Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2020 / Paper 2 / Section A(i) / 13 / i / Solution

The carbon nanotube chiral vector identifies equivalent sites of graphene that become one point on rolling:The primitive vectors meet at $60^{\circ }$, $b$ is the C–C bond length, and one can choose $0\le m\le n$. The carbon nanotube chiral angle satisfies $\tan \theta =\sqrt{3}m/(2n+m)$. Armchair tubes have $(n,n)$ and $\theta =30^{\circ }$; zigzag tubes have $(n,0)$ and $\theta =0$. Other pairs are chiral, with mirror-related left- and right-handed forms. Armchair and zigzag tubes themselves have mirror symmetry.

The circumferential boundary condition requires $\mathbf{k}\cdot {\mathbf{C}}_h=2\pi \ell$. In nanotube zone folding, an allowed wavevector line intersects one of the Dirac valleys in graphene precisely when $n-m$ is divisible by three. Thus all armchair tubes are metallic in this approximation. Zigzag tubes are metallic for $n$ divisible by three and otherwise semiconducting; general chiral tubes follow the same modular rule. Curvature can open small gaps in nominally metallic non-armchair tubes. Strain, interactions and broken symmetries can modify the ideal result; metallic band structure does not imply zero device resistance.