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I have run into a problem trying to calculate angular momenta and I wonder if someone can set me straight.

I think I know that the lowest order rotational mode should have angular momentum of h, and that was the orbital momentum of the electron in Bohr's original hydrogen atom. The kinetic energy of the electron in its orbit was one Rydberg (~13 eV). For the rest of the calculations below it will be convenient to express this as an equivalent temperature of close to 150 000 K (see Wikipedia for the conversion).

I wanted to see if I could understand the rotational modes of diatomic molecules by analogy with the hydrogen atom. So I tried the Hydrogen molecule. Using a bond length of approximately 1.4 Bohr radii (from this table), and mass of 2000 Me (electron masses) for each atom, I got a frequency factor of 2000x compared to the hydrogen atom, which equates to a reduced equivalent temperature of 75 K.

This was not bad according to Table 5.1 from this article which gives an equivalent of 85 degrees for the characteristic rotational mode temperature. Similarly, I get reasonable answers for diatomic oxygen and diatomic Chlorine. But I am way off when I try to calculate HCl.

My source gives a bond length of around 2.2 Bohr radii; because of the mismatch between H and Cl, I assume almost all the momentum is in the hydrogen atom and the calculation gives me a factor of 10 000 on frequency (it goes inversely as mass*r^2). This gives me an equivalent temperature of 15 K; but the table shows a value of 4K.

I seem to be reasonably close on the other diatomic molecules. I wonder why I'm way off on HCl? Any ideas would be appreciated.

EDIT: Okay, Fabian's answer is helpful. I probably shouldn't even have mentioned the hydrogen atom; since Fabian gives us the correct formula for the energy of the rigid rotor (first excited state), I can plug in the values for my diatomic molecules and see what happens. Here is what I get, using masses in AMU's and distances in Bohr radii, and letting h=1:

Energy of Rigid Rotor, first mode

Hydrogen molecule: M1 = M2 = 1, d = 1.4, energy = 1.0 approx.

Oxygen molecule: M1 = M2 = 16, d = 2.2, energy = 0.025

Chlorine molecule: M1 = M2 = 35, d = 3.8, energy = 0.0038

Taken in proportion, these numbers line up pretty well with the equivalent temperatures given in the table: Equivalent Temperatures of Rotational Energies

But when I apply the formula to HCl, I'm way off:

Hydrogen Chloride: M1=1, M2=35, d=2.2, energy = 0.2

Extrapolating from the other molecules, this would give me an equivalent temperature of 15 to 20 degrees K; but the table gives a value of 4.2 K.

I wonder why the formula seems to give consistent values for the H2, O2, and Cl2, but is way out of whack for HCl?

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What equations are you using to solve for the temperature? It might make it clearer if you are making assumptions in the equation that HCl violates. – tpg2114 Dec 12 '12 at 19:12
Ar you sure about the bond length. If that changes by a factor of 2 then the energy is a factor 4 smaller. – Fabian Dec 13 '12 at 14:54
E.g., this page tells us that the bond length should be $.13$nm. – Fabian Dec 13 '12 at 14:56
Yes, that's the same bond length I'm using; I've expressed it as 2.2 Bohr radii, using a value of 53 picometers. – Marty Green Dec 13 '12 at 15:48

I do not understand the relation which you try to put up with the hydrogen atom. In the hyrdogen atom the energy is due to the Coulomb attraction. As far as I understand your question, you are interested in the rotational energy instead.

The rotational energy of a rigid rotor is given by $$E_\text{rot} = \frac{L^2}{2I}$$ with $L$ the angular momentum and $I$ the moment of inertia. In quantum mechanics, we have $L^2= \hbar^2 l (l+1)$ with $l=0,1,2,...$. A diatomic molecule has the inertial moment $$I=\mu d^2$$ with $\mu= M_1 M_2/(M_1+M_2)$ the reduced mass and $d$ the distance between the two atoms.

The energy splitting between the ground state ($l=0$) and the first rotationally excited state ($l=1$) is given by $$\Delta E= \frac{(M_1+M_2)\hbar^2 }{M_1 M_2 d^2}.$$

There are also transitions between higher $l$. In fact, we have $$E_{l+1} - E_l = \Delta E \times (1+l)$$

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