the classification of stellar spectra

Atoms/Ions in stellar atmospheres are excited and ionized primarily by collisions with atoms/ions/electrons (along with a small contribution from the absorption of photons). The temperature dependence of stellar absorption lines reflect how such collisions affect the excitation and ionization of atoms/ions.

The Classification of Stellar Spectra

Ionization of Atoms/Ions in Stellar Atmospheres Atoms/Ions in stellar atmospheres are ionized primarily by collisions between atoms/ions/electrons.

To ionize an atom/ion from a lower to higher excitation state, the colliding atom/ion/electron must have a kinetic energy equal to or greater than the corresponding ionization energy of the other atom/ion.

To understand how the strength of stellar photospheric absorption lines depend on temperature, we not only have to understand how the distribution of atoms/ions in different excited states depend on temperature but also how their distribution in different ionization states depend on temperature.

Learning Objectives Collisional Ionization of Atoms/Ions

HydrogenHelium

Distribution of Ionization StatesPartition FunctionSaha equationDegree of ionization in stellar atmospheres

Strength of Stellar Absorption LinesCombining collisional excitation and ionization

Relative line strengths of different elements Relative Abundance of Different Elements

Collisional Ionization of Atoms/Ions Consider two hydrogen atoms in their ground state; i.e., with their individual

electrons in the n = 1 quantum state. In a collision between these two atoms, a part of their kinetic energy can be

transferred into ionizing one or both atoms; i.e., cause their individual electrons to make a transition from the n = 1 to a free state.

Consider the solar photosphere, which is at a temperature of 5778 K:- vmp = 9779.1 m/s, corresponding to a kinetic energy of 0.50 eV- ‹v› = 11026.4 m/s , corresponding to a kinetic energy of 0.64 eV - vrms = 11950.4 m/s, corresponding to an average kinetic energy of 0.75 eV

An energy of 13.6 eV (4.3 × vrms) is required to ionize a hydrogen atom from the ground state. Thus, in the solar photosphere, only collisions between hydrogen atoms at the tail of the Maxwell-Boltzmann velocity distribution can ionize hydrogen atoms from the ground state. (Recall that an energy of 3.7 × vrms is required to excite a hydrogen atom from the ground to its first excited state.)

Collisional Ionization of Atoms/Ions Because higher excited states are less populated than lower excited states, one

might think that the ionized state is even less populated than excited states. Yet the fact that hydrogen Balmer lines become weaker from A0 → B → O stars imply otherwise. How can the ionized be more populated than excited states?

Collisional Ionization of Atoms/Ions Because higher excited states are less populated than lower excited states, one

might think that the ionized state is even less populated than excited states. Yet the fact that hydrogen Balmer lines become weaker from A0 → B → O stars imply otherwise. How can the ionized be more populated than excited states?

An increasing fraction of hydrogen atoms are in higher excited states at higher temperatures. Collisional excitation of atoms from an excited state to an even higher excitation state requires less energy than collision excitation of atoms from the ground state. Similarly, collisional ionization of atoms at higher excited states require less energy than collisional ionization of atoms at lower excited states.

Recall that a collision between two atoms can result in either the excitation or de-excitation of an atom. The frequent collisions between atoms result in and maintain a Boltzmann distribution of atoms in different excitation states.

Once an atom is ionized, however, collisions between an atom and an ion (or between two ions) do not “de-ionize” the ion. The i+1 ionization stage no longer participates in the distribution of excited states of atoms in the i ionization stage. It takes much longer for an electron to recombine with an ion than for an electron to de-excite by collisions in stellar atmospheres. (Can you guess why?)

Recall that Canon placed O stars, which have visible lines of ionized helium (He II), …

Note that this photograph is a negative, so that bright lines correspond to absorption lines.

Collisional Ionization of Atoms/Ions

Collisional Ionization of Atoms/Ions Recall that Canon placed O stars, which have visible lines of ionized helium (He II), before B stars, which do not have He II lines but have the most intense neutral helium (He I) lines, and both O and B before A stars despite having weaker H lines.

Note that this photograph is a negative, so that bright lines correspond to absorption lines.

