AST 104 LAB 1 Temperature and Luminosity of Stars: Wein s Law and the Stephan-Boltzmann Law Learning Objectives To understand thermal spectra To understand Wien s Law and the Stephan-Boltzmann Law To understand how thermal spectra can be used to evaluate the temperature of a star To understand how temperature and radius of a star determine a star s luminosity Introduction: In this activity we will learn how light from a star can tell us its temperature and how much energy per second the star is emitting its luminosity. Light that is emitted from stars, like light from any hot object made of many closely packed atoms, forms a continuous spectrum called a thermal spectrum. The range of wavelengths emitted and the relative intensity of the light for each wavelength are dependent on the temperature of the object. In this activity we will use a PhET simulation to study the thermal spectra of stars in order to understand the relationship between these spectra and temperature. We will see how this relationship results in two laws that describe these spectra -- Wein s Law and the Stephan-Boltzmann Law. Go to http://phet.colorado.edu/sims/blackbody-spectrum/blackbody-spectrum_en.html As you work through this activity, please answer the questions on a separate page neatly handwritten or typed. Part 1: Understanding the graph This simulation plots a graph that describes the spectrum emitted by a hot object. The horizontal axis indicates the wavelength of the emitted photons and the vertical axis indicates the amount of energy per second (power) emitted by one square meter of the object at a particular wavelength. To investigate this further, set the temperature to 5800 K. Either move the slider or type the numbers in the temperature box. 5800 K is approximately the temperature of the Sun. Note that the horizontal axis specifies the emitted wavelengths ( ) in units of micrometers, µm. One micrometer equals 1000 nanometers. (1µm = 1000 nm.) For example, look at the visible spectrum. The longest wavelength of the visible spectrum is approximately 0.7µm = 0.7x1000nm or 700 nm. 1
(1 point) Question 1. Approximately, what is the shortest visible wavelength? Express your answer in µm and nm. ~.4 µm or 400nm Observe the peak of the curve. The Sun emits the most energy at this wavelength. We will label this wavelength λpeak-- the wavelength where the curve is at its peak. (1 point) Question 2: What color corresponds to λpeak? What color is the Sun? λpeak corresponds to a color of light blue/green. The sun appears white, because it mixes all colors roughly equally. (1 point) Question 3. Approximately, what is the value for λpeak? Express your answer in µm and nm. λpeak is roughly.5µm or 500nm. The vertical axis indicates for each wavelength the amount of energy per second emitted by one square meter of the surface of the emitting object, the Sun in this setting. Recall that energy per second is power, measured in watts, W. MW indicates megawatts. 1MW = 1Million Watts. Thus the vertical units are MW/m 2. Note that the only values on the graph are 0 and 100. You can use the ruler by clicking Show Ruler in the bottom right corner to estimate values in between. For example, to measure value for the peak wavelength, rest the bottom of the ruler on the horizontal axis, and note that 30cm corresponds to 100 MW/m 2. The peak measures 24 cm. So the value at this wavelength = (24/30) x 100 MW/m 2 = 80 MW/m 2. Hit the save button! Enter your values for the Sun in Table 1 on the next page. Part 2: Other Stars at other Temperatures A. Change the temperature to 3800K. Compare this curve to the saved one for the Sun. (1 point) Question 4: In what two ways does this peak differ from that of the Sun? The peak gets lower, to a smaller power ouput per m^2. The peak moves to the right to higher wavelengths. 2
For easier viewing, magnify the curve by pressing the + button on the left. Note that the maximum reading on the vertical axis changes. Make measurements to complete the Table 1 entries for this temperature. B. Change the temperature to 6600K. Compare this curve to the saved one for the Sun. For easier viewing, press the - button on the left. Make measurements to complete the Table 1 entries for this temperature. Sun? (1 point) Question 5: In what two ways does this peak differ from that of the The peak is taller, with a higher peak power output per m^2. The peak is further to the left, towards smaller wavelengths. C. Make measurements to complete the Table 1 entries for a temperature of 8300 K, adjusting the scales as necessary. (3 points) Table 1: Thermal Spectra Characteristics for Different Temperatures Temperature λpeak (nm) λpeak (color, UV, or IR) What color is the star? 3800 K 750 Red/IR Red 9.48 Peak Energy/sec/m 2 (MW/m 2 ) 5800 K (Sun) 500 Light-blue/ White 80 green 6600 K 430 Dark Blue/ Blue 152.7 Light Purple 8300 K 350 UV Blue/ Purple 483.33 Part 3: Wien s Law A. Qualitative discussion (1 point) Question 6: Based on your values in Table 1, how does the value of λpeak change as the temperature increases? As temperature increases, λpeak decreases. 3
