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Angular momentum and Quantum Number l

  • #1
6
0
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!
 

Answers and Replies

  • #2
59
2
Moment of Intertia about center of mass = ∑miri
Are you sure this is correct?
 
  • #4
59
2
Look again, that gives the another formula than what you wrote
 
  • #5
6
0
Well I also tried 1/12 ML2 but that still isn't working.
 
  • #6
59
2
No you have to square the radius. I = mr^2, not mr.
Well I also tried 1/12 ML2 but that still isn't working.
I don;t see that on the page
 
  • #8
6
0
OH that's my mistake, that was a typo it's supposed to be Σmiri2
 
  • #9
59
2
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
 
  • #10
6
0
it's a rigid body.. so the mass of the rod is being ignored
 

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