1. Introduction
Choose One • 10 points
It is mentioned in the week 4 course notes that the understanding of something called blackbody
radiation lead to the formulation of quantum mechanics. This first part of your Physical Science
in Action this week will take a brief look at this phenomenon.
As a reminder, here is a picture of the electromagnetic spectrum. It is oriented in the same
direction as the spectrum you will use in today's simulation.
Have you ever noticed that on a cool winter day that your house feels much cooler than on a
warm summer day even though your thermostat is set to the same temperature?
You are probably also aware that as objects heat up they will begin to give off visible light and
also change color as they get hotter. For example a piece of metal placed in a furnace will begin
to glow deep red, then orange, and then become “white hot”. Why do you think this is?
Both of these questions will be easily answered by learning about blackbody radiation.
Open the following link:
http://phet.colorado.edu/sims/blackbody-spectrum/blackbody-spectrum_en.html
It just so happens that regardless of the material, when objects are heated up they will start to
glow and change colors at near identical temperatures. The plot that you see is called a
blackbody spectrum. This plot tells us the intensity or the “amount” of light that an object will
emit at different wavelengths (or “colors”). The visible wavelengths are marked by their colors
on the plot. To the right of the visible band is lower energy infrared light. To the left of this band
is higher energy ultraviolet (UV) light.
Click the + button that is to the left of the intensity scale (far left side of the screen) such that the
top of the scale is at .001. (in the picture above the top of the scale says 100).
Now use the temperature slider to the right, and take the temperature all the way down to 300
Kelvin (80 Fahrenheit).
Now slowly begin to raise the temperature. At approximately what temperature would a heated
material (metal, wood, etc.) begin to give off visible light at a deep red color?
Note: This will be the temperature where your spectrum first begins to come off of the
wavelength axis in the visible region, and so is giving off a small amount of red light.
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500 K (440 Fahrenheit)
1050 K (1430 Fahrenheit)
1800 K (2780 Fahrenheit)
2500 K (4040 Fahrenheit)
2. Blackbody Spectrum
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Click the - button that is to the left of the intensity scale to zoom out such that the top of the scale
is at 10.
Move the temperature slider to that of a light bulb. The red part of the thermometer on the far
right should just be touching the line marked light bulb. At approximately what temperature does
the filament in a household light bulb operate?
Note: This is written in blue in the simulation.
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660 K (728 F)
1800 K (2780 F)
3000 K (4940 F)
5700 K (9800 F)
3. Blackbody Spectrum
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What type of light does this light bulb produce most (i.e. at what wavelength does the spectrum
have maximum intensity)?
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Infrared light
Red visible light
Violet visible light
Ultraviolet light
4. Blackbody Spectrum
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Click the - button that is to the left of the intensity scale to zoom out such that the top of the scale
is at 100.
Move the temperature slider to that of the Sun. The red part of the thermometer on the far right
should just be touching the line marked Sun. Approximately what temperature is the surface of
the Sun?
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2100 K (3320 F)
4500 K (7640 F)
5700 K (9800 F)
9800 K (17,180 F)
5. Blackbody Spectrum
Choose One • 10 points
Based on the simulation, what type of light does the Sun produce the most?
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Infrared light
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Green visible light
Orange visible light
Ultraviolet light
6. Blackbody Spectrum
Choose One • 10 points
Relative to the peak intensity in the Sun’s spectrum, the Sun emits nearly equal amounts of light
across the entire visible part of the EM-spectrum. This is demonstrated by the star shaped symbol
at the top of the simulation being white. Therefore, if you look at the Sun when it is directly
overhead on a clear day, it will appear white.
Click the - button that is to the left of the intensity scale to zoom out such that the top of the scale
is at 316.
Use the star shaped symbol above your graph and to the right of the blue, green, and red dots to
estimate the temperature at which something will begin to glow blue. At approximately what
temperature does the object gain a faint blue tint?
Note: This will also be the temperature where the max intensity of the objects spectrum is in the
blue portion of the visible spectrum.
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3000 K (4940 F)
6600 K (11,420 F)
7900 K (13,760 F)
Object cannot glow blue at any temperature.
