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Equipartition theorem and motion

  1. Nov 14, 2007 #1
    What is the typical rotational frequency f_rot for a molecule like N_2 (nitrogen) at room temperature (25 C)? Assume that d for this molecule is 1 angstrom = 10^{-10} m. Take the atomic mass of N_2 to be 4.65 * 10^{-26} kg.
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    I know that the rms angular speed which is the root mean square (rms) of the x component of velocity of the gas particles is = sqrt(2k_B(T)/(m(d^2))) where k_B is the Boltzman constant 1.38*10^-23 J/K. Ok so I have all the info I need I believe to solve for velocity but what about the wavelength component that I need to use this equation velocity=Freq*wavelength and to get my answer for frequency. What is this wavelength that I have to calculate for?
     
  2. jcsd
  3. Nov 14, 2007 #2
    i found this equation relating frequency and veolcity --> angular velocity=2(pi)f
    but i DONT UNDERSTAND why i'm not getting the right answer. i get velocity to be 4.2057*10^12 m/s and i SHOULD be able to get frequency by dividing it by 2pi, but f=6.69*10^11 is not right!

    whats wrOong?!
     
  4. Nov 14, 2007 #3

    marcusl

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    Hold on, you're getting confused. You don't want the de Broglie wavelength because this isn't a wave, it's a true classical rotation. The x component of velocity is just that--translational speed, not angular rotation. Also your title mentions the equipartition theorem so you know that will figure in somewhere.

    Take a look at the equipartion theorem in your text. How many degrees of freedom does N2 have? What is the energy of each one? Then set it equal to the energy of a rotating object (what is that?)

    Come back if you need more help.
     
  5. Nov 14, 2007 #4

    marcusl

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    Jaded18, I just saw your other post in Introductory Physics. Please don't double post in the future. (And that forum is the appropriate one for homework problems.)
     
  6. Nov 14, 2007 #5
    I don't understand how the energy relates to frequency!!!!!!!!! Degrees of freedom for diatomic molecules: 5. energy of each one = 0.5k_B(T)= 0.5(1.38*10^-23 J/K)(25+273K)=2.056*10^-21= 0.5I(angular velocity)^2... what is I?. How come what i did in the beginning with the equation i found didn't work out to get the correct answer?
     
  7. Nov 15, 2007 #6

    marcusl

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    Good, now you're on the right track; you properly found the energy of one degree of freedom and you set it equal to the rotational energy of that mode. I is the moment of inertia, which is the sum of (each mass times that mass's distance squared from the center). You can finish the problem now and solve for the angular velocity.

    The reason you didn't get the right answer before is because you used equations that made no sense.
     
  8. Nov 15, 2007 #7
    first of all, i don't need the angular velocity. i need the rotational frequency. and also, the equations are all very real. it turns out that my ans was right and it was a mistake in the ans key.... So yeah won't double post in the future, it's pretty much useless anyways.
     
  9. Nov 16, 2007 #8

    marcusl

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    Well, I'm glad it worked out for you.
     
  10. Feb 2, 2009 #9
    Actually - For others who have to finish this problem there was an error in the above calculations. When dealing with N_2 there are 2 molecules of nitrogen when your solving for just 1. This means you'll need to cut your Mass in half:
    (4.65E-26)/2 = M
    sqrt((2*k_B*T(K)/M*d^2))= ω <---(Remember Kelvins)
    f=(ω/2pi)
     
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