[R1 James Dalessio 2010]
6. Spectra I - PreLab
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Electro-Magnetic Waves
Nerd-speak for light
When we talk about light we usually assume we are talking about visible light, the type of light we see with our eyes. We
know that visible light comes in many different colors. However, light also comes in many different colors we can't see.
Below is a chart showing all of the different “colors” of light. Note that some textbooks refer to visible light as simply
“light” and all types of light as Electro-Magnetic radiation.
For light, “color” is just a measurement of the energy contained by each piece of light. Each piece of light is called a
photon. For example, a blue photon would have more energy than an orange photon. Notice that Ultra-Violet, X-Rays, and
Gamma Rays are all more energetic than visible light. This is why they are all harmful in large quantities. The low energy
types of light are rather harmless, and often pass clear through our bodies with no effect. (think about all the radio waves
that are passing through you right now, no need for a foil hat)
When describing light, scientists often use the units of eV, or electron volts, as a unit of energy. Sometimes instead of
energy, scientists will use wavelength or frequency to describe the color of light. Here are 4 different ways someone might
describe a piece of light....
The light is Yellowish-Green.
Each photon has an energy of 2.25eV
The light has a wavelength of 550nm (10^-9 meters) or 5500 A (10^-10 meters)
The light has a frequency of 5.45E14 Hz
Every one of these statements is identical. The color, energy, frequency, and wavelength of light are all one in the
same.
(2) Question 1: Which has more energy, a photon of Ultra-Violet light or a photon of Radio light?
Daddy, where does light come from?
Well little Jimmy, when two pieces of light really love each other ...
To understand how most visible light is created we have to understand Quantum Mechanics. I'll spare you the gory details,
for they are rather complicated and strange. Lets just use a loose metaphor.
Imagine you are leaning out the window of a skyscraper. If you were to drop a bowling ball from the 3rd floor it would hit
the ground really hard. If you dropped it from the 4th floor, it would hit the ground even harder. The harder it hits, the more
energy we say it had.
In an atom, the electrons are pulled on by the nucleus, the same way the bowling ball is pulled towards the Earth. If an
electron “falls” in closer to the nucleus it should gain energy, just like the bowling ball. However, due to Quantum
Mechanics, the electron isn't allowed to gain any energy. So where does this energy go? It is released as light.
Light is produced when a charged particle (typically an electron) is accelerated. Visible light is typically produced
when electrons get closer or “fall” to/towards the nucleus of an atom.
(2) Question 2: Imagine I drop the bowling ball out of a third story window. When the bowling ball hits the ground the
energy it picked up while falling is usually converted into heat. Suppose I take that heat energy and put it back into the
bowling ball. How high will the bowling ball travel?
Electron Energy Levels in the Atom
Shells and cheese never tasted so good.
You may remember from high school chemistry that electrons live in “shells” around the nucleus. Electrons are only
allowed to be in these shells, never in between. You can think about this like people in a building. People only live on the
first, second, third floor (etc). Here's a digram showing the shells around a nucleus. Don't worry about the names of the
shells.
So imagine you are dropping bowling balls out the window of a building again. When you drop the ball out the first floor
window it hits the ground with a certain amount of energy. When you drop it out the second floor it hits with more energy.
You can imagine that there are only certain amounts of energy the bowling ball can hit the ground with, because there are
only certain places it can fall from. When electrons move into shells closer to the nucleus they release a specific amount of
energy. If an electron drops from the 3rd to 1st shell, the energy from the “falling” electron is released as light. You can
imagine that there are only certain colors of light that this atom can emit. The light generated from an electron falling from
the 2nd to 1st shell might be very red, 3rd to 1st might be an greenish-yellow, and 4th to 1st might be Ultra-Violet.
For a specific type of atom (Hydrogen, Helium, etc), there are only very specific colors of light which it can emit.
Now imagine you are on the ground trying to throw a bowling ball into a window. If you throw the ball a little too hard it
will hit above the window, too soft and it wont go high enough. When a piece of light hits an atom it will pass right through
unless the energy of the light is the perfect amount to knock one of the electrons into a “higher” shell.
For a specific type of atom, there are only very specific colors of light which it can absorb.
(2) Question 3: I fill a large box with a gas of atoms. The atoms in the box like to absorb blue light. If I were to shine a
mixture of red and blue light in a beam through the middle of the box, what color would the beam be when it reached the
other side?
