Ch 8: Stars & the H-R Diagram  Nick Devereux 2006 Revised 9/12/2012

Spectroscopy

Stellar Spectra

Emission & Absorption Lines

The Sun’s Spectrum

The Spectrum of Hydrogen

Electronic Transitions in the Hydrogen Atom

Energy Level Diagram

Two important facts about Hydrogen 1.The ionization potential = 13.6 eV 2.The wavelength of the H  emission line is 6563Å where 1Å = m You can figure out everything about Hydrogen from these two facts and knowing that the energy difference between two electronic states,  E, is proportional to  E  1/n 2 Where n is the principal quantum number

For Example The H  absorption line results from the electron jumping from the n=2 to n=3 level. We can use this fact (#2) to calculate the constant of proportionality, R  E = hc  constant [ 1/ n 1 2 – 1/n 2 2 ] So that, 1  R [ 1/ n 1 2 – 1/n 2 2 ] Substitute n 1 = 2 and n 2 = 3 and  = 6563Å to yield R = x which is known as the Rydberg Constant

Now you can calculate the wavelength for all other electronic transitions Since, 1  x [ 1/ n 1 2 – 1/n 2 2 ] But, remember, that this equation yields  in Å Question: what is the wavelength of the Lyman  transition ? Question: what is the wavelength that corresponds to the ionization potential of hydrogen ?

Ionic Spectra The wavelengths of H like ions, such as He II, Li III, O VIII, can be estimated using the Bohr model of the H atom;  E = -  U/2  E  Z 2 R[ 1/ n 1 2 – 1/n 2 2 ] So, the energy levels scale by Z 2 compared to H. e.g. He has Z=2, Z 2 = 4, so the Lyman  transition in He I may be calculated using; 1  Z 2  x [ 1/ n 1 2 – 1/n 2 2 ] yielding = 304 Å which occurs in the far UV part of the spectrum.

Back to the Sun

The Balmer Discontinuity The Balmer discontinuity is the break in the spectrum at 3646 Å due to the ionization of Hydrogen from the n=2 state; 1  x [ 1/ n 1 2 – 1/n 2 2 ] Substitute n 1 = 2, n 2 = ∞, to yield  = 3646 Å

Summary Stars are made mostly of Hydrogen. So, the stellar continuum exhibits Hydrogen absorption lines which modifies it from a perfect Planck function. The absorption line strengths depend on two things; How many of the H atoms are in each of the excitation levels, n=1, n=2, etc, described by the Boltzmann equation and secondly, how many of the H atoms are completely ionized, for if the H atoms are ionized they can not produce any absorption lines. This is described by the Saha equation.

The Spectral Classification of Stars The relative strength of the absorption lines depends on the temperature of the star. If the star is too hot, all the H is ionized and there are no absorption lines. If the star is too cool, all the H atoms will be in the ground state, for which there are no transitions in the optical part of the spectrum. A star of “medium” temperature will have lots of H absorption lines. The range of H absorption lines strengths described above defines another way to measure stellar temperature.

The Strengths of Absorption Lines

Boltzmann & Saha Equations The strength of the absorption lines can be calculated fairly easily for H and with some difficulty for other species. The calculation is a two step process. Step 1 utilizes the Boltzmann equation to determine the relative number of atoms in the various energy levels with principle quantum level n =1, n=2, etc. Step 2 utilizes the Saha equation to determine the relative number of atoms that are ionized.

Boltmann Equation N 2 = g 2 e –h /kT N 1 g 1 Where: N 1 = number of atoms in n = 1 level N 2 = number of atoms in n = 2 level g 1 = 2 (statistical weight = degeneracy = 2n 2 ) g 2 = 8 T = temperature n = frequency

Application of the Boltzmann Equation For the purposes of this class, we are interested in the Balmer series of H lines that originate from the n=2 level and occur in the optical part of the spectrum. The graph in the textbook, Fig. 8.7, shows the relative number of atoms in the n=2 level, as a function of temperature. The graph shows that very high temperatures are required for significant population of the n=2 level. Why then does the strength of the Balmer lines get weaker as the temperature increases? ie. Why are the Balmer lines so weak in O stars? The answer is, the H atoms are mostly ionized………

Saha Equation N II = 2 A (kT) 3/2 Z II e –h /kT ___ _________________ N I n e Z I Where= N I = number of neutral atoms N II = number of ionized atoms Z I = partition function for neutral atom (= 2 for H in gd. State) Z II = partition function for ionized atom (=1 for H) n e = electron number density ( e/m 3 ) A is a constant = 2  m e /h 3

Application of the Saha Equation The graph in the textbook, Fig. 8.8, shows the fraction of atoms that are ionized as a function of temperature. The graph shows that H becomes ionized at relatively low temperatures corresponding to ~ 10 4 K. The combination of these two equations; Boltzmann and Saha is illustrated in Fig. 8.9 which shows that the population of the n=2 level is sharply peaked at ~ 10 4 K. So, the Balmer absorption lines are strong only in the atmospheres of stars with temperatures ~ 10 4 K. (see previous figure on slide 17).

The Harvard Spectral sequence O, B, A, F, G, K, M The spectral sequence is a much more precise way to measure the temperature of stars than the B-V color index, discussed previously. The reason is because the B-V color index is directly affected by the stellar absorption lines which leads to an “incorrect” temperature, since stars are not perfect blackbodies after all.

The Hertzsprung-Russell Diagram This diagram requires a measurement of the distance to each star ( to get Mv) and a spectrum for each star ( to get the Spectral type ).

Two More Methods to Measure Distances to Stars Spectroscopic Parallaxes The name of this method is a bit deceiving as it incorrectly implies some measure of stellar parallax. Actually, what happens is we obtain a spectrum for a star of unknown distance. We use the spectrum to determine the spectral type, which locates it on the x-axis of the H-R diagram. Now draw a line up to the main sequence, and continue it horizontally to determine it’s absolute magnitude, M v. The absolute magnitude combined with the apparent magnitude, m v, allows the distance to be determined using m v - M v = 5 log d - 5.

The Second Method is called Main Sequence Fitting In this method, we plot a B-V color vs. apparent magnitude diagram for a cluster of stars of unknown distance. This graph is then shifted vertically over a copy of a calibrated H-R diagram until the two main sequences overlap. The difference between the cluster apparent magnitude and the calibrated H-R diagram absolute magnitude is the distance modulus m v - M v which yields the distance to the cluster using m v - M v = 5 log d - 5.

The M-K Luminosity Classification scheme