SOLUTION SET HW5 FALL 2016 PROBLEM 1 - all unites cgs (a) n_N = 7.243E+13 (b) alpha_1 = 9.026E-01 alpha_2 = 9.741E-02 (c) n_1 = 6.537E+13 n_2 = 7.055E+12 (d) n_11 = 9.806E+12 n_21 = 5.557E+13 n_12 = 7.055E+10 n_22 = 2.822E+12 n_32 = 4.163E+12 (e) n_e = 6.671E+13 j,k max = 2,1 ionized hydrogen max contribution to n_e %max = 8.329E+01 83% comes from ionized hydrogen (f) n = 1.391E+14 (g) P_g = 1.921E+02 (h) P_rad = 2.523E+01 beta = 8.839E-01 (i) mu_N = 1.300E+00 mu_e = 1.411E+00 mu = 6.765E-01 (j) rho_N = 1.563E-10 rho_e = 6.077E-14 rho = 1.564E-10 re/rN = 3.888E-04 rho_e is 4 parts in 10,000 of rho_N PROBLEM 2 (a,b) Boltzmann Excitation Ratios T=10,000 K i/1 n_i/n_1 i+1/i n_i+1/n_i 2/1 2.899E-05 2/1 2.899E-05 3/1 7.288E-06 3/2 2.514E-01 4/1 6.016E-06 4/3 8.255E-01 5/1 6.591E-06 5/4 1.096E+00 n_i11/n111 is the ratio of the number density of neutral hydrogen with in excitation state i to the number density of ground state neutral hydrogen. For i=2, this ratio is about 30 parts per million at T=10,000 K. The ratio drops to several parts per million for i>2, with a hint of increasing again for i>=5. n_i+1,11/ni11 is the ratio of the number density of neutral hydrogen with in excitation state i+1 to the number density of neutral hydrogen in excitation state i. For i+1/i=3/2, this ratio is about 1/4 at T=10,000 K, increasing to unity for 5/4. (c) Partition Function T=10,000 K i = 1 U_i11 = 2.000E+00 i = 2 U_i11 = 5.798E-05 i = 3 U_i11 = 1.458E-05 i = 4 U_i11 = 1.203E-05 i = 5 U_i11 = 1.318E-05 total i=1-5 U_11 = 2.000E+00 Note that if you continue the sum to i=infinity that the partition function actually will diverge to infinity. So the question becomes, at what value of i does one truncate the summation. This depends on the gas density and temperature. One determines the level i at which the electron is equally influenced by the gas as it is by the potential well of the nucleus. You can find this worked out in some text books. We didn't cover it in class. (d) Boltzmann Excitation/Total Ratios T=10,000 K i = 1 n_i11/n11 = 9.9995E-01 i = 2 n_i11/n11 = 2.8987E-05 i = 3 n_i11/n11 = 7.2872E-06 i = 4 n_i11/n11 = 6.0159E-06 i = 5 n_i11/n11 = 6.5906E-06 n_i11/n_11 is the ratio of the number density of nuetral hydrogen in excited state i to the number density of neutral hydrogens (accounting for all excitations states). At T=10,000 K most of the neutral hydrogen is in the ground state and only 3 in every 100,000 are in the first excited state (Balmer). Yet is is at T=10,000 K (A stars) the the Balmer lines and break are maximized! So, even in A stars, most of the nuetral hydrogen is int he ground state. PROBLEM 3 Many people did not correctly account for the ionization potential for the different ionization stages of iron. FeII has an ionization potential that is higher than the ionization potential of FeI. Note that you did not require the ionization potential of the final ionization stage, FeIII. Ionization fractions for FeII ne T f_1,26 f_2,26 f_3,26 G2 V 1.000E+14 5.800E+03 1.780E-01 8.220E-01 1.738E-08 G2 Ia 1.000E+12 5.800E+03 2.161E-03 9.978E-01 2.109E-06 A0 V 1.000E+14 1.000E+04 1.233E-04 9.646E-01 3.532E-02 Ratio from G2 V star ne T r_1,26 r_2,26 r_3,26 G2 V 1.000E+14 5.800E+03 1.000E+00 1.000E+00 1.000E+00 G2 Ia 1.000E+12 5.800E+03 1.214E-02 1.214E+00 1.214E+02 A0 V 1.000E+14 1.000E+04 6.928E-04 1.173E+00 2.032E+06 Instead of percent differences, I computed the ratios (so, I didn't grade for the percent difference calculations). Consider the G2 V and G2 Ia star, which have same T but different n_e. The G2 Ia star is more highly ionized than the G2 V star. That is, for a smaller n_e at fixed T, the balance trends toward higher ionization equilibrium (note the Saha ratio is inversely proportional to n_e). Physically, as n_e decreases, collision times increase and recombination rates decrease, which results in a higher ionization balance. Consider the G2 V and A0 V star, which have same n_e but different T The A0 star is hotter than the G2 star. As T goes up for fixed n_e, the balance trends toward higher ionization equilibrium. This is because as T increases, the photon field energy increases, increasing ionizations. As for collisions, the collision times shorten but the dominant effect is more ionizations (even though the recombinations are more common , their cross sections drop rapidly with increasing T. (I didn't expect you to know that last bit).