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Equipartition Theorem and Microscopic Motion

  1. Dec 4, 2008 #1
    What is the typical rotational frequency f_rot for a molecule like {\rm N}_2 at room temperature (25\;^\circ \rm{C})? Assume that d for this molecule is 1\; {\rm \AA} =10^{-10}\; {\rm m}. Take the atomic mass of {\rm N}_2 to be m_{\rm N_2} = 4.65 \times 10^{-26}\; {\rm kg}.
    Express f_rot numerically in hertz, to three significant figures.

    I know that :

    Boltzmann's constant = k_B = 1.38*10^-23 J/K

    angular speed about the x axis is omega = [(k_B*T)/(m*((d^2)/2)]^1/2

    Rotational frequency = omega/ 2pi

    Then I got 6.69*10^6 as the answer, I was wondering if it is correct, since I only have one more try left.

    Or do I need to take the 3 degrees of freedom into consideration, so would the equation for omega then becomes [3(k_B*T)/(m*((d^2)/2)]^1/2, so the final answer for rotational frequency is 2.00*10^8 ?

    Please help.
     
  2. jcsd
  3. Dec 6, 2008 #2
    I think you should consider m - mass of N-2 first, then you apply it to the equation of omega to find omega and frequency.

    Note: m in the omega formula is mass of only 1 atom while m given is the mass of Nitrogen molecule.

    Hope this helps =)
     
  4. Dec 6, 2008 #3
    But the mass of N_2 is already given in the problem.

    m_{\rm N_2} = 4.65 \times 10^{-26}\; {\rm kg}
     
  5. Dec 6, 2008 #4
    That's what I said above. The m given is mass of N-2 molecule which has 2 atoms. The mass m in the omega formula is mass of 1 single atom, not a molecule, which means you have to divide the given m by 2.
     
  6. Dec 10, 2008 #5
    Actually then do I still need to take the degrees of freedom into consideration other than the mass situation?
     
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