6) SPECTRAL LINE FORMATION

Text Reading: Chpt 8 (Sect 4-5)

A) ATOMIC EXCITATION/DE-EXCITATION MECHANISMS
a) GETTING FROM LOWER LEVEL I TO UPPER LEVEL J:

RADIATIVE EXCITATION	COLLISIONAL EXCITATION
Photon absorption such that: DEij=hn	Free e- collision w/ atom such that: Ee+Ei=Ej+Ee'




b) GETTING FROM UPPER LEVEL J TO LOWER LEVELI:

SPONTANEOUS EMISSION	STIMULATED EMISSION	COLLISIONAL EMISSION
e- in an upper state has finite lifetime in state, and can decay spontateously to lower state.	e- in upper state decays to lower state if atom/ion is bombarded by radiation of energy hn	Free e- induces an excited bound e- to decay to lower state, adding energy to free e-.

c) QUICK OVERVIEW OF SELECTION RULES:

i) FIRST, A REVIEW OF QUANTUM NUMBERS FOR A SINGLE ELECTRON
1) QUANTUM NUMBERS: ®n=principle quantum number ® shell ID

n=	1	2	3	4	5	6	...
shell=	K	L	M	N	O	P	...

each shell can contain a maximum of 2n² e^-

2) ORBITAL ANGULAR MOMENTUM QUANTUM NUMBER ®subshell ID

l=	0	1	2	3	4	5	...	(n-1)
subshell=	s	p	d	f	g	h	...

• each subshell can contain a max of 2(2l+1) e^-

• the orbital angular momentum vector can have 2l+1 orientations in a magnetic field, from -l to +l:

-l £ ml £ l

3) SPIN ANGULAR MOMENTUM QUANTUM NUMBER ®spin direction ("up" or "down")

s=1/2

The spin angular momentum vector can have 2s+1=2 orientations in a B-field.

ms=±1/2

4) TOTAL ANGULAR MOMENTUM QUANTUM NUMBER

j=l±s

The total angular momentum vector can have 2j+1 orientations in a B-field:

-j£Mj£j

Note: What exactly the "l" and "s" here represent are described below in (LS coupling).

ii) NOW A REVIEW OF THE QUANTUM NUMBERS DESCRIBING MULTIPLE ELECTRONS:

1) LS COUPLING (RUSSELL-SAUNDERS COUPLING): ®describes how individual orbital angular momentum vectors of e^- within an atom/ion add to produce a net vector:

• are added vectorially: =å

• are added vectorially: =å

• L and S combine vectorially to give the total angular momentum J: |L-S|£J£L+S

2) SPECTROSCOPIC NOTATION FOR A "TERM":

^(2S+1)L_J

L=	0	1	2	3	4	5	...	(n-1)
state=	S	P	D	F	G	H	...

iii) FINALLY, THE TRANSITION RULES:

1) A TRANSITION IS "ALLOWED" IS NONE OF THE FOLLOWING ARE VIOLATED:

• Dl=±1 (parity rule)

• for terms,

  DS=0

  DL=0,±1

  DJ=0,±1

  J=0 to J=0 forbidden

2) RESONANCE LINES: ®Allowed transitions from E=0. (e.g., CaII K at l3934Å goes from 4s ²S_1/2 to 4p ²P_3/2

3) SEMI-FORBIDDEN LINES: ®transitions in which DS or DL rule is violated (e.g., CII]l2325Å)

4) FORBIDDEN LINES: ®if both DS and DL are violated, and/or the transition involves NO change in parity. (e.g., [OIII]l5006Å)

B) FORBIDDEN LINES

Due to quantum mechanical rules (see above), certain transitions are less likely to occur than others. Therefore, once an electron finds itself in one of these "metastable" levels, it takes a long time to return to the ground state. But once it does finally return to the ground state, a forbidden line is produced. A low density gas is where you are most likely to see such a line (can you guess why?).

C) PROPERTIES OF GAS:

a) KIRCHOFF LAW #2:

Hot gas will produce emission lines. Exactly where those emission lines appear (i.e., at what wavelength they will appear) depends on the type of gas present.



b) KIRCHOFF LAW #3:

If a continuous spectrum passes through a gas that is at a lower temperature, absorption features will result:



c) PRESSURE OF IDEAL GAS

P=nkT

where "k" is the Boltzman constant having a value of 1.38x10^-23 J/K = 8.62x10^-5 eV/K, and "T" is the temperature.

D) CALCULATING THE RELATIVE NUMBER OF ATOMS IN A PARTICULAR EXCITED STATE

a) EXCITATION EQUILIBRIUM

Excitation equilibrium means that the number of atoms in a given excitation state will remain constant with time (in other words, every excitation is balanced by a de-excitation). The relative number of atoms in a given excitation state is given by:



where N_B and N_A are the number density of atoms (#atoms per unit volume) in excitation states B and A, resp., g_B and g_A are the "statistical weights" of states B and A, resp., and E_B and E_A are the energies of levels B and A, resp., and "k" is the "Boltzman constant", and "T" is the temperature of the gas. For hydrogen gas, g_n = 2n².

Example: For hydrogen gas, at what temperature will there be just as many atoms in the 2nd excited state (n=3) as in the ground level?

