Homework 3 Solutions

The term in parentheses can be simplified as 241x1021

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Unformatted text preview: ns (number of states) for each ionization state. ZII = 1 because there is no electron left; ZI we’ll take to be 2, assuming most of the electrons are in the ground state (you can show this by solving part (a) for N2/N1). The term in parentheses can be simplified as 2.41x1021 m ­3 T(K)3/2, while the exponential term is  ­1.58x105/T(K). Combining these and the numerical factors in front (except for ne), we get: NII 1 = 1.63 × 1015 m−3 NI ne for the Sun and NII 1 = 1.17 × 1027 m−3 NI ne for λ Orionis. Because we are assuming a pure H gas, all free electrons will pair with free protons, so NII ne = nH = nH NI + NII NI 1+ NII −1 We can relate nH to the total mass density because the protons and H atoms dominate the mass nH = ρ = 6 × 1020 m3 MH Combining these we get: NII NI = C (1 + ) NI NII where C = 2.72x10 ­6 for the Sun and 1.94x105 for λ Orionis. Multiplying both sides by NII/NI yields a quadratic equation whose positive root is: NII C = NI 2 1+ 1+ 4 C which is 1.65x10 ­3 for the Sun (predominantly neutral) and 1.94x105 for λ Orionis (predominantly ionized) (c) [10 pts] H I Balmer α absorption comes from the n = 3→2 transiti...
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This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.

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