In this lab we’ll be looking at how astronomers (or other scientists studying on-Earth stuff) can learn more about what things are made of by taking the light we get and spreading it out into the rainbow.
This lab doesn’t involve any special materials, though there are a couple of additional PDF files that this document will have you look at.
By the end of this lab, you should be able to…
- …sort different types of light by wavelength, frequency, or energy.
- …understand how reflectance spectroscopy works.
- …tell the difference between continuous spectra, absorption spectra, and emission spectra, in terms of what they look like and what causes them.
- …interpret the spectrum of an object.
This lab has a total of 32 points, and is due on Monday October 24.
Part 1: Intro to the Spectrum (8 points)
Arrange these types of light by wavelength, going from shortest wavelengths to longest: red light, green light, yellow light, blue light, violet light, bluish-violet light, infrared light, ultraviolet light, radio waves. (There are 9 items in that list.) [4 points]
Out of blue light, orange light, and ultraviolet light, which has the highest frequency? Which has the lowest frequency? [2 points]
Out of green light, radio waves, and infrared light, which has the most energy per photon? Which has the least energy per photon? [2 points]
Part 2: Reflected Spectra (6 points)
In the “reference reflectance spectra” PDF you’ll find six reference spectra for different minerals (and other things), from the USGS Spectral Library: carnallite, kaolinite, lazurite, lawn grass (probably Kentucky bluegrass), olivine, and parisite. (Be careful with the spelling for that last one. It’s not something that feeds off of a host; it’s a mineral named after someone named J. J. Paris.)
You have a sample that contains one of those six substances. It’s not pure: it’s mixed with a small amount of some other stuff. You send it to the lab for analysis. They shine a light on your sample, convert the light seen from the sample into a spectrum, and compare that with the original light to get the reflectance spectrum seen below. (Note that it’s only from 0.3 µm to 1.5 µm.)
Which of the six substances do you have here? Explain how you determined this. [2 points]
You have another sample, and this time you know that it is mostly carnallite (the first mineral in the reference PDF), mostly kaolinite (the second mineral in the PDF), or mostly a more-or-less equal mixture of both minerals. Your friend at the spectroscopy lab offers to give you a high-quality reflectance spectrum of the sample over any 0.5-µm range of wavelengths: for example, over 0.5-1.0 µm, or 1.0-1.5 µm, or 1.3-1.8 µm.
If you want to figure out whether your sample is mostly carnallite, mostly kaolinite, or mostly a mix of the two, which wavelength range would you request? What would you look for in the resulting spectrum to figure out what you have in the sample? [2 points]
Your friend is eager to get into the mineral identification game. She has a stone she found on a camping trip, and while the outside is boring, she thinks there’s something interesting deep inside it – like a geode or a fossil. Would reflectance spectroscopy be able to tell her what’s inside this stone? Explain your answer. [2 points]
Part 3: Hydrogen (8 points)
A hydrogen discharge lamp creates light by exciting atoms of hydrogen in a sealed tube, producing a rosy pink color. When you send the light from this lamp through a prism or grating, you don’t get a full rainbow – instead you see just a handful of lines:
Images from utk.edu and anonymous.
We get a red line, a blue line (sometimes depicted as blue-green), and a bunch of lines in the violet. Below is a different pic of the spectrum, with a scale added. Read the scale carefully, and report the wavelengths of the four lines at wavelengths greater than 400 nm below. (For example, the line at a wavelength just smaller than 400 nm would be around 397 nm.) [4 points]
the red line:
the blue line:
the violet line:
the other violet line:
(Don’t forget to include units in your answers!)
These four spectral lines’ wavelengths are based on the energy released when the hydrogen atom’s electron falls down from a high energy level down to the second energy level. These four lines in particular are for when the electron falls from the 6th energy level down to the 2nd, from 5 to 2, from 4 to 2, and from 3 to 2.