Collisional Ionization of Atoms/Ions Energy diagram of helium and permitted transitions to the n = 1 and n = 2 states. (In this rendering of the energy diagram, the ground state of parahelium is defined to have an energy of 0 eV).

Parahelium Orthohelium

Cannon must have reasoned that it requires more energy to ionize He than H. Around 1895, Carl Runge and Louis Paschen discovered that helium comprised

Collisional Ionization of Atoms/Ions

a mixture of two gases, named parahelium and orthohelium, that produced different spectral lines. Notice that the lines of both parahelium and orthohelium are weaker in O than in B stars, implying that both species are more highly ionized in B than in O stars. Ionization of parahelium requires more energy than hydrogen.


HydrogenHelium




Distribution of Ionization States Atoms/Ions of a gas can gain energy during a collision in the form of

- kinetic energy- excitation of an electron to a higher energy level- removal of an electron to a free state (ionization)

Atoms/Ions of a gas also can lose energy during a collision in the form of- kinetic energy- de-excitation of an electron to a lower energy level (energy transferred as

kinetic energy to another atom) The collision between an ion and electron can also result in recombination. In a collection of atoms/ions excited and ionized by collisions, the distribution of

excited and ionization states depends on the distribution in kinetic energies of the atoms. In stellar atmospheres, the distribution in speeds of the impacting atoms − given by the Maxwell-Boltzmann velocity distribution − produces a definite distribution of excited and ionization states.

The Partition Function Recall that the ratio of the number of atoms ni with a higher energy Ei to the

number of atoms nj with a lower energy Ej in different states of excitation is given by the Boltzmann equation

Expressed in terms of nj = n1, the number of atoms in the ground state,

The total number of atoms in all states is therefore

Eij = Ei - Ej

Partition Function, Z

Ei = Ei – E1

The Partition Function By defining the partition function as

the number of atoms in each excited state can be written in terms of the partition function

The above equation tells us that the number of atoms in a given excited state i depends on the distribution of atoms in different excited states as encoded in the partition function Z. What is the reason for this dependence?

Ei = Ei – E1



The above equation tells us that the number of atoms in a given excited state i depends on the distribution of atoms in different excited states as encoded in the partition function Z. What is the reason for this dependence? This dependence reflects the excitation/de-excitation of electrons to a given state i from lower and higher excitation states.

Ei = Ei – E1



The partition function encodes the statistical properties (in this case, distribution of excited states) of a system in thermal (more accurately, thermodynamic) equilibrium. As an illustration, if the hydrogen atom had only one possible electron orbital (n = 1), what would the partition function be?

Ei = Ei – E1



The partition function encodes the statistical properties (in this case, distribution of excited states) of a system in thermal (more accurately, thermodynamic) equilibrium. As an illustration, if the hydrogen atom had only one possible electron orbital (n = 1), what would the partition function be? Z = 2, as there are two degenerate levels (electron spin up or spin down).

Ei = Ei – E1

The Partition Function The partition function (returning to the notation used in the textbook)

Let us compute the partition function for hydrogen in the solar photosphere, where the temperature is 5778 K. What would you expect the partition function to be in this case?


Let us compute the partition function for hydrogen in the solar photosphere, where the temperature is 5778 K. Recalling that gj = 2 j2 , E1= -13.6 eV, and Ej = -13.6 eV/ j2, each term is given by 2 j2 exp[-13.6(1-1/j2)/0.5]

- the 1st term, 2 exp(-0) = 2- the 2nd term, 8 exp(-20.4) = 1.1 ×10-8

- the 3rd term, 18 exp(-24.2) = 5.7 ×10-10

-the 4th term, 32 exp(-25.5) = 2.7 ×10-10

Thus Z 2, reflecting that most of the hydrogen atoms are in the ground state, as ≅given by the Boltzmann equation

- N2/N1 = 5.0 × 10-9

- N3/N1 = 2.5 × 10-10

- N4/N1 = 6.0 × 10-11


Let us compute the partition function for hydrogen in the solar photosphere, where the temperature is 5778 K. Recalling that gj = 2 j2 , E1= -13.6 eV, and Ej = -13.6 eV/ j2, each term is given by gj exp[-13.6(1-1/j2) 2.0]

