**Lines near the band origin in the high-resolution infrared spectrum of **^{12}C^{16}O are observed at 2131.58, 2135.50, 2139.38, 2147.04, 2150.81 and

2154.55 cm^{-1}. Assign J_{initial} and J_{final} quantum numbers to each transition

and, hence, determine the rotational constant of this molecule in both the v = 1 and the v = 0 vibrational levels. Determine the band origin.OK so, in the P branch:

2131.58 J=2

J=3

2135.50 J=1

J=2

2139.38 J=0

J=1

ΔE=hν

_{0}+hB[J(J+1)-J'(J'+1)]

ΔE=hν

_{0}+hB[J(J-1)-J(J+1)]

ΔE=hν

_{0}-2hBJ

R branch:

2147.04 J=1

J=0

2150.81 J=2

J=1

2154.55 J=3

J=2

R branch: ΔE=hν

_{0}+hB[J(J+1)−J′(J′+1)]

ΔE =hν

_{0} +hB [(J+1)(J+2)-J(J+1)]

ΔE =hν

_{0}+2hB(J+1)

These are the equations I've found but I'm guessing that they're not right as I don't have v

_{0}, but I can't do the rotational constant from the line spacing because they're all different? Just need a point in the right direction, not the answer, thanks.