This preview shows page 1. Sign up to view the full content.
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...
View
Full
Document
This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.
 Fall '08
 Norman,M
 Physics, Work

Click to edit the document details