[PoM] Most intense line of HCl molecule

• BRN
In summary, In this conversation, an expert summarizer has determined the equilibrium internuclear distance of HCl molecule, knowing that some contiguous lines of the rotational spectrum of 2H35Cl are observed at the wavelengths: 234.1, 186.8, 156.4, 134.1, 117, 5. Using these wavelengths and the energy difference between the rotational spectra lines, the equilibrium internuclear distance is calculated to be 1.2837*10^{-10} [m]. Assuming emission, the most intense line is associated to l=4 or l=5.
BRN
Hi guys, I've another exercise...

1. Homework Statement

Determine the equilibrium internuclear distance ##R_M## of HCl molecule, knowing That some contiguous lines of the rotational spectrum of 2H35Cl are observed at the wavelengths: 234.1, 186.8, 156.4, 134.1, 117, 5 ##\mu m##. If the experiment is conducted with a gaseous sample at T = 410 K, indicates which of these lines is the most intense.

The Attempt at a Solution

The energy difference between the rotational spectra lines is:

##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1)##

For a ##l \rightarrow l-1## transition ##\Rightarrow \Delta E_{rot}=\frac{\hbar^2}{I}l##

and for simplicity, I assume ##l=1 \rightarrow l=0## transition. So:

##\Delta E_{rot}=\frac{\hbar^2}{\mu R_M^2}l##

For example, I take account the first two wavelengths:

##E_1=\frac{hc}{\lambda_1}=8.4852*10^{-22}[J]##

##E_2=\frac{hc}{\lambda_2}=1.0633*10^{-21}[J]##

Then, I have ##\Delta E_{rot}=2.1478*10^{-22}[J]##

also:

##\mu=\frac{35*2}{37} \frac{10^{-3}}{N_A}=3.1415*10^{-27}[kg]##

and:

##R_M=\sqrt{\frac{\hbar^2}{\mu \Delta E_{rot}}}=1.2837*10^{-10} [m]##

This result is OK.

Now, ##I=\mu R_M^2=5.1768*10^{-47}[kg m^2]##

The most intense line is that of the transition that start from the most populated level.
The most populous level ##l_{max}## is calculated as:

from the rotational level population
##P_{rot}=\frac{2l+1}{Z_{rot}}e^{-\beta E_{rot}}=\frac{2l+1}{Z_{rot}}e^{-\beta (\frac{\hbar^2}{2\mu}\frac{l)l+1)}{R_M^2})}##

##\frac{\partial P_{rot}}{\partial l}|l_{max}= 2+(2l+1)^2(-\beta \frac{\hbar^2}{2I})|l_{max}=0##

##\Rightarrow l_{max}=\sqrt{\frac{I}{\beta \hbar^2}}-\frac{1}{2}=\sqrt{\frac{k_BTI}{ \hbar^2}}-\frac{1}{2}=4.63##

therefore, the most intense line is associated to ##l=4 \rightarrow l=5## or ##l=5 \rightarrow l=4##

but already at this point I'm wrong...

Solutions are: ##R_M=129,7 [pm]##; transition ##l=5 \rightarrow l=6## with ##\lambda=156.4[\mu m]##

Where am I wrong?

Last edited:
No one can help me?

BRN said:
from the rotational level population
##P_{rot}=\frac{2l+1}{Z_{rot}}e^{-\beta E_{rot}}=\frac{2l+1}{Z_{rot}}e^{-\beta (\frac{\hbar^2}{2\mu}\frac{l)l+1)}{R_M^2})}##

##\frac{\partial P_{rot}}{\partial l}|l_{max}= 2+(2l+1)^2(-\beta \frac{\hbar^2}{2I})|l_{max}=0##

##\Rightarrow l_{max}=\sqrt{\frac{I}{\beta \hbar^2}}-\frac{1}{2}=\sqrt{\frac{k_BTI}{ \hbar^2}}-\frac{1}{2}=4.63##

therefore, the most intense line is associated to ##l=4 \rightarrow l=5## or ##l=5 \rightarrow l=4##
What is ##P_{rot}(l=5) / P_{rot}(l=4)##?

First, are you dealing with an absorption spectrum or an emission spectrum? You seem to be assuming emission, the answer assumes absorption. Does the question make this clear?
Next, you assume l = 1 for the first line. This is unjustified; you are only told that these lines appear in the spectrum. You only need to know the difference between the successive transition energies, which to a first approximation is independent of the rotational quantum number. That gives you the rotational constant and moment of inertia.
Your expression ##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1) ## refers to the energy of a transition between two rotational energy levels - not to the difference in energy between two rotational transitions.

