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BRN

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Hi guys, I've another exercise...

Determine the equilibrium internuclear distance ##R_M## of HCl molecule, knowing That some contiguous lines of the rotational spectrum of

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?

1. Homework Statement1. Homework Statement

Determine the equilibrium internuclear distance ##R_M## of HCl molecule, knowing That some contiguous lines of the rotational spectrum of

^{2}H^{35}Cl 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?

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