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 :-)
Since 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.
To determine if a carbon nanotube is metallic or semiconducting, you can use the formula n - m, where n and m are the values used to construct the chiral vector.
In this case, the values used are n = 7 and m = 3. So, n - m is equal to 7 - 3, which is 4.
According to the given information, if n - m is a multiple of 3, then the nanotube is metallic. Otherwise, it is considered a semiconductor.
Since 4 is not a multiple of 3, we can conclude that the nanotube with the chiral vector (7,3) is a semiconductor, not a metallic carbon nanotube.