Spectral Classes Worksheet-1

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Clemson University *

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1020

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Physics

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Dec 6, 2023

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Spectral Class Worksheet These lab activities have evolved over years of use in Clemson University’s Department of Physics and Astronomy general astronomy laboratory. Contributors include Tom Collins, Mark Leising, Neil Miller, Peter Milne, Grant Williams, Donna Mullenax, Jessica Crist, Keith Davis, Amber Porter, Lea Marcotulli, and David Connick. Please direct all questions, complaints, and corrections to David Connick (dconnic@clemson.edu) who is responsible for all errors and omissions. Worksheet Name III Example Stars Table 1. Stars Spectral Class Name Temperature (K) B-V Color Index (number) Absolute Magnitude (total) Distance (ly) O Zeta Ophiuch 32000 .02 -5.93 460 B Spica 26200 -.13 -5.95 250 A Sirius A 9400 .02 1.21 8.7 F Muceti 7300 .32 2.11 84.1 G Alpha Centauri A 5790 .69 4.17 4.36 K Gienah 4660 1.03 .25 72 M Menkar 3620 1.63 -3.31 220 1) In which properties do you see a trend as you move down the spectral class? What is the trend? a) As the spectral classes move from left to right, the temperature decreases with each class. In the opposite direction, from right to left, the color index increases with each class. In absolute magnitude, from left to right, the numbers begin small, increase to positive values, then decrease back to negative. In distance, from left to right, the numbers start large, then decrease, then increase again. Table 2. Spectral Class K or G Stars Luminosity Class Name Temperature (K) Absolute Magnitude (total) Distance (ly) I Beta Camel- opardalis 5280 -3.29 1000 II Rastaban 5190 -2.75 362
III HR1129 5850 -2.81 450 IV 12 Comae Berenices 5850 -.18 295 V Achird 5940 4.40 19.4 2) Which property of the star determines its Luminosity Class. How is that property changing as you move from class I to class V? a) The absolute magnitude of the star determines the luminosity of the star. The property changes through the classes because it ranges through negative and positive values. IV Spectral Class Spectra 3) Calculate the wavelength extremes for the visible spectrum of hydrogen (n 1 = 2): Ask your instructor for help with this calculation. Longest wavelength: 6578.95 A Line designation (n 2 ) = 3 Shortest wavelength: 4166.67 A Line designation (n 2 ) = 6 V Spectral Class Ordering 4) Using the numerical designations of the absorption spectrum plots which is the simplest plot (least dips and fluctuations)? a) Graph 1 is the simplest plot. 5) Which absorption spectrum plot is the most complex (most dips and fluctuations)? a) Graph 7 is the most complex plot. 6) In which of the absorption spectrum plots are the hydrogen lines the strongest? a) Graph 7 has the strongest Hydrogen lines. 7) Which of the absorption spectrum plots is from the highest temperature source? a) Graph 1 has the highest temperature source. 8) Which of the absorption spectrum plots is from the lowest temperature source? a) Graph 3 has the lowest temperature source. 9) Focus on the dips in the absorption spectrum plots due only to Hydrogen. Using the numerical designations list the spectrum from strongest (deepest) hydrogen lines to weakest (shallowest) hydrogen lines. a) 7, 6, 5, 4, 3, 2, 1 10) Looking at the calcium emission lines, at what wavelengths do the K and H lines appear at in Angstroms? a) K is located at 3940 angstroms and H is located at 3990 angstroms.
11) In which absorption spectrum plots can you see (both dips about the same size) the K and H calcium dips? a) Charts 2, 3, 4, and 5 have the same K and H calcium dips. When the spectra are ordered by the intensity of the hydrogen lines alone, other spectral lines do not seem to be appropriately ordered. In particular there is a discontinuity (abrupt change) in the intensity of the K and H lines of calcium between adjacent spectra. In fact, this hydrogen ordering is not the best way to organize the physical properties of the stars. Now list your spectra in a way so that there is a smooth transition between the appearances and disappearances of lines of all elements. List them with the spectral sequence designations (O, B, A, F, G, K, M). Make sure that these make sense given what you know about the spectral classes. Hint: consider the temperature and expected continuum spectra of these stellar types. O B A F G K M Chart 1 Chart 6 Chart 7 Chart 5 Chart 4 Chart 2 Chart 3 Now using the Letter designation of the spectral classes answer the following questions. 12) Which is the simplest spectrum? a) Chart 1 is the simplest spectrum. 13) Which is the most complex spectrum? a) Chart 7 is the most complex spectrum. 14) Which is the spectrum of the coolest source? a) Chart 3 has the coolest source. 15) Which is the spectrum of the hottest source? a) Chart 1 has the hottest source. 16) Conclusion - The key stellar property that determines the spectrum, both continuum and absorption line is: a) Temperature 17) Hydrogen is by far the most abundant element in the atmospheres of virtually all "normal" stars. Explain how a star can have very strong lines of other low-abundance elements and only very weak lines of hydrogen. (Hint: Visible hydrogen absorption lines can occur only if the electron starts in the n = 2 orbit.) What basic condition in the star’s atmosphere (other than absence of hydrogen) could make this possible? a) These stars are so hot that the Hydrogen is not in the atomic state. There are no transitions and therefore no absorption lines.
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