1. No category

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.
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
Hydrogen
Helium

Distribution of Ionization States
Partition Function
Saha equation
Degree of ionization in stellar atmospheres

Strength of Stellar Absorption Lines
Combining collisional excitation and ionization
Relative line strengths of different elements

Relative Abundance of Different Elements
Learning Objectives

Collisional Ionization of Atoms/Ions
Hydrogen
Helium

Distribution of Ionization States
Partition Function
Saha equation
Degree of ionization in stellar atmospheres

Strength of Stellar Absorption Lines
Combining 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 deexcitation 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?)
Collisional Ionization of Atoms/Ions

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

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
Collisional Ionization of Atoms/Ions

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
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.
Learning Objectives

Collisional Ionization of Atoms/Ions
Hydrogen
Helium

Distribution of Ionization States
Partition Function
Saha equation
Degree of ionization in stellar atmospheres

Strength of Stellar Absorption Lines
Combining collisional excitation and ionization
Relative line strengths of different elements

Relative Abundance of Different Elements
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
Eij = Ei - Ej
Expressed in terms of nj = n1, the number of atoms in the ground state,
Ei = Ei – E1

The total number of atoms in all states is therefore
Partition Function, Z
The Partition Function

By defining the partition function as
Ei = Ei – E1
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?
The Partition Function

By defining the partition function as
Ei = Ei – E1
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? This dependence
reflects the excitation/de-excitation of electrons to a given state i from lower and
higher excitation states.
The Partition Function

By defining the partition function as
Ei = Ei – E1
the number of atoms in each excited state can be written in terms of the partition
function

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?
The Partition Function

By defining the partition function as
Ei = Ei – E1
the number of atoms in each excited state can be written in terms of the partition
function

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).
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?
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. 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
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. 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
Eq. (8.7) diverges: that is, if we
- the 2nd term, 8 exp(-20.4) = 1.1 ×10-8
include
the contribution by very
-10
the 3rd term, 18 exp(-24.2) = 5.7 ×10
high order terms, Z → ∞! In
-the 4th term, 32 exp(-25.5) = 2.7 ×10-10
practice, based
on physical
the 5th term, 50 exp(-26.1)arguments,
= 2.3 ×10-10
we should not include
th
the 6 term,
72 exp(-26.4)
= 2.4
×10-10
the contribution
by such
high
- terms.theWhy
7th term,
order
not? 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
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
The Saha equation is derived by equating the ionization and recombination rate
for the i and i+1 ionization stage.
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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?
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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?
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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?
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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 ↓ 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?
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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 ↓ 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
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
ionization energy
from ground state
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.

Meghnad Saha,
1893-1956
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.
Collisional Excitation
Collisional Ionization
Learning Objectives

Collisional Ionization of Atoms/Ions
Hydrogen
Helium

Distribution of Ionization States
Partition Function
Saha equation
Degree of ionization in stellar atmospheres

Strength of Stellar Absorption Lines
Combining collisional excitation and ionization
Relative line strengths of different elements

Relative Abundance of Different Elements
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.
Collisional Excitation
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
Total number of
neutral hydrogen atoms
Total number of
hydrogen atoms and ions
(NI + NII)
Compute from
Boltzmann equation
Compute from
Saha equation
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.
Collisional Excitation
Collisional Ionization
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?
Relative Line Strengths of Different Elements

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.
Relative Line Strengths of Different Elements

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
Relative Line Strengths of Different Elements

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.
Relative Line Strengths of Different Elements

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
= 1/200,000,000
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.
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).
Relative Line Strengths of Different Elements

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).
Relative Line Strengths of Different Elements

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.
Relative Line Strengths of Different Elements

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
[
]Ca II
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.
Relative Line Strengths of Different Elements

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.
Relative Line Strengths of Different Elements

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.
Learning Objectives

Collisional Ionization of Atoms/Ions
Hydrogen
Helium

Distribution of Ionization States
Partition Function
Saha equation
Degree of ionization in stellar atmospheres

Strength of Stellar Absorption Lines
Combining collisional excitation and ionization
Relative line strengths of different elements

Relative Abundance of Different Elements
Relative Abundance of Different Elements

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.
Solar Spectrum
Relative Abundance of Different Elements

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.

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