The primitive translational vectors of the hexagonal spacelattice may be taken a

Question

The primitive translational vectors of the hexagonal spacelattice may be taken as a1=(31/2a/2)x+(a/2)y a2=-(31/2a/2)x+(a/2)y a3=cz (a) Show that the volume of the primitive cell is(31/2/2)a2c (b) Show that the primitive translations of the reciprocallattice are b1= (2/3 1/2a)x+(2/a)y b2=-(2/3 1/2a)x+(2/a)y b3=(2/c)z so that the lattice is its own reciprocal, but with a rotationof axes. (c) Describe and sketch the first Brillouin zone of thehexagonal space lattice. The primitive translational vectors of the hexagonal spacelattice may be taken as a1=(31/2a/2)x+(a/2)y a2=-(31/2a/2)x+(a/2)y a3=cz (a) Show that the volume of the primitive cell is(31/2/2)a2c (b) Show that the primitive translations of the reciprocallattice are b1= (2/3 1/2a)x+(2/a)y b2=-(2/3 1/2a)x+(2/a)y b3=(2/c)z so that the lattice is its own reciprocal, but with a rotationof axes. (c) Describe and sketch the first Brillouin zone of thehexagonal space lattice.

Explanation / Answer

Taking surfaces at the same distance fromone element of the lattice and its neighbours, the volume included is thefirst Brillouin zone. Another definition is as the set of points ink-space that can be reached from the origin withoutcrossing any Bragg plane.Equivalently, this is the Voronoi cell around theorigin of the reciprocal lattice.

There are also second, third,etc., Brillouin zones, corresponding to a sequence ofdisjoint regions (all with the same volume) at increasing distancesfrom the origin, but these are used more rarely. As a result, thefirst Brillouin zone is often called simply theBrillouin zone. (In general, the n-th Brillouinzone consist of the set of points that can be reached from theorigin by crossing n 1 Bragg planes.)

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