Remember which types of light are higher-energy than others. Out of the four transitions described above, the one that involves the lowest amount of energy is when the electron falls from level 3 down to level 2. Which of our spectral lines is associated with that transition? [1 point]
Here is a detailed spectrum of the Sun, mainly in visible light. You can see quite a few narrow dips in the spectrum, where certain wavelengths are dimmer than their neighbors:
Are these dips emission features, or absorption features? [1 point]
Nine of the strongest features are traditionally associated with letters, as indicated above. Out of the features labeled A-G, which ones are caused by hydrogen? [2 points]
Part 4: The Ring Nebula (10 points)
The Ring Nebula (also known as Messier 57) is a famous sight for telescopes, located in the constellation Lyra (where you’ll also find the bright star Vega). This nebula is a bunch of gas expelled as part of a star’s dying breath, released as the star collapsed into a tiny dead husk at the center.
In addition to the distinctive shape, the Ring Nebula treats viewers to a beautiful palette of reds, greens, yellows, and even blues.
Let’s look at the spectrum of the Ring Nebula to see where these colors are coming from. Open up the “spectrum of the Ring Nebula” PDF to look at the Ring Nebula’s spectrum from 450 nm to 670 nm. The first page shows the full spectrum over that range, and the following two pages zoom in in different ways to give better views of some of the subtler features.
Are the lines shown in this spectrum emission or absorption? [1 point]
Based on your previous answer, which of these three scenarios is playing out with the light we’re getting from the Ring Nebula?
- The nebula is hot enough to produce its own light, and that light travels to us without getting affected by anything else.
- The nebula is intercepting light from another source, collecting photons at certain wavelengths, and re-emitting those photons in various directions – including toward us.
- The nebula is along our line-of-sight to a light source behind it, and the light we’re getting is colored by the nebula removing some photons from the light we’d otherwise get from that light source.
[1 point] Your answer:
Below is a reference table showing some prominent spectral lines that may or may not be seen in the spectrum. Each one is associated with an element – some elements have multiple lines, as we’ve seen with hydrogen.
In the left column you see some element abbreviations accompanied by Roman numerals. This is astronomy shorthand for whether these atoms have lost electrons (become ionized). A Roman numeral “I” means the atom has all of its electrons. A “II” means it has lost an electron, a “III” means it has lost two electrons, and so on. For example, a helium atom normally has two electrons: it would be “He I”. If the helium atom has gotten enough energy for one of its electrons to escape entirely, it is “He II”. If the atom has lost both of its electrons, it would be written as “He III”. “Si IV” is a silicon atom that has lost three of its electrons.
| nm | element | |
| O II | 372.8 | oxygen |
| Ca II | 393.4 | calcium |
| Ca II | 396.8 | calcium |
| Si IV | 408.9 | silicon |
| He II | 420.0 | helium |
| Ca I | 422.7 | calcium |
| Fe I | 432.5 | iron |
| He I | 447.1 | helium |
| He II | 454.1 | helium |
| O II | 465.0 | oxygen |
| He II | 468.6 | helium |
| O III | 495.9 | oxygen |
| O III | 500.7 | oxygen |
| He I | 587.6 | helium |
| N II | 654.8 | nitrogen |
| N II | 658.3 | nitrogen |
| H I | 397.0 | hydrogen |
| H I | hydrogen | |
| H I | hydrogen | |
| H I | hydrogen | |
| H I | hydrogen |
Go ahead and write in the wavelengths for the four hydrogen lines you examined earlier. They may be useful in the upcoming exercise.
Let’s look at the spectrum of the Ring Nebula and see if we can identify the major lines.
The brightest line in this spectrum of the Ring Nebula is caused by which element? [1 point]
What about the fairly bright line shown just to its left in the graph, at a slightly shorter wavelength and a relative brightness of a little more than 0.3? [1 point]
What about the third brightest line on the left side of the graph, which looks to be about 10% of the brightness of the most prominent line? [1 point]
Which elements are responsible for the close trio of lines on the right side of the spectrum? [2 points]
What about the line between 550 and 600 nm, with a relative brightness of 0.015? [1 point]
When our eyes see the color yellow, it’s due to our red and green receptors both getting light they’re sensitive to. Both receptors can sense yellow photons (wavelengths around 570 nm), but we can also get “yellow” by receiving both red photons (around 650 nm) and green photons (around 510 nm), as demonstrated below.
Some pictures of the Ring Nebula show it with a greenish color, while other pictures make it look more yellow. Based on the Ring Nebula’s spectrum, are we getting a lot of yellow photons from the Ring Nebula, or is the “yellow” really a mix of red and green light? Explain your answer. [2 points]
