Explaining how surface temperature and composition may be obtained from a star’s spectrum
Using the data below and Wien's Law Eq determine the temperature at each voltage.
Compare the blackbody spectrum of the sun to visible light.
Describe the blackbody spectrum of a light bulb. Where is the peak of the spectrum? Why do light bulbs get hot? Do they seem efficient?
Describe what happens to the shape and peak value of the spectral radiance curve as you change the temperature.
Imagine that you see two stars in the sky, one is glowing orange and the other is glowing blue. Which one is hotter?
Determine the relationship between the peak wavelength and temperature of the blackbody
The surface temperature of a black body can be found if the wavelength of the maximum intensity of radiation on a black body curve is known. The effect of temperature on the black body curve can be investigated using the simulation to the right and above (PhET)
The surface temperature of a black body could also be found if the wavelength of the maximum intensity of radiation on a black body curve is known. The effect of temperature on the black body curve can be investigated using the simulation above (HERE).
Power of the Sun
Use the Vernier Spectrometer to record the spectrum of the Sun. The spectrum can be used to identify elements but also to determine the surface temperature of the Sun.
Plug in the USB cable to your laptop and open LoggerPro.
Click on the Experiment tab and select Change Units => Spectrometer:1 => Select Spectrometer Mode...
In the window that appears select Intensity and then close the window.
The graph should look similar to the one below, measured on the 14th of February 2019. For the calculations that follow use EITHER the data from your own spectrum or the one below.
From the data above:
The wavelength of the peak intensity is 497 nm.
The surface temperature of the Sun can be calculated using Wien's law.
Use the wavelength of maximum intensity to show that the surface temperature of the Sun is 5800 K.
The power of the Sun P can be calculated using the Stefan-Boltzmann law.
Use the surface temperature of 5800 K and the radius of the Sun 6.9 x 108 m to show that the power of the Sun is 3.85 x 1026 W. Assume that the Sun is a perfect black body and has an emissivity of 1.
The power of the Sun fluctuates but is considered to be about 3.85 x 1026 W. This means that it radiates 3.85 x 1026 J of energy per second.
The radiation of the Sun spreads out in an ever increasing sphere.
The intensity I of the radiation at any point on the surface of this increasing sphere can be found from
Intensity = Power / Area (I=P/A)
where A is the surface area of the sphere.
The average distance between the Sun and the Earth is 1.5 x 1011 m. Use the power of the Sun 3.85 x 1026 W to show that the intensity of the radiation reaching Earth is 1360 W m-2.
This value is known as the solar constant S. (Not necessarily important but a common question on the IB Exams).
The readings at ACS had a maximum of 904 W m-2. What could account for the loss of power from the sun? (Why didn't we measure the intensity to be 1360 W m-2?)
The radiation reaching the Earth will either be absorbed or reflected. The Earth absorbs radiation only on the side facing the Sun, as shown in the diagram below.
e - emissivity (Black Body = 1) (unitless)
𝜎 - Stefan-Boltzmann Constant (W m-2 K-4)
A - Surface area of radiating body (m2)
T - Temperature (K)
Explaining how the chemical composition of a star may be obtained from the star’s spectrum
■ Radiation emitted from the very hot core of a star has to pass through the cooler outer layers and some wavelengths of the continuous spectrum are absorbed and then re-emitted in random directions by the elements present.
■ The absorption spectra received can be displayed by passing the light from the star through a diffraction grating or prism, as described and explained in Topic 7.1 .
■ As an example, figure 1 shows the absorption spectrum of the Sun from which elements like helium and hydrogen (and others) can be identified.
■ Figure 2 indicates how the continuous intensity–wavelength graph of radiation from a star with a surface temperature of about 5000K (similar to the Sun and shown in Figure 16.11) is changed after passing through its outer layers.
Figure 1
Figure 2
To get a more detailed breakdown of the light from a star, you can use a spectrograph to separate the light from stars into 100 narrow spectral bands. It’s like having 100 filters all at once. That detail will show you the specific shape of a spectrum. You can also see spectral lines, which correspond to particular elements and ions.
Your spectrograph measures the intensity of light at various wavelengths. We will measure the wavelength of light in nanometers (10-9m = 1 nm). The visual spectrum—where we will operate—extends from about 350 (violet) to 700 (red) nm.
Superimposed on the overall shape of the spectrum, you will see dark lines in the spectra of most stars.
These lines are at wavelengths characteristic of the material in the star’s atmosphere.
Hot, energetic photons, coming up from the interior, get absorbed by atoms or molecules, promoting their electrons to some higher level.
This only happens if they are at wavelengths that correspond to
changes in energy levels in the atom or molecule, and
Where there are atoms actually at the lower level, ready to receive!
Since some of the light at that wavelength gets absorbed, there is a little less light at that wavelength. In a spectrum, this looks like a dark line compared to the rest of the light.
These stellar spectra with dark lines are called absorption spectra.
In the lab, you can make emission spectra (bright lines) by heating up gases in glass tubes called discharge tubes. That way you can learn where the spectral lines are.
If the star is moving away from you, all of the light is Doppler shifted to longer (redder) wavelengths. This is called a red shift. If it’s coming towards you, it’s a blue shift.
The ratio of the change in wavelength to the wavelength (i.e., the proportional change) is approximately equal to the ratio of the star’s speed to the speed of light.
(Figure: Sun’s spectrum and blackbody. Notice how complicated the solar spectrum is in the infrared, and below the atmosphere. In Stella, we simplify: no atmosphere, visible only.)
Stellar Evolution
Massive stars have higher-temperature cores, and burn hydrogen into helium faster. So they are brighter and bluer than their low-mass counterparts.
These massive blue stars burn the hydrogen so much faster that even though they are more massive, they run out of fuel sooner.
When a star runs out of hydrogen, its core contracts and gets hotter, and starts burning helium, converting it to carbon. Its “envelope” expands and gets cooler: the star becomes a “red giant.”
In a cluster of stars that were all formed from the same primordial gas cloud at roughly the same time, there is a consistent relationship between temperature and brightness:
Just-born stars lie along the “zero-age main sequence,” with massive stars being hot and bright, and the least massive being cool and dim.
The top, hot end of this pattern evolves first. The specific pattern can be complex, but in general, the star becomes cooler while roughly maintaining its luminosity.
The older the cluster, the more stars have evolved.