1. In the lecture we introduced the idea of mean molecular weight m. For an fully ionized gas, we gave the expression 1/m = 2 X + 3/4 Y + 1/2 Z, where X, Y and Z are the mass fractions of hydrogen, helium, and everything else, respectively. (a) Explain fully how the coefficient 3/4 is obtained for fully ionized helium (consider its atomic weight, and number of electrons). (b) In the same way, explain where the coefficient 1/2 comes from (consider some of the more common species such as C, O and N). At higher atomic weights, does the fraction get larger or smaller? (c) What is the mean molecular weight of a partially ionized helium gas (with no other constituents) in which only one electron, on average, is free while one remains bound? (d) What is the mean molecular weight of the same gas when cooled to the point where all of the helium is molecular (He₂)? Note, there will be no free electrons, and the individual particles are helium molecules.

2. In an earlier lecture we showed that dP/dr = -gr, and in lecture 22 we derived the expression P = rkT/mm_H. (a) Eliminate r in the two equations to find the expression dP/dr = -(mm_Hg/kT) P. (b) Solve this first-order differential equation by separation of variables (that is, multiply both sides by dr/P) and integrate (assume that the quantities in parentheses are constant). You should obtain an expression of the form P = P₀ exp(-r/H), where H is the "scale height" (the height over which the pressure changes by a factor of e). Give the expression for H. (c) The solar corona is nearly isothermal at a very high temperature T ~ 2 x 10⁶ K. What is the pressure scale height in the low corona (you can use the value of g for the surface of the Sun)? Give your value as a fraction of the solar radius. Compare with the scale height in the solar photosphere (T = 5770 K). This explains why the solar corona is so extended compared to the sharp edge of the photosphere.

3. The main energy generation mechanism in the Sun is the fusion of 4 protons into a He nucleus (2 protons and 2 neutrons, bound into a single nucleus). The atomic weight of a proton (in atomic mass units, u = 1.66 x 10^-27 kg) is 1.007825 u. The atomic weight of a He nucleus is 4.002603 u. (a) What is the difference in mass of 4 protons vs. a He nucleus? Expressing the difference as a fraction (4m_p - m_He) / 4m_p, what percentage of mass is converted to energy? (b) Equating mass to energy, using E = mc², how much energy is released through the fusion process? (c) If 10% of the Sun's mass is converted from hydrogen to helium in this process, the energy released would be E = (0.1*M_sun) (0.007) c² = 1.3 x 10⁴⁴ J. With this much energy available, how long can the Sun shine with its current luminosity of L = 3.84 x 10²⁶ W?