1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quantum Atomic States:- Check of workings

  1. May 3, 2009 #1
    This post is simply to check through these problems I have attempted to see if I have understood the question properly and applied the correct physics to each senario. Thankyou in advance for looking through my work;

    1. The simplest model of describe the rotations of a diatomic molecule is the dumbell model. This is two point like massive balls connected by a ridgid massless rod length 'a'.

    Find the rotational energy levels of a hydrogen molecule.

    Applying the Bohr model where the angular momentum is maintained in discrete levels;

    [tex]
    mv_{n}r_{n} = \frac{nh}{2 \pi}
    [/tex]

    where n = 1,2,3,...

    Implies;

    [tex]
    r_{n} = \frac{nh}{2 \pi mv_{n}}
    [/tex]

    So;

    [tex]
    r_{n}^{2} = \frac{n^{2}h^{2}}{4 \pi^{2} m^{2}v_{n}^{2}}
    [/tex]

    Comparing with the moment of inertia: I = mr2 for the rotational kinetic energy formula: 1/2 Iw2 w = angular frequency;

    [tex]
    I = \frac{n^{2}h^{2}}{4 \pi^{2} mv_{n}^{2}}
    [/tex]

    Therefore into the RKE formula;

    [tex]
    E_{RKE} = \frac{1}{2}\frac{n^{2}h^{2}w^{2}}{4 \pi^{2} m^{2}v_{n}^{2}}
    [/tex]

    As v=rw w=v/r w2 = v2/r2 In this case as r = 1/2 a and v=vn;

    [tex]
    E_{RKE} = \frac{1}{2}\frac{n^{2}h^{2}}{4 \pi^{2} m^{2}r_{n}^{2}}
    [/tex]

    [tex]
    E_{RKE} = \frac{1}{2}\frac{n^{2}h^{2}}{4 \pi^{2} m^{2} \frac{1}{4} a^{2}}
    [/tex]

    Hence mulitipliny by two, for both molecules;

    [tex]
    E_{RKE} = \frac{n^{2}h^{2}}{\pi^{2} m^{2} a^{2}}
    [/tex]

    Yay or nay?

    2. Distance of closest approach of a 4.78MeV alpha particle and uranium nucleus at 180 degrees.

    (There are 2 protons in an alpha particle, and 92 in the uranium nucleus 180 degrees means that it's a head on collision.)

    Using the work energy theorum with;

    [tex]
    work = \int \frac{kq_{A}q_{U}}{r^{2}} dr
    [/tex]

    Where the qs are the charges on the respective particles.

    [tex]
    Work = \frac{-kq_{A}q_{U}}{r}
    [/tex]

    Evaluated between infinity (because the question didn't state a possition vector for the alpha particle to be at when it has its 4.78MeV of energy) and the distance of closest aproach.

    Hence;

    [tex]
    E_{A} = \frac{-kq_{A}q_{U}}{r}
    [/tex]

    [tex]
    r = \frac{-kq_{A}q_{U}}{E_{A}}
    [/tex]

    Plug the numbers in. Can ignore the (1.6x10-19) conversion factor because it appears in the numerator and denomiator.

    r = 3.5x105m

    Well, it's a tad large isn't it? Hence why I'm questioning my method, however it is coming from infinity, so perhaps that is why it doesn't get as close as in real life when it would have this kind of energy just mear meters before it feels the electric field of the uranium nucleus.

    Or perhaps I am meant to place the alpha particle not at infinity, but at about 1x10-10, as that is 'within' the orbit of electrons. But as the question doesn't talk about electrons, nor does it say anything about the possitions of those electrons (i.e. would they be attracting the alpha particle towards the nucleus or away, and then what are their movements with respect to the two particles in the question.

    Futhermore, due to the alpha particle being very much more massive, the electrons will be drawn to it, more than the alpha particle being slowed down, by the electrons.

    Eitherways did I do the above calculations correctly.

    Cheers,
    Haths
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Quantum Atomic States:- Check of workings
  1. Atom Trap (Replies: 0)

  2. Atom trap (Replies: 0)

  3. Atomic Radiation (Replies: 0)

Loading...