By D. Kinderlehrer, et al.,
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2, I D 24:6 eV, and Pe D 20 N m one-half of the He I atoms have been ionized when T ' 1:5 104 K. 10 (with Pe D 1000 N m 2 ) with Eq. 9) to evaluate NIII =NII and NII =NI . This results in the graph shown in Fig. 3. One-half of the helium atoms have been twice-ionized when T ' 4 104 K. 4 and the information given in the problem, use ZI D 2, ZII D 1, ne D 6:1 1031 m 3 , and T D 1:57 107 K in the Saha equation, Eq. 8), to find that I D 13:6 eV, NII NII =NI NII D D D 0:709 Nt NI C NII 1 C NII =NI at the center of the Sun.
5, we have ZII D 2:30, ZIII D 1, Pe D 1:5 N m 2 , II D 11:9 eV, and T D 5777 K. Inserting these values into the alternative form of the Saha equation, Eq. 9), we find that ŒNIII =NII Ca D 2:08 10 3. 5 shows that nearly all of the calcium atoms are in the form of Ca II, available for forming the H and K lines. 14 The Saha equation, Eq. 8), is NiC1 2ZiC1 D Ni ne Zi Â 2 me kT h2 Ã3=2 e i = kT : Stars of the same spectral type have the atoms in their atmospheres in the same states of ionization and excitation.
Z cos Â sin Â dÂ C Iin ÂD =2 cos Â sin Â dÂ Iin /: From Eq. 46) into Eq. 51) gives Iout C Iin D 2 ! Z 2 ! 47) into Eq. 43) gives Te4 : Iin D Iout Adding these, we find that Te4 Iout D Â 3 4 v Ã C1 : Subtracting Eq. 5) from Eq. Iout Iin C Iin / D 0:01: Using the above expressions, we have 1 2 h 3 Te4 = Á Te4 2 or 3 so v D 133. 19 From Eq. 53), at the top of a star’s atmosphere where v D 0, T 4 D Te4 =2. 1=2/1=4 D 0:841. If Te D 5777 K, then at the top of the atmosphere T D 4858 K. 20 For a plane-parallel gray atmosphere in LTE, Eq.