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The Decay Process 235U > 231 Th-> 231 Pa (t1/2 = 7.04 108 Y For 235U, T1/2 = Question: In the decay process 235U -> 231Th --> 231 Pa (t1/2 = 7.04 x 108 y for&n In the decay process 235U -> 231Th -> 231Pa (t1/2 = 7.04 108 y for 235U ; t1/2 = 25.5 h for 231Th) plot the 235U and 231 Th activities as a function of time from t=0 to t=100 h. Assume the sample initially consists of 1.0 mCi of 235U Discuss the condition of secular equilibrium in this decay process