# If a star is very hot, the electrons will be freed from the hydrogen atom. (Ionized) Once they are free, they act like particles and emit a continuous.

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If a star is very hot, the electrons will be freed from the hydrogen atom. (Ionized)
Once they are free, they act like particles and emit a continuous spectrum. If the star is hot enough that hydrogen is ionized all the way to the surface, then there will be no hydrogen Balmer lines.

Here is how line strength depends of temperature

Remember, the line strength depends primarily on two parameters
Remember, the line strength depends primarily on two parameters. (1) Surface Temperature and (2) Number of absorbers. If we want learn about the number of absorbers for a given element (say, calcium, iron, gold, etc) then we need to know the temperature of the star. If we know the temperature we can account for its effect and… The line strength will only depend on the Number of Absorbers.

Spectral typing Summary
Spectral typing can be used to find the surface temperature of a star. (Along with color and Wien’s Law) Spectral typing can also be used to find out how much of a given element is in a star. HD has much less of all the elements, other than Hydrogen and Helium, than the Sun. In fact, it has about 0.03 the value of the Sun for all 90 elements. That is 3% the amount in the Sun. The most deficient star known has about 0.001% the Sun. There are also stars with up to 3 times the amount in the Sun.

A little bit more to do. Stefan-Boltzmann Law.
Imagine we have a cube, that is 1meter X 1meter X 1meter. It is made of steel and is being heated from inside. We move a light collector (maybe a huge bundle of fiber optic cables) that is 1meter X 1meter, right up close to one of the surfaces so we can capture all the light being emitted from that side of the cube. The cube begins to glow, and we send the light collected to a grating which makes a spectrum. As the cube heats up we continue to collect data.

Detector in front of cube collecting all the light that is coming from one side of the cube

When the spectrum is recorded, as the cube heats up, this is what you find.

To find the total energy being emitted from one side of the cube every second you would need to add up all the intensity at each wavelength. In other words, find the area underneath the curve.

The result would be: Lfor one side = σT4 where σ is called the Stefan-Boltzman constant

This means that if you increase the temperature (T) by a factor of 2, then L will ???

If the temperature increases by a factor of 2 then the Luminosity will ???
45 30 Increase by a factor of 2 Increase by a factor of 4 Increase by a factor of 16 Decrease by a factor of 4 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

To find the TOTAL luminosity of the cube in our experiment we have to..
30 30 Multiply by 2 Multiply by 6 Divide by 16 Do nothing, we already have the total luminosity 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

To figure out the total luminosity being emitted by the box, we would need to multiply the luminosity of one side by 6. This is because there are six sides of the cube for the energy to escape from.

Stars work the same way. The total Luminosity being emitted through the surface of a star is: L = σT4(4πR2) Where (4πR2) is the surface area of the star and R is the radius of the star.

Let’s consider three stars.
Procyon B, Sirius and Deneb. All three stars have very similar surface temperatures. T ~ 9000 K But the Luminosities are not similar. Deneb has by far the largest, then Sirius, and then Procyon B. Remember that: L = σT4(4πR2)

So we can use the Physics of a radiating object to determine, not only the surface temperature of a star (Wien’s Law) But also the star’s radius. (Stefan-Boltzman Law) Deneb is not more luminous than Sirius because it is hotter, it is more luminous because it has a bigger radius. Likewise, Procyon B is not less luminous because it has a cooler temperature, it is less luminous because it has a smaller radius.

A new distance determination
Now that we know the way things work on the H-R diagram, for stars that we can measure distance to using trigonometric parallax (triangulation) we can make use of these laws to calculate the distance to stars that are too far away to measure distance directly. Here is what we know. B = L/4πd2 and L = σT4(4πR2) So B = σT4(4πR2)/4π d or B = σT4(R2)/d2 If we measure the apparent brightness, the temperature and the radius, we can calculate the distance to the star.

B, the apparent brightness comes from observing the star at a telescope with a detector attached to it. T can come from either, Wien’s law, the color of the star, or from spectral typing the absorption lines. But we have to still get a handle on R.

The strength of an absorption line depends primarily on two variables.
The temperature of the star. And the Number of Absorbers. But there is a secondary effect to the absorption lines that we can use. This effect is called Pressure Broadening. Here is how it works.

The electron is bound to the atom by the electromagnetic force.
Different elements have different number of charged particles. This means that the electromagnetic force is different for each element. This causes the electrons to have different standing waves for different elements. So the transition energies between different excited states are different. The absorption lines occur at a different wavelength for each element.

BUT, what happens when atoms, with electrons attached, are packed really close together?
The electrons from the neighboring atoms can have a small effect on the standing wave of each of the atom’s electrons.

In a supergiant star (luminosity class I) the star has a huge volume
In a supergiant star (luminosity class I) the star has a huge volume. That means the atoms are not close to each other near the surface. They have virtually no effect on the given energy levels. In a giant star (luminosity class III) the star has a large radius but not as large as the supergiant. The atoms near the surface interact more with each other. In a main-sequence star (luminosity class V) the radius is much smaller and atoms are packed closely together near the surface.

Luminosity classes in stars.

The amount of pressure broadening is related to the radius of the star
The amount of pressure broadening is related to the radius of the star. When the pressure is high, (luminsoity class V) the pressure broadening is large and the radius of the star is small. When the pressure is low (luminosity class I) the broadening is small and the radius of the star is big. We can calculate the radius of the star in this fashion.

How do we now find the distance?
We can calculate it from B = σT4(R2)/d2 Here is a visual example of how it works. We know the properties of the H-R diagram for the near by stars.

Quiz #5 There are two stars, star A and star B. Star A is approaching the Earth at 100 km/s and Star B is moving away from the Earth at 200 km/s. Compare the Doppler shift for these two stars by explaining how the spectra will be shifted and by how much. (I am not looking for a number here, just a qualitative comparison)

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