Page 73 - Carbon Nanotubes

to better than 1 x if d = 0.339 nm and G = sible to find long series of successive cylindrical sheets
0.142 nm. This is well illustrated by the sequence with 01 = O", it is clear that one should have, for two
formed from the values rl = 0.678 nm, G = 0.142 successive cylinders with identical helicity:
nm, d = 0.339 nm, and 6r/r = 51.5%, which give
more than 30 consecutive non-helical (a! = 0") sym-
metric cylinders, with characteristics ri (nm) = 0.678 +
(i - 1)d and Pi = (i + 1)lO. with
+
2.4 Symmetrical helical cylinders (ri+l/G)' = (3/16~~)(3Pf+~ Qi",,) (19)
with a = 30"
For the configuration of hexagons in case 11, or
ni,i+l(II) should now be an integral and even multiple
of Iy1 or, using eqn (11), [(ri + d)/G]' = (3/16.rr2) [3(Pi +p)' + (Qi + q)']

= ni,i+,(II)d?G. (15)
This equality is not even approximately satisfied for in which p and q are, respectively, integral increments
small integral values of ni,i+l(II). There is, however, of Pi and Qi. For the pitch angles of the two cylinders
a possibility offered by the distance di,i+3, that is, for to be equal, we must have:
1.02 nm, namely:

ni,i+3 (11) = 2~d,,~+~/&C (16)
and it can be shown that
= 26

to better than 0.2%. In other words, if a (Pi,Qi)
doublet yields a = 30", which is the pitch angle cor- In view of the remarks in Q 2.3, it is legitimate to
responding to case 11, the chances are high that this substitute the value 15 for the round bracket in the
angle will be repeated for the (i + 3)th cylinder. A r.h.s. of eqn (22). Taking then into account the fact
good example of this is given by the series described that the maximum and minimum possible values of p
by the values rl = 0.71 nm, G = 0.1421 nm, d = are obtained when q = 0 and q =p, respectively, eqn
0.339 nm, and 6d/d = 2%, which yield CY = 30" for (22) yields:
i = 1,4, 10, 19, 22, 27, 30, 33, 36, 39, . . ., with most
differences between successive values of i either equal 8.66 ~p I 10. (23)
to 3 or to a multiple of 3, and the values P, = Qi =
I&, 44, 96, 174, 200, 243, 269, 295, 321, 347,. . ., The only doublets consistent with this inequality,
which show the expected differences. with the necessity of identical parity for p and q, and
with values of r, close to 0.34 nm, are given in Table 1
2.5 Symmetric helical tubes and cylinder with (from which the doublet (10,O) can be excluded since
a different from 0" or 30" it is characteristic of a symmetrical, non-helical sheet).
Equality (4) can be rewritten as: Hence, if 6r = 0, the necessary conditions for two suc-
cessive helical cylindrical sheets to have strictly iden-
(ri/G)2 = (3/l6a2)(3Pf + Qf). (17) tical pitch angles are:
Since the first bracket on the right-hand side is a
constant and the second is an integer, it is evident that,
for any particular i, some leeway must exist in the q:p = 7:9, or 2:10, or 4:10, (25)
value of the ratio ri/G for the equality to be satisfied.
Here too, the presence of screw helicity must affect ei- since all other values would give either d < 0.334 nm
ther ri, or G, or both. In view of the fairly small vari- (which is already less than d of graphite and, there-
ations of G allowed if the hybridization of the C atoms
is to remain sp', and since the deformation of the C
orbitals decreases as the radius of the cylindrical sheets Table 1. ( p, q) increments for obtaining identical successive
increases, the distance between successive cylinders pitch angles
must decrease and probably tend towards a value char-
Interlayer
acteristic of turbostratic graphite. distance d (nm) Pitch
The second problem associated with helical sym- (P, 4) angle CY'
metric cylindrical sheets addresses the possibility of (9,7) 0.334 24.2
finding the same helix angle or pitch in two neigh- (10,O) 0.339 0
bouring sheets. Apart from cases similar to the one (103 0.341 6.6
described above (case I, section 2.3) in which it is pos- (10~4) 0.348 13.0