How to Estimate the Energy of a Delocalized π Electron in Benzene?

AI Thread Summary
The discussion focuses on estimating the energy of a delocalized π electron in benzene, modeled as a particle on a ring. The energy formula E = n² * (h-bar)² / (2 * I) is provided, with I defined as the moment of inertia. Participants express confusion about determining the quantum number n and how to fill the energy levels with the six π electrons. It is clarified that n represents angular momentum, with n=0 corresponding to a delocalized electron and higher n values indicating moving electrons with associated de Broglie wavelengths. The importance of understanding nodal planes and their relationship to quantum numbers is emphasized for accurately estimating the energy.
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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)

d0fz4Iw.png


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.
 
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spaghettibretty said:
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:
spaghettibretty said:
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
spaghettibretty said:
(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?
 
spaghettibretty said:

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)

d0fz4Iw.png


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 said:
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?
 
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