Line Formation
Just connect the dots.
When photons pass through some material, that material will tend to absorb certain colors of light. After white light (made
of all colors) passes through this material, an observer will notice that the light is missing that color. This is called
absorption. The lack of light at a certain color is called an absorption line.
If a material is heated up the collisions of atoms can knock elections into higher shells. When these electrons come back
down the atom releases photons. As there are only select color photons the atom can emit, and observer will see the atoms
glow at this color. This is called an emission line.
When atoms absorb light of a certain color, it knocks electrons into higher shells. When these electrons come back down the
atom will re-emit the same light, but in a random direction. This is called scattering.
Why is the Sky Blue?
Read. Process. Impress your friends.
The atoms in our atmosphere tend to absorb and re-emit (scatter) blue light. The Sun emits white light. When the Sun's light
hits the sky, the blue light is absorbed and sent back out in random directions. When we look directly at the Sun, the blue
light is missing. The blue light gets scattered around in the atmosphere and eventually reaches our eyes. Because the light
came from the direction of the sky, the sky looks blue.
(2) Question 4: When the Sun is low on the horizon, the light has to pass through a lot more atmosphere to reach you.
Explain why sunsets appear reddish.
(2) Question 5: If I were to shine a blue and red laser beam through a gas that absorbed red light, what color would the
beam be when it reached the other side? What color would the gas be?
7. Spectra I – Lab
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Section:
Date:
Start the program “Solar Energy”, “Flow of Energy in and Out of the Sun”, or “Solar Energy Transfer”. Click login, and
then OK. (Don't worry about entering your names).
Click on the top menu “simulation” and select “line formation”. This will bring up a window with a light source (left), a gas
(center), and a detector (right). In this exercise a source will be emitting photons through a gas.
The electrons in the atoms are in shells. The energy needed to by the atom to push an electron from the first shell to the
second shell happens to correspond exactly to energy of the pieces of light (photons) we are going to send through.
(2) Question 1: What do you think will happen to these photons as they pass through the gas?
Send at least 15 photons through this gas (by clicking “Run”).
(2) Question 2: What percentage of the photons made it through? Does this agree or disagree with your prediction?
Close that window and select the “continuum” simulation from the pull down menu in the main window. This is exactly as
before except now the individual photons all have some random energy.
(3) Question 3: Do you think many photons will be scattered? Why?
(3) Question 4: What percentage of the photons made it though? Some may have gotten scattered by the gas. Why is this?
Close the “continuum simulation” window and open the “experiment” simulation. Now we have the ability to select the
energy of the photons we are going to send through. Click on the “parameters” menu and select “# of photons”. Enter 10 in
the box and click ok. We are going to cycle through different energy photons and see how many pass through. The material
in the cylinder is hydrogen.
Click on “parameters” and select “change photon energy”. Choose the lowest value, 1.5eV. Click run. Record how many
photons made it through in the table below. Repeat this for each energy in the table below.
Photon Energy (eV)
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
3.1
3.2
Number Through of 10
Plot the results from your table below. Plot energy on the x axis and number of photons through on the y axis.
(2) Question 1: In our experiment, what three colors of light (in eV) does Hydrogen tend to scatter most?
(3) Question 2: A star generates energy in its interior. This energy passes though the atmosphere in the form of light. If we
were to split up the light from a star by color and noticed that there was very little light coming from the star at the colors
from question one, what could we conclude about the atmosphere of the star?
Here are a few example spectra of stars. The light from a star has been split up by color, and the amount of light coming at
each color plotted. Remember that “Wavelength” is the same thing as color. These two stars are very similar...
From this information, lets see if we can determine some properties of these stars. First look at the dips in the spectrum.
These are absorption lines.
(3) Question 3: Given the following lines, identify what substances are in this star's atmosphere.
Helium Lines: 3888A, 4471A, 5875A
Hydrogen Lines: 6562A, 4865A
Mercury: 4360A, 5790A
Iron: 5170A
Oxygen: 6870A
Given a spectrum, a simple law can be used to estimate the temperature of the star. The law is simply stated as...
The temperature of the object (in Kelvin) is equal to .29 divided by the wavelength of the brightest color light (in
cm).
(2) Question 4: Given the above spectrum, estimate the temperature of this star. Remember that 1A=10E-8cm.