Lecture3 Spectra

7/24/2019 Lecture3 Spectra

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Spectra

Original: Michael Balogh, Univ. Waterloo


2/33

Stellar spectraStellar spectra show interesting trends as a function of temperature:

Increasingtemperature


3/33

Review: spectral classesSpectral

Type

Colour Temperature

(K)

Main characteristics Example

O Blue-white >25000 Strong HeII absorption (sometimesemission); strong UV continuum

10 Lacertra

B Blue-white 11000-25000 HeI absorption, weak Balmer lines Rigel

A White 7500-11000 Strongest Balmer lines (A0) Sirius

F Yellow-white 6000-7500 CaII lines strengthen Procyon

G Yellow 5000-6000 Solar-type spectra Sun

K Orange 3500-5000 Strong metal lines Arcturus

M Red


4/33

The HR diagram revisited

Henry Norris original diagram, showing stellar

luminosity as a function of spectral class.

The main sequence is clearly visible

Spectral Class

O B A F G K M

Luminosity


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Luminosity class

The lumino s i ty c lassis assigned a roman numeral: I II III

IV V or VI and is related to the width of the spectral linewhich we will see is related to the stellar luminosity


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Spectroscopic parallax

In principle, you can identify both

the spectral class and theluminosity class from the spectrum.

It is therefore possible to locate

the stars position on the HR

diagram and determine its

absolute magnitude. This can be

used to determine the distance to

the star. This method is known as

spectroscopic parallax


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Example

The star Rigel has a spectral type B8Ia and a magnitude

V=0.14. What is its distance?


8/33

Kirchoffs laws

1. A hot, dense gas or hot solid object

produces a continuous spectrum with no

dark spectral lines

2. A hot, diffuse gas produces bright spectral

emiss ion l ines

3. A cool, diffuse gas in front of a source of acontinuous spectrum produces dark

absorpt ion l inesin the continuous

spectrum


9/33

Review: The hydrogen atom

eV6.132n

En

where n is the principal quantumnumber.

2

1

2

2

21

116.13

nn

eVEEE

2

1

2

2

11

16.91

11

nnnm

The change in energy is

associated with the

absorption or emission of aphoton. For Hydrogen:

Thus the energy difference between twoorbitals n1 and n2 of Hydrogen is givenby:


10/33

The Boltzmann factor

The probability that an electron is

in a given energy level dependson the Boltzmann factor:

kTEE

kT

E

kT

E

a

bab

a

b

e

e

e

sP

sP

)(

)(

kT

E

e

Thus the ratio of the probability that an electron is in state sb to the probabilitythat it is in state sa is just:


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1. n is the principal quantum

number related to the

energy of the orbital

2. The angular momentum is

quantized to have values of

3. The z-component of angularmomentum can only have

values of

where mlis an integer

betweenland l

The quantum atom

Electron probability distributions are described by orbitals that

are specified by three quantum numbers:n l m

l

)1( llL

lz mL


12/33

1. is the principal quantum





3. The z-component of angularmomentum can only have

values of


betweenland l

The quantum atom



l

)1( llL

lz mL


13/33





quantized to have values

of

3. The z-component of angular

momentum can only have

values of


betweenland l

The quantum atom



l

)1( llL

lz mL


14/33






3. The z-component ofangular momentum can

only have values of

where ml

is an integer

between

l and l

The quantum atom



l

)1( llL

lz mL


15/33

Degeneracies

There may be more than one state with the same energyE. For example, for an isolated Hydrogen atom thequantum numbers associated with spin and angularmomentum do not affect the energy

kT

EE

a

bab

esP

sP

)(

)(

n l ml

ms

E (eV)