7. Blackbody Spectrum
Choose One • 10 points
Note that in the above question, although the object still emits all colors of visible light, it
appears blue now instead of white because of the significant difference in the intensity or amount
of blue light radiated versus the amount of red light emitted.
Click the + button that is to the left of the intensity scale to zoom in such that the top of the scale
is at 1. Now slowly decrease the temperature from 5000K down to 300K (room temperature).
Notice how the entire spectrum decreases in intensity and moves to the right into the infrared
region. Even though the spectrum appears completely flat, objects at room temperature and
below also emit their own light. If our eyes could detect infrared light, we would be able to see in
the dark with warmer objects being brighter than others.
In the introduction of this activity, we mentioned the temperature of your home on hot and cold
days. Your body is kept warm in your home primarily by two ways: by direct contact with the air
around you and by absorbing infrared light that is radiated from the walls. As you have seen in
this activity, the light that is radiated from an object depends almost solely on the temperature of
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the object. Based on what you have learned here, what is one reason for feeling warmer in your
house on a summer day versus a winter day even though your thermostat is set the same?
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The walls of the house are warmer during the summer. Therefore, they radiate more
infrared light that can serve to warm our body.
The walls of the house are warmer during the summer. Therefore, they radiate more
visible light that can serve to warm our body.
The walls of the house are warmer during the summer. Therefore, they radiate more
ultraviolet (UV) light that can serve to warm our body.
The temperature of the walls of the house has no effect on the light they radiate.
8. Blackbody Spectrum
Choose One • 10 points
Since we cannot physically collect data from stars and most other objects in the universe, almost
all of the information we obtain from the universe comes from analyzing the light, or spectra,
from those objects.The study of light is known as spectroscopy.
As we have seen in this simulation, every blackbody emits light with an easily identified pattern
known as the blackbody curve. This is the particular way the total light emitted by a blackbody
varies with its frequency. The exact form of the curve depends only on the body’s temperature.
Since we can treat stars as blackbodies, this is incredibly useful in astronomy that shows us that
the color of a star is also indicative of its temperature.
Use the simulation to determine the surface temperature of the following star:
Betelgeuse is a red supergiant star in the constellation Orion.
Adjust the intensity scale so that the top of the scale is 31.6. Then, knowing that Betelgeuse has
peak intensity in the red and infrared wavelengths, move the temperature scale up and down until
you can determine the approximate surface temperature of the star.
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2000 K
3500 K
7700 K
11,000 K
9. Wien's Law
Choose One • 10 points
The relationship between temperature and peak wavelength is given by an equation known as
Wien’s Law:
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In this equation:
λ(max)= peak wavelength (cm)
T = temperature (K)
Based on what you have seen in the simulation and your knowledge of proportionality
relationships learned this month, what is the relationship between temperature and peak
wavelength?
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They are directly proportional.
They are inversely proportional.
They are exponentially proportional.
They are unrelated.
10. Wien's Law
Choose One • 10 points
Use Wien’s Law to calculate the peak wavelength of Betelgeuse, based on the temperature found
in Question #8.
Note: 1 nanometer (nm) = .0000001 centimeters (cm)
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6
208 nm
400 nm
828 nm
1800 nm
1. Scientific Notation
Choose One • 5 points
Because quantum mechanics is physics that describes the interactions of very small
objects (i.e. molecules, atoms, and electrons), this week you will need to know how to
multiply very small numbers. Remember that scientific notation writes very small or
large number in terms of powers of 10. For example, .0008 can be written in scientific
notation as 8 x 10-4 or as 8E-4. The power of 10 (-4 in this case) tells you to take the
number 8.0 and move the decimal 4 places to the left giving us .0008.
Which is a correct representation of .000025 in scientific notation?
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2.5E-4
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2.5E-5
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2.5E-6
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25E-5
2. Scientific Notation
Choose One • 5 points
Let’s now multiply two numbers in scientific notation using Google.
