By Carroll B.W., Ostlie D.A.

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1 In the following, it is easiest to express Boltzmann’s constant as k D 8:6174 temperature of 300 K, 1 kT D 0:0259 eV eV: 40 If kT D 1 eV, then T D 1:16 104 K. If kT D 13:6 eV, then T D 1:58 10 5 eV K 1 . At a room 105 K. 3 The peak of the Maxwell–Boltzmann distribution, Fig. 6, shows that nv =n ' 6:5 10 5 s m 1 where v D vmp . nv =n/ v ' 0:13. 4 The most probable speed, vmp , occurs at the peak of the Maxwell–Boltzmann distribution, Eq. 1). Setting d nv =dv D 0, we find Á d nv m Á3=2 d 2 e mv =2kT v 2 D 0: D4 n dv 2 kT dv This leads to Â Ã mv 3 2 C 2v e mv =2kT D 0; kT p so that vmp D 2kT =m, which is Eq.

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.

From Eq. 25 Heisenberg’s uncertainty principle, Eq. 20), relates E, the uncertainty in the energy of an atomic orbital, to t, the time an electron occupies the orbital before making a downward transition: E t D h 2 t : When an electron makes a downward transition from an initial to a final orbital, the energy of the emitted photon is (Eq. W = /. 594 nm line: These values can then be used with the general curve of growth for the Sun, Fig. 22, to obtain a value of log10 Œf Na . =500 nm/ for each wavelength.