= Solution
The sequence primarily orders <effective temperature> rather than chemical composition. Hot O spectra show strong ionized-helium features and highly ionized metals, with relatively weak neutral-hydrogen lines. B <stars> show neutral <helium> and increasingly prominent <hydrogen>. A <stars> have the strongest Balmer absorption. Through F and G, lower-ionization metal lines and calcium features become prominent. K <stars> have strong low-ionization metal absorption and cool-star features; M <stars> develop strong molecular bands such as TiO, while Balmer absorption is weak.
Line <opacity> depends on how many <atoms> occupy both the relevant <ionization> stage and the particular lower level. Boltzmann excitation controls the latter and Saha <ionization> the former, so equal elemental abundances need not give equal line strengths. <Electron> <pressure>, <surface gravity>, broadening and atmospheric <temperature> gradients also affect the spectrum.
For Balmer absorption the lower state is $n=2$, $10.2$ eV above the <hydrogen> ground state. In cool atmospheres its population is exponentially small. At higher <temperatures> it rises, but at still higher <temperature> <ionization> removes neutral <hydrogen>. The competition gives the maximum in A-type <stars>, with <effective temperatures> around $9000$-$10{,}000$ K. This is an excitation-and-ionization effect, not evidence that A <stars> have the largest <hydrogen> abundance.
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