- the 1st term, 2 exp(-0) = 2- the 2nd term, 8 exp(-20.4) = 1.1 ×10-8

- the 3rd term, 18 exp(-24.2) = 5.7 ×10-10

-the 4th term, 32 exp(-25.5) = 2.7 ×10-10

- the 5th term, 50 exp(-26.1) = 2.3 ×10-10

- the 6th term, 72 exp(-26.4) = 2.4 ×10-

10 - the 7th term, 98 exp(-26.6) = 2.6 ×10-10

-the 100th term, 20000 exp(-27.2) = 3.1 ×10-8

-the 1000th term, 2 × 106 exp(-27.2) = 3.1 ×10-6

- the 100,000th term, 2 × 1010 exp(-27.2) = 0.031- the 1,000,000th term, 2 × 1012 exp(-27.2) = 3.1

Eq. (8.7) diverges: that is, if we include the contribution by very high order terms, Z → ∞! In practice, based on physical arguments, we should not include the contribution by such high order terms. Why not?

Collisional Ionization of Atoms The ratio of the number of atoms in ionization stage i + 1

(i = 0, 1, 2, … depending on the element) to the number of atoms in stage i is given by the Saha equation

E.g., in the case of hydrogen NI is neutral hydrogen, and NII is ionized hydrogen. This equation was first derived by the Indian astrophysicist Meghnad Saha in 1920.

The Saha equation is derived by equating the ionization and recombination rate for the i and i+1 ionization stage.

Meghnad Saha, 1893-1956

ionization energy from ground state




Notice that the number fraction of ions to neutrals depends on - ionization energy; as χ1 ↑, Ni+1/Ni

- temperature; as T ↑, Ni+1/Ni

- ne; as ne ↑, Ni+1/Ni - the partition function for the

atoms/ions; why?






Notice that the number fraction of ions to neutrals depends on - ionization energy; as χ1 ↑, Ni+1/Ni ↓

- temperature; as T ↑, Ni+1/Ni

- ne; as ne ↑, Ni+1/Ni - the partition function for the atoms/ions; why?







- temperature; as T ↑, Ni+1/Ni ↑- ne; as ne ↑, Ni+1/Ni

- the partition function for the atoms/ions; why?







- temperature; as T ↑, Ni+1/Ni ↑- ne; as ne ↑, Ni+1/Ni ↓ since there are more

electrons per volume with which the ions may recombine (i.e., the recombination rate increases with increasing ne) - the partition function for the atoms/ions; why?







- temperature; as T ↑, Ni+1/Ni ↑- ne; as ne ↑, Ni+1/Ni ↓ since there are more

electrons per volume with which the ions may recombine (i.e., the recombination rate increases with increasing ne) - the partition function for the atoms/ions; it takes less energy to ionize an atom

from a higher than lower excited state, and this dependence is encoded in the partition function (which reflects the distribution of excited states)






The Saha equation is often expressed in terms of electron pressure rather than electron density through the ideal gas equation Pe = nekT so that



Degree of Ionization in Stellar Atmospheres Let us compute the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.

In this temperature range, the partition function ZI = 2 (most of the hydrogen atoms are in the ground state, and neglecting high order terms). A hydrogen ion is just a proton, and has no energy levels or degeneracy so that ZII = 1.

Inserting into the Saha equation, we can solve for NII/NI; i.e., the ratio of ionized to neutral hydrogen.

Degree of Ionization in Stellar Atmospheres Let us compute the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.

In this temperature range, the partition function ZI = 2 (most of the hydrogen atoms are in the ground state, and neglecting high order terms). A hydrogen ion is just a proton, and has no energy levels or degeneracy so that ZII = 1.

Inserting into the Saha equation, we can solve for NII/NI; i.e., the ratio of ionized to neutral hydrogen.

To compute the fraction of hydrogen ions to the total number of hydrogen (atoms+ions), we write

Degree of Ionization in Stellar Atmospheres In stellar photospheres, hydrogen atoms are ionized over a narrow range of temperatures around 10,000 K.

Degree of Ionization in Stellar Atmospheres In stellar photospheres, hydrogen atoms are ionized over a narrow range of temperatures around 10,000 K.

