Astronomy Lab Report
Q1: Based upon the properties of the spectra, what can you conclude about the relative temperatures of the two tungsten filaments in the halogen and incandescent bulb? Explain your reasoning.
Based on the spectra, it appears that the filament in the halogen bulb is at a higher temperature than the filament in the incandescent bulb. This is indicated because a) the peak intensity occurs at a shorter wavelength, and b) the spectrum of the incandescent lamp is spread out more into the longer wavelengths (Bennett et al., 2013, p. 154-155; Elvidge et al., 2010, 3963-3969).
Q2: How do the spectra of the two CFL bulbs differ from each other?
The 2700 K CFL emits more radiation in the longer wavelength area of visible light than the 5000 K bulb. Therefore, the 5000 K produces a whiter light similar to daylight, while the 2700 K produces yellow light that is similar to an incandescent bulb (EPA, 2013).
Q3: The color temperature of these CFL bulbs is supposed to mimic the color temperature of a black body that has the same color of the light source. Explain how they do this through emission lines?
A black body will emit radiation according to its temperature and will absorb all other radiation. Thus, it will produce an emission spectrum based solely on temperature. If it is producing yellow light, it absorbs all the other colors. The color can be identified based on the wavelength(s) in the emission spectrum and the temperature (Gearhart, 2009, 39-42).
Q4: Which spectra belongs to the 5000 K color temperature bulb?
The 5000 K color temperature bulb gives off a white light similar to natural daylight; under certain conditions it may be slightly bluish. It has its greatest intensity at a shorter wavelength, about 500 nm (EPA, 2013).
Q5: Why don’t they just make a light bulb with a filament that’s heated to 5000K?
Most metals will melt at temperatures lower than 2700 K; tungsten has a higher melting point, which is one reason that tungsten is the metal used in light bulbs. However, even at 2700 K the bulb must contain a gas such as argon to prevent the filament from burning away. If it were at 5000 K, the filament would burn out quickly (Argonne, 2012).
Q6: One of the two light yellow tubes contains plant chlorophyll which creates a broad absorption feature around 670 nm. By referring to the observed properties of the transmission spectra, which one of the two light yellow fluid tubes contains chlorophyll?
Yellow tube #1 contains chlorophyll. This is because its spectrum has the expected absorption near 670 nm whereas tube #2 does not (Atwell et al., 1999).
Q7: How did you estimate the error in your measurement?
I estimated the error to be 5% on either side of my measurement.
Q8: How does the spectrum of helium compare to that of hydrogen?
The hydrogen emission spectrum produced red, green, and blue lines plus 2 faint violet lines. In comparison, helium gave off a faint red line and two faint violet lines, plus an intense yellow lines and two blue lines. These differences result from the atomic structure of the two elements. Helium has two electrons and therefore more energy levels available, plus combinations of the two electrons’ transitions, while hydrogen has only one electron and is more limited (NASA, 2006).
Q9: From the observed neon emission lines, can you explain why neon lights seen in store front signs look so red?
Neon has at least 8 strong emission lines in the red wavelengths, and they are close together, almost appearing continuous. As a result, neon signs look red (Georgia State University, n.d.; EPA, 2013).
Q10: Examine the strongest lines observed in the spectrum of mercury and record their wavelengths. Then go back and look at the two CFL lights. Do either one or both contain mercury? How do you know? Are you sure?
The strong lines in the red, yellow and green seen with mercury are also present in the CFL spectra, however, it is possible that another substance could cause similar lines. For instance, neon, which has many red lines, could produce the red lines in the CFL spectrum (Georgia State University, n.d.).
Although this experiment did not deal directly with astronomical observations, it is essential to the study of astronomy because spectroscopy is used extensively to identify the temperature and composition of stars. It is impossible to take direct measurements of temperature and composition, even from our own Sun, due to the extreme heat and intense light near a star. However, by making spectroscopic observations and using Wien’s law and the Stefan-boltzmann law, these quantities can be computed even for very distant stars.