Question

The equilibrium reaction fBe+ e 1 C* can be viewed as analogous to the ionization and recombination of hydrogen: C* corresponds to a neutral hydrogen atom while &Be+ e correspond to a proton and an electron, respectively. So, the relative concentrations of the three species at a temperature T can be described by the Saha equation. But there is a difference: the "ionization energy" is negative, χ =-bd. Derive an expression for the ratio of densities of Be and C in the core of a helium-burning star of mass Mc and radius Rc as a function of temperature. Assume that the core is composed entirely out of helium and that the statistical weights corresponding to all the nuclei involved equal 1. Express your answer in termsof oantities defined above and physical constants

Answer 1

2 spin states (up & Down) relative to the proton Now consider the second term of G_r for neutral hydrogen, g_r_2^2 exp (+ elementof_rgl/RT) [Because x_r = - Vertical-bar x__r Vertical-bar ] The atom require E_r, 1 = 10.2 ev to reach the n = 2 state to ground state of neutral hydrogen is g_r1 = 8 (six p state and two & states) The second term thus becomes 8 exp(- 10.2/0.56) 8exp(- 10.2/0.56) = 9.8 times 10^-8, which is so small compared to g_ri that is can safely be ignored so by G_r (T) = g_r, 0 = 2 similarly because the proton has no bound electrons in the ionized state, we have G_r + 1 (T) = g_r + 1, 0 = 1 X_r = 13.6ev: R T = 0.56 ev so ration of densities n^2_r + 1/n^2_r = 2.41 times 10^21 1 times 2/ 2 (6400)^3/2 exp (- 13.6/0.56) n^2_r + 1/ n^2_r = 3.5 times 10^16 m^-3