Phys 315 Lattice Vectors 2012 Homework Solutions - Solutions for problem set 3 1(i The crystal looks like sketched below Suppose we choose one of the Sr

Phys 315 Lattice Vectors 2012 Homework Solutions -...

This preview shows page 1 - 3 out of 6 pages.

Solutions for problem set 3 1. (i) The crystal looks like sketched below. Suppose we choose one of the Sr sites as a latticesite. Since all Sr are in identical environments, all their sites are part of the lattice. The other sitescontain different atoms so they can’t be part of the lattice. It follows that the lattice is simple cubic,with lattice constanta.The basis, then, contains one Sr, one Ti (at the center of the cube) and 3 O on 3 of the mutuallyperpendicular faces. One possible choice is shown below.(ii) The most symmetric choice for lattice vectors isvectora1=aˆx,vectora2=aˆy,vectora3=aˆz, as indicated above.(iii) First, we need to find the reciprocal lattice vectors. Since they satisfyvectorai·vectorbj= 2πδij, andsince the lattice vectorsvectoraiare mutually perpendicular, it follows that the reciprocal lattice vectorsare also mutually perpendicular and of equal magnitude, i.e.vectorb1=2πaˆx,vectorb2=2πaˆy,vectorb3=2πaˆz. (If thisis not obvious, you can always usevectorb1= 2πvectora2×vectora3vectora1·(vectora2×vectora3), etc., to get to the answer).They also span a cubic lattice. Since the Brillouin zone is the Wigner-Seitz unit cell in thisreciprocal lattice, it will also be cubic, spanning all valueskx(πa,πa],ky(πa,πa],kz(πa,πa].2. (i) The formula is A3B2. There are various ways to figure this out, for example if you removehalf the B’s, each remaining B will be in its own “cage” of 3 A. Another way to answer this is to firstanswer (ii) to find the basis, and that tells you the formula (also see the problem above, the name ofthe crystal gave away its basis. This is usually – but not always! – the case).(ii) Let’s choose one of A’s that sits at the common corner of a “up” and a “down” triangle asbelonging to the lattice (these are shown by open circles, below). All other A’s that sit in similarlocations have similar environments, so they all belong to the lattice. The other A’s (from the basesof these triangles) are in different environments so they are not in the lattice.The lattice, then, is triangular with side 2a. A possible choice for the lattice constants is shownbelow, and hasvectora1= 2aˆx,vectora2=ax+y). 1
Image of page 1
Image of page 2
Image of page 3

You've reached the end of your free preview.

Want to read all 6 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture