# Spacing/population of rotational states

1. Jan 7, 2013

### dscot

1. The problem statement, all variables and given/known data

This is a two part question I can do about half of each but get a little lost when trying to finish.

I have written all the values below but just in case the full question is here (sorry about clarity) - http://screencast.com/t/jHQTMFnYOhp [Broken]

λ = 308nm
T = 2000K

population of ground state (j=0) is 3.6*1010 cm

(i) So we are trying to calculate the spacing (in cm-1) between the ground rotational state and the j =6.

(ii) The population of the rotational state j =6

2. Relevant equations

(i)
εj = BJ(J+1)
gj = 2J+1

(ii)
Nj/Nd = gj exp( -εj / KT)

3. The attempt at a solution

(i)
gj = (2*6) + 1 = 13 but this is only degeneracy

can also try:

wave number = 1 /λ = 1/308*10-9 but this is only the wave number of j=0 and not he distance between j = 0 --> 6?

(ii)
use:
Nj/Nd = gj exp( -εj / KT)

We know gj, K, T Just need εj but how as we don't have a value for B?

Sorry if this is confusing and please let me know if I can clear anything up!
Thanks
David

Last edited by a moderator: May 6, 2017
2. Jan 7, 2013

### TSny

Hello, dscot. If you know the bond length of OH then you can calculate B. Then you could get the energy spacing between J = 0 and J = 6.

3. Jan 7, 2013

### dscot

Hi TSny =)

I think I understand what you are saying but it doesn't look like the bond length was provided in the question? I'm just guessing here but maybe a clue is that its an OH radical? I could try google but this was a past exam question so I think there must be a way to figure it out?

Thanks!

4. Jan 7, 2013

### TSny

I can't see a way to get B without knowing the bond length.

I also don't see the relevance of the 308 nm wavelength. That seems to me to be the wavelength corresponding to some electronic excitation of OH. I don't see how to use it to help answer the question.

Maybe someone else can help.

5. Jan 7, 2013

### dscot

Hi TSny,

This does seem to be a tricky one, I'll keep trying and hopefully someone else might know :)

Thanks for trying!
David