By B. Carroll, D. Ostlie

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The electron number density, ne , is given by ne D NII =V because each ionized hydrogen atom contributes one free electron. mp C me / ' V =mp . Here, D 10 6 kg m 3 is the density. The mass of the 44 Chapter 8 The Classification of Stellar Spectra electron is much less than the mass of the proton, and may be safely ignored in the expression for N t . Combining these expressions produces NII ne D N t mp for use in the Saha equation. With these substitutions, the Saha equation becomes  NII N t mp D NI NII 2 me kT h2 Substituting this into the above equation produces " à  N t mp 2 me kT 3=2 NII 1C e Nt NII h2 # I = kT Ã3=2 I = kT e N t mp NII  : 2 me kT h2 Ã3=2 e I = kT : Multiplying each side by NII =N t and rearranging terms gives  NII Nt Ã2  C NII Nt à mp  2 me kT h2 Ã3=2 e mp I = kT  2 me kT h2 Ã3=2 e I = kT D 0: (b) This is a quadratic equation of the form ax 2 C bx C c D 0, where aD1  bD mp cD b: 2 me kT h2 Ã3=2 e I = kT It therefore has the solution NII 1 D Nt 2a bC p b2 Á b p 4ac D 1 C 4=b 2 Á 1 : Evaluating b for a range of temperatures between 5000 K and 25,000 K produces the graph shown in Fig.

B) From Kepler’s third law (Eq. 37), with P D 1:8 yr, M1 D 0:5 Mˇ and M2 D 2:0 Mˇ , a D 2:0 AU. Also, since m1 =m2 D a2 =a1 (Eq. 1) and a D a1 C a2 D 2:0 AU, we have a1 D 1:6 AU and a2 D 0:4 AU. Now, the orbital velocities of the two stars are v1 D 2 a1=P D 26:5 km s 1 and v2 D 2 a2=P D 6:6 km s 1 . Finally, taking into consideration the orbital inclination of 30ı, the maximum observable radial velocities are v1r; max D v1 sin i D 13:2 km s 1 and v2r; max D v2 sin i D 3:351 km s 1 . (c) The eccentricity can be determined by considering the asymmetries in the velocity and/or light curves.

3). 7% of the mass is converted into energy, the timescale for this process would be t' 0:007Nreactmp c 2 7:2 E D D Lˇ Lˇ 3:84 1014 J D2 1026 W 10 12 s: Quantum mechanical tunneling is definitely required! 5 Beginning with Eq. 9) and substituting Eq. 6 From the non-relativistic expression for kinetic energy (E), r r 2E 2 1 vD and dv D dE: m 2 mE Using the relation nv dv D nE dE and making the appropriate substitutions in Eq. 1) leads directly to Eq. 28). 7 From Eq. 22), the gravitational potential energy of the Sun is approximately Ug D 2:3 1041 J.

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An Introduction to Modern Astrophysics 2nd ed - SOLUTIONS MANUAL by B. Carroll, D. Ostlie

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