1 0 0 +1/2 -13.6

1 0 0 -1/2 -13.6

2 0 0 +1/2 -3.4

2 0 0 -1/2 -3.4

2 1 0 +1/2 -3.42 1 0 -1/2 -3.4

2 1 1 +1/2 -3.4

2 1 1 -1/2 -3.4

2 1 -1 +1/2 -3.4

2 1 -1 -1/2 -3.4


16/33

Degeneracies

Therefore the probability that a system will be found in any state with energy Eb,

relative to the probability that it will be found in any state with energy Ea is:

kT

EE

a

b

kT

E

a

kT

E

b

a

bab

a

b

eg

g

eg

eg

EP

EP

)(

)( n l ml ms E(eV)

1 0 0 +1/2 -13.6

1 0 0 -1/2 -13.6

2 0 0 +1/2 -3.4

2 0 0 -1/2 -3.4

2 1 0 +1/2 -3.4

2 1 0 -1/2 -3.4

2 1 1 +1/2 -3.4

2 1 1 -1/2 -3.4

2 1 -1 +1/2 -3.4

2 1 -1 -1/2 -3.4

For the Hydrogen atom only, the degeneracy

depends on the energy level n like:

22)( nng


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The Boltzmann equation

Since the number of atoms in stellar atmospheres is so large, the ratio ofprobabilities is essentially equal to the ratio of atoms in each state. This is the

Bol tzmann equat ion:

kT

EE

a

b

a

bab

eg

g

N

N


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Example

For a gas of neutral Hydrogen atoms at room temperature,

what is the ratio of the number of electrons in the n=2state to the number in the n=1 state?

What temperature do you need to get a significant number

(say 10%) of electrons into the n=2 state (for a neutral

Hydrogen gas)?


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Puzzling

The Balmer sequence of absorption lines is due to the transition from n=2 ton>2.

The strength of the Balmer lines is largest for A0 stars, which havetemperatures ~9520 K

But we just found you need temperatures ~3 times larger than this to get even10% of electrons into the n=2 state; and this n=2 population will increasefurther with increasing temperature.


20/33

Break


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Ionization

So far we have just dealt with neutralatoms. However, if

the temperature gets high enough, electrons can beentirely removed from the atom.

Lets define ci to be the energy required to remove an electron from an

atom. This increases its ionization state from i to i+1

Eg. It takes 13.6 eV of energy to remove an electron in the ground stateof Hydrogen. So cI=13.6 eV.

Ionization states are usually denoted by Roman numerals. So the

neutral hydrogen atom is HI and the first ionization state is HII and so

on.


22/33

Partition functions

We want to compute the number of atoms in ionization state i+1 relative to the

number in ionization state i.

To do this we need to sum over all possible orbital distributions of each state.

i.e. how could the electrons be distributed in each ionization state?

The sum of the number of configurations, weighted by the probability of each

configuration, is thepartition function:

kT

EE

j

j

j

eggZ1

2

1


23/33

The Saha Equation

In 1920, Meghnad Saha derived an equation for the relative number of

atoms in each ionization state. Well just present the result:

kTe

ie

i

i

ii

eh

kTm

Zn

Z

N

N c

2/3

2

11 22

Note:

This depends on the number density of electrons, ne. This is because

as the number of free electrons increases, it is more likely that they can

recombine with an atom and lower the ionization state.

The Boltzmann factor exp(-ci/kT) means it is more difficult to ionize

atoms with high ionization potentials


24/33

Example: typical Hydrogen atmospheres

4

16.1

2/3

4

1

320

171

m101083.4 Te

i

i

i

i

i

eT

n

Z

Z

N

N c

32010

eV6.13

mne

Ic

Evaluate the partition functions


25/33


4

16.1

2/3

4

1

320

171

m101083.4 Te

i

i

i

i

i

eT

n

Z

Z

N

N c

32010

eV6.13

mne

Ic

Evaluate the partition functions

Hydrogen has only one electron, so there is only HI (neutral) and HII (ionized).

