Astronomy 110 - The Solar System Lab 5 - Blackbody Radiation, Spectroscopy, and
ID: 106228 • Letter: A
Question
Astronomy 110 - The Solar System
Lab 5 - Blackbody Radiation, Spectroscopy, and the Doppler effect
Light is the primary tool astronomers use to study the cosmos. Since we cannot travel to distant celestial objects to physically measure samples, we have to use the information from light we observe from distant objects. Fortunately, nature gives us quite a bit of information hidden within the atomic spectra characteristic of the elements.
When we talk about light in science, we talk about the entire spectrum of light (which we also call radiation). The entire spectrum of light includes very small wavelengths such as gamma rays, to the visible spectrum which is the colors we see with our eyes, to much longer wavelengths such as the infrared and radio spectrum. Technology exploits regions of the spectrum in the non-visible in the form of mobile phones (radio waves), microwave ovens (micro waves), and medical uses (x-rays) among countless other technologies.
We won’t be focusing on technology in this laboratory, as we are going to focus on how the properties of light allow us to determine properties of celestial objects without actually physically visiting them.
Blackbody Radiation
When you heat an object or a gas, some of that energy will eventually be released in the form of radiation. This can include radiation that is not visible. For example, if you hold your hand near a hot pan, the warmth you will feel is really radiation in the infrared region of the spectrum. If you heat that pan (hotter than your stove will actually allow), it will start emitting radiation in the visible spectrum. This is essentially what you see when a welder heats metal to the point where you actually start seeing a glow.
The dierent wavelengths that a heated object emits is related to the density and ma- terial of that object. A dense object (such as the bright glowing metal from a welder) will emit many dierent wavelengths. A non-dense object, such as a diuse gas, will only emit specific wavelengths. We’ll find out why later when we talk about spectroscopy.
An idealized dense object that absorbs all the incoming light is called a blackbody because it doesn’t reflect any incoming light. This should make some sense intuitively, if you wear a black t-shirt on a sunny day, you will feel a lot warmer than if you wear a white t-shirt. That is because the black dye absorbs most of the incoming light, while the white material reflects almost all the light that it encounters.
A star can be considered nearly a perfect blackbody. This may sound counterintuitive since a star emits white light. However, it is a very dense object and it does not reflect any light that falls onto it. The peak intensity of a star is related to how hot the star is.
In this exercise we will be using Wien’s Law to estimate the temperature of the hypo- thetical blackbody curves in Figure 1. Wien’s law gives a mathematical relationship between the peak intensity wavelength, max, and the temperature of the blackbody, T
max = 0.0029K · m, (1) T
where the quantity 0.0029 is a constant with units of Kelvin-meters (K · m). Notice that since T is in Kelvin, the units of max will be in meters as it should be because it is a length. The energy of a photon is related through the speed of light, c, Plank’s constant, h, and
the wavelength of a photon, given by
E = (2)
where hc has the value 1.989 1025J · m and E is in units of Joules (J).
1) Use Figure 1, Eq.(1), and Eq.(2) to fill in Table 1. Note that you will have to solve Eq.(1) for T and that your max has to be converted from nm to m when you calculate T and E. For the “Color” column, indicate the color that the star would appear in the night sky (Hint: the peak wavelength).
Curve
max (nm)
Temperature (K)
Emax (J)
Color
A
B
C
Table 1: Table of values calculated from the blackbody curves in Figure 1.
2) Looking at your table, what is the relationship of the temperature of a star and the peak intensity wavelength emitted by the object?
3) Looking at your table, what is the relationship between the temperature of the object and the energies of the emitted photons?
4) You look up in the night sky and see a star with a red hue and a star with a blue hue, which one is hotter?
Figure 1: Hypothetical blackbody curves for three dierent stars.
Spectroscopy
When light interacts with matter, the atomic structure of atoms and molecules only permits specific energies to be absorbed and re-emitted. When the light from a gas consist- ing of specific atoms or molecules is separated using a prism, only the discrete wavelengths of the permitted energies will appear on a screen. We call this type of spectra emission spectra. Light from a dense heated object (as discussed earlier) such as a filament of a light bulb will produce a continuous spectra consisting of all wavelengths. Starlight will produce an almost continuous spectrum, but the gas on the surface of the star will absorb the spe- cific wavelengths characteristic of the atoms or molecules. These lines will be missing in the spectrum and appears as the inverse of emission spectra. This type of spectra is called an absorption spectra. The missing lines in the visible part of the spectra are called Balmer lines.
Figure 2: The three dierent types of atomic spectra.
Procedure
Go to out class Sapling Learning page. Click on “Resources.” Click “Spectroscopy Interac- tive Lab”. This resource allows you to simulate a laboratory experiment to produce emission spectra for various sources.
5) Click on the Sun. What type of spectra is produced on the screen, continuous, ab- sorption, or emission spectra?
6) Click on the light bulb. What type of spectra is produced on the screen, continuous, absorption, or emission spectra?
Often the background incandescent light will appear in photographs of real spectra. Figure 3 shows images of real spectra for each of the elements included in the interactive lab. Your job will be to identify each and the peak intensity wavelength, peak, of each element. To identify the peak wavelength, look at the plot on the lower left screen and click on the largest peak. Some elements will have more than one peak wavelength. If this happens, list them both in Table 2.
Figure 3: Real images of atomic spectra.
7) Use the Spectroscopy Interactive Lab from Sapling Learning to identify the elements from Figure 3 and fill in Table 2.
Spectra
Element
peak (nm)
A
B
C
D
E
Table 2: Identification of emission spectra from Figure 3.
