The diameter of a single-walled carbon nanotube is given by d = a√(n² + nm + m²)/π, using the graphene lattice constant a = 0.246 nm. Which value is closest to the diameter of a nanotube with chiral indices (10, 10)?

Answer: B. About 1.36 nm

n² + nm + m² = 100 + 100 + 100 = 300, so √300 ≈ 17.32. The length of the chiral vector (the circumference) is 0.246 × 17.32 ≈ 4.26 nm, and dividing by π gives a diameter d ≈ 1.36 nm. About 4.26 nm forgets to divide by π and takes the circumference as the diameter. About 0.78 nm uses the C–C bond length of 0.142 nm instead of the lattice constant a = 0.246 nm, confusing the relation a = √3 × 0.142 nm. (10, 10) is an armchair tube and so is metallic. Most single-walled carbon nanotubes have diameters of about 1–2 nm.