At a gas temperature of 10,000 K: -vmp = 12840.3 m/s, corresponding to a kinetic energy of 0.86 eV- ‹v› = 14488.7 m/s , corresponding to a kinetic energy of 1.10 eV - vrms = 15726.2.4 m/s, corresponding to an average kinetic energy of 1.29 eV

An energy of 13.6 eV (3.2 × vrms) is required to ionize a hydrogen atom from the ground state. Thus, collisions that result in ionization involve hydrogen atoms at the tail of the Maxwell-Boltzmann velocity distribution, as well as between hydrogen atoms with lower speeds and excited hydrogen atoms.

Degree of Ionization in Stellar Atmospheres Notice that the fraction of ionized hydrogen rises significantly just as the fraction of hydrogen atoms in the first excited state starts to rise. Why?

Collisional Excitation Collisional Ionization

Degree of Ionization in Stellar Atmospheres Notice that the fraction of ionized hydrogen rises significantly just as the fraction of hydrogen atoms in the first excited state starts to rise. Why?

It takes 10.2 eV to excite hydrogen from the ground to the first excited state, but only another 3.4 eV to ionize hydrogen from the first excited state. Thus, once there is a significant fraction of hydrogen atoms in the first excited state, it becomes relatively easy to ionize hydrogen.



HydrogenHelium




Combining Collisional Excitation and Collisional Ionization The strength of absorption lines in stellar spectra – i.e., the number of atoms/ions in the ground or excited state involved in the transition that produces a particular absorption line – depends on the effects of both collisional excitation and collisional ionization.


Combining Collisional Excitation and Collisional Ionization The strength of absorption lines in stellar spectra – i.e., the number of atoms/ions in the ground or excited state involved in the transition that produces a particular absorption line – depends on the effects of both collisional excitation and collisional ionization.

E.g., consider the Balmer absorption lines of hydrogen in stellar spectra, produced by hydrogen atoms in the first excited state. As the temperature increases, the fraction of hydrogen atoms in the first excited state to that in the ground state increases, and so the strength of Balmer absorption lines also increases. As the temperature increases, however, an increasing fraction of hydrogen atoms become ionized, leaving fewer neutral hydrogen atoms available to produce (Balmer) absorption lines. Thus, the combination of both collisional excitation and collisional ionization determines how the strength of absorption lines depend on gas temperature, and therefore how the strength of absorption lines in stellar spectra depend on stellar effective temperature.

Combining Collisional Excitation and Collisional Ionization Consider the Balmer absorption lines of hydrogen in stellar atmospheres, produced by the absorption of photons by hydrogen atoms in the first excited state (i.e., electron in the n = 2 orbital).

The strength of the Balmer lines depends on N2/Ntotal, the fraction of all hydrogen particles that are in the first excited state.

Because nearly all hydrogen atoms are in the ground state and most of the remainder in the first excited state so that N1 + N2 ≅ NI, we can write

Compute from Boltzmann equation

Compute from Saha equation

Total number of neutral hydrogen atoms

Total number of hydrogen atoms and ions

(NI + NII)

Plot of N2/Ntotal in stellar atmospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2.

Combining Collisional Excitation and Collisional Ionization


Plot of N2/Ntotal in stellar atmospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2. Do you now understand why the strength of Balmer absorption lines reaches a maximum at spectral type A0?

Combining Collisional Excitation and Collisional Ionization

In the spectra of the Sun and other G-type stars, the absorption lines of calcium are far stronger than the Balmer lines of hydrogen. Why is this the case, given that the cosmic abundance of calcium is far smaller than that of hydrogen?

Relative Line Strengths of Different Elements

Let us first compute the fraction of ionized to neutral hydrogen. The electron pressure in the solar photosphere is 1.5 N m-2.

From the Saha equation, the ratio of ionized to neutral hydrogen

Thus, there is only one hydrogen ion (H II) for every 13,000 hydrogen atoms (H I). Almost all the hydrogen in the solar photosphere is neutral.


This is in agreement with the previous plot (redisplayed below) showing the degree of ionization in stellar photospheres over the temperature range 5,000 K to 25,000 K, assuming for simplicity pure hydrogen gas and a constant electron pressure of Pe = 20 N m-2


Let us now compute the fraction of hydrogen atoms in the first excited state (capable of producing the Balmer absorption lines) to the ground state.

