Ch 8: Stars & the H-R Diagram

Nick Devereux 2006Revised 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 eV2. The wavelength of the H emission line is 6563Å

where 1Å = 10-10 m

You can figure out everything about Hydrogen from these twofacts and knowing that the energy difference between two electronic states, E, is proportional to

E 1/n2

Where n is the principal quantum number

For Example

The Habsorption line results from the electron jumping fromthe n=2 to n=3 level. We can use this fact (#2) to calculatethe constant of proportionality, R

E = hc constant [ 1/ n1

2 – 1/n22 ]

So that, 1R [ 1/ n1

2 – 1/n22 ]

Substitute n1= 2 and n2 = 3 and = 6563Å

to yield R = 1.097 x 10-3

which is known as the Rydberg Constant

Now you can calculate the wavelength for all other

electronic transitions

Since, 11.097 x 10-3 [ 1/ n1

2 – 1/n22 ]

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 SpectraThe 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 Z2 R[ 1/ n12 – 1/n2

2 ]

So, the energy levels scale by Z2 compared to H.

e.g. He has Z=2 , Z2 = 4, so the Lyman transition in He I

may be calculated using;

1Z21.097 x 10-3 [ 1/ n12 – 1/n2

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=2state;

11.097 x 10-3 [ 1/ n12 – 1/n2

2 ]

Substitute n1 = 2, n2 = ∞, to yield = 3646 Å

Summary

Stars are made mostly of Hydrogen. So, the stellar continuumexhibits Hydrogen absorption lines which modifies it from aperfect 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 theH 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 thetemperature of the star. If the star is too hot, all the H is ionizedand 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 definesanother 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 EquationN2 = g2 e –h/kT

N1 g1

Where:

N1 = number of atoms in n = 1 level

N2 = number of atoms in n = 2 level

g1 = 2 (statistical weight = degeneracy = 2n2)

g2 = 8

T = temperature

= 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 EquationNII = 2 A (kT)3/2 ZII e –h/kT

___ _________________

NI ne ZI

Where=

NI = number of neutral atoms

NII = number of ionized atoms

ZI = partition function for neutral atom (= 2 for H in gd. State)

ZII = partition function for ionized atom (=1 for H)

ne= electron number density ( e/m3)

A is a constant = 2 me/h3

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 ~ 104 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 ~ 104 K. So, the Balmer absorption lines are strong only in the atmospheres of stars with temperatures ~ 104 K. (see previous figure on slide 17).

The Harvard Spectral sequenceO, 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 distanceto 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 incorrectlyimplies some measure of stellar parallax. Actually, what happensis we obtain a spectrum for a star of unknown distance. We use thespectrum 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, Mv. The absolute magnitude combined with the apparent magnitude, mv, allows the distance to be determined using

mv - Mv = 5 log d - 5.

The Second Method is calledMain Sequence Fitting

In this method, we plot a B-V color vs. apparent magnitudediagram 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 sequencesoverlap.

The difference between the cluster apparent magnitudeand the calibrated H-R diagram absolute magnitude isthe distance modulus mv - Mv which yields the distance tothe cluster using mv - Mv = 5 log d - 5.

The M-K Luminosity Classification scheme