By B. Carroll, D. Ostlie
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22, to obtain a value of log10 Œf Na . =500 nm/ for each wavelength. 26. log10 Œf hNa i. W = / 4:58 4:69 3:90 4:02 Now calculate log10 Œf . 9 nm line: The average value of log10 Na for this problem is hNa i D 20:18, so hNa i D 1:51 1020 m 2 . This is the value of the number of neutral hydrogen atoms per unit area with an electron in the n D 3 orbit. (b) The Boltzmann equation (Eq. 6) allows us to find the relative numbers of neutral hydrogen atoms in the n D 3, n D 2, and n D 1 states. From Eq.
3), ! Z 2 Z Z 2 Z =2 1 hI i D Iout sin Â dÂ d C Iin sin Â dÂ d 4 D0 ÂD0 D0 ÂD =2 ! Iout C Iin /: 2 52 Chapter 9 Stellar Atmospheres From Eq. Iout Z Iout cos Â sin Â dÂ d C =2 ÂD0 Z 2 D0 ÂD =2 Iin cos Â sin Â dÂ d ! 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.
L/ D j L: Thus you will see emission lines at wavelengths where j is large. 22 If the source function does not vary with position, then Eq. 1 e ;0 /: We consider two cases. ;0 1: In this case, the source function is equal to the Planck function: S D B (assuming thermodynamic equilibrium). 0/ D S D B ; and you will see blackbody radiation. 1 ;0 / DI ;0 L (Eq. 17). I ;ı S /: 54 Chapter 9 Stellar Atmospheres If I ;0 > S , then you will see absorption lines superimposed on the incident spectrum (I ;0 ).