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Vibrational spectrum of HCl

  1. Jun 30, 2014 #1
    1. The problem statement, all variables and given/known data

    Chlorine has two naturally occurring isotopes, Cl-35 and Cl-37. Show that the vibrational spectrum of HCl should consist of closely spaced doublets, with a splitting given by ##\Delta \nu = 7.51x10^{-4}\nu ## where ##nu## is the frequency of the emitted photon. Hint: think of it as an harmonic oscillator, with ##\omega = \sqrt{k/\mu} ## where ##mu## is the reduced mass, and k is the same for both isotopes.

    2. Relevant equations
    To put it into to context this comes after an explanation of how a two-particle quantum system can be reduced to a central-force problem.

    3. The attempt at a solution
    I am at a loss here. I'm not even sure what "vibrational spectrum" or "splitting" mean. Are they the range of the frequency and the space between consecutive values?

    My guess is that I should compute the energies for the 35 and 37 excited states of an harmonic oscillator; their difference should be the energy of a photon emitted for that transition. However I can't understand when is the photon emitted. Does the chlorine spontaneously change from Cl-53 to Cl-37?
     
  2. jcsd
  3. Jun 30, 2014 #2

    ehild

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    Homework Helper
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    You can consider the H and Cl atoms in the HCl molecule as if they were connected by a spring and vibrate with respect to the centre of mass of the molecule. It can be shown that the vibration frequency is the same as that of a single particle connected to the same spring, when
    the mass of the particle is the reduced mass of the two-atomic system: [tex]μ= \frac{1}{\frac{1}{m_H}+\frac{1}{m_{Cl}}}[/tex]

    You need only know about vibrational spectra that the molecule emits such photons which energy is the same as the vibrational frequency of the molecule. If you scan the emitted light intensity in terms of frequency, you get maximum intensity at the frequency equal to the vibrational frequency of the HCl.

    Chlorine has got two isotopes, with mass numbers 35 and 37. The reduced mass of the molecules are different for the molecules containing Cl35 and Cl37
    Determine the difference between the vibration frequencies relative to the average frequency.


    ehild
     
  4. Nov 29, 2016 #3
    Bump. I have the same question. I would assume the change in vibration frequencies would be equal to Δf=(n/2π)√(k/Δμ) but when I look at the solution, the Δμ is in the numerator and there's a factor of -1/(2μ3/2). How does that work?

    Also, when finding Δμ, the way I've worked it out only works if you subtract the chlorine masses in the numerator but not in the denominator. Reduced mass being μ=(mHmCl)/(mH+mCl). I can't figure it out, can someone point me in the right direction?
     
  5. Nov 29, 2016 #4

    DrClaude

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    $$\sqrt{a}-\sqrt{b} \neq \sqrt{a -b}$$

    Please show your work.
     
  6. Nov 29, 2016 #5
    Well by the definition of reduced mass, μ=(mHmCl)/(mH+mCl). If were to make it (mHmCl')/(mH+mCl') - (mHmCl)/(mH+mCl) I can reduce it down to (m2HΔmCl)/((mH+mCl')((mH+mCl')). Which is equal to the correct answer if ((mH+mCl')(mH+mCl))=(mH+mCl)2 (The denominator) but I don't understand why the prime is negligible in the denominator for the Mass of the Chlorine. I understand for Hydrogen because there is clearly only one Hydrogen isotope in the question. Does the denominator deal with average mass?
     
  7. Nov 29, 2016 #6

    DrClaude

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    There is surely an error somewhere, as the entire point is that the mass of Cl changes. Looking at your other post, which value of μ is supposed to be in the factor?

    In any case, I think it is better to calculate the change in frequency starting from ##\delta \nu## and replacing μ by the corresponding masses.
     
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