Midterm Solutions

# 04x1018 m2 therefore n2n0 104x1018 m232 e3 ev05 ev

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Unformatted text preview: transition occurs at 1.24 eV µm/2.25 eV = 552 nm – visible =&gt; 2 ­&gt;3 transition occurs at 1.24 eV µm/0.42 eV = 2.95 µm – infrared (b) [10 pts] Assume the Sun’s atmosphere contains Nubulium. What is the ratio of neutral Nubulium in the n = 2 to n = 1 states? This makes use of the Boltzmann Equation: N2/N1 = g2/g1 e ­(ΔE21)/kT = (2x22)/(2x12) e ­[2.25 eV/(8.62x10 ­5 eV/K x 5800 K)] = 4e ­4.5 = 0.044 Page 10 of 16 Physics 160 Fall 2013 Midterm Exam (c) [15 pts] If the cosmic fraction of Nubulium to Hydrogen atoms is 10 ­8 (about the same as Lithium to Hydrogen), what are the number densities of Hydrogen and Nubulium atoms in the Sun’s photosphere (τ = 2/3)? For simplicity, assume the Sun’s atmosphere is pure Hydrogen, has a mean opacity κ = 0.1 m2/kg, and the surface pressure P = 0 at τ = 0. nNu = 10 ­8 nH = 10...
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