[bmu123]

Really stuck on this, any help would be appreciated!
(i) So is this asking me to work out the wavenumber for the J=0 J=2 transition?
Using wavenumber = Te + 1/2We' - 1/4WeXe' - 5/2We" + 25/4WeXe"
And putting the A values in for We' and WeXe' and the X ones in for We" and WeXe". I thought Te would be Te'-Te" so 14068.3 and from this I get a wavenumber of 13300.8425 cm^-1

(ii) So would this be the same thing for P(4) working it out for J=3 J=4
Te + 7/2We' - 49/4WeXe' - 9/2We" + 81/4WeXe" = 13414.7475 cm^-1
and R(3) for J=3 J=2
Te + 7/2We' - 49/4WeXe' - 5/2We" + 25/4WeXe" = 14416.6575 cm^-1

(iii) Is the band head in the R branch and the band with the highest wavenumber? How do I know the value of J"?

(iv) weaker because J=0 to J=0 transitions are not allowed. The vector angular momentum must change by one unit in a electronic transition, j=0 to j=0 can't happen because there is no total angular momentum to re-orientate to get a change of 1.

[mjc123]

You need to clearly distinguish between vibrational and rotational quantum numbers. (2,0) and (0,0) in parts i and iv refer to the vibrational quantum numbers in the two electronic states. (I'm not quite sure which refers to the ground and which the excited state, but if it's an absorption band I would expect n = 0 in the ground state.) Your formula looks OK (and I assume your calculation, though I haven't checked it) on the assumption that n = 2 in the ground state. Check your notes or textbook for the notational convention.

ii and iii are asking about rotational levels in the above electronic-vibrational transition. You should be alerted at once by the fact that both your P line and your R line have wavenumbers higher than the band origin! In this case your formulas are incorrect - you are applying rotational quantum numbers to a vibrational expression. The energy of a rovibronic state is
Te + (n+1/2)We - (n+1/2)²WeXe + BJ(J+1)
Use that to work out the difference between your initial and final states.

[bmu123]

Thanks! Yes I think (i) might be the other way around.
I did this for the P(4) branch first and I must still be going wrong somewhere, so I plug the values in (I think n=0 for both) from the A row for the initial and X for the final
14068.3 + (0+1/2)255.5 - (0+1/2)²1.58 + 0.4948x4(5) = 14205.551
0 + (0+1/2)351.4 - (0+1/2)²2.61 + 0.6691x3(4) = 183.0767
Then the difference between these is 14022.4743 which is still higher than the band origin.

[mjc123]

In part ii you are still dealing with the (2,0) transition.

[bmu123]

OK so I used n=0 for the J'' value (initial) and n=2 for the J' value (final) aand I got 14514.68 cm^-1 for P(4) and 14340.911 cm^-1 for R(3). Looks right so really hoping it is! The next question about the band head, I've read somewhere it's in the R branch but can't find a reason why, and how would I know which J'' value it's for?

[mjc123]

OK so I used n=0 for the J'' value (initial) and n=2 for the J' value (final)

This is either wrong or, if right, confusing. n is the vibrational quantum number; J is the rotational quantum number. You can't substitute one value for the other. If you mean you used n = 0 for the initial state, the state for which the J value is denoted J'', then say that.
Please specify, for my sake if nothing else, which electronic state (A or X) is the initial and which the final state, what is the value of n in each state, and the value of J in each state.
Assuming X is the initial state, with n = 0, and A (n=2) is the final state, I calculate a band origin of 14522.13 cm^-1 and I agree with your value for P(4) (J = 4 3). I'm at a loss how you get a lower value for R(3). Where did that come from?
The band head arises because of the difference between the B values in the two states, which leads to a correction to the transition energy which is quadratic in J, and (depending on the sign of ΔB) will eventually cause either the P band frequencies to start increasing, or the R band frequencies to start decreasing.
For example, let us express the R band frequencies as {band origin} + B'J'(J'+1) - B''J''(J''+1)
= {band origin} + (B''+ΔB)(J''+1)(J''+2) - B''J''(J''+1)
= {band origin} + 2B''(J''+1) + ΔB(J''+1)(J''+2)
If ΔB is negative, the frequency will at some point start to decrease with increasing J''. I leave it to you to work out where this point is.