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Ch.12 - Solids and Modern Materials

Chapter 12, Problem 30a

The unit cell of a compound containing potassium, aluminum, and fluorine is shown here. (a) What type of lattice does this crystal possess (all three lattice vectors are mutually perpendicular)?

Diagram of a simple cubic unit cell with dimensions and angles labeled.

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Video transcript

Welcome back everyone. We need to consider the unit cell of a compound containing potassium, nitrogen and oxygen and identify the type of lattice structure. So we're going to define three different labels for the faces of our unit. Sell our first face, we can call a which has the value of 9.7-6 angstrom. We represent face a. By the front and back faces of our unit cell here. Now we move on to face B. Of our unit cell which we would recognize as the These two faces that I'm outlining the side faces of our unit cell which have a measurement of 4.562 extremes. And then we have faced sea of our unit cell which has a measurement of 7.267 Angstrom defining the top and bottom faces of our unit cell. We are also given angles where we would recognize that our angle for alpha corresponds to the degree of 70.498° where alpha is defining our side. Or sorry, our face of A. Which is our front and back of our unit cell. As we stated, Having an angle of 70.498 whereas were given to other measurements for angles beta and gamma, which have a measurement of 90° here. So we would see that we can confirm from this key here that we have two angles that are equal to the same degree. We can also confirm based on what we've outlined that A B. And see I'm sorry this is an A. Here. So A B and C we would determine are not equal because they all have different measures in Angstrom, as we labeled in our diagram Now because we have two angles that are equal to 90° being our beta and gamma angles. We would recognize or recall from our notes that this corresponds to a hexagonal or Mono clinic lattice. But we need to pick one and sorry, this is mono clinic here. So we need to pick between these two Where we want to recognize that our last angle, as we stated is the 70.498° defined by alpha. And because we see that 70.498 Is not 1:20 And it's not 1 or sorry, not 90. We would then be able to rule out the fact that the lattice could be hexagonal. And so therefore we would say that our lattice structure is Mono Clinic. So for our final answer, we have been successfully able to identify the type of lattice structure which is Mono Clinic. So this is our final answer. I hope everything I explained was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video