Discussion
In rotational fine structure of vibration-electronic transitions of a diatomic molecule, the band head will appear in R-branch if :
Answer: A
A band head forms in the R branch if B′ \(\lt\) B″.
In the rotational fine structure of an electronic band, the line positions follow
\[ \begin{aligned} \bar{\nu} &= \bar{\nu}_{0} + (B'+B'')m \\ &+ (B'-B'')m^{2} \end{aligned} \]
with m = J + 1 for the R branch (m positive) and m = −J for the P branch. If B′ \(\lt\) B″, which is common because a bond usually lengthens on excitation, the m2 term is negative. In the R branch the lines first move to higher wavenumber, then turn back once the negative quadratic term wins: they pile up at a band head, and the band is degraded (shaded) to the red.
If B′ \(\gt\) B″, the head forms in the P branch instead. If B′ = B″, there is no quadratic term and no head, and B′ = −B″ is physically impossible, since rotational constants are positive.
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