Wien s Law states: Hotter objects emit photons with a higher average energy, which means a higher frequency and, therefore, a shorter wavelength. (1 point) Question 7: Do the values in Table 1 support this law or not? Explain your choice. Yes, the table does support this law. As the temperature increases, the average wavelength shrinks, and so the average energy increases! B. Quantitative discussion Quantitatively, Wien s law reflects this inverse relationship between temperature and peak wavelength. It states that if wavelength is measured in nm and temperature is measured in Kelvin, then T 2,900,000 peak Use the values you measured for λpeak and Wien s law to determine T and compare to the values from the PhET. Enter the values in Table 2. (3 points) Table 2: Comparison with Wien s Law Temperature λpeak (nm) Temperature % error (From PhET) (Wien s Law) 3800 K 750 3866.67 K 1.75% 5800 K (Sun) 500 5800 K 0% 6600 K 430 6744.19 K 2.18% 8300 K 350 8285.71 K.17% 4
Part 4: The Stephan-Boltzmann Law A. Qualitative discussion The picture above is a screen shot of the 5800K and 6600K thermal spectra. (1 point) Question 8: Which radiates more MW/m 2 at infrared wavelengths? The 6600 K star. (1 point) Question 9: Which radiates more MW/m 2 at visible wavelengths? The 6600 K star. (1 point) Question 10: Which radiates more MW/m 2 at ultraviolet wavelengths? The 6600 K star. The Stephan-Boltzmann Law states: Each square meter of a hotter object s surface emits more light at all wavelengths. (1 point) Question 11: Do your observations of this figure support this law or not? Explain your choice. Yes, our obserations match this law. Even though it doesn t peak at all of them, the 6600K star emits more of every wavelength than the 5800K star! 5
B. Quantitative discussion The Stephan-Boltzmann law can be expressed mathematically as follows: The energy per second (power) emitted by one square meter of a star s surface = T 4 where the temperature T is measured in K and the constant = 5.7 x 10-8 watt/(m 2 K 4 ). We will not be putting numbers into this equation. We are most interested in noting the following: The energy emitted per second varies as the fourth power of the temperature so that, for example, doubling the temperature results in 2x2x2x2 = 16 times as much energy per second. This only tells us the energy per second emitted by one square meter of the star s surface. Part 5: Luminosity Luminosity, L, is the total power (energy per second) that a star radiates into space. The Stephan-Boltzmann law tells us how much energy per second one square meter of the star radiates into space, so to obtain the total energy radiated we must multiply by the surface area of the star, 4πR 2. L = T 4 x 4πR 2 where T is in Kelvin, R, the radius of the star is in m (meters) and L is in watts. (1 point) Question 12. What two properties of a star determine its luminosity? The temperature and size (Radius) determine its luminosity. (1 point) Question 13. Can two stars have the same temperature and different luminosities? Explain. Yes, two stars can have the same temperature and different luminosities. If they are different sizes but the same temperature, their luminosities will be different (1 point) Question 14. Can two stars have the same luminosity and different temperatures? Explain. Yes, two stars can have the same luminosity and different temperatures. If their sizes are also different so that the product of T and R^2 is the same,_ they will have the same luminosity but different temperatures. 6
(5 points) In conclusion Summarize what you learned from this activity by: a. listing the key scientific terms encountered in the activity. b. constructing a bulleted list that utilizes these key words (and, as needed, words from previous activities) in summarizing what you learned. a.) Thermal spectrum: The spectrum of light emitted by objects based on their temperatures, such as stars. Stefan-Boltzman law: Hotter objects emit more photons at all wavelengths than colder objects. Wein s Law: Hotter objects emit more photons with higher average energy, and hence higher average frequency and lower average wavelength. Luminosity: The total power emitted by a hot object such as a star, determined by the objects size and temperature. Peak wavelength: The wavelength of light at which an objects emits the most power/m^2 b.) Hot stars have a higher peak power output and a smaller peak wavelength than cold stars. Hot stars also emit more photons of all types of light than cold stars. Luminosity, the total power outputted by hot objects, depends on the temperature and size of the object. Two stars could have the same temperature and different luminosities, or the same luminosity but different temperatures, depending on their size. Overall grading for this lab: The 14 questions are each worth 1 point. The 2 tables are each worth 3 points. The final summary is worth 5 points. The total score should be 25 points. Make sure Luminosity includes temperature AND size, not one or the other! 7