Morgan-Keenan Spectral Classification (O,B,A,F,G,K,M, hi to lo T)

Based on the appearance of the spectra of stars, a spectral classification scheme was devised in the late 1800s and the early 1900s at Harvard in a group led by Annie Jump Cannon. The criteria used to define the sequence were based primarily on the strengths of the hydrogen Balmer lines but other features were also considered. Today, other criteria are used and so the ordering is rather more obscure. The ordering is O, B, A, F, G, K, M . This is a temperature sequence starting from the hottest stars at O going to the coolest stars at M.

Note that different spectral lines are seen and that the strengths of the lines are quite different. Below is a rough description of the spectral classes:

Spectral Class	Temperature	Comments
O	28,000-50,000 K	ionized atoms, especially helium
B	10,000-28,000 K	neutral helium, some hydrogen
A	7,500-10,000 K	strong hydrogen, some ionized metals
F	6,000-7,500 K	hydrogen and ionized metals, such as calcium and iron
G	5,000-6,000 K	ionized calcium and both neutral and ionized metals
K	3,500-5,000 K	neutral metals
M	2,500-3,500 K	strong molecular lines, e.g., titanium oxide, and some neutral calcium

• a graphical representation of the variation in the strengths of the spectral lines.

We know that the chemical compositions of most stars are roughly the same, why do stars show such different spectra? To clear up this point, we discuss the hydrogen lines.

The hydrogen lines are weak in cool stars, increase in strength as the temperature increases, reach a peak around the A stars, and then weaken at higher temperatures.

Question: Why don't all stars show strong hydrogen lines if hydrogen is the most abundant element in the Universe?

The reason has to do with the energy level structure of hydrogen and the temperatures of the stars.

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• Because electrons are a lazy sort, they like to reside in the lowest energy level. If they are in a higher state (excited state), they quickly return to the lowest state (the ground state). The only way to populate an excited state is for collisions with high energy particles (other atoms or electrons) or absorption of photons to occur rapidly enough to overcome the tendency for all things to reside in the ground state.

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• For cooler stars, the electrons in most hydrogen atoms are in the ground state. Therefore, if an absorption occurs, a Lyman line is produced. Because Lyman lines correspond to large transitions, the lines fall in the ultraviolet and will not appear in the optical portion of the spectrum. As a result, in cool stars one does not see strong hydrogen lines, because not enough electrons reside in the n = 2 level.

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• For warmer stars, the number of electrons which can be pushed into the n = 2 level increases and consequently, the number of absorptions by electrons in the n = 2 level increases (stronger Balmer lines are produced). Balmer transitions are less energetic than Lyman transitions and fall in the optical portion of the spectrum.

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• The strength of the hydrogen lines does not increase forever, however. Why not? Well, once stars start to get hot enough, the collisions between hydrogen atoms and photons (and perhaps with some particles) in their atmospheres are so energetic that they will liberate electrons, that is, the atoms become ionized. Since lines are produced by electrons changing energy states, if an atom has no electrons then it cannot produce lines! Because of this, in the very hottest stars, hydrogen lines are not seen.

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The competition between excitation and ionization leads to the maximum hydrogen line strength in stars with T roughly 8,000-10,000 K, A stars.

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Similar arguments can be made for the other lines seen in stellar spectra.

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