Electrons inside a carbon nanotube - Quatum Mechanics

[Feodalherren]

Homework Statement

Electrons inside a carbon nanotube can be approximated as a one dimensional "particle in a box". If the nanotube is 3 micrometers long, what is the minimum speed of an electron inside the tube?

Homework Equations

The Attempt at a Solution

The minimum occurs as n=1 so therefore

$K=\frac{h^{2}}{8M_{e^{-}}(3E-6)}= 2.01E-32J$

If I use 1/2 mV^2 to find V I get the wrong answer. Why can't I use it and what should I use instead?

 

[eprparadox]

The lowest energy state in a particle in a box is

$$E_n = \frac{h^2}{8m_eL^2}$$

If you used the equation as you wrote it, you didn't square your L.

Then you can use the

$$E = \frac{1}{2}mv^2$$

formula to solve for v.

 

[Feodalherren]

Gah it's always something stupid like that. Thanks.

 

[eprparadox]

No worries! I've done 28328348238 stupid things like that over the years.

 

[HalfLostAndSearching]

I would like to clarify that the concept of "minimum speed" for an electron inside a carbon nanotube is not well-defined. In quantum mechanics, the speed of an electron is not a measurable quantity, as it is described by a wave function that represents the probability of finding the electron at a certain position and energy. Therefore, instead of minimum speed, we can talk about the minimum kinetic energy of the electron inside the nanotube.

To calculate this minimum kinetic energy, we can use the equation for the energy of a particle in a one-dimensional box:

E_n = \frac{n^2h^2}{8mL^2}

where n is the quantum number, h is the Planck's constant, m is the mass of the electron, and L is the length of the nanotube.

Substituting the given values, we get:

E_1 = \frac{1^2(6.626E-34 J.s)^2}{8(9.11E-31 kg)(3E-6 m)^2} = 2.07E-32 J

This is the minimum energy of the electron inside the nanotube, which is equivalent to its kinetic energy. To find the minimum speed, we can use the equation for kinetic energy:

K = \frac{1}{2}mv^2

Rearranging this equation, we get:

v = \sqrt{\frac{2K}{m}}

Substituting the minimum kinetic energy we found above, and the mass of the electron, we get:

v = \sqrt{\frac{2(2.07E-32 J)}{9.11E-31 kg}} = 5.79E5 m/s

Therefore, the minimum speed of an electron inside a 3 micrometer long carbon nanotube is approximately 5.79E5 m/s. However, as mentioned before, this speed is not a measurable quantity and should be interpreted as the minimum speed needed for the electron to have the minimum kinetic energy required to be confined inside the nanotube.