Can the Vibrational Spectrum of HCl Be Explained by Isotopic Differences?

In summary: It is easier to see what happens.In summary, the conversation is discussing the vibrational spectrum of HCl and how it is affected by the two naturally occurring isotopes of chlorine. The formula for the splitting of the spectrum is given as ##\Delta \nu = 7.51x10^{-4}\nu ## where ##nu## is the frequency of the emitted photon. The conversation also mentions the reduced mass and how it is calculated for a two-particle quantum system. There is confusion about the calculation of the reduced mass and the use of primes for the masses of chlorine isotopes. The solution is suggested to be found by calculating the change in frequency starting from ##\delta \nu## and replacing the reduced mass with the corresponding
  • #1
carllacan
274
3

Homework Statement



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.

Homework 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.

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?
 
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  • #2
carllacan said:

Homework Statement



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.

Homework 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.

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?

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
 
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  • #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?
 
  • #4
danmel413 said:
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?
$$\sqrt{a}-\sqrt{b} \neq \sqrt{a -b}$$

danmel413 said:
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?
Please show your work.
 
  • #5
DrClaude said:
$$\sqrt{a}-\sqrt{b} \neq \sqrt{a -b}$$Please show your work.

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?
 
  • #6
danmel413 said:
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?
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.
 

Related to Can the Vibrational Spectrum of HCl Be Explained by Isotopic Differences?

1. What is the vibrational spectrum of HCl?

The vibrational spectrum of HCl is a graphical representation of the different vibrational energy levels that the molecule can occupy. It shows the various frequencies at which the molecule can absorb or emit infrared radiation.

2. How is the vibrational spectrum of HCl measured?

The vibrational spectrum of HCl is measured using infrared spectroscopy, which involves passing infrared light through a sample of HCl and measuring the amount of light absorbed at different frequencies. This data is then plotted to create the vibrational spectrum.

3. What does the vibrational spectrum of HCl tell us about the molecule?

The vibrational spectrum of HCl provides information about the bond between the hydrogen and chlorine atoms in the molecule. The peaks in the spectrum correspond to different vibrational modes, which can give insights into the bond strength, bond length, and other properties of the molecule.

4. What factors affect the vibrational spectrum of HCl?

The vibrational spectrum of HCl can be affected by various factors such as temperature, pressure, and the presence of other molecules. These factors can alter the vibrational energy levels and lead to changes in the spectrum.

5. Why is studying the vibrational spectrum of HCl important?

Studying the vibrational spectrum of HCl is important for several reasons. Firstly, it helps us understand the chemical and physical properties of the molecule, which can have practical applications in fields such as medicine and materials science. Additionally, the vibrational spectrum can be used to identify and analyze different molecules, making it a valuable tool in chemical analysis and research.

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