Rotation Spectrum of a diatomic molecule (QM)

joker_900
Messages
53
Reaction score
0

Homework Statement


In the rotation spectrum of 12C16O the line arising from the transition l = 4 -> 3 is at 461.04077GHz, while that arising from l = 36 -> 35 is at 4115.6055GHz. Show from these data that in a non-rotating CO molecule the intra-nuclear distance is s ~ 0.113 nm, and that the electrons provide a spring between the nuclei that has force constant ~1904Nm−1. Hence show that the vibrational frequency of CO should lie near 6.47×10^13Hz (measured value 6.43 × 1013 Hz). Hint: show from classical mechanics that the distance of O from the centre of mass is (3/7) s and that the molecule’s moment of inertia is (48/7) ms^2. Recall also the classical relation L = Iw.


Homework Equations


f = j h/((2pi)^2*I)

Where f is the frequency of the emitted photon and j(j+1) is the eigenvalue of J^2 (I think j is the same as l in this question?)

The Attempt at a Solution


I can derive the two classical bits in the hint and the above formula. I then tried taking one of the frequencies given, working out I using the above formula, and plugging it into the second classical expression to get s. It didn't work!

Thanks
 
Physics news on Phys.org
It's not (2pi)^2 on the bottom, it's just 2pi. Plugging in the numbers with the modified formula gives the correct value.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top