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Homework Help: Particle on a Ring Applications

  1. Nov 19, 2014 #1
    1. The problem statement, all variables and given/known data

    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)


    2. Relevant equations

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

    3. 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.
    Last edited: Nov 19, 2014
  2. jcsd
  3. Nov 21, 2014 #2


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    Staff: Mentor

    There is just one n to determine to answer the question:
    To do that, you need to use the hint
    So, in the particle in a box, what is the relation between n and the nodes?
  4. Jan 9, 2015 #3

    Quantum Defect

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    Homework Helper
    Gold Member

    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?
  5. Jan 9, 2015 #4

    Quantum Defect

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    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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