“You miss 100% of the shots you don't take “You miss 100% of the shots you don't take.” Wayne Gretzky HW2 is due on Tuesday. Question?

The formation of spectral lines Continuing with our light discussion. Only loosely covered in your book. The formation of spectral lines

Velocity distribution The Maxwell-Boltzmann distribution: The number of gas particles per unit volume having a speed between v and v+Dv is: Where n is the number density, m is the particle's mass, k is Boltzmann's constant.

The velocity distribution also affects excited states of atoms.

The ratio of the number Nb of atoms with energy Eb to the number Na of atoms with energy Ea is the same function (called the Maxwell equation): k=8.6174x10-5eV/K, For H: gn=2n2 En= -13.6/n2 eV What is En? What is gb?

We found that for a H gas at 10,000 K the ratio of atoms in the first excited state (n=2) to that of the ground state (n=1) is very small. 1 in 34,600 atoms!

But ionization needs to be accounted for. We found that for a H gas at 10,000 K the ratio of atoms in the first excited state (n=2) to that of the ground state (n=1) is very small. 1 in 34,600 atoms! But ionization needs to be accounted for.

e.g. to ionize H from the ground state (HI to HII) would be cI=13.6eV. Ionization states The Boltzmann equation also enters into the expression for the relative number of atoms in different stages of ionization. Let ci be the ionization energy needed to remove an electron from an atom in the ground state, going from ionization state i to (i+1). e.g. to ionize H from the ground state (HI to HII) would be cI=13.6eV.

To ionize H from the ground state (HI to HII) would be cI=13.6eV. Ionization states To ionize H from the ground state (HI to HII) would be cI=13.6eV. However, the atoms may not all be in the ground state. An average must be taken over the orbital energies to allow for the possible partitioning of the atom's electrons among its orbitals. The partition function, Z, is the weighted sum of the number of ways the atom can arrange its electrons, with different energey configurations weighted by the Boltzmann factor.

Partition functions If Ej is the energy of the jth energy level and gj is the degeneracy of that level, then Notes: Every Z has the ground state. Usually one energy level dominates, simplifying the equation.

Using partition functions of Zi and Z(i+1) the ratio of the number of atoms in ionization stage (i+1) to the number of atoms in ionization stage i is Which is known as the Saha equation, after Meghnad Saha who first derived it in 1920. In this equation, ne is the number of free electrons per unit volume and the 2 in front of the Z(i+1) is for the 2 spin states of the free electron (up and down)

Later we will discuss how to find Pe. Electron pressure (from the ideal gas law), Pe=nkT, is used, as it's more easily measured. Then the Saha equation becomes: Later we will discuss how to find Pe.

NOTE on the Boltzmann constant: When you use energies in eV (in the exponential here), use: k = 8.617x10-5 eV/K When you use MKS units (the first two factors in this equation), use k = 1.381x10-23 m2kg.s-2K-1

What fraction of H is ionized at 10,000K? (Our question from before.)

What fraction of H is ionized at 10,000K? (Our question from before.) First we need the partition functions: Ionized H is just a proton, with only 1 state. So ZII=1. At 10,000K, we also know that for each atom in the first excited state, there are >104 atoms in the ground state. So the partition function is: ZI~g1=2.

Something like Log Pe=1.3 so Pe~20 Pa (MKS units) Models for the Sun Something like Log Pe=1.3 so Pe~20 Pa (MKS units)

Balmer lines We saw that at 10,000 K, most of the H was still in the ground state. To get more H in the first excited state (to make Balmer lines), you want to heat up the material. We calculated that at about 85,000 K, there are equal amounts of H in the ground and first excited state. But at 10,000 K, already most of the H is ionized. If you heat the material more, you ionize it more.

What fraction of H is ionized at 10,000K? (Our question from before.) ZII=1 and ZI~g1=2. Assume that Pe=20 Pa (MKS units), h=6.626x10-34m2kg/s, me=9.109x10-31kg, k=1.38x10-23m2kg/s2K. Units: e-c/kT can be in eV

This gives us the ration of HII to HI, but we really want the overall fraction of HII: HII/HTotal

We really want the overall fraction of HII

This shows the HII/HTot fraction This shows the HII/HTot fraction. Between 8,000 and 11,000 K, there is a small region called the partial ionization zone. So you can see that the Balmer lines will peak at the hottest temperature for which there are still sufficient HI atoms.

We can combine our previous formulae to get one that will tell us when the Balmer lines will peak. The solution to this problem comes out to be about 9,900 K (very close to our 10,000 K value!).

The solution, plotted.

Spectral classification: These examples tell us something very important about H within line-forming regions of stars. At Teff>10,000 K, most H is ionized and therefore not available to form lines (unless pressure is greatly increased!) At Teff <~7,500K, the number of H atoms in the 1st excited state is quite small.

The spectrum of our Sun has stronger CaII lines than Balmer lines The spectrum of our Sun has stronger CaII lines than Balmer lines. Let's investigate why. Our Sun has about 1 Ca atom for every 500,000 H atoms.

Start with the fraction of H is ionized at 5777 K? ZII=1 and ZI~g1=2. This time Pe=1.5 Pa (MKS units), h=6.626x10-34m2kg/s, me=9.109x10-31kg, k=1.38x10-23m2kg/s2K.

Start with the fraction of H is ionized at 5777 K? Note these values! ZII=1 and ZI~g1=2. This time Pe=1.5 Pa (MKS units), h=6.626x10-34m2kg/s, me=9.109x10-31kg, k=1.38x10-23m2kg/s2K.

Start with the fraction of H is ionized at 5777 K? ZII=1 and ZI~g1=2. This time Pe=1.5 Pa (MKS units), h=6.626x10-34m2kg/s, me=9.109x10-31kg, k=1.38x10-23m2kg/s2K. I get 7.7x10-5 (1 in 13,000)