BRN
mjc123 said:
First, are you dealing with an absorption spectrum or an emission spectrum? You seem to be assuming emission, the answer assumes absorption. Does the question make this clear?
Not exactly. How do I understand this from exercise text?

mjc123 said:
Next, you assume l = 1 for the first line. This is unjustified; you are only told that these lines appear in the spectrum. You only need to know the difference between the successive transition energies, which to a first approximation is independent of the rotational quantum number. That gives you the rotational constant and moment of inertia.
Your expression ##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1) ## refers to the energy of a transition between two rotational energy levels - not to the difference in energy between two rotational transitions.
Yes, you are right. In this part of exercise my process is right, but I made an unjustified assumption... Thanks for the clarification.

DrClaude said:
What is ##P_{rot}(l=5) / P_{rot}(l=4)##?
Ok, thi is the next step. With the population defined as:

## P_{rot}=\frac{2l+1}{Z_{rot}}e^{-\beta E_{rot}}##

I have:

##\frac{P_{rot}(l=5) }{P_{rot}(l=4)}=0.98##

and then the most populous level should be ##l=4##, with a transition ##l=4 \rightarrow l=5##. But, this is wrong...

BRN said:
Not exactly. How do I understand this from exercise text?
If the question is exactly as quoted, you can't tell. You have assumed the lines are l → l-1 transitions, while the given answer assumes they are l → l+1.

BRN said:
Yes, you are right. In this part of exercise my process is right, but I made an unjustified assumption... Thanks for the clarification
Your process is not right. Your answer is coincidentally correct because you chose l = 1; any other l would have given you the wrong answer. Read my post again.

BRN said:
Prot(l=5)Prot(l=4)=0.98Prot(l=5)Prot(l=4)=0.98\frac{P_{rot}(l=5) }{P_{rot}(l=4)}=0.98

My book reports that in a rotovibrational spectra, the lines are equally spaced in accord with ##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1)##

At this point I'm very confused... Could you explain in more detail this thing?

mjc123 said:

I checked my calculations and I get 0.99. How I get it, I wrote it in the first and 5th post.

BRN said:
My book reports that in a rotovibrational spectra, the lines are equally spaced in accord with ##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1)##
Yes, but that is the spacing between rotational energy levels l and l+1, hence the energy of the transition between them. It is not, as you originally wrote, the spacing between lines in the rotational spectrum. That is E(l+1 → l+2) - E(l → l+1), which equals ħ2/I, independent of l. The difference in energy between the first two lines is not ΔErot as you defined it. You accidentally got the right answer because you assumed that the first line was l=1 → l=0, which is unjustified and in fact untrue.

I got ##\frac{P(5)}{P(4)} = \frac{11}{9}exp(-\frac{E(5)-E(4)}{kT}) = \frac{11}{9}exp(-\frac{30B - 20B}{kT}) ##
where B = ħ2/2I. I used a value of 1.05*10-22 J as an average from the given line spacings. Then I get
##\frac{11}{9}exp(-\frac{10*1.05e-22}{1.38e-23*410}) = 1.015## (Perhaps I got 1.017 last time, not 1.17. Or just made a mistake!)

I try to recap.

I convert the wavelengths in energy by relation ##E_i=\frac{hc}{\lambda_i}##

then:

##\lambda_1=234.1 [\mu m] \rightarrow E_1=8.4852*10^{-22}[J]##
##\lambda_2=186.8 [\mu m] \rightarrow E_2=1.0633*10^{-21}[J]##
##\lambda_3=156.4 [\mu m] \rightarrow E_3=1.2700*10^{-21}[J]##
##\lambda_4=134.1 [\mu m] \rightarrow E_4=1.4812*10^{-21}[J]##
##\lambda_5=117.5 [\mu m] \rightarrow E_5=1.6905*10^{-21}[J]##

The difference between these transition energy are:

##|E_1-E_2|=2.1478*10^{-22}[J]##
##|E_2-E_3|=2.067*10^{-22}[J]##
##|E_3-E_4|=2.112*10^{-22}[J]##
##|E_4-E_5|=2.093*10^{-22}[J]##

the average value is ##\Delta =2.1049*10^{-22}[J]##

The spectra lines are spaced, in energy, by ##\Delta = \Delta E_{rot}(l_i) - \Delta E_{rot}(l_i-1)=\frac{\hbar^2}{I}=\frac{\hbar^2}{\mu R_M^2}##
So, the equilibrium internuclear distanece is:

##R_M=\sqrt{\frac{\hbar^2}{\mu \Delta}}=1.2967*10^{-10} [m]##

Now it's OK?

For 2nd part nothing changes for me...

##I=\mu R_M^2=5.2822*10^{-47}[kg m^2]##

##l_{max}=\sqrt{\frac{I}{\beta \hbar^2}}-\frac{1}{2}=\sqrt{\frac{k_BTI}{ \hbar^2}}-\frac{1}{2}=4.68##

##\frac{P_{rot}(l=5) }{P_{rot}(l=4)}=0.99##

I don't understand what is wrong...

