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Topic: rotational spectroscopy HCl  (Read 3805 times)

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Offline Crazy_girl

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rotational spectroscopy HCl
« on: April 17, 2011, 10:44:45 AM »
I have to calculate the wavenumber of the fundamental vibrational transiton (w) for HCl. I know that this transition is from
v0--> v1
which is equal to E1-E0 = 3/2 We-9/4XeWe-1/2+1/4XeWe
                                =We-2XeWe


This is the only data that was given : The first overtone wavenumber is 5668 cm-1
Ro =2906.24 P1 = 2865.10
R1 = 2925. 90 P2 = 2843.62

but with out knowing either Xe or Wei dont see how i am supposed to do this



Offline iolzizlyi

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Re: rotational spectroscopy HCl
« Reply #1 on: May 01, 2011, 10:03:39 PM »
First, note that the R(0) transition is the fundamental plus 2B since we are going from v'' = 0, J = 0 to v' = 1, J = 1.  Likewise the P(1) transition is the fundamental minus 2B.  So you have:

we - 2wexe + 2B = 2906.24
we - 2wexe - 2B = 2865.1

Then solve for B to get 10.29 cm-1.

Now that you have B, note that you also have the first overtone which means you know the transition that corresponds to v'' = 0, J = 0 to v' = 2, J = 0.  Set up the energy equation for the difference between these two vibrational levels and you should get

2we - 6wexe = 5668

along with

we - 2wexe = 2906.24 - 2B = 2885.66

Then solve the system for the unknowns.

Offline iolzizlyi

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Re: rotational spectroscopy HCl
« Reply #2 on: May 02, 2011, 05:03:20 PM »
I was reflecting, and I realized I forgot something important.  The rotational constant for the ground vibrational level is not the same as for the first excited vibrational level.  What you would really want to do is find the rotational constant B for both vibrational levels, B0 and B1.  You can find B0 by noting that R(0) and P(2) both end in the same vibrational and rotational state (v' = 1, J = 1).  However, these transitions differ in the rotational level in which they begin, i.e. P(2) starts in v" = 0, J = 2 and R(0) starts in v" = 0, J = 0.  Thus the energy difference between them is 6B0 and we can find B0 by noting that

6B0 = 2906.24 - 2843.62 = 62.62

B0 = 10.437 cm-1

Similarly, we find B1 by using the R(1) and P(1) transitions which start in the same ground states but finish in different rotational states.  So we have

6B1 = 2925.9 - 2865.1 = 60.8

B1 = 10.133 cm-1

The rotational constants are different because B is proportional to 1/re where re is the equilibrium bond length.  re shifts to the right as vibrational level increases due to anharmonicity and so as re increases, B is expected to decrease.

Now you can use the overtone equation above with the equations for any of the four transitions (with the appropriate B values) to determine the fundamental frequency and anharmonicity.  For instance

2we - 6wexe = 5668

along with

we - 2wexe = 2906.24 - 2B1 = 2885.974

Sorry if anyone was confused.

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