Doppler Eect
When an object is moving toward us, the original wavelength of the light to become shorter. When an object is moving away from us, the original wavelength of the light will be longer. A schematic of this eect is shown in Figure 4.
Figure 4: The green wavelength, o, corresponds to an object at rest or moving transversely with respect to the observer. If the object is moving away from the observer, the original wavelength will be elongated and appear longer (red). If the object is moving toward the observer, the wavelength will be compressed (blue).
This is known as the Doppler Eect. Police use this eect with radar guns to determine your speed. Meteorologists use this eect to track the motion of air masses. While the actual velocity of the object can be calculated by the change in wavelength, we will only focus on how we can tell if a celestial object is moving toward or away from us from the absorption spectra.
When an object is blue-shifted (moving towards us), all the wavelengths emitted by the object will be shifted toward the blue spectrum and conversely for a red-shifted object (mov- ing away). As a consequence, the characteristic Blamer lines within the absorption spectra shift toward the blue or red end of the visible spectrum.
Figure 5: The top absorption spectrum corresponds to object at rest or moving transversely relative to an oberver, while the two below corresponds to moving objects.
8) In Figure 5, which absorption spectrum corresponds to an object moving away from an observer, A or B?
9) In Figure 5, which absorption spectrum corresponds to an object moving toward an observer, A or B?
Curve
max (nm)
Temperature (K)
Emax (J)
Color
A
B
C
In this exercise we will be using Wien's Law to estimate the temperature of the hypo- thetical blackbody curves in Figure 1. Wien's law gives a mathematical relationship between the peak intensity wavelength, Amar, and the temperature of the black body, T 0.0029 where the quantity 0.0029 is a constant with units of Kelvin-meters (K.m). Notice that since T in the units of will be in meters as it should be because it is a length. The energy of the a is related through the speed of light, c wavelength of a photon, A by (2) 1) Use Figure 1, Eq.(), and Eq. (2) to fill in Table 1. Note that you will have to solve Eq for T and that your be from nm to m when you T and E. For the column, indicate the color that the star would appear in the night sky (Hint: the peak wavelength) Curve Amar (nm) Temperature (K) EA (J) ColorExplanation / Answer
1) Use Figure 1, Eq.(1), and Eq.(2) to fill in Table 1. Note that you will have to solve Eq.(1) for T and that your max has to be converted from nm to m when you calculate T and E. For the “Color” column, indicate the color that the star would appear in the night sky (Hint: the peak wavelength).
Curve
max (nm)
Temperature (K)
Emax (J)
Color
A
0.125
34000
1.60*10-15
Blue
B
600
5800
3.31-10-19
Yellow
C
966
3900
3*10-20
Orange
2) Looking at your table, what is the relationship of the temperature of a star and the peak intensity wavelength emitted by the object?
Since stars emit light with different wavelengths, they have different colors. Stars do not just emit one wavelength of electromagnetic radiation, but a range of wavelengths. The wavelength at which a star emits the most light is called the star's peak wavelength. Wien's displacement law states that the black body radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature. The shift of that peak is a direct consequence of the Planck radiation law which describes the spectral brightness of black body radiation as a function of wavelength at any given temperature.
3) Looking at your table, what is the relationship between the temperature of the object and the energies of the emitted photons?
To estimate the surface temperature of a star, we can use the known relationship between the temperature of a blackbody, and the wavelength of light where its spectrum peaks. That is, as you increase the temperature of a blackbody, the peak of its spectrum moves to shorter (bluer) wavelengths of light. The temperature of an object is a direct measurement of the energy of motion of atoms and/or molecules. The faster the average motion of those particles (which can be rotational motion, vibrational motion, or translational motion), the higher the temperature of the object. A blackbody is an object that absorbs all of the radiation that it receives (that is, it does not reflect any light, nor does it allow any light to pass through it and out the other side). The energy that the blackbody absorbs heats it up, and then it will emit its own radiation. The only parameter that determines how much light the blackbody gives off, and at what wavelengths, is its temperature. There is no object that is an ideal blackbody, but many objects (stars included) behave approximately like blackbodies. We can see the similar in the table values.
4) You look up in the night sky and see a star with a red hue and a star with a blue hue, which one is hotter?
The temperature of an object is a measurement of the amount of random motion (the average speed) exhibited by the particles that make up the object; the faster the particles move, the higher the temperature we will measure. They create electromagnetic radiation (light) since some of the particles within an object are charged, any object with a temperature above absolute zero (0 K or –273 degrees Celsius) will contain moving charged particles, so it will emit light. A blackbody, which is an “ideal” or “perfect” emitter (that means its emission properties do not vary based on location or the composition of the object), emits a spectrum of light with the following properties:
The hotter the blackbody, the more light it gives off at all wavelengths. That is, if you were to compare two blackbodies, regardless of what wavelength of light you observe, the hotter blackbody will give off more light than the cooler one.
The spectrum of a blackbody is continuous (it gives off some light at all wavelengths), and it has a peak at a specific wavelength. The peak of the blackbody curve in a spectrum moves to shorter wavelengths for hotter objects. If you think in terms of visible light, the hotter the blackbody, the bluer the wavelength of its peak emission. For example, the sun has a temperature of approximately 5800 Kelvin. A blackbody with this temperature has its peak at approximately 500 nanometers, which is the wavelength of the color yellow. A blackbody that is twice as hot as the sun (about 12000 K) would have the peak of its spectrum occur at about 250 nanometers, which is in the UV part of the spectrum.
Curve
max (nm)
Temperature (K)
Emax (J)
Color
A
0.125
34000
1.60*10-15
Blue
B
600
5800
3.31-10-19
Yellow
C
966
3900
3*10-20
Orange
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