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I'm looking into nanotubes and I thought I'd assure myself that the basic geometry equations are indeed correct. No problems for the radius, I quickly found the known formula:

$$R = \sqrt{3(n^2+m^2+nm)}\frac{d_{CC}}{2\pi}$$

if $d_{CC}$ is the carbon bond length.

For the chiral angle, however, the equation should be

$$\tan{\theta} = \frac{m\sqrt{3}}{2n+m}$$

but I got

$$\tan{\theta} = \frac{m\sqrt{3}}{2n-m},$$

a sign difference. I thought I'd look for the mistake and quickly correct it, but I cannot for the life of me find it. I'll show my derivation below. The starting point is of course the nanotube structure:

Nanotube structure

I'll simplify this picture to make it more clear what I did:

Simplified picture

From the above picture, it's easy to see that

$$\sin{\theta} = \frac{h}{na_1}$$

and

$$\sin{\left(120^{\circ} - \theta\right)} = \frac{h}{ma_2}.$$

This can be rewritten using the trigonometric identity

$$\sin{\left(\alpha-\beta\right)} = \sin{\alpha}\cos{\beta} - \cos{\alpha}\sin{\beta}$$

where $\alpha = 120^{\circ}$ and $\beta = \theta$, giving rise to

$$\frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta} = \frac{h}{ma_2}.$$

Now, using the fact that $a_1 = a_2 (=\sqrt{3}d_{CC})$, we find the following equations:

$$\left\{\begin{array}{rcl}n\sin{\theta} & = & \frac{h}{a_1}\\ m\left[\frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta}\right] & = & \frac{h}{a_1}\end{array}\right.$$

which obviously allow for combination:

$$\begin{eqnarray} \frac{n}{m}\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} + \frac{1}{2}\sin{\theta} \\ \left(\frac{n}{m}-\frac{1}{2}\right)\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} \\ \frac{2n-m}{2m}\sin{\theta} & = & \frac{\sqrt{3}}{2}\cos{\theta} \\ \tan{\theta} & = & \frac{m\sqrt{3}}{2n-m}. \end{eqnarray}$$

This differs from the known expression in that it has a minus sign in front of $m$ in the denominator, but I fail to see my mistake. It could be really silly and I'm just being blind... Thanks for taking a look at it.

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The angle at $C$ is not $120^\circ-\theta$, it's $180^\circ-(120^\circ+\theta)=60^\circ-\theta$. The rest of the algebra is right, and you can quickly see where the sign comes from.

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  • $\begingroup$ Ah yes! Of course, I knew it was going to be some blind spot I was having. Thank you :) $\endgroup$ – Wouter Feb 24 '13 at 0:53
  • $\begingroup$ Sure thing. I think we've all had those kind of days, and to be fair I completely missed it on the first read through as well. $\endgroup$ – wsc Feb 24 '13 at 2:35

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