\question
{Shoichi Takahata}
{02/03/12}
{physics2070@yahoo.co.jp}
{2-15}
%question
{
Calculate the diameter for an ({\it n,m}) SWNT.
What are the values of the diameters for (5,5), (9,0), (10,5) SWNTs?
What are possible ({\it n,m}) values for SWNTs having a $1.5\pm0.02\,{\rm nm}$ diameter?
}
%answer
{
A.Calculate the diameter for an ({\it n,m}) SWNT.

B.What are the values of diameters for (5,5), (9,0), (10,5) SWNTs?

C.What are possible ({\it n,m}) values for SWNTs having a $1.5\pm0.02\,{\rm nm}$?


\vspace{1zw}\textbf{The parameters}

$a=0.246\,{\rm nm}$\,:\,Graphene lattice constant

${\bf a_1}=\Big( \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}},
\frac{{\displaystyle 1}}{{\displaystyle 2}} \Big) a$, 
${\bf a_2}=\Big( \frac{{\displaystyle \sqrt{3}}}{{\displaystyle 2}},
-\frac{{\displaystyle 1}}{{\displaystyle 2}} \Big) a$ : Graphene unit vectors

({\it n,m})\,:\,integers such that $0 \le m \le n$

${\mbox{\boldmath $C_h$}}=n{\bf a_1}+m{\bf a_2}$ : Nanotube chiral vector

\vspace{1zw}\textbf{The ansewer for A}

The diameter of a nanotube $d_{\rm t}$ 
can be calculated by $|{\mbox{\boldmath {$C_h$}}}|/\pi$ 
whose formula is given as follows:

\begin{equation}
d_{\rm t}
=\frac{a\sqrt{n^2+nm+m^2}}{\pi} 
\hspace{1zw} {\rm units:nm} \label{eq:d_t} \,.
\end{equation}

\vspace{1zw}\textbf{The answer for B}

Substituting $a=0.246\,{\rm nm}$ and each ({\it n,m}) value 
into Eq.\,(\ref{eq:d_t}), 
we can obtain the corresponding values of $d_{\rm t}$ 
as follows:

\begin{equation}
d_t=
\left\{\begin{array}{rl}
0.678\,{\rm nm} & (n,m)=(5,5)  \\
0.705\,{\rm nm} & (n,m)=(9,0)  \,\,\,\,\,.\\
1.04\,{\rm nm}  & (n,m)=(10,5) 
\end{array}
\right.
\end{equation}

\vspace{1zw}\textbf{The answer for C}

From Eq.\,(\ref{eq:d_t}) and 
the condition that $1.48 \le d_{\rm t} \le 1.52$, 
we can obtain the following inequality:

\begin{equation}
\Bigl(\frac{1.48\times \pi}{0.246}\Bigr)^2 
\le n^2+nm+m^2 \le 
\Bigl(\frac{1.52\times \pi}{0.246}\Bigr)^2 \, .
\end{equation}


Since $\{(1.48\times \pi)/0.246\}^2=357.2$, 
$\{(1.52\times \pi)/0.246\}^2=376.8$ 
and $n^2+nm+m^2$ is integer, 
({\it n,m}) values for SWNTs having a $1.5\pm0.02\,{\rm nm}$ diameter 
should satisfy the condition as follows:

\begin{equation}
358 \le n^2+nm+m^2 \le 376 \,\,\,\,{\rm and} \,\,\,\, 0 \le m \le n \,, 
\end{equation}

by which we can obtain the proper ({\it n,m}) values as follows:

\begin{equation}
(n,m) = (19,0),\,(18,2),\, (17,4),\, (16,5),\, (14,8),\, (13,9),\, (12,10),\, (11,11) \, .
\end{equation}
}