band line corresponds toky = 0and it touches the band above at a certainkxpoint. If we increase
kya little bit what happens is that we miss this point where the two bands meet and then we get
something that looks like a semiconductor. But this (10,10) armchair tube is metallic because we went
through the points where the two solutions meet.
From the dispersion relationship we can calculate the density of states. For a metallic tube the disper-
sion relationship is constant around the fermi energy and that gives a constant density of states like in
fig. 43. The next allowed value ofkyhas a maximum and that results in a Van Hove singularity in the
Figure 43: Dispersion relationship and density of states for both a metallic and a semiconducting
carbon nanotube.
density of states (because there are a lot of states in this energy range). If we have a semiconducting
nanotube (as seen on the right in fig. 43) the bands do not meet and the density of states is zero,
there is a gap. But we also have this Van Hove singularity on both sides because of the maxima and
the minima in the bands.
The formulam−n= 3Z (for a(m,n)-nanotube) says that if m− 3 n is an integer the tube will be
metallic, otherwise semiconducting (for example a (10,0) nanotube: 10 −0 = 10can not be divided
by three so this should be a semiconducting tube).
Other Nanotubes
Carbon is not the only material which makes tubes, others are:
- Tungsten disulfide
- Boron nitride
- Silicon
- Titanium dioxide
- Molybdenum disulfide
Boron nitride for example has the same bond lengths in all directions and so all the directions are the
same, therefore we get get the same phase factors and band structures with the tight binding model
as for graphene (as we see in fig. 44). But because there are two different types of atoms per unit