# [PoM] Most intense line of HCl molecule

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1. Jan 29, 2017

### BRN

Hi guys, I've another exercise...

1. The problem statement, all variables and given/known data

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.

3. 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: Jan 29, 2017
2. Jan 31, 2017

### BRN

No one can help me?

3. Jan 31, 2017

### Staff: Mentor

What is $P_{rot}(l=5) / P_{rot}(l=4)$?

4. Jan 31, 2017

### mjc123

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.

5. Feb 1, 2017

### BRN

Not exactly. How do I understand this from exercise text?

Yes, you are right. In this part of exercise my process is right, but I made an unjustified assumption... Thanks for the clarification.

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...

6. Feb 1, 2017

### mjc123

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.

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.

7. Feb 1, 2017

### BRN

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?

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

8. Feb 2, 2017

### mjc123

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!)

9. Feb 2, 2017

### BRN

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...

10. Feb 2, 2017

### mjc123

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.

11. Feb 2, 2017

### BRN

[/QUOTE]

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

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

12. Feb 2, 2017

### mjc123

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

13. Feb 2, 2017

### BRN

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

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?

14. Feb 2, 2017

### mjc123

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.

15. Feb 2, 2017

### BRN

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.

16. Feb 3, 2017

### mjc123

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!)

17. Feb 3, 2017

### BRN

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

Now I post another exercise.