 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.