Thursday

July 31, 2014

July 31, 2014

Posted by **Lewis** on Saturday, July 12, 2008 at 11:25am.

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 1:13pmSince n-m 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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