From the Boltzmann equation, the ratio of hydrogen atoms in the first excited state to the ground state

Thus, there is only one hydrogen atom in the first exited state to every 198,000,000 hydrogen atoms in the ground state. Almost all the hydrogen atoms in the solar photosphere are in the ground state.


Distribution of Excited States This is in agreement with the earlier plot (redisplayed below) of the relative

number of hydrogen atoms in the first exited state to the total number of hydrogen atoms in the ground and first excited states.

Combining the Botzmann and Saha equations, the fraction of hydrogen atoms in the first excited state (capable of producing the Balmer absorption lines) to the total number of hydrogen (atoms+ions) (employing N1 + N2 ≅ NI) is

Thus, there is only one hydrogen atom in the first excited state (capable of producing Balmer absorption) to every 200,000,000 hydrogen particles (mostly neutral and in the ground state) in the solar photosphere.


= 1/200,000,000

Distribution of Excited States This is in agreement with the earlier plot (redisplayed below) of the relative

number of hydrogen atoms in the first exited state compared to the total number of hydrogen (atoms and ions).

Let us compute the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium atoms in the solar photosphere.

The ionization energy of neutral calcium (to single-ionized calcium) is 6.11 eV, only about half of the 13.6 eV ionization energy of hydrogen.

From the Saha equation, the ratio of singly-ionized to neutral calcium is (ZI = 1.32 and ZII = 2.30 for calcium)

Practically all of the calcium atoms in the solar photosphere are singly-ionized (in the form of Ca II), with only one of every 918 calcium atoms remaining neutral (in the form of Ca I). Note: the ionization energy of Ca II to Ca III is 11.9 eV, and so there are very few doubly-ionized calcium atoms (Ca III).


Let us now compute the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium atoms.

The first excited state of Ca II is E2 - E1 = 3.12 eV above the ground state.

From the Boltzmann equation, the ratio of Ca II atoms in the first excited state to the ground state

Thus, there is only one Ca II atom in the first exited state to every 264 Ca II atoms in the ground state. Almost all of the Ca II atoms in the solar photosphere are in the ground state.


Combining the Boltzmann and Saha equations, the fraction of singly-ionized calcium atoms (Ca II) in the ground state (capable of producing the Ca II H and K absorption lines) to the total number of calcium (atoms and ions) is

Thus, 995 out of 1000 calcium atoms are singly ionized and in the ground state. Almost all calcium atoms in the solar photosphere are singly-ionized and in the ground state.


[ ]Ca II

There are 500,000 hydrogen atoms to every calcium atom in the solar photosphere. However, the fraction of hydrogen atoms in the first excited state to all hydrogen is only 1/200,000,000. By comparison, virtually all calcium atoms in the solar photosphere are singly-ionized and in the ground state. Thus, for every hydrogen atom in the first excited state capable of producing Balmer absorption line, there are 400 singly-ionized calcium atoms in the ground state capable of producing the Ca II H and K lines.

This is the reason why the Ca II Ha and K lines in the spectrum of the Sun is so much stronger than the Balmer lines of hydrogen.


Plot of the strength of various spectral lines in stellar atmospheres as a function of stellar effective temperature as calculated from the Boltzmann and Saha equations.



HydrogenHelium




How do we know the cosmic abundance of the elements? Until the mid-1920’s, the accepted wisdom was that the

relative abundance of different elements in the Sun was the same as that on the Earth, ~65% iron and ~35% hydrogen.

Relative Abundance of Different Elements

Solar Spectrum

If we know the relative abundances of different elements and how their line strengths depend on temperature (through the Boltzmann and Saha equations), we can compute the relative line strengths of different elements.

Similarly, if we know the relative line strengths of different elements (from stellar spectra) and how their line strengths depend on temperature (through the Boltzmann and Saha equations), we can compute the relative abundances of different elements.

Relative Abundance of Different Elements

The first person to do so was Cecilia Payne, who (using the Boltzmann and Saha equations) in 1925 calculated the abundance of 18 elements in the stellar photospheres in her PhD thesis. She showed for the first time that hydrogen is the most abundant element in the Universe.

Cecilia Payne, 1900-1979

the classification of stellar spectra

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