homework3sol - Key for AST3018 HMW 3 25 Points Leah Simon 8...

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Key for AST3018 HMW 3 25 Points Leah Simon 8 Oct 2009 3.1 (3 pts) Here we simply need to use the Boltzmann Distribution: N 2 N 1 = g 1 g 2 e − E 2 E 1 / kT (1) where g n is the degeneracy in the nth state. For Hydrogen this is g n = 2n 2 , and E n = -13.6 eV * 1/n 2 a.) n 2 /n 1 = 1, g 2 /g 1 = 4, therefore, using k = 8.617 × 10 −5 eV K −1 yields T = 8.54 × 10 4 K. b.) n 3 /n 1 = 1, g 3 /g 1 = 9, therefore T = 6.38 × 10 4 K. Q 3.2 (1 pts) We get this from the original classi±cation method where the stars were given a letter designation based on their Hα strength. As such strongest to weakest would be: A, B, F, G, K, M, O. Q 3.9 (2 pts) When the temperature increases, first the average time between collisions is decreased, increasing the chance for a collisional excitation, that is, the collisional cross section is increased. Next, because the thermal energy of the gas goes as ≈ kT , there is more energy available per collision for excitation. Q 3.12 (2 pts)
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homework3sol - Key for AST3018 HMW 3 25 Points Leah Simon 8...

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