Determine the Transition Wavelength

[Samuel Williams]

Hi all, the question is as follows:

1. Homework Statement

From the energy level diagram for OH in the study guide, it can be seen that the first rotationally
excited level of OH (²∏₃₌₂ J =5\2) lies 120 K above the groundstate. What is the wavelength of radiation associated with a transition between the groundstate and the first excited level?

Homework Equations

This is an energy level diagram that contains the relevant wavelengths, the numbers in blue are the transition wavelengths, which is what I require.

I would just like to know how to actually calculate the wavelengths.

The Attempt at a Solution

The difference in frequencies is, for example from J = 5/2 to J = 3/2, 4365MHz where the main line frequencies of 1665MHz and 6030MHz are used. However, this corresponds to a wavelength of 0.068m, which is far off of the 0.000119m actual value. I suspect that I am calculating it incorrectly, I am unsure how to actually do it.

 

[mjc123]

What do you mean by the "mainline frequencies"? Are they the 1.6 GHz and 6 GHz marked on the diagram? These are not the energies of the 3/2 and 5/2 levels; they are the energy differences between the + and - sub-levels of these levels. A wavelength of 119.2 µm corresponds to 2517 GHz.

What is meant by the statement "the first rotationally excited level lies 120 K above the groundstate"? How might you use this information to calculate the answer?

 

[Samuel Williams]

What do you mean by the "mainline frequencies"? Are they the 1.6 GHz and 6 GHz marked on the diagram? These are not the energies of the 3/2 and 5/2 levels; they are the energy differences between the + and - sub-levels of these levels. A wavelength of 119.2 µm corresponds to 2517 GHz.

What is meant by the statement "the first rotationally excited level lies 120 K above the groundstate"? How might you use this information to calculate the answer?

Yes, the mainline frequencies are the corresponding 1.6GHz and 6 GHz as seen. They are referred to as mainline because F = constant for their transitions. Are you referring to thermal excitation, or something with regards to blackbody radiation such as Wien's law? I don't think that is applicable, although I could be wrong. I am frequently afterall

 

[nrqed]

Hi all, the question is as follows:

1. Homework Statement

From the energy level diagram for OH in the study guide, it can be seen that the first rotationally
excited level of OH (²∏₃₌₂ J =5\2) lies 120 K above the groundstate. What is the wavelength of radiation associated with a transition between the groundstate and the first excited level?

Homework Equations

This is an energy level diagram that contains the relevant wavelengths, the numbers in blue are the transition wavelengths, which is what I require.

View attachment 232114

I would just like to know how to actually calculate the wavelengths.

The Attempt at a Solution

View attachment 232115

The difference in frequencies is, for example from J = 5/2 to J = 3/2, 4365MHz where the main line frequencies of 1665MHz and 6030MHz are used. However, this corresponds to a wavelength of 0.068m, which is far off of the 0.000119m actual value. I suspect that I am calculating it incorrectly, I am unsure how to actually do it.

EDIT: I just realized that mcj123 had already made the same comments.
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You cannot combine the 1665 MHz and the 6030 MHz to get the transition J=5/2 to J=3/2. The 1665 MHz gives the splitting between the J=3/2 lines (the more precise values are given in the graph) while the 6030 MHZ gives the splitting between the J=5/2 lines. This tells us nothing about the transition 5/2 to 3/2. The energy of transition between the J=5/2 and the J=3/2 is approximately $85 cm^{-1}$ if we use the vertical axis to estimate the value, which gives a wavelength of $0.000118$ cm. But it is hard to read precisely on the graph since no graduation is shown. However, it seems that the wavelength in $\mu m$ is directly written on the graph, it is the 119.2 number written there.

Note that if you convert 1GHz to the unit $cm^{-1}$, you will get something tiny. The graph is not drawn to scale, the splittings wishing the J=5/2 lines or within the J=3/2 lines are tiny tiny compared to the separation between J=3/2 and J=5/2.

 

[mjc123]

I assume "the first rotationally excited level lies 120 K above the groundstate" means that 120K is the temperature at which E = k_BT. Hence you can work out E, and hence the wavelength.