Well here:

g₃=2(3)²=18 and g₁=2(1)²=2

Also:

E₁=-13.6 eV/1² and E₃=-13.6 eV/3²

Thus, after plugging into the Boltzmann equation, we get (after forcing N₃/N₁=1) that T=6.4x10⁴ K.

b) IONIZATION EQUILIBRIUM

Ionization equilibrium means that the number of electrons that are ionized equals the number of electrons that recombine with ions. The relative number of atoms in a given ionization state is given by:



where N_i+1 and N_i are the number density of atoms (#atoms per unit volume) at ionization levels i+1 and i, resp., A is a "constant" that depends on the element being considered, N_e is the free-electron number density, and X_i is the ionization energy for an atom in ionization state "i".

Example: What fraction of hydrogen gas is ionized, if gas is at a temperature of 10000 K? Assume that A/N_e=1.027x10²⁵?

Using the Saha equation, we get that N_II/N_I=2.34. But this is how much singly ionized atoms there are relative to neutral atoms, NOT how many ionized atoms there are, relative to the total number of hydrogen atoms. However, if we have the fraction N_II/N_I, we can get N_II/N_total

Rewriting,

N_II/N_total=N_II/(N_I+N_II) = ( N_I/N_II+1)^-1

Thus, N_II/N_total=0.7=70%

Note that the ionization of hydrogen is very sensitive to temperature. The ionization fraction goes from 0 to 100% in temperature range of only ~300 K. This explains why Balmer lines are strongest in A-type stars (T = 10000K), despite the fact that a temperature that is 8 times higher is required in order to get substantial excitation into the n=3 and higher energy levels.

c) SAHA-BOLTZMANN EQUATION

This equation calculates the number of atoms that are available to make a particular transition for a spectral line:

N_i,s/N=N_i,s/(N_i-1+N_i+N_i+1)

where N_i,s is the number of atoms in the excited state i and ionization level s, and N is the total number of atoms and ions. This equation makes the approximation that only 3 states of ionization need to be considered.

This equation can be re-written as:



Note that the Boltzmann equation can be used to evaluate some of the above fractions, and that the Saha equation can be used to evaluate the others.

Example: Estimate the relative strengths of the absorption lines due to hydrogen (Balmer lines) and those due to calcium (CaII H and K lines) in the Sun.

The strategy here will be to use the Saha equation to determine the degree of ionization, and the Boltzmann equation to determine the distribution of e^_- between the ground and the 1st excited states.

First, let's do hydrogen, where A=7.89x10²⁵. For the Sun, T=5770 K.

(N_II/N_I)_hydrogen=7.5x10^-5. This means that almost none of the hydrogen is ionized. Now let's determine the relative number of atoms that are in excited states:

Since g₂=2(2)² and g₁=2(1)², then (N₂/N₁)_HI=5x10^-9

This means that very few atoms are in the 1st excited, where they need to be, in order to produce Balmer lines!

N₂/N_total= 5x10^-9

Now do Calcium, where A=2.75x10²⁶ and evergy of ionization=6.11 eV:

(N_II/N_I)_Ca=903

So hardly any of the atoms remain neutral.

For the CaII H line: l=3933Å, E₂-E₁=3.12 eV, g₁=2 and g₂=4.

Thus, (N₂/N₁)_CaII=3.77x10^-3

This means that most of the CaII ions are in the ground state, and capable of producing H & K spectral lines.

Thus:



Therefore, CaII lines are stronger than the Balmer lines, even though there are 5 orders of magnitude more hydrogen atoms than Ca atoms in the Sun.

A plot of the Saha equation for an A-star looks like:



So, given our discussion on the roles of the Boltzmann and Saha equations, can you explain why the Saha-Boltzmann equation for an A-star looks like:



E) SPECTRAL LINE BROADENING

a) NATURAL BROADENING: QUANTUM EFFECT

Consider the Heisenberg uncertainty principle:



where DE is the uncertainty in the energy of a particular energy level (this is sort of the "thickness" of that energy level), and Dt is the amount of time that an electron will stay in a particular energy level before spontaneously decaying to a lower energy state (this is called the "lifetime" of that transition). Let's calculate the width of of an Ha absorption line. (This transition has a lifetime of 10^-8 sec and l=6563Å).



b) THERMAL DOPPLER BROADENING

Consider a Maxwellian velocity distribution:



Where V_p is the "most probable velocity", and is mathematically expressed as:



Now consider the Doppler shift, expressed as:



Thus, the thickness of a spectral feature due to thermal broadening is:



For example, for the Sun's photosphere, where T=5770, the amount of thermal broadening in the H-alpha absorption line is expected to be 0.427Å

c) COLLISIONAL BROADENING

Here, energy levels within single atoms are modified due to the electric field of nearby atoms and ions ("Stark effect"). The greater the density of atoms, the greater the effect of this kind of broadening. This broadening can be expressed as:



where s is the collisional cross section, m is the mass of the atom, and n is the number density of atoms.

d) ZEEMAN EFFECT

Here, the atomic energy levels separate into 3 or more sublevels when an atom is placed in a magnetic field. This phenomena occurs because of the electron's magnetic dipole moment. These multiple energy levels "blur" into one thick level, if we don't have enough spectral resolution in our observing instruments that look at the spectrum.

e) TURBULENT BROADENING

Large-scale random motions at the surface of a star causes Doppler shifts.

f) ROTATIONAL BROADENING

F) HOW ALL THIS AFFECTS STELLAR SPECTRA ...