Enter .0008 into Google exactly as it was written above as:
We could now multiply it by .000056 by typing:
Note that we have separated our two numbers by putting them inside parentheses, and
the * symbol (SHIFT+8) is used as the multiplication sign. We could have done a
division instead of multiplying by separating the two numbers by a forward slash
Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in
scientific notation?
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6.78E-11
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3.33E-12
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2.40E-12
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2.33E-11
3. Electron Transitions
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As mentioned in this week’s notes on page 4, the electrons of an atom can occupy
different energy shells within the atom (similar to how the planets all occupy different
orbits around the Sun). Electrons prefer to be in the lowest energy shell possible (the
ground state); however, they can gain energy and jump to a higher shell by absorbing
light or being excited by an electric current. In accordance with the conservation of
energy, if an electron drops from a higher energy level to a lower one, this must emit a
photon (particle of light) with energy equal to the energy difference of the shells.
A Balmer series transition is any transition of an electron from some higher energy shell
down to the second lowest energy shell (n=2) in hydrogen.
Looking at image (b) above, what is the wavelength of a photon emitted during the
Balmer transition from the n=3 shell in hydrogen? (remember nm is short for a
nanometer, for example 656 nm = 656 x 10-9 meters)
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656E-9 meters
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486E-9 meters
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434E-9 meters
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410E-9 meters
4. Momentum
Choose One • 5 points
Use the momentum equation for photons found in this week's notes, the wavelength you
found in #3, and Plank’s constant (6.63E-34) to calculate the momentum of this photon:
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1.0E-27 kgm/s
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1.8E-27 kgm/s
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2.0E-27 kgm/s
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3.0E-27 kgm/s
5. Frequency
Choose One • 5 points
Use the equation from week 3:
frequency=wavespeedwavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember
the speed of light is 3E8 m/s):
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7.6E14 Hz
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6.0E14 Hz
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4.6E14 Hz
6. Energy
Choose One • 5 points
Use the energy equation from this week’s notes, your answer from #5, and Plank’s
constant(6.63E-34) to find the approximate energy of this photon:
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4.8E-19 Joules
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3.0E-19 Joules
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3.0E-17 Joules
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1.21 Gigawatts
7. Atomic Spectra
Choose One • 5 points
A glass tube is filled with hydrogen gas. An electric current is passed through the tube, and
the tube begins to glow a pinkish/purple color (this is how fluorescent bulbs and neon
signs produce light). If you were to pass this pink light through a prism to separate the
individual light frequencies, you would see that this pink light is composed of four distinct
colors: violet, green, blue, and red. Notice the similarity between image (b) above and
image (b) from question #3.
Which is the best description of why this occurs?
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The electrons within the hydrogen atoms gain energy from the current
causing them to jump to higher energy orbitals. When they fall back to a lower
energy orbital they release a single proton. These protons have discrete energies
equal to the difference in energy of the two orbitals.
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Atoms contain continuous energy orbitals, meaning that the light the
hydrogen atoms produce can be of any energy. Depending on the type of prism
used, when the light reaches it, the prism will only allow specific light energies
(frequencies) to pass through.
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The light spectrum from any source contains all colors (frequencies) because
the Planck, constant, h, is so small.
8. Momentum and Energy
Choose One • 5 points
The lights used by Mark Watley (played by Matt Damon) during the film The
Martian seem to be Metal Halide lamps. Metal Halide lamps are filled with vaporized
mercury and metal-halogen compounds. When an electric current is passed through
the lamp, the tube begins to glow a bright white/blue color. If you were to pass this light
through a prism to separate the individual light frequencies, you would see a rainbow
just as you would if using natural sunlight because of the complexity of the metal halide
gas and the vast amount of possible electron transitions.
(The study of light in this way is known as spectroscopy and allows astronomers to
know exactly what atoms compose distant stars, simply by looking at the light they
emit. The spectral lines an atom produces uniquely identifies that atom just like a
fingerprint uniquely identifies a person.
The momentum equation and energy equation that we have used above can be
combined to give the following equation:
c=Ep
where again p is the phonon momentum, E is the photon energy and c is the speed of
light. When you divide the photon energy found in #6 by the photon momentum found
in #4, do you get the speed of light?
