[PoM] Most intense line of HCl molecule

[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?

 

[BRN]

No one can help me?

 

[DrClaude]

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

 

[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.

 

[BRN]

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?

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.

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...

 

[mjc123]

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.

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.

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

I get 1.17. How did you get your answer?

 

[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 get 1.17. How did you get your answer?

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

 

[mjc123]

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 ΔE_rot 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!)

 

[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...

 

[mjc123]

First part is fine. Second part, l_max 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.

 

[BRN]

[/QUOTE]

First part is fine.

Ok, thanks. I return to see the light!

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

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

 

[mjc123]

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

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.

 

[BRN]

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...

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?

 

[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.

 

[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.

 

[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!)

 

[BRN]

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

Now I post another exercise.