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Angular momentum and Quantum Number l

  1. Feb 1, 2015 #1
    • Move here from another forum, so no homework template.
    The question: Consider two masses of 0.1 gm each, connected by a rigid rod of length 0.5 cm, rotating about their center of mass with an angular frequency of 800 rad/s. a.) What is the value of l corresponding to this situation? b.) What is the energy difference between adjacent l-values for the l you have just calculated?

    Relevant equations:
    Moment of Intertia about center of mass = ∑miri
    L=I * ω
    L= ħ[l(l+1)]1/2

    My attempt at part a of the problem:
    r= 0.0025m
    mass= 0.0001 kg

    I = 2 * (0.0001 kg) * (0.0025 m2) = 1.25 * 10-9 kg m2

    L = 1.25 * 10-9 kg m2 * 800 rad/s

    L/ħ = 9.48*1027 = [l(l+1)]1/2

    Which this is the answer, but when I solve for l(the quantum number for angular momentum) I get a number with order of magnitude of 1055 am I completely overlooking something here? I can't think of any other ways to solve this problem. Thank you in advance for your help!
  2. jcsd
  3. Feb 1, 2015 #2
    Are you sure this is correct?
  4. Feb 1, 2015 #3
  5. Feb 1, 2015 #4
    Look again, that gives the another formula than what you wrote
  6. Feb 1, 2015 #5
    Well I also tried 1/12 ML2 but that still isn't working.
  7. Feb 1, 2015 #6
    No you have to square the radius. I = mr^2, not mr.
    I don;t see that on the page
  8. Feb 1, 2015 #7
  9. Feb 1, 2015 #8
    OH that's my mistake, that was a typo it's supposed to be Σmiri2
  10. Feb 1, 2015 #9
    If you don't know the mass of the ord you can't use the second equation because M is the mass of the rod
  11. Feb 1, 2015 #10
    it's a rigid body.. so the mass of the rod is being ignored
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