HII is just a proton: there is only one state, so Z II=1

We saw that, for T


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4/11777.152/344.3

T

I

II eT

N

N

T(K) T4 NII/NI NII/(NI+NII)

5500 0.55 3.410-6 3.410-6

8000 0.8 0.047 0.0459000 0.9 0.50 0.33

10000 1 3.4 0.77

15000 1.5 1201 0.9992

20000 2 25640 0.99996

where T4 =T/(10,000K)


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In the interior of stars, temperature decreases from the core to the surface. The narrowregion inside a star where Hydrogen is partially ionized is called the hydrogen part ial

ionizat ion zone

Ionization fraction as a

function of temperaturefor 3 different electron

densities:

ne=1021 m-3

ne=1020 m-3

ne=1019 m-3


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Balmer line formation

Calculate the relative strength of the Balmer absorption

lines as a function of temperature

IIItotal NNNNNN

/1)1/(

1

21

2

T N1/N2 NII/NI N2/Ntotal

4000 1.6x1012 4.3x10-11 6.2x10-13

5000 4.4x10

9

1.6x10

-7

2.3x10

-10

9000 1.2x105 0.50 5.5x10-6

15000 652 1203 1.3x10-6

20000 91 2.6x104 4.2x10-7


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Balmer line formation

This shows why Balmer lines are strongest at ~9000 K.

They quickly get weaker at higher temperatures becausethe ionization fraction increases.


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Boltzmann and Saha equations: applicability

Saha equation depends on electron density

10% of the atoms in a real star are Helium. Ionized Heliumincreases ne, and therefore decreases the ionization fraction of

H at a fixed temperature

These equations only apply if the gas is in thermal

equilibrium

The energy levels of H were calculated for an isolated

atom. If the gas density gets too large (~1 kg/m3) this is

no longer a good approximation, and the ionization

energy becomes lower than 13.6 eV.

(recall the average density in the Sun is ~1410 kg/m3)


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Example: Calcium lines in the Sun

The Calcium absorption lines are ~400 times stronger than theHydrogen lines. How much Calcium is there, relative to Hydrogen?

Assume ne=1.88x1019 m-3 and T=5770 K

First, lets look at hydrogen. The Balmer lines arise due to transitionsfrom the n=2 level of neutral H.

Ca H+K

5

777.15

2/3

4

1

320

7

1051.7

m101083.4 4

Te

I

II

I

II eTn

Z

Z

N

N

So almost all the Hydrogen is

neutral.

Ha

9

1

1

2

1

577.0

777.152

11777.152

1

2

1096.4

1

2 22

21

224

1

2

e

en

n

N

NnnT

n

n

So only 1 of every ~200 million H atoms is in the

first excited state (and capable of creating aBalmer absorption line).


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The Calcium absorption lines are ~400 times stronger than the Hydrogen lines.

How much Calcium is there, relative to Hydrogen? Assume ne=1.88x1019 m-3

and T=5770 K

For Hydrogen:

The partition functions are more complicated to compute, so Ill just give them to

you: ZI=1.32 and ZII=2.3

The =393.3 nm Calcium line in the sun come from the n=12

transition of singly-ionized calcium.

51051.7 I

II

N

N 91096.4

1

2 n

n

N

N

Now consider Calcium, for which cI=6.11 eV

For the Boltzmann equation, I tell you that for singly ionized Calcium 4and2 21 gg


33/33


The Calcium absorption lines are ~400 times stronger than the Hydrogen lines. How

much Calcium is there, relative to Hydrogen? Assume ne=1.88x1013 and T=5770 K

There is only one Ca atom for every 520,000 H atoms.

The difference: Ca is easier to ionize than H (cI=6.11 instead of 13.6), and

Ca has more than 1 electron, so is able to emit radiation in the singly-

ionized state!

We have found that only 4.8x10-9 of H atoms are in the first excited state (to produce

Balmer lines), whereas 99.5% of Ca atoms are in the ground, singly-ionized state (to

produce the solar absorption lines).

Since the Ca lines are ~400 times stronger, this means:

HCa

H

Ca

NN

N

N

6

9

100.2

4001096.4

995.0

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