= Solution
For an armchair <single-walled carbon nanotube>, $|\mathbf C_h|=3bn$. Hence
$$
d_n=\frac{0.426n}{\pi}\ {\rm nm},\qquad
\frac{\pi}{0.426}\le n\le\frac{1.7\pi}{0.426}.
$$
The integer bounds give \b[$(n,m)=(8,8),(9,9),(10,10),(11,11),(12,12)$].
\Table[
|| Indices
|| Exact diameter in nm
|| Rounded diameter in nm
| $(8,8)$
| $3.408/\pi$
| $1.08480$
| $(9,9)$
| $3.834/\pi$
| $1.22040$
| $(10,10)$
| $4.260/\pi$
| $1.35600$
| $(11,11)$
| $4.686/\pi$
| $1.49160$
| $(12,12)$
| $5.112/\pi$
| $1.62720$
]
{title=Armchair nanotube indices and atom-center diameters}
The neighboring values $d_7\simeq0.9492\ {\rm nm}$ and $d_{13}\simeq1.7628\ {\rm nm}$ lie outside the interval. “Exact” here means exact within the stipulated cylindrical model and input bond length; the decimals are rounded. No van der Waals wall thickness is added because the diameter is measured between atomic centers.
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