Ph 444 Solutions for Problem Set 8

1. (Ryden 10.1)

With τ_n= 89 s, instead of 890 s.,

N_n/N_p = e^-200/89 / (5 + (1 – e^-200/89)) = 0.018 at T_nuc=200s.

Then, Y_max = 2f/(1+f)= 0.0353

2. (Ryden 10.2)

With Q_n = 0.129 Mev, at t_freeze ,

N_n,i/N_p,i = e ^–Q_n^/kT = e^-0.129/0.8 = 0.851

At T_nuc = 200 s, N_n/N_p = N_n,i e^-200/890 / (N_p,i +N_n,i (1-e^-200/890))

= e^-200/890 / (1/0.851 + (1- e^-200/890)) = .7987/(1.175+(1-0.7987))

= 0.58

Then, Y_max = 1.16/1.58 = 0.73