(If not, check your work for questions #4 through #6).
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Yes
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No
9. Light
Choose One • 5 points
All visible light (light that our eyes can detect) has wavelength between 400-700
nanometers. Wavelengths just smaller than 400 nm are Ultraviolet light. Wavelengths just
larger than 700 nanometers are infrared light. What type of light is the Balmer series light
that we have consider so far?
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Visible
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Ultraviolet
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Infrared
10. Light
Choose One • 5 points
The solar panels used by Mark function because of the photoelectric effect. Light shines on
the cells causing electrons to be ejected from the metal, which produces an electric
current. At night on Mars, no light will fall on the solar cells and no electric current will be
generated. According to your notes, what type of light is typically needed to cause the
photoelectric effect?
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Visible
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Ultraviolet
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Infrared
11. Balmer Series
Choose One • 5 points
If we were to illuminate them only with light from the Balmer transition considered above,
would the solar panels produce a current?
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Yes
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No
12. Balmer Series
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Starting with only the Balmer series light (visible light), how could we ensure that the solar
panels generate a current that Mark can use for his power station? (It may help to look at
the electromagnetic spectrum from week 3):
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By gradually increasing the brightness (amount) of light that we shine on it.
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By gradually increasing the frequency of the light we shine on it.
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By gradually increasing the wavelength of the light that we shine on it.
13. Special Relativity
Choose One • 5 points
Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf
ball at 100 mph from the deck of the yacht. You see the ball move away from you at
100mph, while a person standing on a near by beach would observe your golf ball traveling
at 120 mph (20 mph + 100 mph).
Now imagine you are aboard the Hermes spacecraft traveling at 0.1c (1/10 the speed of
light) past Mars and shine a laser from the front of the ship. You would see the light
traveling at c (the speed of light) away from your ship. According to Einstein’s special
relativity, how fast will a person on Mars observe the light to be traveling?
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0.1c (1/10 the speed of light)
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c (the speed of light)
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1.1c (c+0.1c)
14. Stellar Evolution
Choose One • 5 points
Note: The following questions are unrelated to the Balmer series or The
Martian. Please refer to your course notes.
A Sun-sized star will spend most of its lifetime as a:
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White Dwarf
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Red Giant
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Protostar
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Main-Sequence Star
15. Stellar Evolution
Choose One • 5 points
Our Sun will eventually:
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explode in a supernova.
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become a white dwarf star.
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become a black hole.
16. Stellar Evolution
Choose One • 5 points
A main sequence star does not expand or contract due to the balance between the
internal heat pushing outward and the weight of the material pressing inward due to
gravity. This state of maintaining a constant size is known as:
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hydrostatic equilibrium
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thermal equilibrium
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dynamic equilibrium
17. Stellar Remnants
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Neutron stars are:
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Low density star remnants with many neutrons, which mass is less than the
mass of the Sun.
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Incredibly small remnants of super massive stars where the gravitational collapse
is stop by neutron degeneracy.
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Incredibly big and massive star remnants which expelled all its neutrons in a
supernova explosion.
18. Stellar Remnants
Choose One • 5 points
Black holes are:
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Star remnants from super massive stars which gravitational collapse can not be
halt by electron or neutron degeneracy and gravity is so strong in their vicinity
that not even light can escape.
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Regions of the universe with space empty of matter or radiation that becomes so
dark that forbids us from investigating it.
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Regions of space where matters is not sufficiently hot to radiate in the visible
spectrum.
19. Newton vs. Einstein
Choose One • 5 points
Which of the following states that all matter tends to "warp" space in its vicinity and that
objects react to this warping by changing their paths?
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Newton's Universal Law of Gravitation
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Einstein's General Relativity
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Einstein's Special Relativity
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Newton's First Law of Motion
20. Quantum Mechanics
Choose One • 5 points
Wave-particle duality tells us that wave and particle models apply to all objects
whatever the size, so why don't we observe wave properties in macroscopic objects?
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Because their particle properties forbid us from observing their wave properties.
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Because their wavelength is extremely long (undetectable).
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Because their wavelength is extremely short (undetectable).
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