Particle on a Ring Applications

[spaghettibretty]

Homework Statement

Six of the electrons from benzene C₆H₆ form a delocalized conjugated π-bond. We will model it as a "particle on a ring" with ring radius a, particle (electron) mass m, and "moment of inertia" I = ma². After obtaining the energy diagram, we will fill in these 6 electrons. Please estimate the energy (in Joules) of a single delocalized π electron in the specific state shown above. Assume the radius a = 0.15 nanometers. (Hint: Just like the "quantum square" or particle in a 1-D box" problem, one can estimate the quantum number or which state from the number and structure of the nodes)

Homework Equations

E = n² * (h-bar)² / (2 * I)
h-bar = reduced Planck's constant = h / (2 * pi)
I = ma²

The Attempt at a Solution

I'm not sure where to go with this problem, I don't really understand this problem since this is one of those "extended knowledge" problems not generally covered. I know, after solving the energy formula, that each energy state n after 0 can hold 4 electrons since there is a positive and negative energy state n, and each n holds 2 electrons. Therefore on the energy level diagram, n=0 holds 2 electrons and n>0 holds four electrons. I'm pretty sure you just plug in numbers to the energy equation, but I have no idea how to determine the n's.

 

[DrClaude]

I know, after solving the energy formula, that each energy state n after 0 can hold 4 electrons since there is a positive and negative energy state n, and each n holds 2 electrons. Therefore on the energy level diagram, n=0 holds 2 electrons and n>0 holds four electrons. I'm pretty sure you just plug in numbers to the energy equation, but I have no idea how to determine the n's.

There is just one n to determine to answer the question:

Please estimate the energy (in Joules) of a single delocalized π electron in the specific state shown above.

To do that, you need to use the hint

(Hint: Just like the "quantum square" or particle in a 1-D box" problem, one can estimate the quantum number or which state from the number and structure of the nodes)

So, in the particle in a box, what is the relation between n and the nodes?

 

[Quantum Defect]

Homework Statement

Six of the electrons from benzene C₆H₆ form a delocalized conjugated π-bond. We will model it as a "particle on a ring" with ring radius a, particle (electron) mass m, and "moment of inertia" I = ma². After obtaining the energy diagram, we will fill in these 6 electrons. Please estimate the energy (in Joules) of a single delocalized π electron in the specific state shown above. Assume the radius a = 0.15 nanometers. (Hint: Just like the "quantum square" or particle in a 1-D box" problem, one can estimate the quantum number or which state from the number and structure of the nodes)

Homework Equations

E = n² * (h-bar)² / (2 * I)
h-bar = reduced Planck's constant = h / (2 * pi)
I = ma²

The Attempt at a Solution

I'm not sure where to go with this problem, I don't really understand this problem since this is one of those "extended knowledge" problems not generally covered. I know, after solving the energy formula, that each energy state n after 0 can hold 4 electrons since there is a positive and negative energy state n, and each n holds 2 electrons. Therefore on the energy level diagram, n=0 holds 2 electrons and n>0 holds four electrons. I'm pretty sure you just plug in numbers to the energy equation, but I have no idea how to determine the n's.

Remember, the n's are quantum numbers. These are integers.

The first relevant equation uses unfortunate choices for quantum numbers. The particle on a ring is an interesting example to show the quantization of angular momentum. In the first equation, the quantum number "n" is actually a measure of the electron-bead's angular momentum. If you imagine an electron-bead living/spinning around on the ring, what might its angular momentum be? You could have a ground-state (non-moving) electron, delocalized over the entire ring (n=0). With a moving electron/bead (n<>0) how might it be moving? As you note, n can be positive or negative, and n=-1 and n=1 states have the same energy (you alluded to "positive and negative energy states" which is not quite right -- all of the energies in this >= 0.

You will have an "energy level" with n=0, whose energy is E0 = 0; a degenerate pair of |n| = 1 levels, whose energy you can calculate; a degenerate pair of |n| = 2 levels, whose energy you can caclulate; etc. You have six pi electrons to fill these levels -- you do this just like you did in atoms (Aufbau + Pauli) in general chemistry. The total energy is the sum of the energies of the individual electrons.

What important interaction is this problem neglecting?

 

[Quantum Defect]

Remember, the n's are quantum numbers. These are integers.

The first relevant equation uses unfortunate choices for quantum numbers. The particle on a ring is an interesting example to show the quantization of angular momentum. In the first equation, the quantum number "n" is actually a measure of the electron-bead's angular momentum. If you imagine an electron-bead living/spinning around on the ring, what might its angular momentum be? You could have a ground-state (non-moving) electron, delocalized over the entire ring (n=0). With a moving electron/bead (n<>0) how might it be moving? As you note, n can be positive or negative, and n=-1 and n=1 states have the same energy (you alluded to "positive and negative energy states" which is not quite right -- all of the energies in this >= 0.

You will have an "energy level" with n=0, whose energy is E0 = 0; a degenerate pair of |n| = 1 levels, whose energy you can calculate; a degenerate pair of |n| = 2 levels, whose energy you can caclulate; etc. You have six pi electrons to fill these levels -- you do this just like you did in atoms (Aufbau + Pauli) in general chemistry. The total energy is the sum of the energies of the individual electrons.

What important interaction is this problem neglecting?

Doh! Forgot the point of the problem. The n=0 electron is delocalized over the whole ring (no nodes). The |n|=1 electrons are moving, they have a de Broglie wavelenth. How could you fit something with a wavelength into the ring (this is like the Bohr model of the hydrogen atom)? If you draw the picture of the wave, are there any nodal planes? How many? A higher |n| will be for an electron with a shorter de Broglie wavelength. What do these waves look like? How do the number of nodal planes change as the de Broglie wavelength decreases?

 

[DeeNos]

I would suggest breaking this problem down into smaller components and using known principles to solve it. First, let's define the variables given in the problem:

- Ring radius, a = 0.15 nanometers
- Particle (electron) mass, m = mass of an electron
- "Moment of inertia", I = ma^2

Using these values, we can calculate the moment of inertia, I, for the "particle on a ring" system. Next, we can use the energy equation provided in the problem, E = n^2 * (h-bar)^2 / (2 * I), to calculate the energy of a single delocalized π electron in the specific state shown above.

To determine the quantum number, n, we can use the information given in the problem about the number and structure of the nodes. The number of nodes in a wave function corresponds to the quantum number, n. For example, a wave function with one node has a quantum number of n = 1, a wave function with two nodes has a quantum number of n = 2, and so on. Using this information, we can estimate the quantum number for the specific state shown in the problem.

Once we have calculated the energy for a single delocalized π electron in the specific state, we can then use this value to calculate the total energy for all six delocalized π electrons in the benzene molecule.

In summary, as a scientist, I would approach this problem by breaking it down into smaller components and using known principles to solve it. By using the given variables, the energy equation, and the information about the number and structure of the nodes, we can estimate the energy of a single delocalized π electron in the specific state shown and then use this value to calculate the total energy for all six electrons in the benzene molecule.