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Homework Help: Rotational energy, find minimum, for nitrogen molecule

  1. Apr 17, 2009 #1
    i made a mistake, and fixed it, but I'm still wrong. I totally thought I had this one. Little help here.

    1. The problem statement, all variables and given/known data
    What is the minimum (nonzero) rotational energy of the <sup>14</sup>N_2 molecule?

    2. Relevant equations
    E_rot = L * (L + 1) * E_0 * (m_e) / (4*M)
    M = (M_1 + M_2) / (M_1 * M_2)

    3. The attempt at a solution

    first off, is E_0 = 13.6 eV? that's what it was in the similar problem on molecular hydrogen. If it isn't...well, how do I calculate it for this situation?

    Ok, so the minimal non-zero E has to be when L=1, since L=0 yields E=0. With L=1, we get

    E_rot = (E_0 * m_e) / (2 * M)

    From the far-above equation for reduced mass M, I plug in 7 times the mass of a proton for each of the nitrogen nuclei, and get M = (7*m_p*7*m_p)/(7*m_p + 7*m_p). After simplifying, M = (7*m_p)/2

    Plugging in E_0, m_e = .00091*10^(-27) kg, m_p = 1.672*10^(-27) kg, and M into the above equation for E_rot, I get

    E_rot = (13.6 * .00091*10^(-27) * 2) / (2 * 7 * 1.672*10^(-27)), and after simplifying

    E_rot = 0.00105742 eV. But this is wrong.

    I thought I was pretty careful. Can anyone see a calculation error, or a mistake in my understanding of how to use these equations with this particular molecular nitrogen situation?

    Many thanks.
  2. jcsd
  3. Apr 17, 2009 #2


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    I am about to log off, and don't have time to read through your post in detail. But, wouldn't the minimum (nonzero) rotational energy occur when the angular momentum is h-bar?
  4. Apr 18, 2009 #3
    I'm not sure how to respond to your response, since the approach of using angular momentum isn't quite comprehensible to me. in the book, they say A is angular momentum, and it's defined in this context as A^2 = L * (L + 1) * h-bar^2. Well, I guess this is the numerator of my new equation for E_rot, shown below. I just didn't think L could be anything but non-negative integers, so I don't know how A would be h-bar as you suggest. Can you see a way to do this, or explain what I don't get about angular momentum here?

    back to my own approach, so I think I was wrong on a few points, my chemistry is super rough.

    I switched equations first off for E_rot, using E_rot = (L * (L + 1) * h-bar^2)/(2*M*r_0^2).

    I re-calculated r_0, using r_0=2*a_0, where a_0 is just like hydrogen but swapping out the term m_e for 7*m_e, since nitrogen has 7 electrons instead of the single electron for hydrogen. so now a_0 = 7.57143*10^(-12) m, making r_0 = 1.51429*10^(-11) m. r_0 is what I'll actually use below.

    Also the reduced mass figure, M. I was forgetting to include the neutrons, in addition to the protons. So it ultimately doubles the result, M = 7 * m_p.

    Plugging in these new values for M and r_0 into the new E_rot equation, (and also using L=1 since it has to be an integer and L=0 causes E_rot to be 0), I get

    E_rot = (L * (L+1) * h-bar^2) / (2*M*r_0^2), I get 0.0258943 eV. But I submitted this and it was wrong.

    It's frustrating, 'cause the problem w/ answer in the book is for the easiest thing ever, the hydrogen molecule. When I do this problem for hydrogen using this new method, I totally get the correct answer of 0.00739841 eV. So switching between hydrogen and nitrogen, something is changing that I'm not accounting for.

    I'm beat. The more I figure out for this problem, the less I feel I know.

    Tips, suggestions? Thanks much.
  5. Apr 18, 2009 #4


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    Ah, you're right. Yes, A2 is what you say above, and L=1 would give the lowest nonzero value.

    This might be where things are going awry. r0 should be something you would look up for nitrogen -- does your text book have a table of atomic radii? I really don't think it's possible to calculate r0 using another molecule and substituting masses.

    Another useful relation is the kinetic energy of a rotating object:

    E = ½ A2 / I

    For I, use the moment of inertia of two point masses (the nitrogen nuclei) separated by a distance ... is it r0 or 2r0 (not sure how you're defining it)?
  6. Apr 18, 2009 #5
    well, you were right about the atomic radius. What I calculated before from modifying hydrogen's radius, and what I found on the internet about the atomic radius for nitrogen, were totally different.

    so using the new correct atomic radius for nitrogen (56 pm), I get 0.00100192 eV, which is also wrong. There's not much else to fiddle with in this problem, except for the M. It's possible that the reduced mass calculation is different for nitrogen, but I don't see how.

    Maybe finding the minimum is found when L is not equal to 1.

    Or maybe I can't use this new formula for E_rot, and instead have to use the formula you stated in terms of I and A. But it was my impression that the two forms were equivalent.

    I've only got one chance left to get the right answer, so I'm gonna stall on this problem and work on some others. Thanks for your help. I feel pretty close to getting this problem, but proximity to the answer doesn't count for much unfortunately.
  7. Apr 18, 2009 #6


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    Good luck. It's definitely the case that L=1 here -- if it could be anything, then the energy could be arbitrarily close to 0 and there would be no answer to this problem.

    The trick is in getting the moment of inertia I. It can be tricky to work in terms of reduced mass, so I recommend using the simple two-point-masses picture to figure out I.

    Also, is r0 the atomic radius, or is it the distance between nitrogen nuclei? Those two things differ by a factor of 2, so be careful there.
  8. Apr 19, 2009 #7
    I think you're making this way more complicated than it has to be, Yroyathon. I believe we have the same book, because most of the problems you ask questions on, I have been assigned as well. In this case, I used your original

    E_rot = [l(l+1)*E_0*m_e]/(4*M)

    The value of E_0 is 13.6 eV, as the book states on the top of the page, saying it is just a simplification of their expanded notation. The use of l=1 is also correct, and the value for m_e is 9.10938*10^-31 kg.

    Thus, the only unknown was the M, which you initially forgot to solve including both protons and neutrons. If you include both, meaning 7 protons and 7 neutrons, your M becomes
    M=2.34328e-26 kg.

    Plug this value in to the initial equation and solve, looks like it should be
    E_rot = 2.643462e-4 eV.

    Hope this helps, I know your posts have helped me this semester so far.
  9. Apr 19, 2009 #8
    icysoul, you give me a tiny bit of confidence I might finish this problem.

    but for one thing.

    when calculating the reduced mass M, M = (M_1 * M_2) / (M_1 + M_2). Using M_1=M_2=7*m_p, I get M=7*m_p=1.1704*10^(-26) kg. this is different than what you wrote, M=2.34328*10^(-26) kg.

    this would lead to a slightly different result, E_rot = 0.000529231.

    do you agree? or am I misunderstanding something about reduced mass calculations?

  10. Apr 19, 2009 #9
    You are completely correct about M, all I did was add up 7p and 7n apparently, haha. I blame 2 AM. But yes, you are right, and after recalculating I also got
    E_rot = 5.28692e-4 eV,
    with the values of m_p and m_n being the ever-so-slightly different values given in the book. Either way, your answer should be within the correct answer tolerance, good catch.
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