Download 1 AST 104 LAB 1 Temperature and Luminosity of Stars: Wein`s Law

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.
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(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/m2. 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/m2. The peak measures 24 cm. So the value at this wavelength
= (24/30) x 100 MW/m2 = 80 MW/m2.
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.______________________
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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.
(1 point) Question 5: In what two ways does this peak differ from that of the
Sun?
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)
3800 K
750
Red/IR
5800 K (Sun)
500
6600 K
430
8300 K
350
Light-blue/
green
Dark Blue/
Light Purple
UV
What color is Peak
the star?
Energy/sec/m2
(MW/m2)
Red
9.48
White
80
Blue
152.7
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.______________________________
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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
(From PhET)
3800 K
λpeak (nm)
% error
750
Temperature
(Wien’s Law)
3866.67 K
5800 K (Sun)
500
5800 K
0%
6600 K
430
6744.19 K
2.18%
8300 K
350
8285.71 K
.17%
1.75%
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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/m2 at infrared wavelengths?
The 6600 K star.___________________________________________________
(1 point) Question 9: Which radiates more MW/m2 at visible wavelengths?
The 6600 K star.___________________________________________________
(1 point) Question 10: Which radiates more MW/m2 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!_________________________________
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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/(m2 K4).

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, 4πR2.
L =  T4 x 4πR2
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.____________
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(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!
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