ECE : Nanoelectronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University

VM Ayres, ECE , F13 Lecture 18, 29 Oct 13 Carbon Nanotubes and Graphene Carbon nanotube/Graphene physical structure Carbon bond hybridization is versatile : sp 1, sp 2, and sp 3 sp 2 : origin of mechanical and electronic structures Carbon nanotube/Graphene electronic ‘structure’ R. Saito, G. Dresselhaus and M.S. Dresselhaus Physical Properties of Carbon Nanotubes Imperial College Press, London, 1998.

VM Ayres, ECE , F13 CNT Structure The Basis Vectors: a 1 and a 2 The Chiral Vector: C h The Chiral Angle: cos(  The Translation Vector: T The Unit Cell of a CNT – Headcount of available  electrons

VM Ayres, ECE , F13

Lec 17: The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 where magnitude a = |a 1 | = |a 2 |

VM Ayres, ECE , F13 Lec 17: The Basis Vectors 1.44 Angstroms is the carbon-to- carbon distance in individual ring

VM Ayres, ECE , F13 Lec 17: The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 Magnitude a= 2 [ (1.44 Angstroms)cos(30) ] = 2.49 Angstroms 1.44 A 120 o a

VM Ayres, ECE , F13 The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 where magnitude a = 2.49 Ang Example: label the vectors shown in red

VM Ayres, ECE , F13 The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 where magnitude a = 2.49 Ang Example: label the vectors shown in red Answer: as shown a1a1 a2a2

VM Ayres, ECE , F13 The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 where magnitude a = 2.49 Ang Example: what is a 1 a 2 ? a1a1 a2a2

VM Ayres, ECE , F13 The Basis Vectors a 1 = √3 a x + 1 a y 2 2 a 2 = √3 a x - 1 a y 2 2 where magnitude a = 2.49 Ang Example: what is a 1 a 2 ? Answer: non-orthogonal in a i system a1a1 a2a2

VM Ayres, ECE , F13 The Chiral Vector C h

VM Ayres, ECE , F13 The Chiral Vector C h Basic definition Vector: C h = n a 1 + m a 2 C h = (n, m) Magnitude of vector: |C h | = a √ n 2 + m 2 + mn

VM Ayres, ECE , F13 Lec 17: Introduction Many different types of wrapping result in a seamless cylinder. But The particular cylinder wrapping dictates the electronic and mechanical properties. Buckyball endcaps

VM Ayres, ECE , F13 The Chiral Vector C h Basic definition Example: Prove that the magnitude of |C h | is a√n 2 + m 2 + mn

VM Ayres, ECE , F13 The Chiral Vector C h Basic definition Answer:

VM Ayres, ECE , F13 The Chiral Vector C h Basic definition Example: Evaluate |C h | for a (10,10) SWCNT

VM Ayres, ECE , F13 The Chiral Vector C h Basic definition Example: Evaluate |C h | for a (10,10) SWCNT Answer:

VM Ayres, ECE , F13 The Chiral Vector C h A number associated with the Chiral Vector : the Greatest Common Divisor d (or gcd) Definition of d: IfC h = n a 1 + m a 2 Thend = the greatest common divisor of n and m.

VM Ayres, ECE , F13 The Chiral Vector C h The Greatest Common Divisor (d, or gcd) Example: Find d for the following SWCNTs: (10,10), (9,9), (9,0), (7,4), and (8, 6) Definition of d: IfC h = n a 1 + m a 2 Thend = the greatest common divisor of n and m.

VM Ayres, ECE , F13 The Chiral Vector C h The Greatest Common Divisor (d, or gcd) Example: Find d for the following SWCNTs: (10,10), (9,9), (9,0), (7,4), and (8, 6) Answer: (10,10): d=10; (9,9): d = 9; (9,0): d=9; (7,4): d=1; and (8, 6): d=2 Definition of d: IfC h = n a 1 + m a 2 Thend = the greatest common divisor of n and m.

VM Ayres, ECE , F13 The Chiral Vector C h Define the CNT tube diameter d t ( Note that diameter d t is different than greatest common divisor d! ) |C h | =  diameter =  d t d t = |C h |/  = a √ n 2 + m 2 + mn / 

VM Ayres, ECE , F13 The Chiral Vector C h CNT diameter d t |C h | =  diameter =  d t d t = |C h |/  = a √ n 2 + m 2 + mn /  Example: Find the nanotube diameter for a (10, 10) CNT.

