Posted by Lewis on Saturday, July 12, 2008 at 11:25am.
Hi,
Can anyone please check ....
If the values of n and m that have been used to construct the chiral vector equal ( n, m ) = ( 7,3 ).
n  m is 7  3 = 4
For a given (n,m) nanotube, if n − m is a multiple of 3, then the nanotube is metallic, otherwise the nanotube is a semiconductor. Thus all armchair (n=m) nanotubes are metallic, and nanotubes (5,0), (6,4), (9,1),
Is it right to assume that the above is a semi conducting nanotube rather than a metallic carbon nanotube ...
Thank you :)

Physics .. critique  drwls, Saturday, July 12, 2008 at 1:13pm
Since nm is not a multple of 3 for the nanotubes with chiral vectors (n,m) = (5,0), (6,4)and (9,1), they are all semiconductors.
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