Question

1. Carbon nanotubes are large cylindrical molecules made up of a network of covalently bound carbon atoms. The electrons in carbon nanotubes can be considered as particles in a one-dimensional box. For a nanotube of length 30 nm what is the probability of an electron being within 10 nm of the centre of nanotube if the electron is in the state n 3? a. b. What wavelength of light is needed to excite the electron from the n-2 to n -6 energy level? vibrational frequency of NO is measured as 1904 cm, while the vibrational frequency of NOt is 2376 cm Calculate the force constant of the NO bond in N m1 for each molecule. Remember that in cm the frequency is 1/2 and that ω-2πλ. The mass of 14N is 14.003074 amu. 1 kg s-2-1 N mi-l

Answer 1

a) Energy of electron in one dimension box

En = n²h² / 8 m L²  , where n= number of energy level =3 (given) , h = plank constant 6.63 * 10^-34  m² k.g. / s ,

m =mass of electron , 9.10 *10^-34 k.g. ,  L=10*10^{-9 m}

electron 거 3 n de 내 of , pr lability a{ 2 2 En9rma 2- วา 3 x 3 3

b) wavelength required for n=2 to n= 6

The question wants to determine the energy that the incoming photon must have in order to allow the electron that absorbs it to jump from n =2 to n =6.

A good starting point here will be to calculate the energy of the photon emitted when the electron falls from n =6 to n =2 by using the Rydberg equation.

1/ λ=R⋅(1/n₁ ² - 1/n₂ ²)

• λ is  the wavelength of the emitted photon
• R is the Rydberg constant, equal to 1.097m−1

• 1/λ=1.097 *( 1/2² - 1/6² )

• 1/λ=2.4378⋅106.m−1

  This means we have

  λ=4.10 * 10^-7 m

3) Vibrational frequency given and force constant solved as

Fron the pro lem, te knouw that 2. ヒ (2.x x 3.oxio ) x 14x16 14 +16 14 x 16 3 0 2.0 132x132x10-14X/6 3 0