nb-4.5wt%C alloy. Niobium carbide has a face centred cubic lattice. The specimen has been deep-etched to remove the surrounding matrix chemically and reveal the dendrite. The dendrite has 3 sets of "arms" which are orthogonal to one another (one set pointing out of the plane of the image, the other two sets, to a good approximation, lying in the plane of the image), and each arm has a pyramidal shape at its end. It is known that the crystallographic directions along the dendrite arms correspond to the < 100 > lattice directions, and that the direction **ab ** labelled on the micrograph is [10 ].

1) If point **c ** (not shown) lies on the axis of this dendrite arm, what is the direction **cb **? Index face C , marked on the micrograph.

The diagram shows the [10**ab**, and can be used as **cb**. Of the < 100 > type directions, we could also have used [00 ].

Using a right handed set of axes, we then have z-axis pointing out of the plane of the image, the x-axis pointing along the direction **cb**, and the y-axis pointing towards the top left of the image.

Face C must contain the direction **cb**, and its normal must point out of the plane of the image. Therefore face C is a (001) plane.

2) The four faces which lie at the end of each dendrite arm have normals which all make the same angle with the direction of the arm. Observing that faces A and B marked on the micrograph both contain the direction **ab **, and noting the general directions along which the normals to these faces point, index faces A and B .

Both faces A and B have normals pointing in the positive x and z directions, i.e. positive h and l indices. Face A has a positive k index, and face B has a negative k index.

The morphology of the ends of the arms is that of half an octahedron, suggesting that the faces are (111) type planes. This would make face A, in green, a (111) plane, and face B, in blue, a (1

1) plane. As required, they both contain the [10 ] direction, in red.## Example B

1) Work out the common direction between the (111) and (001) in a triclinic unit cell.

The relation derived from the Weiss zone law in the section Vectors and planes states that:

The direction, [*UVW*], of the intersection of (*h*_{1}*k*_{1}*l*_{1}) and (*h*_{2}*k*_{2}*l*_{2}) is given by:

\[U = k_{1}l_{2} − k_{2}l_{1}\]

\[V = l_{1}h_{2} − l_{2}h_{1}\]

\[W = h_{1}k_{2} − h_{2}k_{1}\]

We can use this relation as it applies to all crystal systems, including the triclinic system that we are considering.

We have *h*_{1} = 1, *k*_{1} = 1, *l*_{1} = 1

and *h*_{2} = 0, *k*_{2} = 0, *l*_{2} = 1

Therefore

\[U = (1 × 1) - (0 × 1) = 1\]

\[V = (1 × 0) - (1 × 1) = −1\]

\[W = (1 × 0) - (0 × 1) = 0\]

So the common direction is:

[1

0].This is shown in the image below:

If we had defined the (001) plane as (*h*_{1}*k*_{1}*l*_{1}) and the (110) plane as (*h*_{2}*k*_{2}*l*_{2}) then the resulting direction would have been, [ 10] i.e. anti-parallel to [1 0].

2) Use the Weiss zone law to show that the direction [1

0] lies in the (111) plane.We have *U = *1, *V = *−1, *W = *0,

and *h* = 1, *k* = 1, *l* = 1.

\[hU + kV + lW = (1 × 1) + (1 × −1) + (1 × 0) = 0\]

Therefore the direction [1

0] lies in the plane (111).