First part is fine. Second part, lmax is obtained by differentiating as a continuous function what is in fact a discrete function. I don't see how you get from there to a value for P(5)/P(4). Do as I did, show what formula you used for P(5)/P(4), what numbers you used, and the calculation.

[/QUOTE]
mjc123 said:
First part is fine.

For 2nd part:
Now, ##I=\mu R_M^2=5.2822*10^{-47}[kg m^2]##

The most intense line is that of the transition that start from the most populated level.
The most populous level ##l_{max}## is calculated as:

from the rotational level population
##P_{rot}=\frac{2l+1}{Z_{rot}}e^{-\beta E_{rot}}=\frac{2l+1}{Z_{rot}}e^{-\beta (\frac{\hbar^2}{2\mu}\frac{l)l+1)}{R_M^2})}##

##\frac{\partial P_{rot}}{\partial l}|l_{max}= 2+(2l+1)^2(-\beta \frac{\hbar^2}{2I})|l_{max}=0##

##\Rightarrow l_{max}=\sqrt{\frac{I}{\beta \hbar^2}}-\frac{1}{2}=\sqrt{\frac{k_BTI}{ \hbar^2}}-\frac{1}{2}=4.68##

therefore, the most intense line is associated to ##l=4 \rightarrow l=5## or ##l=5 \rightarrow l=4##

To determine what is the most populous level, I do the ratio between the two populations. If the result is > 1, the most populous level will have ##l = 5##, if the result is < 1, the most populous level will have ##l = 4##.
With the population defined as:

##P_{rot}=\frac{2l+1}{Z_{rot}}e^{-\beta E_{rot}}##

I have

##\frac{P_5}{P_4}=\frac{2(5+1)}{2(4+1)}\frac{e^{-\beta\frac{\hbar^2}{2I}5(5+1)}}{e^{-\beta\frac{\hbar^2}{2I}4(4+1)}}=\frac{12e^{-\beta\frac{\hbar^2}{I}15}}{10e^{-\beta\frac{\hbar^2}{I}10}}=0.99##

The multiplicity is 2J + 1, not 2(J+1). So the fraction you want at the beginning is 11/9, not 12/10.
BRN said:
therefore, the most intense line is associated to ##l=4 \rightarrow l=5 or l=5 \rightarrow l=4##
It's an absorption spectrum (the answer gives us to understand); all the lines are l → l+1. So either 4→5 or 5→6.

mjc123 said:
The multiplicity is 2J + 1, not 2(J+1). So the fraction you want at the beginning is 11/9, not 12/10.

Oh DAMN! I want to die! Thanks, I did not realize...

mjc123 said:
It's an absorption spectrum (the answer gives us to understand); all the lines are l → l+1. So either 4→5 or 5→6.
Ok, but the exercise text tells me only "some contiguous lines of the rotational spectrum of 2H35Cl are observed". How can I know if it is an emission spectrum or an absorption spectrum if I don't have the solution?

A priori you can't, without being told. But either way, the spacing between the lines is 2B (to a first approximation), so that gives you the moment of inertia.

I forgot a step.
with ##l_{max}=5##, if I assume an absorption spectrum, the spacing between rotational energy levels 5 and 6 is:

##\Delta E_{rot}=\frac{\hbar^2}{I}(l_i+1)=\frac{\hbar^2}{I}6=1.2630*10^{-21}[J]##

close to the wavelength ##\lambda=156.4 [\mu m]##

But, then, if I had considered the transition 5→4, the solution would be equally correct, I think.

No, for a l → l-1 transition the energy is ħ2/I*l. (Just treat it as l-1 → l using your formula, the energy difference is the same!)

Yes, I said something stupid.
Thanks for your help, now everything is clear.

Now I post another exercise.

1. What is the most intense line of HCl molecule?

The most intense line of HCl molecule is the R(0) line, which corresponds to the transition from the lowest rotational energy level (J=0) of the ground state to the first excited rotational energy level (J=1).

2. Why is the R(0) line of HCl molecule the most intense?

This line is the most intense because it is the most probable transition to occur due to the Boltzmann distribution of molecular rotational energy levels. In other words, the J=0 to J=1 transition has the highest transition probability.

3. What is the significance of the most intense line in the HCl spectrum?

The most intense line in the HCl spectrum provides valuable information about the rotational energy levels and the rotational constants of the molecule. It is also used in spectroscopic studies to determine the concentration and temperature of HCl in a sample.

4. Can the intensity of the R(0) line be affected by external factors?

Yes, the intensity of the R(0) line can be affected by factors such as temperature, pressure, and the presence of other molecules in the sample. These factors can alter the rotational energy levels and therefore change the transition probabilities.

5. How is the intensity of the R(0) line measured?

The intensity of the R(0) line can be measured using a spectrometer, which detects and measures the intensity of light emitted or absorbed by the HCl molecules. The intensity is then compared to a reference spectrum to determine the concentration and other properties of the HCl sample.

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