VM Ayres, ECE , F13 The Chiral Vector C h CNT diameter d t |C h | =  diameter =  d t d t = |C h |/  = a √ n 2 + m 2 + mn /  Example: Find the nanotube diameter for a (10, 10) CNT. Answer: d t = Ang/  = Ang

VM Ayres, ECE , F13 The Chiral Angle 

VM Ayres, ECE , F13 The Chiral Angle  Direction cosine of a 1 C h : a 1 C h = |a 1 | |C h | cos  cos  = a 1 C h |a 1 | |C h | = ( n + m/2) √ n 2 + m 2 + mn Defines the tilt of C h with respect to a 1

VM Ayres, ECE , F13 The Chiral Angle  Example: Prove that cos  = ( n + m/2) √n 2 + m 2 + mn

VM Ayres, ECE , F13 The Chiral Angle  Answer:

VM Ayres, ECE , F13 The Translation Vector T

VM Ayres, ECE , F13 The Translation Vector T Note that T and C h are perpendicular. Therefore TC h = 0 Let T = t 1 a 1 + t 2 a 2 Take TC h = 0 Solve for t 1 and t 2

VM Ayres, ECE , F13 The Translation Vector T Vector: T = t 1 a 1 + t 2 a 2 t 1 = 2m + n/ d R t 2 = - (2n + m) /d R Magnitude: | T | = √3 |C h |/d R What is d R ?

VM Ayres, ECE , F13 The Translation Vector T d R is a NEW greatest common divisor: d R = the greatest common divisor of 2m + n and 2n+ m d R = d if n-m is not a multiple of 3d

VM Ayres, ECE , F13 The Unit Cell of a CNT (single wall)

VM Ayres, ECE , F13 The Unit Cell of a CNT (single wall) Note that T and C h are perpendicular. Therefore T X C h = the area of the CNT Unit Cell

VM Ayres, ECE , F13 The Primitive Cell of a CNT (single wall) Also a 1 x a 2 is the area of a single basic or primitive cell. Therefore: the number of hexagons N per CNT Unit Cell is: N = | T X C h | | a 1 x a 2 | = 2(m 2 + n 2 +nm)/d R

VM Ayres, ECE , F13 The Unit Cell of a CNT (single wall) Each primitive cell contains two C atoms. There is one p z -orbital per each C atom. Therefore there are 2N p z orbitals available per CNT Unit Cell

VM Ayres, ECE , F13 Example Problems: CNT Structure 1.Find C h, | C h |, cos , , T, | T |, and N for a (10,10) SWCNT 2.What does N count? 3.What type of CNT is this? 4.Find C h, | C h |, cos , , T, | T |, and N for a (9,0) SWCNT 5.What type of SWCNT is this? 6.Find C h, | C h |, cos , , T, | T |, and N for a (7,4) SWCNT 7.What type of SWCNT is this?

VM Ayres, ECE , F13 For use in Example Problems: C h = n a 1 + m a 2 |C h | = a√n 2 + m 2 + mn d t = |C h |/  cos  = a 1 C h |a 1 | |C h | T = t 1 a 1 + t 2 a 2 t 1 = (2m + n)/ d R t 2 = - (2n + m) /d R d R = the greatest common divisor of 2m + n and 2n+ m |T| = √ 3 |C h | / d R N = | T X C h | | a 1 x a 2 | = 2(m 2 + n 2 +nm)/d R

VM Ayres, ECE , F13 1.Find C h, | C h |, cos , , T, | T |, and N for a (10,10) SWCNT 2.What does N count? 3.What type of CNT is this? 4.Find C h, | C h |, cos , , T, | T |, and N for a (9,0) SWCNT 5.What type of SWCNT is this? 6.Find C h, | C h |, cos , , T, | T |, and N for a (7,4) SWCNT 7.What type of SWCNT is this?

VM Ayres, ECE , F13 Answer 1:

VM Ayres, ECE , F13 Answer 1 continued: -

VM Ayres, ECE , F13 | T | = √3 |C h |/d R = 0.25 nm Answers 2 and 3: 2. 3.

VM Ayres, ECE , F13 1.Find C h, | C h |, cos , , T, | T |, and N for a (10,10) SWCNT 2.What does N count? 3.What type of CNT is this? 4.Find C h, | C h |, cos , , T, | T |, and N for a (9,0) SWCNT 5.What type of SWCNT is this? 6.Find C h, | C h |, cos , , T, | T |, and N for a (7,4) SWCNT 7.What type of SWCNT is this?

VM Ayres, ECE , F13 Answer 4:

VM Ayres, ECE , F13 Answers 4 and 5: 4. 5.

VM Ayres, ECE , F13 1.Find C h, | C h |, cos , , T, | T |, and N for a (10,10) SWCNT 2.What does N count? 3.What type of CNT is this? 4.Find C h, | C h |, cos , , T, | T |, and N for a (9,0) SWCNT 5.What type of SWCNT is this? 6.Find C h, | C h |, cos , , T, | T |, and N for a (7,4) SWCNT 7.What type of SWCNT is this?

VM Ayres, ECE , F13 Answer 6:

VM Ayres, ECE , F13 Answers 6 and 7: 6. 7.